(All 5 Parts Combined) Understanding Physics JEE Main & Advanced - DC Pandey 9789313190554

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Understanding Physics

JEE Main & Advanced

MECHANICS Volume 1

Understanding Physics

JEE Main & Advanced

MECHANICS Volume 1

DC PANDEY [B.Tech, M.Tech, Pantnagar, ID 15722]

ARIHANT PRAKASHAN (Series), MEERUT

Understanding Physics

JEE Main & Advanced

ARIHANT PRAKASHAN (Series), MEERUT All Rights Reserved

© SARITA PANDEY No part of this publication may be re-produced, stored in a retrieval system or by any means, electronic, mechanical, photocopying, recording, scanning, web or otherwise without the written permission of the publisher. Arihant has obtained all the information in this book from the sources believed to be reliable and true. However, Arihant or its editors or authors or illustrators don’t take any responsibility for the absolute accuracy of any information published and the damage or loss suffered thereupon. All disputes subject to Meerut (UP) jurisdiction only.

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Understanding Physics

JEE Main & Advanced

PREFACE The overwhelming response to the previous editions of this book gives me an immense feeling of satisfaction and I take this an opportunity to thank all the teachers and the whole student community who have found this book really beneficial. In the present scenario of ever-changing syllabus and the test pattern of JEE Main & Advanced. The NEW EDITION of this book is an effort to cater all the difficulties being faced by the students during their preparation of JEE Main & Advanced. Almost all types and levels of questions are included in this book. My aim is to present the students a fully comprehensive textbook which will help and guide them for all types of examinations. An attempt has been made to remove all the printing errors that had crept in the previous editions. I am very thankful to (Dr.) Mrs. Sarita Pandey, Mr. Anoop Dhyani and Mr. Nisar Ahmad Comments and criticism from readers will be highly appreciated and incorporated in the subsequent editions.

DC Pandey

Understanding Physics

JEE Main & Advanced

CONTENTS 1. BASIC MATHEMATICS

1-12

1.1 Basic Mathematics

2. MEASUREMENT AND ERRORS

13-31

2.1 Errors in Measurement and Least Count 2.2 Significant Figures 2.3 Rounding off a Digit 2.4 Algebraic Operations with Significant Figures 2.5 Error Analysis

3. EXPERIMENTS 3.1 Vernier Callipers 3.2 Screw Gauge 3.3 Determination of ‘g’ using a Simple Pendulum 3.4 Young’s Modulus by Searle’s Method 3.5 Determination of Specific Heat 3.6 Speed of Sound using Resonance Tube 3.7 Verification of Ohm’s Law using Voltmeter and Ammeter 3.8 Meter Bridge Experiment 3.9 Post Office Box 3.10 Focal Length of a Concave Mirror using u-v method 3.11 Focal Length of a Convex Lens using u-v method

33-78

Understanding Physics

JEE Main & Advanced

79-96

4. UNITS AND DIMENSIONS 4.1 Units 4.2 Fundamental and Derived Units 4.3 Dimensions 4.4 Uses of Dimensions

5. VECTORS

97-125

5.1 Vector and Scalar Quantities 5.2 General Points regarding Vectors 5.3 Addition and Subtraction of Two Vectors 5.4 Components of a Vector 5.5 Product of Two Vectors

6. KINEMATICS

127-214

6.1 Introduction to Mechanics and Kinematics 6.2 Few General Points of Motion 6.3 Classification of Motion 6.4 Basic Definition 6.5 Uniform Motion 6.6 One Dimensional Motion with Uniform Acceleration 6.7 One Dimensional Motion with Non-uniform Acceleration 6.8 Motion in Two and Three Dimensions 6.9 Graphs 6.10 Relative Motion

7. PROJECTILE MOTION

215-259

7.1 Introduction 7.2 Projectile Motion 7.3 Two Methods of Solving a Projectile Motion 7.4 Time of Flight, Maximum Height and Horizontal Range of a Projectile 7.5 Projectile Motion along an Inclined Plane 7.6 Relative Motion between Two Projectiles

Understanding Physics

JEE Main & Advanced

8. LAWS OF MOTION

261-359

8.1 Types of Forces 8.2 Free Body Diagram 8.3 Equilibrium 8.4 Newton’s Laws of Motion 8.5 Constraint Equations 8.6 Pseudo Force 8.7 Friction

9. WORK, ENERGY AND POWER

361-427

9.1 Introduction to Work 9.2 Work Done 9.3 Conservative and Non-conservative Forces 9.4 Kinetic Energy 9.5 Work-Energy Theorem 9.6 Potential Energy 9.7 Three Types of Equilibrium 9.8 Power of a Force 9.9 Law of Conservation of Mechanical Energy

10. CIRCULAR MOTION

429-476

10.1 Introduction 10.2 Kinematics of Circular Motion 10.3 Dynamics of Circular Motion 10.4 Centrifugal Force 10.5 Motion in a Vertical Circle

Hints & Solutions JEE Main & Advanced Previous Years' Questions (2018-13)

477-608 1-18

Understanding Physics

JEE Main & Advanced

SYLLABUS JEE Main Physics and Measurement Physics, Technology and Society, SI units, Fundamental and derived units, Least count, Accuracy and Precision of measuring instruments, Errors in measurement, Dimensions of physical quantities, Dimensional analysis and its applications.

Kinematics Frame of reference, Motion in a straight line, Position-time graph, Speed and velocity, Uniform and non-uniform motion, Average speed and instantaneous velocity, Uniformly accelerated motion, Velocity-time and position-time graphs, Relations for uniformly accelerated motion, Scalars and Vectors, vectors addition and subtraction, Zero vector, scalar and vector products, Unit vector, resolution of a vector, Relative velocity, motion in plane, Projectile motion, Uniform circular motion.

Laws of Motion Force and inertia, Newton's first law of motion, Momentum, Newton's second law of motion, impulse, Newton's third Law of motion, law of conservation of linear momentum and its applications, Equilibrium of concurrent forces. Static and kinetic friction, Laws of friction, rolling friction. Dynamics of uniform circular motion, centripetal force and its applications.

Work, Energy and Power Work done by a constant force and a variable force, Kinetic and potential energies, Work energy theorem, power. Potential energy of a spring, Conservation of mechanical energy, Conservative and non-conservative forces, Elastic and inelastic collisions in one and two dimensions.

Centre of Mass Centre of mass of a two particle system, Centre of mass of a rigid body.

Experimental Skills Vernier Callipers and its use to measure internal and external diameter and depth of a vessel. Screw gauge its use to determine thickness/diameter of thin sheet/wire.

Understanding Physics

JEE Main & Advanced

JEE Advanced General Physics Units and dimensions, Dimensional analysis, Least count, Significant figures, Methods of measurement and error analysis for physical quantities pertaining to the following experiments, Experiments based on vernier callipers and screw gauge (micrometer).

Kinematics Kinematics in one and two dimensions (Cartesian coordinates only), Projectiles, Uniform circular motion, Relative velocity.

Laws of Motion Newton's laws of motion, Inertial and uniformly accelerated frames of reference, Static and dynamic friction.

Work, Energy and Power Kinetic and potential energy, Work and power, Conservation of linear momentum and mechanical energy.

Centre of Mass and Collision System of particles, Centre of mass and its motion, Impulse, Elastic and inelastic collisions.

Understanding Physics

JEE Main & Advanced

This book is dedicated to my honourable grandfather

(Late) Sh. Pitamber Pandey a Kumaoni poet and a resident of Village

Dhaura (Almora), Uttarakhand

01 Basic Mathematics Chapter Contents 1.1

Basic Mathematics

2 — Mechanics - I

1.1 Basic Mathematics The following formulae are frequently used in Physics. So, the students who have just gone in class XI are advised to remember them first.

Logarithms (i) e ≈ 2.7183 (iii) If 10 x = y, then x = log 10 y (v) log ( ab) = log ( a ) + log ( b)

(ii) If e x = y, then x = log e y = ln y (iv) log 10 y = 0.4343 log e y = 0.4343 ln y a (vi) log   = log ( a ) − log ( b)  b

(vii) log a n = n log ( a )

Trigonometry (i) sin 2 θ + cos 2 θ = 1 (ii) 1 + tan 2 θ = sec 2 θ (iii) 1 + cot 2 θ = cosec 2 θ (iv) sin 2 θ = 2 sin θ cos θ (v) cos 2 θ = 2 cos 2 θ − 1 = 1 − 2 sin 2 θ = cos 2 θ − sin 2 θ (vi) sin ( A ± B ) = sin A cos B ± cos A sin B (vii) cos ( A ± B ) = cos A cos B + sin A sin B C − D C + D (viii) sin C + sin D = 2 sin    cos   2   2  C + D C − D (ix) sin C − sin D = 2 sin    cos   2   2  C +D C −D (x) cos C + cos D = 2 cos cos 2 2 D −C C +D (xi) cos C − cos D = 2 sin sin 2 2 2 tan θ (xii) tan 2 θ = 1 − tan 2 θ tan A ± tan B (xiii) tan ( A ± B ) = 1 + tan A tan B (xiv) sin (90° + θ ) = cos θ (xv) cos (90° + θ ) = − sin θ (xvi) tan (90° + θ ) = − cot θ (xvii) sin (90° − θ ) = cos θ (xviii) cos (90° − θ ) = sin θ (xix) tan (90° − θ ) = cot θ (xx) sin (180° − θ ) = sin θ

Chapter 1

Basic Mathematics — 3

(xxi) cos (180° − θ ) = − cos θ (xxii) tan (180° − θ ) = − tan θ (xxiii) sin (180° + θ ) = − sin θ (xxiv) cos (180° + θ ) = − cos θ (xxv) tan (180° + θ ) = tan θ (xxvi) sin ( − θ ) = − sin θ (xxvii) cos ( − θ ) = cos θ (xxviii) tan (− θ ) = − tan θ

Differentiation (i) (iii) (v) (vii) (ix) (xi)

(xii) (xiii)

d d (ii) (constant) = 0 ( x n ) = nx n −1 dx dx d d 1 d (iv) (log e x ) or (ln x ) = (sin x ) = cos x dx dx x dx d d (vi) (cos x ) = − sin x (tan x ) = sec 2 x dx dx d d 2 (viii) (cot x ) = − cosec x (sec x ) = sec x tan x dx dx d d x (x) (cosec x ) = − cosec x cot x (e ) = e x dx dx d d d { f 1 ( x ) . f 2 ( x )} = f 1 ( x ) f 2 (x ) + f 2 (x ) f (x ) dx dx dx 1 d d f 2 (x ) f 1 (x ) − f 1 (x ) f (x ) d f 1 (x ) dx dx 2 = dx f 2 (x ) { f 2 (x )}2 d d f ( ax + b) = a f ( X ) , where X = ax + b dx dx

Integration (i) ∫ x n dx =

x n +1 + c ( n ≠ −1) n +1

(iii) ∫ sin x dx = − cos x + c (v) ∫ e x dx = e x + c (vii) ∫ cosec 2 x dx = − cot x + c

(ii)

∫

dx = log e x + c or x

(iv) ∫ cos x dx = sin x + c (vi) ∫ sec 2 x dx = tan x + c (viii) ∫ sec x tan x dx = sec x + c

(ix) ∫ cosec x cot x dx = − cosec x + c 1 f ( X ) dX , where X = ax + b a∫ Here, c is constant of integration.

(x) ∫ f ( ax + b) dx =

ln x + c

4 — Mechanics - I Graphs Following graphs and their corresponding equations are frequently used in Physics. (i) y = mx, represents a straight line passing through origin. Here, m = tan θ is also called the slope of line, where θ is the angle which the line makes with positive x-axis, when drawn in anticlockwise direction from the positive x-axis towards the line. The two possible cases are shown in Fig. 1.1. In Fig. 1.1 (i), θ < 90° .Therefore, tan θ or slope of line is positive. In Fig. 1.1 (ii), 90° < θ < 180° . Therefore, tan θ or slope of line is negative. y

y θ

θ

x

(i)

x

(ii)

Fig. 1.1

Note That y = mx or y ∝ x also means that value of y becomes 2 times if x is doubled. Or it will remain 41 th if x becomes

1 4

times.

(ii) y = mx + c, represents a straight line not passing through origin. Here, m is the slope of line as discussed above and c the intercept on y-axis. y

y y c = +ve

θ

c = +ve

θ

x

θ

x

x

c = –ve

(i)

(iii)

(ii)

Fig. 1.2

In figure (i) : slope and intercept both are positive. In figure (ii) : slope is negative but intercept is positive and In figure (iii) : slope is positive but intercept is negative. Note That in y = mx + c , y does not become two times if x is doubled.

1 2 or y = etc., represents a rectangular hyperbola in first and third quadrants. The shape of x x rectangular hyperbola is shown in Fig. 1.3(i).

(iii) y ∝

y

y

x

x

(ii)

(i)

Fig. 1.3

Chapter 1

Basic Mathematics — 5

From the graph we can see that y → 0 as x → ∞ or x → 0 as y → ∞. 4 Similarly, y = − represents a rectangular hyperbola in second and fourth quadrants as shown in x Fig. 1.3(ii). Note That in case of rectangular hyperbola if x is doubled y will become half.

(iv) y ∝ x 2 or y = 2x 2 , etc., represents a parabola passing through origin as shown in Fig. 1.4( i). y

y y∝x

2

x ∝ y2 x

x

(ii)

(i)

Fig. 1.4

Note That in the parabola y = 2 x or y ∝ x , if x is doubled, y will become four times. 2

2

Graph x ∝ y 2 or x = 4 y 2 is again a parabola passing through origin as shown in Fig 1.4 (ii). In this case if y is doubled, x will become four times. (v) y = x 2 + 4 or x = y 2 − 6 will represent a parabola but not passing through origin. In the first equation ( y = x 2 + 4), if x is doubled, y will not become four times. (vi) y = Ae − Kx ; represents exponentially decreasing graph. Value of y decreases exponentially from A to 0. The graph is shown in Fig. 1.5. y A

x

Fig. 1.5

From the graph and the equation, we can see that y = A at x = 0 and y → 0 as x → ∞. (vii) y = A (1 − e − Kx ), represents an exponentially increasing graph. Value of y increases exponentially from 0 to A. The graph is shown in Fig. 1.6. y A

x

Fig. 1.6

From the graph and the equation we can see that y = 0 at x = 0 and y → A as x → ∞.

6 — Mechanics - I Maxima and Minima Suppose y is a function of x. Or y = f ( x ). Then we can draw a graph between x and y. Let the graph is as shown in Fig. 1.7. y P x Q Fig. 1.7

dy   Then from the graph we can see that at maximum or minimum value of y slope  or  to the graph is  dx  zero. dy Thus, = 0 at maximum or minimum value of y. dx d2 y dy By putting = 0 we will get different values of x. At these values of x, value of y is maximum if 2 dx dx (double differentiation of y with respect to x) is negative at this value of x. Similarly y is minimum if d2 y is positive. Thus, dx 2 d2 y = −ve for maximum value of y dx 2 d2 y and = +ve for minimum value of y dx 2 Note That at constant value of y also V

dy d2y = 0 but in this case 2 is zero. dx dx

Example 1.1 Differentiate the following functions with respect to x (a) x3 + 5x 2 − 2 Solution

(a)

(b) x sin x

(c) ( 2x + 3) 6

x sin x

d 3 d 3 d 2 d ( x + 5x 2 − 2) = (x ) + 5 (x ) − ( 2) dx dx dx dx = 3x 2 + 5( 2x ) − 0 = 3x 2 + 10x

(b)

(d)

d d d ( x sin x ) = x (sin x ) + sin x . (x ) dx dx dx

= x cos x + sin x (1) = x cos x + sin x d d (c) ( 2x + 3) 6 = 2 ( X ) 6 , where dx dX

X = 2x + 3

= 2{6X 5 } = 12X 5 = 12( 2x + 3) 5

(e) e (5 x + 2)

Chapter 1 sin x

d  x  (d)  = dx  sin x

(e) V

d d (x ) − x (sin x ) dx dx (sin x ) 2

(sin x )(1) − x (cos x )

=

Basic Mathematics — 7

2

sin x

=

sin x − x cos x sin 2 x

d (5x + 2) d X e =5 e , where X = 5x + 2 = 5e X = 5e 5x + 2 dx dX

Example 1.2 Integrate the following functions with respect to x (a)

∫ ( 5x

(c)

∫ 4x + 5

2

+ 3x − 2) dx

dx

Solution

(a)

∫ ( 5x

2

2   4 sin x −  dx  x

(b)

∫

(d)

∫ (6x + 2)

3

dx

+ 3x − 2) dx = 5 ∫ x 2 dx + 3 ∫ x dx − 2 ∫ dx

5x 3 3x 2 + − 2x + c 3 2 dx dx = 4 ∫ sin x dx − 2 ∫ x =

(b)

(c)

∫ ∫

  4 sin x − 

2  x

= − 4 cos x − 2 ln x + c dx 1 dX , where X = 4x + 5 = 4x + 5 4 ∫ X

1 1 ln X + c1 = ln ( 4x + 5) + c 2 4 4 1 (d) ∫ ( 6x + 2) 3 dx = ∫ X 3 dX , where X = 6x + 2 6 =

= V

( 6x + 2) 4 1 X 4 + c2   + c1 = 6 4  24

Example 1.3 Draw straight lines corresponding to following equations (a) y = 2x Solution

(b) y = − 6x

(c) y = 4x + 2

(d) y = 6x – 4

(a) In y = 2x, slope is 2 and intercept is zero. Hence, the graph is as shown below. y

θ

Fig. 1.8

tan θ = slope = 2 x

8 — Mechanics - I (b) In y = − 6x, slope is − 6 and intercept is zero. Hence, the graph is as shown below. y

θ

tan θ = – 6 x

Fig. 1.9

(c) In y = 4x + 2, slope is + 4 and intercept is 2. The graph is as shown below. y 2 tan θ = 4 x

θ

Fig. 1.10

(d) In y = 6x − 4, slope is + 6 and intercept is – 4. Hence, the graph is as shown below. y

θ

tan θ = 6 x

–4

Fig. 1.11 V

Example 1.4 Find maximum or minimum values of the functions (a) y = 25x 2 + 5 − 10x Solution

(a) For maximum and minimum value, we can put

dy = 0. dx

dy = 50x − 10 = 0 dx 1 x= 5

or ∴

d2 y

Further, or

(b) y = 9 − ( x − 3) 2

dx 2

d2 y

= 50

1 1 has positive value at x = . Therefore, y has minimum value at x = . 5 5 dx 2

Chapter 1 Substituting x =

Basic Mathematics — 9

1 in given equation, we get 5 2

 1  1 y min = 25   + 5 − 10   = 4  5  5 (b) y = 9 − ( x − 3) 2 = 9 − x 2 − 9 + 6x y = 6x − x 2

or

dy = 6 − 2x dx

∴

For minimum or maximum value of y we will substitute

dy =0 dx

or 6 − 2x = 0 or x = 3 To check whether value of y is maximum or minimum at x = 3 we will have to check whether d2 y is positive or negative. dx 2 d2 y dx 2 or

d2 y dx 2

= −2

is negative at x = 3. Hence, value of y is maximum. This maximum value of y is, y max = 9 − ( 3 − 3) 2 = 9

Exercises Subjective Questions Trigonometry 1. Find the value of (a) (c) (e) (g)

(b) sin 240° (d) cot 300° (f) cos (− 60° ) (h) cos (− 120° )

cos 120° tan (− 60° ) tan 330° sin (− 150° )

2. Find the value of (a) sec2 θ − tan 2 θ (c) 2 sin 45° cos 15°

(b) cosec2 θ − cot2 θ − 1 (d) 2 sin 15° cos 45°

Calculus 3. Differentiate the following functions with respect to x (a) x4 + 3x2 − 2x (c) (6x + 7)4 (1 + x) (e) ex

(b) x2 cos x (d) ex x5

4. Integrate the following functions with respect to t (a) (c)

∫ (3t − 2t ) dt −4 ∫ (2t − 4) dt 2

(b) (d)

∫ (4 cos t + t dt ∫ (6t − 1)

2

) dt

5. Integrate the following functions 2

(a)

∫0 2t dt

(c)

∫4

(e)

∫1(2t − 4) dt

10

dx x

π /3

(b)

∫π / 6 sin x dx

(d)

∫0 cos x dx

π

2

6. Find maximum/minimum value of y in the functions given below (a) y = 5 − (x − 1)2 (c) y = x3 − 3x (e) y = (sin 2x − x), where −

π π ≤x≤ 2 2

(b) y = 4x2 − 4x + 7 (d) y = x3 − 6x2 + 9x + 15

Graphs 7. Draw the graphs corresponding to the equations (a) y = 4x (c) y = x + 4 (e) y = 2x − 4

(b) y = − 6x (d) y = − 2x + 4 (f) y = − 4x − 6

Basic Mathematics — 11

Chapter 1 8. For the graphs given below, write down their x-y equations y

y

y

y 4

45°

x

x

30°

(a)

135°

30°

x

2

(c)

(b)

x

(d)

9. For the equations given below, tell the nature of graphs. (a) y = 2x2 (c) y = 6e−4x 4 (e) y = x

(b) y = − 4x2 + 6 (d) y = 4(1 − e−2x ) 2 (f) y = − x

10. Value of y decreases exponentially from y = 10 to y = 6. Plot a x-y graph. 11. Value of y increases exponentially from y = − 4 to y = + 4. Plot a x-y graph. 12. The graph shown in figure is exponential. Write down the equation corresponding to the graph. y 12

4 x

13. The graph shown in figure is exponential. Write down the equation corresponding to the graph. y 6 x –4

Answers Subjective Questions 1. (a) −

1 2

(b) −

3 2

(c) − 3

 3 + 1  2. (a) 1 (b) 0 (c)   2  

(d) −

1 3

(e) −

1 3

(f)

1 2

(g) −

(b) 4 sin t +

(h) −

1 2

 3 − 1  (d)   2   

3. (a) 4x3 + 6x − 2 (b) 2x cos x − x 2 sin x (c) 24 (6x + 7)3 4. (a) t 3 − t 2 + C

1 2

t3 +C 3

(c) −

1 +C 6 (2t − 4)3

(d) 5e x x 4 + e x x 5 (d)

(e) − xe − x

1 ln (6t − 1) + C 6

12 — Mechanics - I

6. (a) ymax

( 3 − 1)

(c) ln (5 /2) (d) Zero (e) − 1 2 = 5 at x = 1 (b) ymin = 6 at x = 1 /2 (c) ymin = − 2 at x = 1 and ymax = 2 at x = − 1

5. (a) 4 (b)

 3 π (d) ymin = 15 at x = 3 and ymax = 19 at x = 1 (e) ymin = −  −  at x = − π / 6 and 2 6    3 π ymax =  −  at x = π /6 6  2 7.

y

y

y

x

(b)

y

(c)

(d)

y

x

x

(e)

8. (a) y = x (b) y = −

(f)

x 3

(c) y =

x + 4 (d) y = − x + 2 3

9. (a) parabola passing through origin (b) parabola not passing through origin (c) exponentially decreasing graph (d) exponentially increasing graph (e) Rectangular hyperbola in first and third quadrant (f) Rectangular hyperbola in second and fourth quadrant y

y

10

10.

x

x

x

(a)

y

11.

4

6

x

x

12. y = 4 + 8e − Kx Here, K is a positive constant 13. y = − 4 + 10 (1 − e − Kx ) Here, K is positive constant

–4

02 Measurement and Errors Chapter Contents 2.1

Errors in Measurement and Least Count

2.2

Significant Figures

2.3

Rounding off a Digit

2.4

Algebraic Operations with Significant Figures

2.5

Error Analysis

2.1 Errors in Measurement and Least Count To get some overview of error, least count and significant figures, let us have some examples. V

Example 2.1 Let us use a centimeter scale (on which only centimeter scales are there) to measure a length AB. A

B 0

1

2

3

4

5

6

7

8

9

10

Fig. 2.1

From the figure, we can see that length AB is more than 7 cm and less than 8 cm. In this case, Least Count (LC) of this centimeter scale is 1 cm, as it can measure accurately upto centimeters only. If we note down the length ( l ) of line AB as l = 7 cm then maximum uncertainty or maximum possible error in l can be 1 cm ( = LC ) , because this scale can measure accurately only upto 1 cm. V

Example 2.2 Let us now use a millimeter scale (on which millimeter marks are there). This is also our normal meter scale which we use in our routine life. From the figure, we can see that length AB is more than A B 3.3 cm and less than 3.4 cm. If we note down the length, 1 2 3 4 l = AB = 3.4 cm Fig. 2.2 Then, this measurement has two significant figures 3 and 4 in which 3 is absolutely correct and 4 is reasonably correct (doubtful). Least count of this scale is 0.1 cm because this scale can measure accurately only upto 0.1 cm. Further, maximum uncertainty or maximum possible error in l can also be 0.1 cm.

INTRODUCTORY EXERCISE

2.1

1. If we measure a length l = 6.24 cm with the help of a vernier callipers, then (a) What is least count of vernier callipers ? (b) How many significant figures are there in the measured length ? (c) Which digits are absolutely correct and which is/are doubtful ?

2. If we measure a length l = 3.267 cm with the help of a screw gauge, then (a) What is maximum uncertainty or maximum possible error in l ? (b) How many significant figures are there in the measured length ? (c) Which digits are absolutely correct and which is/are doubtful ?

2.2 Significant Figures From example 2.2, we can conclude that: "In a measured quantity, significant figures are the digits which are absolutely correct plus the first uncertain digit".

Chapter 2

Measurement and Errors — 15

Rules for Counting Significant Figures Rule 1

All non-zero digits are significant. For example, 126.28 has five significant figures.

Rule 2 The zeros appearing between two non-zero digits are significant. For example, 6.025 has four significant figures.

Trailing zeros after decimal places are significant. Measurement l = 6.400 cm has four significant figures. Let us take an example in its support. Rule 3

Table 2.1 Measurement

Accuracy

l lies between (in cm)

Significant figures

l = 6.4 cm

0.1 cm

6.3 − 6.5

Two

Remarks

l = 6.40 cm

0.01 cm

6.39 − 6.41

Three

closer

l = 6.400 cm

0.001 cm

6.399 − 6.401

Four

more closer

Thus, the significant figures depend on the accuracy of measurement. More the number of significant figures, more accurate is the measurement. The powers of ten are not counted as significant figures. For example, 1.4 × 10 −7 has only two significant figures 1 and 4. Rule 4

Rule 5 If a measurement is less than one, then all zeros occurring to the left of last non-zero digit are not significant. For example, 0.0042 has two significant figures 4 and 2. Rule 6 Change in units of measurement of a quantity does not change the number of significant figures. Suppose a measurement was done using mm scale and we get l = 72 mm (two significant figures). We can write this measurement in other units also (without changing the number of significant figures) : ` 7.2 cm → Two significant figures. 0.072 m → Two significant figures. 0.000072 km → Two significant figures. 7 7.2 ×10 nm → Two significant figures Rule 7 The terminal or trailing zeros in a number without a decimal point are not significant. This also sometimes arises due to change of unit.

For example, 264 m = 26400 cm = 264000 mm All have only three significant figures 2, 6 and 4. All trailing zeros are not significant. Zeroes at the end of a number are significant only if they are behind a decimal point as in Rule-3. Otherwise, it is impossible to tell if they are significant. For example, in the number 8200, it is not clear if the zeros are significant or not. The number of significant digits in 8200 is at least two, but could be three or four. To avoid uncertainty, use scientific notation to place significant zeros behind a decimal point 8.200 ×10 3 has four significant digits.

16 — Mechanics - I 8.20 ×10 3 has three significant digits. 8.2 ×10 3 has two significant digits. Therefore, if it is not expressed in scientific notations, then write least number of significant digits. Hence, in the number 8200, take significant digits as two. Exact measurements have infinite number of significant figures. For example, 10 bananas in a basket 46 students in a class speed of light in vacuum = 299, 792, 458 m /s (exact ) 22 π = (exact) 7 All these measurements have infinite number of significant figures. Rule 8

V

Example 2.3

Table 2.2

Measured value

Number of significant figures

Rule number

12376 cm

5

1

6024.7 cm

5

2

0.071 cm

2

5

4100 cm

2

7

2.40 cm

3

3

1.60 × 1014 km

3

4

INTRODUCTORY EXERCISE

2.2

1. Count total number of significant figures in the following measurements: (a) 4.080 cm

(b) 0.079 m

(c) 950

(d) 10.00 cm

(e) 4.07080

(f) 7.090 × 105

2.3 Rounding Off a Digit Following are the rules for rounding off a measurement : Rule 1 If the number lying to the right of cut off digit is less than 5, then the cut off digit is retained as such. However, if it is more than 5, then the cut off digit is increased by 1. For example, x = 6.24 is rounded off to 6.2 to two significant digits and x = 5.328 is rounded off to 5.33 to three significant digits. Rule 2 If the digit to be dropped is 5 followed by digits other than zero, then the preceding digit is increased by 1. For example, x =14.252 is rounded off to x =14.3 to three significant digits. Rule 3 If the digit to be dropped is simply 5 or 5 followed by zeros, then the preceding digit it left unchanged if it is even. For example, x = 6.250 or x = 6.25 becomes x = 6.2 after rounding off to two significant digits.

Chapter 2

Measurement and Errors — 17

If the digit to be dropped is 5 or 5 followed by zeros, then the preceding digit is raised by one if it is odd. For example, x = 6.350 or x = 6.35 becomes x = 6.4 after rounding off to two significant digits. Rule 4

V

Example 2.4 Measured value

Table 2.3 After rounding off to three significant digits

Rule

7.364

7.36

1

7.367

7.37

1

8.3251

8.33

2

9.445

9.44

3

9.4450

9.44

3

15.75

15.8

4

15.7500

15.8

4

INTRODUCTORY EXERCISE

2.3

1. Round off the following numbers to three significant figures : (a) 24572

(b) 24.937

(c) 36.350

(d) 42.450 × 109

2. Round 742396 to four, three and two significant digits.

2.4 Algebraic Operations with Significant Figures The final result shall have significant figures corresponding to their number in the least accurate variable involved. To understand this, let us consider a chain of which all links are strong except the one. The chain will obviously break at the weakest link. Thus, the strength of the chain cannot be more than the strength of the weakest link in the chain.

Addition and Subtraction Suppose, in the measured values to be added or subtracted the least number of digits after the decimal is n. Then, in the sum or difference also, the number of digits after the decimal should be n. V

Example 2.5 1.2 + 3.45 + 6.789 = 11.439 ≈ 11.4 Here, the least number of significant digits after the decimal is one. Hence, the result will be 11.4 (when rounded off to smallest number of decimal places).

V

Example 2.6 12.63 − 10.2 = 2.43 ≈ 2.4

Multiplication or Division Suppose in the measured values to be multiplied or divided the least number of significant digits be n. Then in the product or quotient, the number of significant digits should also be n. V

Example 2.7 1.2 × 36.72 = 44.064 ≈ 44 The least number of significant digits in the measured values are two. Hence, the result when rounded off to two significant digits become 44. Therefore, the answer is 44.

18 — Mechanics - I 1101 ms −1

= 107.94117647 ≈ 108

V

Example 2.8

V

Example 2.9 Find, volume of a cube of side a = 1.4 × 10 −2 m.

10.2 ms −1

Solution Volume V = a 3

= (1.4 × 10−2 ) × (1.4 × 10−2 ) × (1.4 × 10−2 ) = 2.744 × 10−6 m 3 Since, each value of a has two significant figures. Hence, we will round off the result to two significant figures. V = 2.7 × 10−6 m 3

∴ V

Ans.

Example 2.10 Radius of a wire is 2.50 mm. The length of the wire is 50.0 cm. If mass of wire was measured as 25 g, then find the density of wire in correct significant figures. [Given, π = 3.14, exact] Solution Given,

r = 2.50 mm = 0.250 cm

(three significant figures) (three significant figures)

Note Change in the units of measurement of a quantity does not change the number of significant figures.

Further given that, l = 50.0 cm m = 25 gm π = 3.14 exact m m ρ= = 2 V πr l =

(three significant figures) (two significant figures) (infinite significant figures)

25 ( 3.14 )( 0.250)( 0.250)( 50.0)

= 2.5477 g /cm 3 But in the measured values, least number of significant figures are two. Hence, we will round off the result to two significant figures. ∴

ρ = 2.5 g /cm 3

INTRODUCTORY EXERCISE

2.4

1. Round to the appropriate number of significant digits (a) 13.214 + 234.6 + 7.0350 + 6.38 (b) 1247 + 134.5 + 450 + 78

2. Simplify and round to the appropriate number of significant digits (a) 16.235 × 0.217 × 5 (b) 0.00435 × 4.6

Ans.

Chapter 2

Measurement and Errors — 19

2.5 Error Analysis We have studied in the above articles that no measurement is perfect. Every instrument can measure upto a certain accuracy called Least Count (LC).

Least Count The smallest measurement that can be measured accurately by an instrument is called its least count. Instrument

Its least count

mm scale

1 mm

Vernier callipers

0.1 mm

Screw gauge

0.01 mm

Stop watch Temperature thermometer

0.1 sec 1°C

Permissible Error due to Least Count Error in measurement due to the limitation (or least count) of the instrument is called permissible error. Least count of a millimeter scale is 1 mm. Therefore, maximum permissible error in the measurement of a length by a millimeter scale may be 1 mm. If we measure a length l = 26 mm. Then, maximum value of true value may be (26 + 1) mm = 27 mm and minimum value of true value may be (26 − 1) mm = 25 mm. Thus, we can write it like, l = (26 ± 1) mm If from any other instrument we measure a length = 24.6 mm, then the maximum permissible error (or least count) from this instrument is 0.1 mm. So, we can write the measurement like, l = (24.6 ± 0.1) mm

Classification of Errors Errors can be classified in two ways. First classification is based on the cause of error. Systematic error and random errors fall in this group. Second classification is based on the magnitude of errors. Absolute error, mean absolute error and relative (or fractional) error lie on this group. Now, let us discuss them separately.

Systematic Error Systematic errors are the errors whose causes are known to us. Such errors can therefore be minimised. Following are few causes of these errors : (a) Instrumental errors may be due to erroneous instruments. These errors can be reduced by using more accurate instruments and applying zero correction, when required. (b) Sometimes errors arise on account of ignoring certain facts. For example in measuring time period of simple pendulum error may creep because no consideration is taken of air resistance. These errors can be reduced by applying proper corrections to the formula used. (c) Change in temperature, pressure, humidity, etc., may also sometimes cause errors in the result. Relevant corrections can be made to minimise their effects.

20 — Mechanics - I Random Error The causes of random errors are not known. Hence, it is not possible to remove them completely. These errors may arise due to a variety of reasons. For example the reading of a sensitive beam balance may change by the vibrations caused in the building due to persons moving in the laboratory or vehicles running nearby. The random errors can be minimized by repeating the observation a large number of times and taking the arithmetic mean of all the observations. The mean value would be very close to the most accurate reading. Thus, a + a 2 +… + a n a mean = 1 n Absolute Error The difference between the true value and the measured value of a quantity is called an absolute error. Usually the mean value a m is taken as the true value. So, if am =

a1 + a 2 + … + a n n

Then by definition, absolute errors in the measured values of the quantity are, ∆a1 = a m − a1 ∆a 2 = a m − a 2 … … … ∆a n = a m − a n Absolute error may be positive or negative.

Mean Absolute Error Arithmetic mean of the magnitudes of absolute errors in all the measurements is called the mean absolute error. Thus, ∆a mean =

| ∆ a1 | + | ∆ a 2 | +… + | ∆a n | n

The final result of measurement can be written as, a = a m ± ∆a mean This implies that value of a is likely to lie between a m + ∆a mean and a m − ∆a mean .

Relative or Fractional Error The ratio of mean absolute error to the mean value of the quantity measured is called relative or fractional error. Thus, Relative error =

∆a mean am

Relative error expressed in percentage is called as the percentage error, i.e. Percentage error =

∆a mean × 100 am

Chapter 2 V

Measurement and Errors — 21

Example 2.11 The diameter of a wire as measured by screw gauge was found to be 2.620, 2.625, 2.630, 2.628 and 2.626 cm. Calculate (a) mean value of diameter (b) absolute error in each measurement (d) fractional error (c) mean absolute error (e) percentage error (f) Express the result in terms of percentage error Solution (a) Mean value of diameter 2.620 + 2.625 + 2.630 + 2.628 + 2.626 am = 5 = 2.6258cm (rounding off to three decimal places) = 2.626cm (b) Taking a m as the true value, the absolute errors in different observations are, ∆a1 = 2.626 – 2.620 = + 0.006cm ∆a 2 = 2.626 – 2.625 = + 0.001cm ∆a 3 = 2.626 – 2.630 = – 0.004 cm ∆a 4 = 2.626 – 2.628 = – 0.002 cm ∆a 5 = 2.626 – 2.626 = 0.000cm (c) Mean absolute error, | ∆a1 | + | ∆a 2 | + | ∆a 3 | + | ∆a 4 | + | ∆a 5 | ∆a mean = 5 0.006 + 0.001+ 0.004 + 0.002 + 0.000 = 5 = 0.0026 = 0.003 ∆a mean ± 0.003 (d) Fractional error = ± = = ± 0.001 am 2.626

(rounding off to three decimal places)

(e) Percentage error = ± 0.001 × 100 = ± 0.1% (f) Diameter of wire can be written as, d = 2.626 ± 0.1%

Combination of Errors Errors in Sum or Difference Let x = a ± b Further, let ∆a is the absolute error in the measurement of a, ∆b the absolute error in the measurement of b and ∆x is the absolute error in the measurement of x. Then, x + ∆ x = ( a ± ∆ a ) ± ( b ± ∆ b) = ( a ± b) ± ( ± ∆ a ± ∆ b) = x ± ( ± ∆ a ± ∆ b) or ∆x = ± ∆a ± ∆b The four possible values of ∆x are ( ∆a − ∆b), ( ∆a + ∆b), ( −∆a − ∆b) and ( −∆a + ∆b).

22 — Mechanics - I Therefore, the maximum absolute error in x is, ∆ x = ± ( ∆ a + ∆ b) i.e. the maximum absolute error in sum and difference of two quantities is equal to sum of the absolute errors in the individual quantities. V

Example 2.12 The volumes of two bodies are measured to be V1 = (10.2 ± 0.02) cm3 and V2 = (6.4 ± 0.01) cm3 . Calculate sum and difference in volumes with error limits. Solution V1 = (10.2 ± 0.02) cm 3

V2 = (6.4 ± 0.01) cm 3

and

∆V = ± ( ∆V1 + ∆V2 ) = ± (0.02 + 0.01) cm 3 = ± 0.03 cm 3 V1 + V2 = (10.2 + 6.4 ) cm 3 = 16.6 cm 3 and

V1 − V2 = (10.2 − 6.4 ) cm 3 = 3.8 cm 3

Hence, sum of volumes = (16.6 ± 0.03) cm 3 and difference of volumes = (3.8 ± 0.03) cm 3

Errors in a Product Let x = ab Then, or or or

( x ± ∆ x ) = ( a ± ∆ a ) ( b ± ∆ b)  ∆x   ∆a   ∆b  x 1 ±  = ab 1 ±  1 ±    x  a   b ∆x ∆b ∆a ∆a ∆b 1± =1 ± ± ± ⋅ x b a a b ∆x ∆a ∆b ∆a ∆b ± =± ± ± ⋅ x a b a b

(as x = ab)

∆a ∆b is a small quantity, so can be neglected. ⋅ a b ∆x ∆a ∆b Hence, ± =± ± x a b

Here,

Possible values of

∆x  ∆a ∆b   ∆a ∆b   ∆a ∆b   ∆a ∆b  are  + − + −  and  − . ,  , −  a  a x b  a b  a b b

Hence, maximum possible value of ∆x  ∆a ∆b  =± +   a x b Therefore, maximum fractional error in product of two (or more) quantities is equal to sum of fractional errors in the individual quantities.

Chapter 2

Measurement and Errors — 23

Errors in Division a b a ± ∆a x ± ∆x = b ± ∆b x=

Let Then,

 ∆x  x 1 ± =  x 

or

 ∆a  a 1 ±   a   ∆b  b 1 ±   b

 ∆x   ∆a   ∆b  1 ±  = 1 ±  1 ±   x   a   b

or

–1

a   as x =   b

∆b < T2 (c) T1 < T2

(b) T1 = T2 (d) nothing can be said

3. The length of a seconds hand in watch is 1 cm. The change in velocity of its tip in 15 s is π cm/s 30 2 π 2 (d) cm/s 30

(a) zero (c)

(b)

π cm/s 30

4. When a ball is thrown up vertically with velocity v0, it reaches a maximum height of h. If one wishes to triple the maximum height then the ball should be thrown with velocity (a)

3 v0

(b) 3 v0

(c) 9 v0

(d)

3 v0 2

Chapter 6

Kinematics — 191

5. During the first 18 min of a 60 min trip, a car has an average speed of 11 ms−1. What should be the average speed for remaining 42 min so that car is having an average speed of 21 ms−1 for the entire trip?

(a) 25.3 ms −1 (c) 31 ms −1

(b) 29.2 ms −1 (d) 35.6 ms −1

8t3 where 3 x is in metres and t in seconds. Find the acceleration of the particle at the instant when particle is at rest.

6. A particle moves along a straight line. Its position at any instant is given by x = 32t −

(a) − 16 ms −2 (c) 32 ms −2

(b) − 32 ms −2 (d) 16 ms −2

7. The acceleration of a particle is increasing linearly with time t as bt. The particle starts from the origin with an initial velocity v0. The distance travelled by the particle in time t will be

1 3 bt 6 1 (c) v0t + bt 2 3

1 3 bt 3 1 (d) v0t + bt 2 2

(a) v0t +

(b) v0t +

8. Water drops fall at regular intervals from a tap 5 m above the ground. The third drop is leaving the tap, the instant the first drop touches the ground. How far above the ground is the second drop at that instant. ( g = 10 ms−2 ) (a) 1.25 m

(b) 2.50 m

(c) 3.75 m

(d) 4.00 m

9. A stone is dropped from the top of a tower and one second later, a second stone is thrown

vertically downward with a velocity 20 ms−1. The second stone will overtake the first after travelling a distance of ( g = 10 ms−2 ) (b) 15 m (d) 19.5 m

(a) 13 m (c) 11.25 m

10. A particle moves in the x-y plane with velocity vx = 8t − 2 and v y = 2 . If it passes through the point x = 14 and y = 4 at t = 2 s, the equation of the path is

(a) x = y2 − y + 2 (c) x = y2 + y − 6

(b) x = y2 − 2 (d) None of these

11. The horizontal and vertical displacements of a particle moving along a curved line are given by

x = 5t and y = 2t 2 + t. Time after which its velocity vector makes an angle of 45° with the horizontal is

(a) 0.5 s

(b) 1 s

(c) 2 s

(d) 1.5 s

12. A ball is released from the top of a tower of height h metre. It takes T second to reach the ground. What is the position of the ball in T/3 second? h metre from the ground 9 (c) (8h / 9) metre from the ground (a)

(b) (7h / 9) metre from the ground (d) (17h / 18) metre from the ground

13. An ant is at a corner of a cubical room of side a. The ant can move with a constant speed u. The minimum time taken to reach the farthest corner of the cube is 3a u 5a (c) u

(a)

3a u ( 2 + 1) a (d) u (b)

192 — Mechanics - I 14. A lift starts from rest. Its acceleration is plotted against time. When it comes to rest its height above its starting point is a (ms–2) 2 8 0

12

4

t (s)

–2

(a) 20 m

(b) 64 m

(c) 32 m

(d) 36 m

15. A lift performs the first part of its ascent with uniform acceleration a and the remaining with uniform retardation 2a. If t is the time of ascent, find the depth of the shaft. (a)

at 2 4

(b)

at 2 3

(c)

at 2 2

(d)

at 2 8

16. Two objects are moving along the same straight line. They cross a point A with an acceleration a, 2a and velocity 2u , u at time t = 0. The distance moved by the object when one overtakes the other is (a)

6u 2 a

(b)

2u 2 a

(c)

4u 2 a

(d)

8u 2 a

17. A cart is moving horizontally along a straight line with constant speed 30 ms−1. A particle is to be fired vertically upwards from the moving cart in such a way that it returns to the cart at the same point from where it was projected after the cart has moved 80 m. At what speed (relative to the cart) must the projectile be fired? (Take g = 10 ms−2) (a) 10 ms −1 40 (c) ms −1 3

(b) 10 8 ms −1 (d) None of these

18. The figure shows velocity–time graph of a particle moving along a

v (m/s)

straight line. Identify the correct statement. (a) The particle starts from the origin

10

(b) The particle crosses it initial position at t = 2 s (c) The average speed of the particle in the time interval, 0 ≤ t ≤ 2 s is zero

t (s) 0

1

2

3

–10

(d) All of the above –20

19. A ball is thrown vertically upwards from the ground and a student gazing out of the window

sees it moving upward past him at 10 ms −1. The window is at 15 m above the ground level. The velocity of ball 3 s after it was projected from the ground is [Take g = 10 ms−2 ]

(a) 10 m/s, up (c) 20 ms −1, down

(b) 20 ms −1, up (d) 10 ms −1, down

20. A body starts moving with a velocity v0 = 10 ms−1. It experiences a retardation equal to 0.2v 2. Its velocity after 2s is given by

(a) + 2 ms −1 (c) − 2 ms −1

(b) + 4 ms −1 (d) + 6 ms −1

Chapter 6

Kinematics — 193

21. Two trains are moving with velocities v1 = 10 ms−1 and v2 = 20 ms−1 on the same track in opposite directions. After the application of brakes if their retarding rates are a1 = 2 ms−2 and a2 = 1 ms−2 respectively, then the minimum distance of separation between the trains to avoid collision is (b) 225 m (d) 300 m

(a) 150 m (c) 450 m

22. Two identical balls are shot upward one after another at an interval of 2s along the same vertical line with same initial velocity of 40 ms−1. The height at which the balls collide is

(a) 50 m (c) 100 m

(b) 75 m (d) 125 m

23. A particle is projected vertically upwards and reaches the maximum height H in time T . The height of the particle at any time t (< T ) will be

(a) g (t − T )2 1 (c) g (t − T )2 2

(b) H − g (t − T )2 1 (d) H − g (T − t )2 2

x2 t2 . Here x varies with time as x = . Where x and y are 2 2 measured in metres and t in seconds. At t = 2 s, the velocity of the particle (in ms−1) is

24. A particle moves along the curve y = (a) 4$i + 6$j (c) 4$i + 2$j

(b) 2$i + 4$j (d) 4$i + 4$j

25. If the displacement of a particle varies with time as x = t + 3 (a) (b) (c) (d)

velocity of the particle is inversely proportional to t velocity of particle varies linearly with t velocity of particle is proportional to t initial velocity of the particle is zero

26. The graph describes an airplane’s acceleration during its take-off run. The airplane’s velocity when it lifts off at t = 20 s is a (ms–2) 5 3

0

(a) 40 ms −1 (c) 90 ms −1

10

20

t (s)

(b) 50 ms −1 (d) 180 ms −1

27. A particle moving in a straight line has velocity-displacement equation as v = 5 1 + s. Here v is in ms−1 and s in metres. Select the correct alternative. (a) Particle is initially at rest (b) Initially velocity of the particle is 5 m/s and the particle has a constant acceleration of 12.5 ms −2 (c) Particle moves with a uniform velocity (d) None of the above

194 — Mechanics - I 28. A particle is thrown upwards from ground. It experiences a constant resistance force which can produce a retardation of 2 ms−2. The ratio of time of ascent to time of descent is ( g = 10 ms−2 ) (a) 1 : 1

(b)

2 3

(d)

(c)

2 3 3 2

29. A body of mass 10 kg is being acted upon by a force 3t 2 and an opposing constant force of 32 N. The initial speed is 10 ms −1. The velocity of body after 5 s is

(a) 14.5 ms −1 (c) 3.5 ms −1

(b) 6.5 ms −1 (d) 4.5 ms −1

30. A stone is thrown vertically upwards. When stone is at a height half of its maximum height, its speed is 10 ms−1; then the maximum height attained by the stone is ( g = 10 ms−2 )

(a) 25 m (c) 15 m

(b) 10 m (d) 20 m

Subjective Questions 1. (a) What does

dv d|v| and represent? dt dt

(b) Can these be equal?

2. The coordinates of a particle moving in x-y plane at any time t are ( 2 t , t 2 ). Find (a) the trajectory of the particle, (b) velocity of particle at time t and (c) acceleration of particle at any time t.

3. A farmer has to go 500 m due north, 400 m due east and 200 m due south to reach his field. If he takes 20 min to reach the field. (a) (b) (c) (d)

What distance he has to walk to reach the field ? What is the displacement from his house to the field ? What is the average speed of farmer during the walk ? What is the average velocity of farmer during the walk ?

4. A rocket is fired vertically up from the ground with a resultant vertical acceleration of 10 m/ s2. The fuel is finished in 1 min and it continues to move up.(a) What is the maximum height reached? (b) After how much time from then will the maximum height be reached? (Take g = 10 m/s2 )

5. A particle is projected upwards from the roof of a tower 60 m high with velocity 20 m/s. Find (a) the average speed and (b) average velocity of the particle upto an instant when it strikes the ground. Take g = 10 m/s 2.

6. A block moves in a straight line with velocity v for time t0. Then, its velocity becomes 2v for next t0 time. Finally, its velocity becomes 3v for time T . If average velocity during the complete journey was 2.5 v, then find T in terms of t0.

7. A particle starting from rest has a constant acceleration of 4 m/ s2 for 4 s. It then retards uniformly for next 8 s and comes to rest. Find during the motion of particle (a) average acceleration (b) average speed and (c) average velocity. 21 8. A particle moves in a circle of radius R = m with constant speed 1m/s. Find, 22 (a) magnitude of average velocity and (b) magnitude of average acceleration in 2 s.

Chapter 6

Kinematics — 195

9. Two particles A and B start moving simultaneously along the line joining them in the same direction with acceleration of 1 m/s2 and 2 m/s2 and speeds 3 m/s and 1 m/s respectively. Initially, A is 10 m behind B. What is the minimum distance between them?

10. Two diamonds begin a free fall from rest from the same height, 1.0 s apart. How long after the first diamond begins to fall will the two diamonds be 10 m apart? Take g = 10 m/ s2.

11. Two bodies are projected vertically upwards from one point with the same initial velocity v0. The second body is projected t0 s after the first. How long after will the bodies meet?

12. Displacement-time graph of a particle moving in a straight line is as shown in figure. c

s b

a

d O

t

(a) Find the sign of velocity in regions oa,ab, bc and cd. (b) Find the sign of acceleration in the above region.

13. Velocity-time graph of a particle moving in a straight line is shown in figure. In the time interval from t = 0 to t = 14 s, find v (m/s) 20 10 0

10 12 2

4

14

6

t (s)

–10

(a) average velocity and (b) average speed of the particle.

14. A person walks up a stalled 15 m long escalator in 90 s. When standing on the same escalator, now moving, the person is carried up in 60 s. How much time would it take that person to walk up the moving escalator? Does the answer depend on the length of the escalator?

15. Figure shows the displacement-time graph of a particle moving in a straight line. Find the signs of velocity and acceleration of particle at time t = t1 and t = t2. s

t1

t2

t

16. Velocity of a particle moving along positive x-direction is v = ( 40 − 10 t ) m/s. Here, t is in seconds.

At time t = 0, the x coordinate of particle is zero. Find the time when the particle is at a distance of 60 m from origin.

196 — Mechanics - I 17. Velocity-time graph of a particle moving in a straight line is shown in figure. Plot the corresponding displacement-time graph of the particle if at time t = 0, displacement s = 0. v (m/s) C

20 10

A

B

2

4

O

6

D 8

t (s)

18. Acceleration-time graph of a particle moving in a straight line is as shown in figure. At time t = 0, velocity of the particle is zero. Find a (m/s2) 20 10 10 12 4

14

6

t (s)

–10

(a) average acceleration in a time interval from t = 6 s to t = 12 s, (b) velocity of the particle at t = 14 s.

19. A particle is moving in x-y plane. At time t = 0, particle is at (1m, 2m) and has velocity ( 4$i + 6$j) m/ s. At t = 4 s, particle reaches at (6m, 4m) and has velocity ( 2 $i + 10 $j) m/ s. In the given time interval, find (a) average velocity, (b) average acceleration and (c) from the given data, can you find average speed?

20. A stone is dropped from the top of a tower. When it crosses a point 5 m below the top, another stone is let fall from a point 25 m below the top. Both stones reach the bottom of the tower simultaneously. Find the height of the tower. Take g = 10 m/ s2.

21. A point mass starts moving in a straight line with constant acceleration. After time t0 the acceleration changes its sign, remaining the same in magnitude. Determine the time T from the beginning of motion in which the point mass returns to the initial position. 22. A football is kicked vertically upward from the ground and a student gazing out of the window sees it moving upwards past her at 5.00 m/s. The window is 15.0 m above the ground. Air resistance may be ignored. Take g = 10 m/ s2. (a) How high does the football go above ground? (b) How much time does it take to go from the ground to its highest point?

23. A car moving with constant acceleration covered the distance between two points 60.0 m apart in 6.00 s. Its speed as it passes the second point was 15.0 m/s. (a) What is the speed at the first point? (b) What is the acceleration? (c) At what prior distance from the first was the car at rest?

Chapter 6

Kinematics — 197

24. A particle moves along the x-direction with constant acceleration. The displacement, measured

from a convenient position, is 2 m at time t = 0 and is zero when t = 10 s. If the velocity of the particle is momentary zero when t = 6 s, determine the acceleration a and the velocity v when t = 10 s. 25. At time t = 0, a particle is at (2m, 4m). It starts moving towards positive x-axis with constant acceleration 2 m/ s2 (initial velocity = 0 ). After 2 s, an additional acceleration of 4 m/ s2 starts acting on the particle in negative y-direction also. Find after next 2 s. (a) velocity and

(b) coordinates of particle.

26. A particle starts from the origin at t = 0 with a velocity of 8.0$j m/ s and moves in the x-y plane

with a constant acceleration of ( 4.0 $i + 2.0 $j) m/ s2. At the instant the particle’s x-coordinate is 29 m, what are (a) its y-coordinate and

(b) its speed ?

27. The velocity of a particle moving in a straight line is decreasing at the rate of 3 m/s per metre of displacement at an instant when the velocity is 10 m/s. Determine the acceleration of the particle at this instant.

28. A particle moves along a horizontal path, such that its velocity is given by v = ( 3t 2 − 6t ) m/ s, where t is the time in seconds. If it is initially located at the origin O, determine the distance travelled by the particle in time interval from t = 0 to t = 3.5 s and the particle’s average velocity and average speed during the same time interval.

29. A particle travels in a straight line, such that for a short time 2 s ≤ t ≤ 6 s, its motion is described

by v = ( 4 / a ) m/ s, where a is in m/ s2. If v = 6 m/ s when t = 2 s, determine the particle’s acceleration when t = 3 s.

30. If the velocity v of a particle moving along a straight line decreases linearly with its displacement from 20 m/s to a value approaching zero at s = 30 m, determine the acceleration of the particle when s = 15 m.

31. Velocity-time graph of a particle moving in a straight line is shown in figure. At time t = 0, s = − 10 m. Plot corresponding a-t and s-t graphs. v (m/s) 10 6 2

8

10

t (s)

4

–10

32. Velocity-time graph of a particle moving in a straight line is shown in figure. At time t = 0, s = 20 m. Plot a-t and s-t graphs of the particle. v (m/s) 20 4

6

8 10 12

–20

14

t (s)

198 — Mechanics - I 33. A particle of mass m is released from a certain height h with zero initial velocity. It strikes the ground elastically (direction of its velocity is reversed but magnitude remains the same). Plot the graph between its kinetic energy and time till it returns to its initial position.

34. A ball is dropped from a height of 80 m on a floor. At each collision, the ball loses half of its speed. Plot the speed-time graph and velocity-time graph of its motion till two collisions with the floor. [Take g = 10 m/ s2]

35. Figure shows the acceleration-time graph of a particle moving along a straight line. After what time the particle acquires its initial velocity? a (m/s2) 2

O

t (s)

2

1

36. Velocity-time graph of a particle moving in a straight line is shown in figure. At time t = 0, displacement of the particle from mean position is 10 m. Find v (m/s) 10 8 2

4

10

6

t (s)

–10

(a) acceleration of particle at t = 1 s , 3 s and 9 s. (b) position of particle from mean position at t = 10 s. (c) write down s-t equation for time interval (i) 0 ≤ t ≤ 2 s, (ii) 4 s ≤ t ≤ 8 s

37. Two particles 1 and 2 are thrown in the directions shown in figure simultaneously with velocities 5 m/s and 20 m/s. Initially, particle 1 is at height 20 m from the ground. Taking upwards as the positive direction, find 1 5 m/s 20 m

+ve 20 m/s 2

(a) acceleration of 1 with respect to 2 (b) initial velocity of 2 with respect to 1 (c) velocity of 1 with respect to 2 after time t = (d) time when the particles will collide.

1 s 2

Chapter 6

Kinematics — 199

38. A ball is thrown vertically upward from the 12 m level with an initial velocity of 18 m/s. At the same instant an open platform elevator passes the 5 m level, moving upward with a constant velocity of 2 m/s. Determine ( g = 9.8 m/ s2 ) (a) when and where the ball will meet the elevator, (b) the relative velocity of the ball with respect to the elevator when the ball hits the elevator.

39. An automobile and a truck start from rest at the same instant, with the automobile initially at some distance behind the truck. The truck has a constant acceleration of 2.2 m/ s2 and the automobile has an acceleration of 3.5 m/ s2. The automobile overtakes the truck when it (truck) has moved 60 m. (a) How much time does it take the automobile to overtake the truck ? (b) How far was the automobile behind the truck initially ? (c) What is the speed of each during overtaking ?

40. Given| vbr| = 4 m/ s = magnitude of velocity of boatman with respect to river, vr = 2 m/ s in the direction shown. Boatman wants to reach from point A to point B. At what angle θ should he row his boat? B

River

θ

45° A

41. An aeroplane has to go from a point P to another point Q, 1000 km away due north. Wind is blowing due east at a speed of 200 km/h. The air speed of plane is 500 km/h. (a) Find the direction in which the pilot should head the plane to reach the point Q. (b) Find the time taken by the plane to go from P to Q.

42. A train stopping at two stations 4 km apart takes 4 min on the journey from one of the station to the other. Assuming that it first accelerates with a uniform acceleration x and then that of 1 1 + = 2. x y

uniform retardation y, prove that

LEVEL 2 Objective Questions Single Correct Option 1. When a man moves down the inclined plane with a constant speed 5 ms−1which makes an angle of 37° with the horizontal, he finds that the rain is falling vertically downward. When he moves up the same inclined plane with the same speed, he finds that the rain makes an angle  7 θ = tan−1   with the horizontal. The speed of the rain is  8 (a) 116 ms −1 (c) 5 ms −1

(b) (d)

32 ms −1 73 ms −1

200 — Mechanics - I dv = − 4v + 8, where v is the velocity in ms−1 and t is the time in dt second. Initial velocity of the particle was zero. Then,

2. Equation of motion of a body is (a) (b) (c) (d)

the initial rate of change of acceleration of the particle is 8 ms −2 the terminal speed is 2 ms −1 Both (a) and (b) are correct Both (a) and (b) are wrong

3. Two particles A and B are placed in gravity free space at ( 0, 0, 0) m and ( 30, 0, 0) m respectively.

Particle A is projected with a velocity ( 5i$ + 10$j + 5k$ ) ms−1, while particle B is projected with a velocity (10 i$ + 5$j + 5k$ ) ms−1 simultaneously. Then,

(a) (b) (c) (d)

they will collide at (10, 20, 10) m they will collide at (10, 10, 10) m they will never collide they will collide at 2 s

4. Velocity of the river with respect to ground is given by v0. Width of the river is d. A swimmer swims (with respect to water) perpendicular to the current with acceleration a = 2t (where t is time) starting from rest from the origin O at t = 0. The equation of trajectory of the path followed by the swimmer is Y v0 d X

0

(a) y =

x3 3v30

(b) y =

x2 2v02

(c) y =

x v0

(d) y =

x v0

5. The relation between time t and displacement x is t = αx 2 + βx, where α and β are constants. The retardation is (a) 2 αv3 (c) 2 αβv3

(b) 2 βv3 (d) 2 β 2v3

6. A street car moves rectilinearly from station A to the next station B (from rest to rest) with an

acceleration varying according to the law f = a − bx, where a and b are constants and x is the distance from station A. The distance between the two stations and the maximum velocity are

2a a , vmax = b b a b (c) x = , vmax = 2b a

b , vmax = 2a a (d) x = , vmax = b

(a) x =

(b) x =

a b a b

7. A particle of mass m moves on positive x-axis under the influence of force acting towards the

origin given by − kx 2$i. If the particle starts from rest at x = a, the speed it will attain when it crosses the origin is

(a)

k ma

(b)

2k ma

(c)

ma 2k

(d) None of these

Kinematics — 201

Chapter 6

v

8. A particle is moving along a straight line whose velocity-displacement

graph is as shown in the figure. What is the magnitude of acceleration when displacement is 3 m ? 4 ms–1

(a) 4 3 ms −2 (c)

3 ms

(b) 3 3 ms −2 4 (d) ms −2 3

−2

60°

3m

s

9. A particle is falling freely under gravity. In first t second it covers distance x1 and in the next t second, it covers distance x2, then t is given by (a)

x2 − x1 g

(b)

x2 + x1 g

(c)

2 (x2 − x1 ) g

(d)

10. A rod AB is shown in figure. End A of the rod is fixed on the

B

ground. Block is moving with velocity 2 ms −1 towards right. The velocity of end B of rod at the instant shown in figure is

(a) 3 ms −1 (c) 2 3 ms −1

(b) 2 ms −1 (d) 4 ms −1

2 (x2 + x1 ) g

A

v = 2 ms–1

30°

11. A thief in a stolen car passes through a police check post at his top speed of 90 kmh−1. A motorcycle cop, reacting after 2 s, accelerates from rest at 5 ms−2. His top speed being 108 kmh−1. Find the maximum separation between policemen and thief. (a) 112.5 m

(b) 115 m

(c) 116.5 m

(d) None of these

12. Anoop (A) hits a ball along the ground with a speed u in a direction which makes an angle 30° with the line joining him and the fielder Babul (B). Babul runs to intercept the ball with a speed 2u . At what angle θ should he run to intercept the ball ? 3 A

30°

B θ

u 2u/3

 3 (a) sin −1    2 

2 (b) sin −1   3 

3 (c) sin −1   4 

4 (d) sin −1   5 

13. A car is travelling on a straight road. The maximum velocity the car can attain is 24 ms−1. The maximum acceleration and deceleration it can attain are 1 ms−2 and 4 ms−2 respectively. The shortest time the car takes from rest to rest in a distance of 200 m is, (a) 22.4 s

(b) 30 s

(c) 11.2 s

(d) 5.6 s

14. A car is travelling on a road. The maximum velocity the car can attain is 24 ms−1 and the maximum deceleration is 4 ms−2. If car starts from rest and comes to rest after travelling 1032 m in the shortest time of 56 s, the maximum acceleration that the car can attain is

(a) 6 ms −2 (c) 12 ms −2

(b) 1.2 ms −2 (d) 3.6 ms −2

15. Two particles are moving along two long straight lines, in the same plane with same speed equal to 20 cm/ s. The angle between the two lines is 60° and their intersection point is O. At a certain moment, the two particles are located at distances 3m and 4m from O and are moving towards O. Subsequently, the shortest distance between them will be (a) 50 cm (c) 50 2 cm

(b) 40 2 cm (d) 50 3 cm

202 — Mechanics - I More than One Correct Options 1. A particle having a velocity v = v0 at t = 0 is decelerated at the rate|a| = α v , where α is a positive constant. 2 v0 α (b) The particle will come to rest at infinity (a) The particle comes to rest at t =

2v30/ 2 α 2v30/ 2 (d) The distance travelled by the particle before coming to rest is 3α (c) The distance travelled by the particle before coming to rest is

2. At time t = 0, a car moving along a straight line has a velocity of 16 ms−1. It slows down with an acceleration of − 0.5 t ms−2, where t is in second. Mark the correct statement (s).

(a) (b) (c) (d)

The direction of velocity changes at t = 8 s The distance travelled in 4 s is approximately 58.67 m The distance travelled by the particle in 10 s is 94 m The speed of particle at t = 10 s is 9 ms −1

3. An object moves with constant acceleration a. Which of the following expressions are also constant ? (a)

d | v| dt

(c)

d (v2) dt

d v (b)   dt   v d   | v| (d) dt

4. Ship A is located 4 km north and 3 km east of ship B. Ship A has a velocity of 20 kmh−1 towards the south and ship B is moving at 40 kmh−1 in a direction 37° north of east. X and Y -axes are along east and north directions, respectively (a) Velocity of A relative to B is (− 32 $i − 44 $j) km/h (b) Position of A relative to B as a function of time is given by rAB = [(3 − 32t )i$ + (4 − 44t )$j] km (c) Velocity of A relative to B is (32$i − 44$j) km/h (d) Position of A relative to B as a function of time is given by (32 t$i − 44 t$j) km

5. Starting from rest a particle is first accelerated for time t1 with constant acceleration a1 and then stops in time t2 with constant retardation a2. Let v1 be the average velocity in this case and s1 the total displacement. In the second case it is accelerating for the same time t1 with constant acceleration 2a1 and come to rest with constant retardation a2 in time t3 . If v2 is the average velocity in this case and s2 the total displacement, then

(a) v2 = 2 v1 (c) s2 = 2 s1

(b) 2 v1 < v2 < 4v1 (d) 2 s1 < s2 < 4s1

6. A particle is moving along a straight line. The displacement of the particle becomes zero in a certain time ( t > 0). The particle does not undergo any collision.

(a) (b) (c) (d)

The acceleration of the particle may be zero always The acceleration of the particle may be uniform The velocity of the particle must be zero at some instant The acceleration of the particle must change its direction

Chapter 6

Kinematics — 203

7. A particle is resting over a smooth horizontal floor. At t = 0, a horizontal force starts acting on it.

Magnitude of the force increases with time according to law F = αt, where α is a positive constant. From figure, which of the following statements are correct ? Y 2

1

X

O

(a) (b) (c) (d)

Curve 1 can be the plot of acceleration against time Curve 2 can be the plot of velocity against time Curve 2 can be the plot of velocity against acceleration Curve 1 can be the plot of displacement against time

8. A train starts from rest at S = 0 and is subjected to an acceleration as shown in figure. Then, a (ms–2) 6

S (m)

30

(a) (b) (c) (d)

velocity at the end of 10 m displacement is 20 ms −1 velocity of the train at S = 10 m is 10 ms −1 The maximum velocity attained by train is 180 ms −1 The maximum velocity attained by the train is 15 ms −1

9. For a moving particle, which of the following options may be correct? (a) |Vav| < vav

(b) |Vav| > vav

(c) Vav = 0 but vav ≠ 0

(d) Vav ≠ 0 but vav = 0

Here, Vav is average velocity and vav the average speed.

10. Identify the correct graph representing the motion of a particle along a straight line with constant acceleration with zero initial velocity. v

v

(a)

x

(b)

(c)

t 0

x

(d)

t 0

t 0

t 0

11. A man who can swim at a velocity v relative to water wants to cross a river of width b, flowing with a speed u. (a) The minimum time in which he can cross the river is

b v

(b) He can reach a point exactly opposite on the bank in time t = (c) He cannot reach the point exactly opposite on the bank if u > v (d) He cannot reach the point exactly opposite on the bank if v > u

b v − u2 2

if v > u

204 — Mechanics - I 12. The figure shows the velocity ( v ) of a particle plotted against time (t). (a) (b) (c) (d)

The particle changes its direction of motion at some point The acceleration of the particle remains constant The displacement of the particle is zero The initial and final speeds of the particle are the same

v T O

2T

t

13. The speed of a train increases at a constant rate α from zero to v and then remains constant for

an interval and finally decreases to zero at a constant rate β. The total distance travelled by the train is l. The time taken to complete the journey is t. Then,

(a) t =

l(α + β ) αβ

(b) t =

(c) t is minimum when v =

2lαβ (α − β )

l v  1 1 +  +  v 2  α β

(d) t is minimum when v =

2lαβ (α + β )

14. A particle moves in x-y plane and at time t is at the point ( t 2 , t3 − 2 t ), then which of the following is/are correct? (a) At t = 0, particle is moving parallel to y-axis (b) At t = 0, direction of velocity and acceleration are perpendicular 2 (c) At t = , particle is moving parallel to x-axis 3 (d) At t = 0, particle is at rest

15. A car is moving with uniform acceleration along a straight line between two stops X and Y . Its speed at X and Y are 2 ms−1 and 14 ms−1, Then

(a) (b) (c) (d)

its speed at mid-point of XY is 10 ms −1 its speed at a point A such that XA : AY = 1 : 3 is 5 ms −1 the time to go from X to the mid-point of XY is double of that to go from mid-point to Y the distance travelled in first half of the total time is half of the distance travelled in the second half of the time

Comprehension Based Questions Passage 1

(Q.Nos. 1 to 4)

An elevator without a ceiling is ascending up with an acceleration of 5 ms−2. A boy on the elevator shoots a ball in vertical upward direction from a height of 2 m above the floor of elevator. At this instant the elevator is moving up with a velocity of 10 ms−1 and floor of the elevator is at a height of 50 m from the ground. The initial speed of the ball is 15 ms−1 with respect to the elevator. Consider the duration for which the ball strikes the floor of elevator in answering following questions. ( g = 10 ms−2 )

1. The time in which the ball strikes the floor of elevator is given by (a) 2.13 s

(b) 2.0 s

(c) 1.0 s

(d) 3.12 s

2. The maximum height reached by ball, as measured from the ground would be (a) 73.65 m

(b) 116.25 m

(c) 82.56 m

(d) 63.25 m

3. Displacement of ball with respect to ground during its flight would be (a) 16.25 m

(b) 8.76 m

(c) 20.24 m

(d) 30.56 m

4. The maximum separation between the floor of elevator and the ball during its flight would be (a) 12 m

(b) 15 m

(c) 9.5 m

(d) 7.5 m

Chapter 6

Kinematics — 205

Passage 2 (Q.Nos. 5 to 7) A situation is shown in which two objects A and B start their motion from same point in same direction. The graph of their u velocities against time is drawn. u A and u B are the initial B velocities of A and B respectively. T is the time at which their velocities become equal after start of motion. You cannot use the data of one question while solving another question of the same uA set. So all the questions are independent of each other. 5. If the value of T is 4 s, then the time after which A will meet B is (a) 12 s (c) 8 s

Velocity of A

Velocity of B t T

(b) 6 s (d) data insufficient

6. Let vA and vB be the velocities of the particles A and B respectively at the moment A and B meet after start of the motion. If u A = 5 ms−1 and u B = 15 ms−1, then the magnitude of the difference of velocities vA and vB is

(a) 5 ms −1 (c) 15 ms −1

(b) 10 ms −1 (d) data insufficient

7. After 10 s of the start of motion of both objects A and B, find the value of velocity of A if u A = 6 ms−1 , u B = 12 ms−1 and at T velocity of A is 8 ms−1 and T = 4 s

(a) 12 ms −1 (c) 15 ms −1

(b) 10 ms −1 (d) None of these

Match the Columns 1. Match the following two columns : Column I

Column II

a

(p) speed must be increasing

(a) t

a t

(b)

(q) speed must be decreasing

s

(c)

t

(r) speed may be increasing

s

(d)

t

(s) speed may be decreasing

2. Match the following two columns : Column I

Column II

(a) v = − 2i$ , a = − 4$j (b) v = 2$i , a = 2i$ + 2$j

(p) speed increasing

(c) v = − 2i$ , a = + 2i$ (d) v = 2$i , a = − 2i$ + 2$j

(r) speed constant

(q) speed decreasing

(s) Nothing can be said

206 — Mechanics - I 3. The velocity-time graph of a particle moving along X-axis is shown in figure. Match the entries of Column I with the entries of Column II. v C B D

A

t

E

Column I (a)

Column II

For AB, particle is

(p) Moving in +ve X-direction with increasing speed

(b)

For BC, particle is

(q) Moving in +ve X-direction with decreasing speed

(c)

For CD, particle is

(r) Moving in −ve X-direction with increasing speed

(d)

For DE, particle is

(s) Moving in −ve X-direction with decreasing speed

4. Corresponding to velocity-time graph in one dimensional motion of a particle as shown in figure, match the following two columns. v (m/s) 10 4

6

8 t (s)

2 – 10

Column I

Column II

(a) Average velocity between zero second and 4 s

(p) 10 SI units

(b) Average acceleration between 1 s and 4 s

(q) 2.5 SI units

(c) Average speed between zero seccond and 6 s

(r) 5 SI units

(d) Rate of change of speed at 4 s

(s) None of the above

5. A particle is moving along x-axis. Its x-coordinate varies with time as : x = − 20 + 5t 2

For the given equation match the following two columns : Column I (a) Particle will cross the origin at

Column II (p) zero second

(b) At what time velocity and acceleration are equal

(q) 1 s

(c) At what time particle changes its direction of motion

(r) 2 s

(d) At what time velocity is zero

(s) None of the above

Chapter 6

Kinematics — 207

6. x and y-coordinates of a particle moving in x -y plane are, x = 1 − 2t + t 2 and y = 4 − 4t + t 2

For the given situation match the following two columns : Column I

Column II

(a) y-component of velocity when it crosses the y-axis

(p) + 2 SI unit

(b) x-component of velocity when it crosses the x-axis

(q) − 2 SI units

(c)

(r) + 4 SI units

Initial velocity of particle

(d) Initial acceleration of particle

(s) None of the above

Subjective Questions 1. To test the quality of a tennis ball, you drop it onto the floor from a height of 4.00 m. It rebounds to a height of 2.00 m. If the ball is in contact with the floor for 12.0 ms, what is its average acceleration during that contact? Take g = 98 m/ s2.

2. The acceleration-displacement graph of a particle moving in a straight line is as shown in figure, initial velocity of particle is zero. Find the velocity of the particle when displacement of the particle is s = 12 m. a (m/s2) 4 2 2

8

10 12

s (m)

3. At the initial moment three points A, B and C are on a horizontal straight line at equal distances from one another. Point A begins to move vertically upward with a constant velocity v and point C vertically downward without any initial velocity but with a constant acceleration a. How should point B move vertically for all the three points to be constantly on one straight line. The points begin to move simultaneously.

4. A particle moves in a straight line with constant acceleration a. The displacements of particle from origin in times t1 , t2 and t3 are s1 , s2 and s3 respectively. If times are in AP with common ( s1 − s3 )2 difference d and displacements are in GP, then prove that a = . d2

5. A car is to be hoisted by elevator to the fourth floor of a parking garage, which is 14 m above the ground. If the elevator can have maximum acceleration of 0.2 m/ s2 and maximum deceleration of 0.1 m/ s2and can reach a maximum speed of 2.5 m/s, determine the shortest time to make the lift, starting from rest and ending at rest.

6. To stop a car, first you require a certain reaction time to begin braking; then the car slows under the constant braking deceleration. Suppose that the total distance moved by your car during these two phases is 56.7 m when its initial speed is 80.5 km/h and 24.4 m when its initial speed is 48.3 km/h. What are (a) your reaction time and (b) the magnitude of the deceleration?

208 — Mechanics - I 7. An elevator without a ceiling is ascending with a constant speed of 10 m/s. A boy on the elevator shoots a ball directly upward, from a height of 2.0 m above the elevator floor. At this time the elevator floor is 28 m above the ground. The initial speed of the ball with respect to the elevator is 20 m/s. (Take g = 9.8 m/ s2 ) (a) What maximum height above the ground does the ball reach? (b) How long does the ball take to return to the elevator floor?

8. A particle moves along a straight line and its velocity depends on time as v = 3t − t 2.Here, v is in m/s and t in second. Find (a) average velocity and (b) average speed for first five seconds.

9. The acceleration of particle varies with time as shown. a (m/s2)

t (s)

1

–2

(a) Find an expression for velocity in terms of t. (b) Calculate the displacement of the particle in the interval from t = 2 s to t = 4 s. Assume that v = 0 at t = 0.

10. A man wishes to cross a river of width 120 m by a motorboat. His rowing speed in still water is 3 m/s and his maximum walking speed is 1 m/s. The river flows with velocity of 4 m/s. (a) Find the path which he should take to get to the point directly opposite to his starting point in the shortest time. (b) Also, find the time which he takes to reach his destination.

11. The current velocity of river grows in proportion to the distance from its bank and reaches the maximum value v0 in the middle. Near the banks the velocity is zero. A boat is moving along the river in such a manner that the boatman rows his boat always perpendicular to the current. The speed of the boat in still water is u. Find the distance through which the boat crossing the river will be carried away by the current, if the width of the river is c. Also determine the trajectory of the boat.

12. The v-s graph for an airplane travelling on a straight runway is shown. Determine the acceleration of the plane at s = 50 m and s = 150 m. Draw the a-s graph. v (m/s) 50 40

100

200

s (m)

Chapter 6

Kinematics — 209

13. A river of width a with straight parallel banks flows due north with speed u. The points O and A are on opposite banks and A is due east of O. Coordinate axes Ox and Oy are taken in the east and north directions respectively. A boat, whose speed is v relative to water, starts from O and v crosses the river. If the boat is steered due east and u varies with x as : u = x( a − x ) 2 . Find a (a) (b) (c) (d)

equation of trajectory of the boat, time taken to cross the river, absolute velocity of boatman when he reaches the opposite bank, the displacement of boatman when he reaches the opposite bank from the initial position.

14. A river of width ω is flowing with a uniform velocity v. A boat starts moving from point P also with velocity v relative to the river. The direction of resultant velocity is always perpendicular to the line joining boat and the fixed point R. Point Q is on the opposite side of the river. P, Q and R are in a straight line. If PQ = QR = ω , find (a) the trajectory of the boat, (b) the drifting of the boat and (c) the time taken by the boat to cross the river. R Q

P

15. The v-s graph describing the motion of a motorcycle is shown in figure. Construct the a-s graph

of the motion and determine the time needed for the motorcycle to reach the position s = 120 m. Given ln 5 = 1.6. v (m/s)

15

3 60

120

s(m)

16. The jet plane starts from rest at s = 0 and is subjected to the acceleration shown. Determine the speed of the plane when it has travelled 60 m. a (m/s2) 22.5

150

s(m)

17. A particle leaves the origin with an initial velocity v = ( 3.00 $i ) m/s and a constant acceleration a = ( −1.00 $i − 0.500 $j ) m/ s2. When the particle reaches its maximum x coordinate, what are (a) its velocity and

(b) its position vector?

18. The speed of a particle moving in a plane is equal to the magnitude of its instantaneous velocity,

210 — Mechanics - I v = | v|= vx2 + vy2 . dv = (vx a x + vy a y ) / vx2 + vy2 . dt dv (b) Show that the rate of change of speed can be expressed as = v ⋅ a /v , and use this result to dt dv is equal to a t the component of a that is parallel to v. explain why dt (a) Show that the rate of change of the speed is

19. A man with some passengers in his boat, starts perpendicular to flow of river 200 m wide and flowing with 2 m/s. Speed of boat in still water is 4 m/s. When he reaches half the width of river the passengers asked him that they want to reach the just opposite end from where they have started. (a) Find the direction due which he must row to reach the required end. (b) How many times more time, it would take to that if he would have denied the passengers?

20. A child in danger of drowning in a river is being carried downstream by a current that flows uniformly at a speed of 2.5 km/h. The child is 0.6 km from shore and 0.8 km upstream of a boat landing when a rescue boat sets out. If the boat proceeds at its maximum speed of 20 km/h with respect to the water, what angle does the boat velocity v make with the shore? How long will it take boat to reach the child?

21. A launch plies between two points A and B on the opposite banks of a

B

river always following the line AB. The distance S between points A and B is 1200 m. The velocity of the river current v = 1.9 m/ s is constant over the entire width of the river. The line AB makes an angle α = 60° with the direction of the current. With what velocity u and at what angle β to the line AB should the launch move to cover the distance AB and back in a time t = 5 min? The angle β remains the same during the passage from A to B and from B to A.

u v β α A

22. The slopes of wind screen of two cars are α1 = 30° and α2 = 15° respectively. At what ratio v1 / v2 of the velocities of the cars will their drivers see the hail stones bounced back by the wind screen on their cars in vertical direction? Assume hail stones fall vertically downwards and collisions to be elastic.

23. A projectile of mass m is fired into a liquid at an angle θ 0 with an initial

y

velocity v0 as shown. If the liquid develops a frictional or drag resistance on the projectile which is proportional to its velocity, i.e. F = − kv where k is a positive constant, determine the x and y components of its velocity at any instant. Also find the maximum distance xmax that it travels?

24. A man in a boat crosses a river from point A. If he rows

perpendicular to the banks he reaches point C ( BC = 120 m ) in 10 min. If the man heads at a certain angle α to the straight line AB (AB is perpendicular to the banks) against the current he reaches point B in 12.5 min. Find the width of the river w, the rowing velocity u, the speed of the river current v and the angle α. Assume the velocity of the boat relative to water to be constant and the same magnitude in both cases.

v0 θ0

B

x

C

v α A

w

Answers Introductory Exercise 6.1 1. Both downwards

2. (a) −2 m2 /s3 , (b) obtuse, (c) decreasing

Introductory Exercise 6.2 1. One dimensional with constant acceleration 3. No

2. Two dimensional with non-uniform acceleration

Introductory Exercise 6.3 π 2 2 cm/s, cm/s 15 15 5. (a) Yes, in uniform circular motion (b) No, yes (projectile motion), yes 6. (a) 25.13 s (b) 1 cm/s, 0.9 cm/s, 0.23 cm/s 2

1. False

2. True

3. g (downwards)

4.

Introductory Exercise 6.4 1. 5.2 m/s

2.

11 m/s 3

Introductory Exercise 6.5 2. See the hints 3. Always g 4. Acceleration 1 7. True 6. u + at 2 9. (a) 6.0 m, (b) 10 s, (c) 50 m 10. 125 m, (b) 5 s, (c) approximately 35 m/s

5. 60 m, 100 m 8. 25 m/s (downwards)

Introductory Exercise 6.6 1. (a) 1 m/s 2

(b) 43.5m

3. (a) x = 1.0m, v = 4 m/s, a = 8 m/s 2 , (b) zero

2. (a) 60 cm/s 2 , (b) 1287 cm 4. (a) x = 2.0 m

(b) zero (c) 26 ms −2

5. s ∝ t 7 /4 and a ∝ t −1/4

Introductory Exercise 6.7 1. 2 7 m/s, 4 3 m

2. (2$j) m / s 2 , (2i$ + $j) m, yes

$ 7 1 3. v = (3$i + j ) m/s, co-ordinates =  m, m 3 4 

Introductory Exercise 6.8 1. (a) Particle A starts at t = 0 from x = 10 m. Particle B starts at t = 4 s from x = 0. (b) vA = + 2.5 m/s , vB = + 7.5 m/s, (c) They strike at x = 30 m and t = 8 s 2. 80 m, 2.5 m/s 2 3. (a) 0.6 m/s 2 , (b) 50 m, (c) 50 m 4. (a) 10 m/s, (b) 20 m/s, zero, 20 m/s, −20 m/s 5. 100 m, zero

212 — Mechanics - I Introductory Exercise 6.9 1. −2 m/s

2. zero

1 4. (a) sin−1   east of the line AB  15 

3. (a) 40 s (b) 50 min

(b) 80 m

5. (a) 200 m, (b) 20 m/min, (c) 12 m/min

6. (a) 10 s, (b) 50 m

Exercises LEVEL 1 Assertion and Reason 1.(d)

2.(d)

11.(a)

12.(d)

3.(a)

4.(d)

5.(c)

6.(d)

7.(d)

8.(d)

9.(d)

10.(a or b)

Single Correct Option 1.(b)

2.(c)

3.(d)

4.(a)

5.(a)

6.(b)

7.(a)

8.(c)

9.(c)

10.(a)

11.(b)

12.(c)

13.(c)

14.(b)

15.(b)

16.(a)

17.(c)

18.(b)

19.(d)

20.(a)

21.(b)

22.(b)

23.(d)

24.(b)

25.(b)

26.(c)

27.(b)

28.(b)

29.(b)

30.(b)

Subjective Questions 1. (a) Magnitude of total acceleration and tangential acceleration, (b) equal in 1-D motion 2. (a) x 2 = 4 y (b) (2$i + 2t$j) units (c) (2$j) units 3. (a) 1100 m, (b) 500 m, (c) 55 m/min, (d) 25 m/min, 4. (a) 36 km (b) 1 min 5. (a) 16.67 ms −1 (b) 10 ms −1 (downwards) 7. (a) zero (b) 8 ms −1 (c) 8 ms −1 9. 8 m

10. 1.5 s

6. T = 4 t 0 21 3 3 ms −1 (b) ms −2 44 2 v t 11. 0 + 0 g 2 8. (a)

12. (a) positive, positive, positive, negative (b) positive, zero, negative, negative 50 14. 36 s, No 13. (a) ms −1 (b) 10 ms −1 7 15. vt1 , at1 and at 2 are positive while vt 2 is negative 16. 2 s, 6 s, 2(2 +

7) s

17. See the hints 18. (a) −5 ms −2 (b) 90 ms −1 $ $ 19. (a) (125 . $i + 0.5 j ) ms −1(b) (− 0.5$i + j ) ms −2 20. 45 m 21. (3.414) t 0 23. (a) 5 ms

−1

22. (a) 16.25 m (b) 1.8 s (b)1.67 ms

−2

(c) 7.5 m

25. (a) (8 ^i − 8 ^j) ms −1 (b) (18 m , − 4m) 27. − 30 ms

−2

29. 0.603 ms −2

(c) No

24. 0.2 ms −2 , 0.8 ms −1 26. (a) 45 m

(b) 22 ms −1

28. 14.125 m, 1.75 ms −1, 4.03 ms −1 20  −2 30.  −  ms  3 

Chapter 6 31. 5

4

10

8

t (s)

16

2

–5

a (ms–2)

32.

a (ms–2)

20

33.

t (s)

s (m) 4 6 8 10 12 14

t (s)

–60 –80 –100 –120

t (s)

–10

14 4 6 8 12

–10

s (m) 30 20 10

Kinematics — 213

34.

KE

Speed (m/s)

Velocity (m/s) 40

2h t0 = g

mgh

20 40 4

20 t0

35. (2 +

2t0

6

8 time (s)

–20

t

4

6

8

time (s)

3) s

36. (a) 5 ms −2 , zero, 5 ms −2 37. (a) zero (b) 25 ms

−1

(b) s = 30 m (c) (i) s = 10 + 2.5 t 2

(c) − 25 ms

−1

(ii) s = 40 + 10 (t − 4) − 2.5 (t − 4)2

(d) 0.8 s

38. (a) 3.65 s, at 12.30 m level (b) 19.8 ms −1 (downwards) 39. (a) 7.39 s (b) 35.5 m (c) automobile 25.9 ms −1, truck 16.2 ms −1 1  40. 45°− sin−1   ≈ 24.3°  2 2 10 41. (a) at an angle θ = sin−1(0.4) west of north (b) h 21

LEVEL 2 Single Correct Option 1.(b)

2.(b)

3.(c)

4.(a)

5.(a)

11.(a)

12.(c)

13.(a)

14.(b)

15.(d)

6.(a)

7.(d)

8.(a)

9.(a)

10.(d)

6.(b,c)

7.(a,b)

8.(b,c)

9.(a,c)

10.(a,d)

6.(b)

7.(d)

More than One Correct Options 1.(a,d) 11.(a,b,c)

2.(all) 12.(all)

3.(b)

4.(a,b)

13.(b,d)

14.(a,b,c)

5.(a,d) 15.(a,c)

Comprehension Based Questions 1.(a)

2.(c)

3.(d)

4.(c)

5.(c)

214 — Mechanics - I Match the Columns 1. (a) → r,s

(b) → r,s

(c) → p

(d) → q

2. (a) → r

(b) → p

(c) → q

(d) → q

3. (a) → p

(b) → p

(c) → q

(d) → r

4. (a) → r 5. (a) → r 6. (a) → q

(b) → s (b) → q (b) → p

(c) → r (c) → s (c) → s

(d) → r (d) → p (d) → s

Subjective Questions 2. 4 3 ms −1

1. 126 . × 103 ms −2 (upward)

v a and downward acceleration − 5. 20.5 s 2 2 7. (a) 76 m (b) 4.2 s 8. (a) − 0.833 ms −1 (b) 2.63 ms −1

3. B moves up with initial velocity 6. (a) 0.74 s (b) 6.2 ms −2

10. (a) 90° + sin−1(3/5) from river current (b) 2 min 40 s

9. (a) v = t 2 − 2t (b) 6.67 m 11.

cv0 ucx , y2 = 2u v0

12. 8 ms −2 , 4.5 ms −2 , For the graph see the hints ^

x2 x3 a aj ^ (b) (c) v (due east) (d) a i + − 2a 3a 2 v 6 1.317 ω 14. (a) circle (b) 3 ω (c) 15. 12.0 s, For the graph see the hints v 13. (a) y =

^

16. 46.47 ms −1

^

17. (a) (− 15 . j ) ms −1 (b) (4.5 i − 2.25 j ) m ^

1 19. (a) At an angle (90° + 2 θ) from river current (upstream). Here : θ = tan−1   2 v 22. 1 = 3 20. 37°, 3 min 21. u = 8 ms −1, β = 12° v2 23. vx = v0 cos θ 0 e − kt/m , vy =

m k

 k  −   v0 sinθ 0 + g  e    m

24. 200 m, 20 m /min, 12 m /min, 36°50.

kt m

 mv cos θ 0 − g , xm = k 

(b)

4 3

07 Projectile Motion Chapter Contents 7.1

Introduction

7.2 Projectile Motion 7.3

Two Methods of Solving a Projectile Motion

7.4

Time of Flight, Maximum Height and Horizontal Range of a Projectile

7.5 Projectile Motion along an Inclined Plane 7.6 Relative Motion between Two Projectiles

7.1 Introduction Motion of a particle under constant acceleration is either a straight line (one-dimensional) or parabolic (two-dimensional). Motion is one dimensional under following three conditions : (i) Initial velocity of the particle is zero. (ii) Initial velocity of the particle is in the direction of constant acceleration (or parallel to it). (iii) Initial velocity of the particle is in the opposite direction of constant acceleration (or antiparallel to it). For small heights acceleration due to gravity (g) is almost constant. The three cases discussed about are as shown in the Fig. 7.1. u=0 g

g

u g u

Case-(i)

Case-(ii)

Case-(iii)

Fig. 7.1

In all other cases when initial velocity is at some angle (≠ 0° or 180°) with constant acceleration, motion is parabolic as shown below. u g θ

Fig. 7.2

This motion under acceleration due to gravity is called projectile motion.

7.2 Projectile Motion As we have seen above, projectile motion is a two-dimensional motion (or motion in a plane) with constant acceleration (or acceleration due to gravity for small heights). The different types of projectile motion are as shown below. u u u

(a)

(b)

(c)

Chapter 7

Projectile Motion — 217

u

u u

(d)

(e)

(f)

Fig. 7.3

The plane of the projectile motion is a vertical plane.

7.3 Two Methods of Solving a Projectile Motion Every projectile motion can be solved by either of the following two methods: Method 1 Projectile motion is a two dimensional motion with constant acceleration. Therefore, we can use the equations 1 v = u + at and s = ut + at 2 2 For example, in the shown figure ^ Vertical (j)

u

α ^ Horizontal (i)

O g

Fig. 7.4

u = u cos α$i + u sin α$j and

a = − g$j

Now, suppose we want to find velocity at time t. v = u + at = ( u cos α$i + u sin α$j ) − gt$j or v = u cos α$i + ( u sin α − gt ) $j Similarly, displacement at time t will be 1 2 1 at = ( u cos α$i + u sin α$j ) t − gt 2 $j 2 2 1   = ut cos α$i +  ut sin α − gt 2  $j   2

s = ut +

Note In all problems, value of a ( = g ) will be same only u will be different.

218 — Mechanics - I V

Example 7.1 A particle is projected with a velocity of 50 m/s at 37° with horizontal. Find velocity, displacement and co-ordinates of the particle (w.r.t. the starting point) after 2 s. Given, g = 10 m/ s 2 , sin 37° = 0.6 and cos 37° = 0.8 Solution In the given problem, u = ( 50cos 37° )$i + ( 50sin 37° )$j = ( 40i$ + 30$j ) m/s

y

a = ( − 10$j ) m/s 2 50 m/s

t = 2s v = u + at = ( 40$i + 30$j ) + ( − 10$j ) ( 2) = ( 40$i + 10$j ) m/s

g = 10 m/s2

37°

x

O

Ans.

Fig. 7.5

1 s = u t + at 2 2 1 = ( 40i$ + 30$j ) ( 2) + ( − 10$j ) ( 2) 2 2 = ( 80i$ + 40$j )m Coordinates of the particle are x = 80 m and V

y = 40 m

Ans.

Example 7.2 A particle is projected with velocity u at angle θ with horizontal. Find the time when velocity vector is perpendicular to initial velocity vector. Solution

y

u

90° v

θ

x

O

Fig. 7.6

Given, v ⊥ u ⇒ v⋅ u = 0 ⇒ ( u + at ) ⋅ u = 0 Substituting the proper values in Eq. (i), we have [{( u cos θ )$i + ( u sin θ )$j} + ( − g$j ) t ] ⋅ [( u cos θ )$i + ( u sin θ )$j] = 0 ⇒ ⇒

u 2 cos 2 θ + u 2 sin 2 θ − ( ug sin θ )t = 0 u 2 (sin 2 θ + cos 2 θ ) = ( ug sin θ ) t

……(i)

Chapter 7

Projectile Motion — 219

Solving this equation, we get t=

u cosec θ u = g sin θ g

Ans.

Alternate method a=g Angle between u and a is α = 90° + θ Now, Eq. (i) can be written as

u θ

u ⋅ u + u ⋅ at = 0 or

u 2 + ( ug cos α )t = 0

or

u 2 + [ ug cos ( 90° + θ )] t = 0

α a=g

Fig. 7.7

Solving this equation, we get t=

u cosec θ u = g sin θ g

Ans.

In this method, select two mutually perpendicular directions x and y and find the two components of initial velocity and acceleration along these two directions, i.e. find ux , u y , a x and a y . Now apply the appropriate equation (s) of the following six equations :  v x = ux + a x t  1 sx = ux t + a x t 2 → Along x - axis 2  2 2 v x = ux + 2a x sx 

Method 2

  1 2 s y = u y t + a y t → Along y - axis  2  2 2 v y = u y + 2a y s y  v y = uy + a y t

and

Substitute v x , ux , a x , sx ,v y , u y , a y and s y with proper signs but choosing one direction as positive and 1 other as the negative along both axes. In most of the problems s = ut + at 2 equation is useful for time 2 calculation. Under normal projectile motion, x-axis is taken along horizontal direction and y-axis along vertical direction. In projectile motion along an inclined plane, x-axis is normally taken along the plane and y-axis perpendicular to the plane. Two simple cases are shown below. y

y

x u

a=g u

α

θ O

x (a)

β β a=g (b)

Fig. 7.8

220 — Mechanics - I In Fig. 7.8 (a) ux = u cos θ, u y = u sin θ, a x = 0, a y = − g In Fig. 7.8 (b) ux = u cos α, u y = u sin α, a x = − g sin β, a y = − g cos β V

Example 7.3 A projectile is fired horizontally with velocity of 98 m/s from the top of a hill 490 m high. Find (a) the time taken by the projectile to reach the ground, (b) the distance of the point where the particle hits the ground from foot of the hill and (c) the velocity with which the projectile hits the ground. (g = 9.8 m/s 2 ) Solution Here, it will be more convenient to choose x and y directions as shown in figure. Here, u x = 98 m/s, a x = 0, u y = 0 and a y = g (a) At A, s y = 490 m . So, applying sy = u y t + ∴

490 = 0 +

∴

t = 10s

(b) BA = sx = u x t +

1 2

1 2

1 2

O

u = 98 m/s x

y

A

B

vx β

vy

Fig. 7.9

ayt 2

(9.8) t 2 Ans.

ax t 2

or

BA = ( 98)(10) + 0

or

BA = 980 m

Ans.

(c) v x = u x + a x t = 98 + 0 = 98 m/s v y = u y + a y t = 0 + (9.8) (10) = 98 m/s ∴ and ∴

v = v x2 + v 2y = ( 98) 2 + ( 98) 2 = 98 2 m/s tan β =

vy vx

=

98 =1 98

β = 45°

Thus, the projectile hits the ground with velocity 98 2 m/s at an angle of β = 45° with horizontal as shown in Fig. 7.9. V

Example 7.4 A body is thrown horizontally from the top of a tower and strikes the ground after three seconds at an angle of 45° with the horizontal. Find the height of the tower and the speed with which the body was projected. Take g = 9.8 m/ s 2 .

Chapter 7 Solution

Projectile Motion — 221

As shown in the figure of Example 7.3. u y = 0 and a y = g = 9.8 m/s 2 , sy = u y t + 1 a y t 2 2 s y = 0 × 3 + 1 × 9.8 × ( 3) 2 2 = 44.1m

Thus, height of the tower is 44.1 m. Further, v y = u y + a y t = 0 + ( 9.8) ( 3) = 29.4 m/s As the resultant velocity v makes an angle of 45° with the horizontal, so vy 29.4 or 1= tan 45° = vx vx v x = 29.4 m/s vx = u x + a x t ⇒ 29.4 = u x + 0 or u x = 29.4 m/s Therefore, the speed with which the body was projected (horizontally) is 29.4 m/s.

INTRODUCTORY EXERCISE

7.1

1. Two particles are projected from a tower horizontally in opposite directions with velocities 10 m/s and 20 m/s. Find the time when their velocity vectors are mutually perpendicular. Take g = 10 m /s 2.

2. Projectile motion is a 3-dimensional motion. Is this statement true or false? 3. Projectile motion (at low speed) is uniformly accelerated motion. Is this statement true or false? 4. A particle is projected from ground with velocity 50 m/s at 37° from horizontal. Find velocity and 3 displacement after 2 s. sin 37° = . 5

5. A particle is projected from a tower of height 25 m with velocity 20 2 m /s at 45°. Find the time when particle strikes with ground. The horizontal distance from the foot of tower where it strikes. Also find the velocity at the time of collision.

Note

In question numbers 4 and 5, $i is in horizontal direction and $j is vertically upwards.

7.4 Time of Flight, Maximum Height and Horizontal

Range of a Projectile Fig. 7.10 shows a particle projected from the point O with an initial velocity u at an angle α with the horizontal. It goes through the highest point A and falls at B on the horizontal surface through O. The point O is called the point of projection, the angle α is called the angle of projection, the distance OB is

222 — Mechanics - I called the horizontal range (R) or simply range and the vertical height AC is called the maximum height ( H ). The total time taken by the particle in describing the path OAB is called the time of flight ( T ). y g

A

u

H

α

B

O

C R

x

Fig. 7.10

Time of Flight (T ) Refer Fig. 7.10. Here, x and y-axes are in the directions shown in figure. Axis x is along horizontal direction and axis y is vertically upwards. Thus, ux = u cos α, u y = u sin α, a x = 0 and ay = − g At point B, s y = 0. So, applying 1 s y = u y t + a y t 2 , we have 2 1 0 = ( u sin α ) t − gt 2 2 2u sin α ∴ t = 0, g 2u sin α correspond to the situation where s y = 0.The time t = 0 corresponds to point g 2u sin α O and time t = corresponds to point B. Thus, time of flight of the projectile is g

Both t = 0 and t =

T = tOAB

or

T=

2u sin α g

Maximum Height (H ) At point A vertical component of velocity becomes zero, i.e. v y = 0. Substituting the proper values in v y2 = u y2 + 2a y s y we have, ∴

0 = ( u sin α ) 2 + 2( −g )(H ) H=

u 2 sin 2 α 2g

Projectile Motion — 223

Chapter 7

Horizontal Range (R ) Distance OB is the range R. This is also equal to the displacement of particle along x-axis in time t = T . Thus, applying sx = ux t +

1 2

a x t 2 , we get

 2u sin α  R = ( u cos α )   +0  g 

t =T =

a x = 0 and

as

2u sin α g

2u 2 sin α cos α u 2 sin 2 α u 2 sin 2 α or = R= g g g Following are given two important points regarding the range of a projectile (i) Range is maximum where sin 2 α = 1 or α = 45° and this maximum range is u2 (at α = 45° ) R max = g (ii) For given value of u range at α and range at 90° − α are equal, although times of flight and maximum heights may be different. Because y u 2 sin 2 ( 90° − α ) R 90° − α = g

∴

R=

=

u 2 sin (180° − 2α ) g

=

u sin 2α = Rα g

u u

2

30°

x

O

R 30° = R 60° R 20° = R 70°

So, or This is shown in Fig. 7.11.

60°

Fig. 7.11

Extra Points to Remember ˜

Formulae of T, H and R can be applied directly between two points lying on same horizontal line. u y O u

Q

α

M P

Q

Tower α O

P

M

x N

S

Fig. 7.12

For example, in the two projectile motions shown in figure, 2u sinα u 2 sin2 α u 2 sin2α , PQ = H = and OM = R = t OQM = T = g 2g g For finding t OQMS or distance NS method-2 discussed in article 7.3 is more useful.

224 — Mechanics - I ˜

˜

As we have seen in the above derivations that a x = 0, i.e. motion of the projectile in horizontal direction is uniform. Hence, horizontal component of velocity u cos α does not change during its motion. Motion in vertical direction is first retarded then accelerated in opposite direction. Because u y is upwards and ay is downwards. Hence, vertical component of its velocity first decreases from O to A and then increases from A to B. This can be shown as in Fig. 7.13. y A u uy O

α ux

B

x

Fig. 7.13 ˜

The coordinates and velocity components of the projectile at time t are x = s x = u xt = (u cos α ) t 1 y = sy = u y t + ay t 2 2 1 = (u sin α )t − gt 2 2 v x = u x = u cos α and

v y = u y + ay t = u sin α − gt

Therefore, speed of projectile at time t is v =

v 2x + v y2 and the angle made by its velocity vector with positive

x-axis is v  θ = tan−1  y   vx  ˜

Equation of trajectory of projectile ∴

x = (u cos α ) t x t = u cos α

Substituting this value of t in, y = (u sin α ) t −

1 2 gt , we get 2

y = x tan α − or

gx2 2u cos 2 α 2

y = x tan α − y = x tan α −

gx2 2u 2

gx2 2u 2

sec 2α

(1 + tan2 α )

These are the standard equations of trajectory of a projectile. The equation is quadratic in x. This is why the path of a projectile is a parabola. The above equation can also be written in terms of range (R) of projectile as: x y = x  1 –  tan α  R

Chapter 7 V

Example 7.5 Find the angle of projection of a projectile for which the horizontal range and maximum height are equal. R=H

Solution Given,

∴

u sin 2α u 2 sin 2 α = g 2g 2

2sin α cos α =

sin 2 α 2

tan α = 4

α = tan −1 ( 4 )

∴

Ans.

Example 7.6 Prove that the maximum horizontal range is four times the maximum height attained by the projectile; when fired at an inclination so as to have maximum horizontal range. Solution

V

or

sin α = 4 or cos α

or

V

Projectile Motion — 225

For θ = 45°, the horizontal range is maximum and is given by R max =

u2 g

Maximum height attained

H max =

u 2 sin 2 45° u 2 R max = = 2g 4g 4

or

R max = 4 H max

Proved.

Example 7.7 For given value of u, there are two angles of projection for which the horizontal range is the same. Show that the sum of the maximum heights for these two angles is independent of the angle of projection. Solution There are two angles of projection α and 90° − α for which the horizontal range R is same. Now, Therefore,

H1 =

u 2 sin 2 α 2g

and

H1 + H 2 = =

H2 =

u 2 sin 2 ( 90° − α ) u 2 cos 2 α = 2g 2g

u2 (sin 2 α + cos 2 α ) 2g u2 2g

Clearly the sum of the heights for the two angles of projection is independent of the angles of projection. V

Example 7.8 Show that there are two values of time for which a projectile is at the same height. Also show mathematically that the sum of these two times is equal to the time of flight. Solution

For vertically upward motion of a projectile, 1 2 1 y = ( u sin α ) t − gt 2 or gt − ( u sin α ) t + y = 0 2 2

226 — Mechanics - I This is a quadratic equation in t. Its roots are t1 = and

t2 =

u sin α − u 2 sin 2 α − 2gy g u sin α + u 2 sin 2 α − 2gy g

2u sin α t1 + t 2 = =T g

∴

INTRODUCTORY EXERCISE

(time of flight of the projectile)

7.2

1. A particle is projected from ground with velocity 40 2 m/s at 45°. Find (a) velocity and (b) displacement of the particle after 2 s. ( g = 10 m /s 2 )

2. Under what conditions the formulae of range, time of flight and maximum height can be applied directly in case of a projectile motion?

3. What is the average velocity of a particle projected from the ground with speed u at an angle α with the horizontal over a time interval from beginning till it strikes the ground again?

4. What is the change in velocity in the above question? 5. A particle is projected from ground with initial velocity u = 20 2 m /s at θ = 45° . Find (a) R, H and T, (b) velocity of particle after 1 s (c) velocity of particle at the time of collision with the ground (x-axis).

6. A particle is projected from ground at angle 45° with initial velocity

y u θ

x

Fig. 7.14

20 2 m /s. Find (a) change in velocity, (b) magnitude of average velocity in a time interval from t = 0 to t = 3 s.

7. The coach throws a baseball to a player with an initial speed of 20 m/s at an angle of 45° with the horizontal. At the moment the ball is thrown, the player is 50 m from the coach. At what speed and in what direction must the player run to catch the ball at the same height at which it was released? ( g = 10 m /s 2 )

8. A ball is thrown horizontally from a point 100 m above the ground with a speed of 20 m/s. Find (a) the time it takes to reach the ground, (b) the horizontal distance it travels before reaching the ground, (c) the velocity (direction and magnitude) with which it strikes the ground.

9. A bullet fired at an angle of 30° with the horizontal hits the ground 3.0 km away. By adjusting its angle of projection, can one hope to hit a target 5.0 away? Assume the muzzle speed to be fixed and neglect air resistance.

10. A particle moves in the xy-plane with constant acceleration a directed along the negative y-axis. The equation of path of the particle has the form y = bx − cx 2, where b and c are positive constants. Find the velocity of the particle at the origin of coordinates.

Projectile Motion — 227

Chapter 7

7.5 Projectile Motion along an Inclined Plane Here, two cases arise. One is up the plane and the other is down the plane. Let us discuss both the cases separately.

Up the Plane In this case direction x is chosen up the plane and direction y is chosen perpendicular to the plane. Hence, y

u

x

B A g sin β

β

α O

β

O

C

g cos β

g

β

Fig. 7.15

u x = u cos (α − β ) , a x = − g sin β u y = u sin (α − β) and a y = − g cos β Now, let us derive the expressions for time of flight (T) and range (R) along the plane.

Time of Flight At point B displacement along y-direction is zero. So, substituting the proper values in sy = u y t +

1 2

a y t 2 , we get 0 = ut sin (α − β) +

1 2

( −g cos β) t 2

t = 0, corresponds to point O and t = T=

⇒ ∴ t = 0 and

2u sin (α − β) g cos β

2u sin (α − β) corresponds to point B. Thus, g cos β 2u sin (α − β) g cos β

Note Substituting β = 0, in the above expression, we get T =

2u sin α which is quite obvious because β = 0 is g

the situation shown in Fig. 7.16. Y u α X

Fig. 7.16

228 — Mechanics - I Range Range (R) or the distance OB can be found by following two methods: Method 1 Horizontal component of initial velocity is uH = u cos α ∴ OC = uH T ( u cos α ) 2u sin (α − β) = g cos β 2 2u sin (α − β) cos α = g cos β OC ∴ R = OB = cos β 2u 2 sin (α − β) cos α = g cos 2 β C − D C + D Using, sin C − sin D = 2 sin   cos  ,  2   2  Range can also be written as, u2 R= [sin (2 α − β) − sin β] g cos 2 β

(as a H = 0)

This range will be maximum when 2α − β = and

R max =

π 2

or u2

g cos 2 β

α=

π β + 4 2

[1 − sin β]

π or α = 45° 4 u2 u2 = ( 1 − sin 0 ° ) = g g cos 2 0°

Here, also we can see that for β = 0, range is maximum at α = and

R max

Range (R) or the distance OB is also equal to the displacement of projectile along x-direction in time t = T . Therefore,

Method 2

R = sx = ux T +

1 2

a xT 2

Substituting the values of ux , a x and T, we get the same result. (ii) Down the Plane Here, x and y-directions are down the plane and perpendicular to plane respectively as shown in Fig. 7.17. Hence, ux = u cos (α + β), a x = g sin β u y = u sin (α + β), a y = − g cos β

Chapter 7

Projectile Motion — 229

Proceeding in the similar manner, we get the following results : T=

2u sin (α + β) u2 , R= [sin (2α + β) + sin β] g cos β g cos 2 β y u

x α β

g sin β

β

β

O

β

g cos β

g

Fig. 7.17

From the above expressions, we can see that if we replace β by −β, the equations of T and R for up the plane and down the plane are interchanged provided α (angle of projection) in both the cases is measured from the horizontal not from the plane. V

Example 7.9 A man standing on a hill top projects a stone horizontally with speed v0 as shown in figure. Taking the co-ordinate system as given in the figure. Find the co-ordinates of the point where the stone will hit the hill surface. y

v0

x

(0, 0)

θ

Fig. 7.18

Solution

Range of the projectile on an inclined plane (down the

plane) is,

q

R= Here, ∴

v0

u2 g cos 2 β

[sin ( 2α + β ) + sin β ]

u = v 0 , α = 0 and β = θ 2v 2 sin θ R= 0 2 g cos θ 2v 02 tan θ g

Now,

x = R cos θ =

and

y = − R sin θ = −

2v 02 tan 2 θ g

R

q

Fig. 7.19

230 — Mechanics - I

7.6 Relative Motion between Two Projectiles Let us now discuss the relative motion between two projectiles or the path observed by one projectile of the other. Suppose that two particles are projected from the ground with speeds u1 and u2 at angles α1 and α2 as shown in Fig. 7.20(a) and (b). Acceleration of both the particles is g downwards. So, relative acceleration between them is zero because y

y

u2

u1 α2

α1

x

O

(a)

x

g (b)

Fig. 7.20

a 12 = a 1 − a 2 = g − g = 0 i.e. the relative motion between the two particles is uniform. Now, u1x = u1 cos α 1 , u2x = u2 cos α 2 u1 y = u1 sin α 1 and u2 y = u2 sin α 2 Therefore, u12x = u1x − u2x = u1 cos α 1 − u2 cos α2 and u12 y = u1 y − u2 y = u1 sin α 1 − u2 sin α 2 u12x and u12 y are the x and y components of relative velocity of 1 with respect to 2. Hence, relative motion of 1 with respect to 2 is a straight line at an angle  u12 y   with positive x-axis. θ = tan −1   u12x 

y

u12y

u12 θ

O

u12x

a12 = 0 x

Fig. 7.21

Now, if u12x = 0 or u1 cos α 1 = u2 cos α 2 , the relative motion is along y-axis or in vertical direction (as θ = 90°). Similarly, if u12 y = 0 or u1 sin α 1 = u2 sin α 2 , the relative motion is along x-axis or in horizontal direction (as θ = 0°). Note Relative acceleration between two projectiles is zero. Relative motion between them is uniform. Therefore, condition of collision of two particles in air is that relative velocity of one with respect to the other should be along line joining them, i.e., if two projectiles A and B collide in mid air, then v AB should be along AB or v BA along BA. V

Example 7.10 A particle A is projected 50 m/s 60 m/s with an initial velocity of 60 m/s at an angle 30° to the horizontal. At the same time α a second particle B is projected in opposite A 30° B direction with initial speed of 50 m/s from a 100 m point at a distance of 100 m from A. If the Fig. 7.22 particles collide in air, find (a) the angle of projection α of particle B, (b) time when the collision takes place and (c) the distance of P from A, where collision occurs. ( g = 10 m/ s2 )

Chapter 7

Projectile Motion — 231

(a) Taking x and y-directions as shown in figure. Here, a A = − g$j

Solution

y

a B = − g$j u Ax = 60 cos 30° = 30 3 m/s

x

u Ay = 60 sin 30° = 30 m/s

uAB

u Bx = − 50cos α and

Fig. 7.23

u By = 50 sin α

Relative acceleration between the two is zero as a A = a B . Hence, the relative motion between the two is uniform. It can be assumed that B is at rest and A is moving with u AB . Hence, the two particles will collide, if u AB is along AB. This is possible only when u Ay = u By i.e. component of relative velocity along y-axis should be zero. or

30 = 50 sin α

∴

α = sin −1 ( 3/ 5) = 37°

Ans.

| u AB | = u Ax − u Bx

(b) Now,

= ( 30 3 + 50 cos α ) m/s =  30 3 + 50 × 

4  m/s 5

= ( 30 3 + 40) m /s Therefore, time of collision is t=

AB 100 = | u AB | 30 3 + 40 t = 1.09 s

or

Ans.

(c) Distance of point P from A where collision takes place is 1 d = ( u Ax t ) 2 +  u Ay t − gt 2    2

2

1 = ( 30 3 × 1.09) 2 +  30 × 1.09 − × 10 × 1.09 × 1.09   2 or

d = 62.64 m

2

Ans.

232 — Mechanics - I

INTRODUCTORY EXERCISE

7.3

1. Find time of flight and range of the projectile along the inclined plane as shown in figure. ( g = 10 m /s 2 ) 20 2 m/s

30° 45° Fig. 7.24

2. Find time of flight and range of the projectile along the inclined plane as shown in figure. ( g = 10 m /s 2 ) 20 2 m/s

45°

30°

Fig. 7.25

3. Find time of flight and range of the projectile along the inclined plane as shown in figure. ( g = 10 m /s 2 ) 20 m/s 30°

30°

Fig. 7.26

4. Passenger of a train just drops a stone from it. The train was moving with constant velocity. What is path of the stone as observed by (a) the passenger itself, (b) a man standing on ground?

5. A particle is projected upwards with velocity 20 m/s. Simultaneously another particle is projected with velocity 20 2 m /s at 45°. ( g = 10 m /s 2 ) (a) What is acceleration of first particle relative to the second? (b) What is initial velocity of first particle relative to the other? (c) What is distance between two particles after 2 s?

6. A particle is projected from the bottom of an inclined plane of inclination 30°. At what angle α (from the horizontal) should the particle be projected to get the maximum range on the inclined plane.

Chapter 7

Projectile Motion — 233

Final Touch Points 1. In projectile motion speed (and hence kinetic energy) is minimum at highest point. Speed = (cos θ ) times the speed of projection and

kinetic energy = (cos 2 θ ) times the initial kinetic energy θ = angle of projection

Here,

2. In projectile motion it is sometimes better to write the equations of H, R andT in terms of u x and u y as T =

2u y g

, H=

u y2 2g

and R =

2u x u y g

3. If a particle is projected vertically upwards, then during upward journey gravity forces (weight) and air drag both are acting downwards. Hence, |retardation| > |g|. During its downward journey air drag is upwards while gravity is downwards. Hence, acceleration < g.. Therefore we may conclude that, v

Gravity

Air drag Air drag

Gravity

v

ime t of ascent < time of descent

Exercise : In projectile motion, if air drag is taken into consideration than state whether the H, R and T will increase, decrease or remain same. Ans. T will increase, H will decrease and R may increase, decrease or remain same.

4. At the time of collision coordinates of particles should be same, i.e. x1 = x 2, Similarly

y1 = y 2

( for a 2 -D motion)

x1 = x 2, y1 = y 2 and z1 = z 2

(for a 3-D motion)

and

Two particles collide at the same moment. Of course their time of journeys may be different, i.e. they may start at different times (t1 and t 2 may be different). If they start together then t1 = t 2.

Solved Examples ‘

TYPED PROBLEMS Type 1. Based on the concept that horizontal component of velocity remains unchanged This type can be better understood by the following example. V

Example 1 A particle is projected from ground with velocity 40m/s at 60° from horizontal. (a) Find the speed when velocity of the particle makes an angle of 37° from horizontal. (b) Find the time for the above situation. (c) Find the vertical height and horizontal distance of the particle from the starting point in the above position. Take g = 10 m/ s2. Solution In the figure shown,

y

ux = 40 cos 60° = 20 m/s uy = 40 sin 60° = 20 3 m/s a x = 0 and a y = − 10 m/s 2 (a) Horizontal component of velocity remains unchanged. or ⇒ ∴ (b) Using, vy = uy + a y t ⇒ t =

vx = ux v cos 37° = 20 v (0.8) = 20 v = 25 m/s. vy − uy

40 m/s

t2 37°

h

v

60° x

O x1 x2

ay

t1 =

v sin 37° − 20 3 (25) (0.6) − 20 3 = = 1.96 s − 10 − 10

For t2,

t2 =

− v sin 37° − 20 3 − (25) (0.6) − 20 3 = = 4.96 s . − 10 − 10

∴

37°

h

For t1,

(c) Vertical height Let us calculate at t1

v

t1

h = sy

.

(at t1or t2)

1 1 a y t12 = (20 3 ) (1.96) + (− 10) (1.96)2 2 2 = 48.7 m.

h = uy t1 +

Horizontal distances

Similarly,

x1 = ux t1 = (20) (1.96) = 39.2 m x2 = ux t2 = (20) (4.96) = 99.2 m.

(as a x = 0)

Chapter 7

Projectile Motion — 235

Type 2. Situations where the formulae, H, R and T cannot be applied directly.

Concept As discussed earlier also, formulae of H, R and T can be applied directly between two points lying on the same horizontal line.

How to Solve? l

In any other situation apply component method (along x and y-axes). In most of the problems it is advisable 1 to first find the time, using the equation, s = ut + at 2 in vertical (or y) direction. 2

V

Example 2 In the figures shown, three particles are thrown from a tower of height 40 m as shown in figure. In each case find the time when the particles strike the ground and the distance of this point from foot of the tower. 20√2 m/s 20 m/s

45°

45° 40 m

40 m

40 m (i)

20√2 m/s (iii)

(ii)

Solution For time calculation, apply sy = uy t + in vertical direction. In all figures, In first figure In second figure, In third figure, For horizontal distance,

1 ay t2 2

s = − 40 m, a y = − 10 m/s 2 uy = + 20 2 cos 45° = + 20 m/s uy = 0 uy = − 20 2 cos 45° = − 20 m/s

x = ux t ux = 20 m/s in all cases Substituting the proper values and then solving we get, In first figure Time t1 = 5.46 s and horizontal distance x1 = 109.2 m In second figure Time t2 = 2.83 s and the horizontal distance x2 = 56.6 m In third figure Time t3 = 1.46 s and the horizontal distance x3 = 29.2 m

y x

236 — Mechanics - I Type 3. Horizontal projection of a projectile from some height

Concept O

O

u

u

h P

Q

v

θ

P

u

Q

vnet

v

θ

u vnet

After falling h Fig. (ii)

After time t Fig. (i)

Suppose a particle is projected from point O with a horizontal velocity ‘u’ as shown in two figures. Then, In Fig. (i) or after time t Suppose the particle is at point P, then Horizontal component of velocity = u Vertical component of velocity, v = gt If g = 10 m/ s, then v = 10 t (downwards) 1 Horizontal distance QP = ut and vertical height fallen OQ = gt 2 2 2 If g = 10 m/ s , then OQ = 5t 2 If Fig. (ii) or after falling a height ‘h’ Suppose the particle is at point P, then Horizontal component of velocity = u (downwards) Vertical component of velocity v = 2gh Time taken in falling a height h is 2h 1 2  t=  as h = g t    g 2 The horizontal distance QP = ut Note In both figures, net velocity of the particle at point P is, vnet = v 2 + u 2 and the angle θ of this net velocity with horizontal is v or tan θ = u

V

v θ = tan −1    u

Example 3 A ball rolls off the edge of a horizontal table top 4 m high. If it strikes the floor at a point 5 m horizontally away from the edge of the table, what was its speed at the instant it left the table?

Chapter 7 1 2 gt , we have 2 1 2 hAB = gtAC 2 2hAB tAC = g h=

Solution Using

or

=

A

v=

or

4m C

B 5m

2× 4 = 0.9 s 9.8 BC 5.0 = = 5.55 m/s tAC 0.9

Ans.

Example 4 An aeroplane is flying in a horizontal direction with a velocity 600 km/h at a height of 1960 m. When it is vertically above the point A on the ground, a body is dropped from it. The body strikes the ground at point B. Calculate the distance AB. 1 2 gt 2 2hOA 2 × 1960 = = = 20 s g 9.8

h=

Solution From we have,

tOB

Horizontal distance

AB = vtOB 5   = 600 × m/s (20 s )   18

O

v

h

= 3333.33 m = 3.33 km V

v

BC = vtAC

Further,

V

Projectile Motion — 237

A

B

Ans.

Example 5 In the figure shown, find O

20m/s

45° Q

P 45°

(a) the time of flight of the projectile over the inclined plane (b) range OP Solution (a) Let the particle strikes the plane at point P at time t, then OQ =

1 g t 2 = 5t 2 2

QP = 20 t In ∆ OPQ, angle OPQ is 45°. ∴ ∴ (b)

OQ = QP or 5t 2 = 20t t =4s OP = QP s 45° = (20t ) ( 2 )

Ans.

Substituting t = 4 s, we have OP = 80 2 m

Ans.

238 — Mechanics - I V

Example 6 In the figure shown, find (a) the time when the particle strikes the ground at P (b) the horizontal distance QP (c) velocity of the particle at P Take g = 10 m/ s2 Solution

40 m/s

O

45 m

2h 2 × 45 (a) t = = =3s g 10

Tower P Q

(b) Horizontal distance QP = 40 t = 40 × 3 = 120 m (c) Horizontal component of velocity at P = 40 m/s Vertical compound of velocity = gt = 10t = 10 × 3 = 30 m/s

(downwards)

40 m/s

θ

v

30 m/s

Net velocity v = (40)2 + (30)2 = 50 m/s

Ans.

30 tan θ = 40

or

3 4

 3 θ = tan −1   = 37°  4

∴

Ans.

Ans.

Type 4. Based on trajectory of a projectile

Concept We have seen that equation of trajectory of projectile is gx 2 gx 2 y = x tan θ − = x tan θ − (1 + tan 2 θ) 2u 2 cos 2 θ 2u 2 In some problems, the given equation of a projectile is compared with this standard equation to find the unknowns. V

Example 7 A particle moves in the plane xy with constant acceleration 'a' directed along the negative y-axis. The equation of motion of the particle has the form y = px − qx 2 where p and q are positive constants. Find the velocity of the particle at the origin of co-ordinates. Solution Comparing the given equation with the equation of a projectile motion, gx2 (1 + tan 2 θ ) 2u 2 a g = a , tan θ = p and (1 + tan 2 θ ) = q 2u 2 u = velocity of particle at origin y = x tan θ −

We find that ∴

=

a (1 + tan 2 θ ) = 2q

a (1 + p2) 2q

Projectile Motion — 239

Chapter 7

Type 5. Based on basic concepts of projectile motion.

Concept Following are given some basic concepts of any projectile motion : (i) Horizontal component of velocity always remains constant. (ii) Vertical component of velocity changes by 10 m/s or 9.8 m/s in every second in downward direction. For example, if vertical component of velocity at t = 0 is 30 m/s then change in vertical component in first five seconds is as given in following table: Time (in sec)

Vertical component (in m/s)

Direction

0

30

upwards

1

20

upwards

2

10

upwards

3

0

-

4

10

downwards

5

20

downwards

In general we can use, vy = u y + ayt (iii) At a height difference ‘h’ between two points 1 and 2, the vertical components v1 and v 2 are related as, v 2 = ± v12 ± 2 gh In moving upwards, vertical component decreases. So, take − 2 gh in the above equation if point 2 is higher than point 1. (iv) Horizontal displacement is simply, (in the direction of u x ) sx = u x t (v) Vertical displacement has two components (say s1 and s 2 ), one due to initial velocity u y and the other due to gravity. s1 = u y t = displacement due to initial component of velocity u y. This s1 is in the direction of u y (upwards or downwards) 1 (if g = 10 m/ s 2 ) s 2 = g t 2 = 5t 2 2 This s 2 is always downwards. Net vertical displacement is the resultant of s1 and s 2 . V

Example 8 In the figure shown, find 20√2 m/s P y 45° 37° O

Q

(a) time of flight of the projectile along the inclined plane. (b) range OP

x

240 — Mechanics - I Solution (a) Horizontal component of initial velocity, ux = 20 2 cos 45° = 20 m/s Vertical component of initial velocity uy = 20 2 sin 45° = 20 m/s Let the particle strikes at P after time t, then horizontal displacement OQ = ux t = 20t 1 In vertical displacement, uy t or 20t is upwards and gt 2 or 5t 2 is downwards. But net 2 displacement is upwards, therefore 20 t should be greater than 5t 2 and QP = 20t − 5t 2 In ∆OPQ, tan 37° =

QP OQ

3 20t − 5t 2 = 4 20t

or, Solving this equation, we get

t =1s

Ans.

(b) Range OP = OQ sec 37°  5 = (20t )    4 Substituting t = 1 s, we have OP = 25 m V

Ans.

Example 9 In the shown figure, find 20√2 m/s y 45° O x P 37°

37° Q

(a) time of flight of the projectile along the inclined plane (b) range OP Solution (a) Horizontal component of initial velocity, ux = 20 2 cos 45° = 20 m/s Vertical component of initial velocity, uy = 20 2 sin 45° = 20 m/s Let the particle, strikes the inclined plane at P after time t, then horizontal displacement QP = ux t = 20t 1 In vertical displacement, uy t or 20t is upwards and gt 2 or 5t 2 is downwards. But net vertical 2 displacement is downwards. Hence 5t 2 should be greater than 20t and therefore, OQ = 5t 2 − 20t

Projectile Motion — 241

Chapter 7 In ∆OQP, tan 37° =

OQ QP

3 5t 2 − 20t = 4 20t

or Solving this equation, we get

t = 7s

Ans.

(b) Range, OP = (PQ )sec37°  5 = (20t )    4 Substituting the value of t, we get OP = 175 m V

Ans.

Example 10 At a height of 45 m from ground velocity of a projectile is, v = ( 30$i + 40$j) m/ s Find initial velocity, time of flight, maximum height and horizontal range of this projectile. Here $i and $j are the unit vectors in horizontal and vertical directions. Solution Given, vx = 30 m/s and vy = 40 m/s Horizontal component of velocity remains unchanged. ∴ ux = vx = 30 m/s Vertical component of velocity is more at lesser heights. Therefore uy > vy or

uy =

y vy vx 45 m

uy

x

vy2

+ 2 gh

= (40)2 + (2) (10) (45) = 50 m/s

O

ux

uy

u

Initial velocity of projectile, u = ux2 + uy2 = (30)2 + (50)2 = 10 34 m/s uy 50 5 tan θ = = = ux 30 3 ∴ Time of flight,

 5 θ = tan −1    3

θ ux

Ans.

Ans.

2u sin θ 2uy = g g 2 × 50 = 10

T=

= 10 s

Ans.

242 — Mechanics - I H=

Maximum height,

=

2 u 2 sin 2 θ uy = 2g 2g

(50)2 = 125 m 2 × 10

Ans.

Horizontal range R = uxT = 30 × 10 = 300 m

Ans.

Miscellaneous Examples V

Example 11 A particle is thrown over a triangle from one end of a horizontal base and after grazing the vertex falls on the other end of the base. If α and β be the base angles and θ the angle of projection, prove that tan θ = tan α + tan β. Solution

The situation is shown in figure. R = Range

Y

A(x, y)

θ O

y α x

β R–x

X

From figure, we have tan α + tan β =

y y + x R−x

tan α + tan β =

yR x(R − x)

…(i)

Equation of trajectory is

or,

x  y = x tan θ 1 −  R  yR tan θ = x(R − x)

…(ii)

From Eqs. (i) and (ii), we have tan θ = tan α + tan β V

Hence Proved.

Example 12 The velocity of a projectile when it is at the greatest height is 2 /5 times its velocity when it is at half of its greatest height. Determine its angle of projection. Solution Suppose the particle is projected with velocity u at an angle θ with the horizontal. Horizontal component of its velocity at all height will be u cos θ.

Chapter 7

Projectile Motion — 243

At the greatest height, the vertical component of velocity is zero, so the resultant velocity is v1 = u cos θ At half the greatest height during upward motion, y = h /2, a y = − g, uy = u sin θ Using vy2 − uy2 = 2a y y h we get, vy2 − u 2 sin 2 θ = 2 (− g ) 2 or

vy2 = u 2 sin 2 θ − g ×

or

vy =

u 2 sin 2 θ u 2 sin 2 θ = 2g 2

 u 2 sin 2 θ  Q h = 2 g  

u sin θ 2

Hence, resultant velocity at half of the greatest height is v2 = vx2 + vy2 = u 2 cos 2 θ + v1 2 = v2 5

Given, ∴

or

u 2 sin 2 θ 2

u 2 cos 2 θ 2 = 2 2 u sin θ 5 u 2 cos 2 θ + 2 1 2 = 1 1 + tan 2 θ 5 v12 = v22

2

or or

2 + tan 2 θ = 5 or

∴ V

tan 2 θ = 3

tan θ = 3 θ = 60°

Example 13 A car accelerating at the rate of 2 m/s 2 from rest from origin is carrying a man at the rear end who has a gun in his hand. The car is always moving along positive x-axis. At t = 4 s, the man fires a bullet from the gun and the bullet hits a bird at t = 8 s. The bird has a $ . Find velocity of projection position vector 40$i + 80$j + 40k of the bullet. Take the y-axis in the horizontal plane. ( g = 10 m/s 2 ) Solution Let velocity of bullet be, v = vx $i + vy $j + vz k$ At t = 4 s, x-coordinate of car is x-coordinate of bird is x b = 40 m ∴ or ∴

xc =

1 2 1 at = × 2 × 16 = 16 m 2 2 x b = xc + vx (8 − 4) 40 = 16 + 4vx vx = 6 m/s

Ans.

244 — Mechanics - I Similarly, or or

yb = yc + vy (8 − 4) 80 = 0 + 4vy vy = 20 m/s

and

zb = zc + vz (8 − 4) − 40 = 0 + 4vz −

or

1 g (8 − 4)2 2

1 × 10 × 16 2 vz = 30 m/s

or ∴ Velocity of projection of bullet

v = (6$i + 20$j + 30k$ ) m/s V

Example 14 Two inclined planes OA and OB having inclinations 30° and 60° with the horizontal respectively intersect each other at O, as shown in figure. A particle is projected from point P with velocity u = 10 3 m/ s along a direction perpendicular to plane OA. If the particle strikes plane OB perpendicular at Q. Calculate x

y B

v u

Q

A P h 60°

30°

O

(a) (b) (c) (d)

time of flight, velocity with which the particle strikes the plane OB, height h of point P from point O, distance PQ. (Take g = 10 m/s 2 )

Solution Let us choose the x and y directions along OB and OA respectively. Then, ux = u = 10 3 m/s, uy = 0 a x = − g sin 60° = − 5 3 m/s 2 and a y = − g cos 60° = − 5 m/s 2 (a) At point Q, x-component of velocity is zero. Hence, substituting in vx = ux + a x t or (b) At point Q, ∴

0 = 10 3 − 5 3t 10 3 t= = 2s 5 3

Ans.

v = vy = uy + a y t v = 0 − (5)(2) = − 10 m/s

Here, negative sign implies that velocity of particle at Q is along negative y-direction.

Ans.

Chapter 7

Projectile Motion — 245

(c) Distance PO = | displacement of particle along y-direction | =|sy| 1 Here, sy = uy t + a y t 2 2 1 = 0 − (5)(2)2 = − 10 m 2 ∴ Therefore,

PO = 10 m  1 h = PO sin 30° = (10)    2

or h =5m (d) Distance OQ = displacement of particle along x-direction = sx Here,

sx = ux t +

1 2

Ans.

ax t2

= (10 3 )(2) −

1 2

(5 3 )(2)2 = 10 3 m

or

OQ = 10 3 m

∴

PQ = (PO )2 + (OQ )2 = (10)2 + (10 3 )2

∴

= 100 + 300 = 400 PQ = 20 m

Ans.

Exercises LEVEL 1 Assertion and Reason Directions Choose the correct option. (a) If both Assertion and Reason are true and the Reason is correct explanation of the Assertion.

(b) If both Assertion and Reason are true but Reason is not the correct explanation of Assertion. (c) If Assertion is true, but the Reason is false. (d) If Assertion is false but the Reason is true.

1. Assertion : A particle follows only a parabolic path if acceleration is constant. Reason : In projectile motion path is parabolic, as acceleration is assumed to be constant at low heights.

2. Assertion : Projectile motion is called a two dimensional motion, although it takes place in space. Reason : In space it takes place in a plane.

3. Assertion : If time of flight in a projectile motion is made two times, its maximum height will become four times. Reason : In projectile motion H ∝ T 2, where H is maximum height and T the time of flight.

4. Assertion : A particle is projected with velocity u at angle 45° with ground. Let v be the velocity of particle at time t ( ≠ 0), then value of u ⋅ v can be zero. Reason : Value of dot product is zero when angle between two vectors is 90°.

5. Assertion : A particle has constant acceleration is x - y plane. But neither of its acceleration components ( ax and a y ) is zero. Under this condition particle cannot have parabolic path. Reason : In projectile motion, horizontal component of acceleration is zero. v − v1 6. Assertion : In projectile motion at any two positions 2 always remains constant. t2 − t1 Reason : The given quantity is average acceleration, which should remain constant as acceleration is constant.

7. Assertion : Particle A is projected upwards. Simultaneously particle B is projected as projectile as shown. Particle A returns to ground in 4 s. At the same time particle B collides with A. Maximum height H attained by B B would be 20 m. ( g = 10 ms−2 )

H A

Reason : Speed of projection of both the particles should be same under the given condition.

8. Assertion : Two projectiles have maximum heights 4H and H respectively. The ratio of their horizontal components of velocities should be 1 : 2 for their horizontal ranges to be same. Reason : Horizontal range = horizontal component of velocity × time of flight.

Chapter 7

Projectile Motion — 247

9. Assertion : If g = 10 m/ s2 then in projectile motion speed of particle in every second will change by 10 ms−1.

Reason : Acceleration is nothing but rate of change of velocity.

10. Assertion : In projectile motion if particle is projected with speed u, then speed of particle at height h would be u 2 − 2gh . Reason : If particle is projected with vertical component of velocity u y . Then vertical component at the height h would be ± u 2y − 2gh

Objective Questions Single Correct Option 1. Identify the correct statement related to the projectile motion. (a) It is uniformly accelerated everywhere (b) It is uniformly accelerated everywhere except at the highest position where it is moving with constant velocity (c) Acceleration is never perpendicular to velocity (d) None of the above

2. Two bodies are thrown with the same initial velocity at angles θ and ( 90° − θ ) respectively with the horizontal, then their maximum heights are in the ratio (a) 1 : 1

(b) sin θ : cos θ

(c) sin 2 θ : cos 2 θ

(d) cos θ : sin θ

3. The range of a projectile at an angle θ is equal to half of the maximum range if thrown at the same speed. The angle of projection θ is given by (a) 15°

(b) 30°

(c) 60°

(d) data insufficient

4. A ball is projected with a velocity 20 ms−1 at an angle to the horizontal. In order to have the maximum range. Its velocity at the highest position must be (a) 10 ms −1

(b) 14 ms −1

(c) 18 ms −1

^

^

(d) 16 ms −1 ^

^

5. A particle has initial velocity, v = 3 i + 4 j and a constant force F = 4 i − 3 j acts on it. The path of the particle is (a) straight line

(b) parabolic

(c) circular

(d) elliptical

6. A body is projected at an angle 60° with the horizontal with kinetic energy K. When the velocity makes an angle 30° with the horizontal, the kinetic energy of the body will be (a) K/2

(b) K/3

(c) 2 K/ 3

(d) 3 K/ 4

7. If T1 and T2 are the times of flight for two complementary angles, then the range of projectile R is given by (a) R = 4 gT1T2

(b) R = 2 gT1T2

(c) R =

1 gT1T2 4

(d) R =

1 gT1T2 2

8. A gun is firing bullets with velocity v0 by rotating it through 360° in the horizontal plane. The maximum area covered by the bullets is (a)

πv02 g

(b)

π 2v02 g

(c)

πv04 g2

(d)

π 2v04 g

9. A grass hopper can jump maximum distance 1.6 m. It spends negligible time on ground. How far can it go in 10 2 s ? (a) 45 m

(b) 30 m

(c) 20 m

(d) 40 m

248 — Mechanics - I 10. Two stones are projected with the same speed but making different angles with the horizontal.

π and the maximum height 3 reached by it is 102 m. Then the maximum height reached by the other in metres is

Their horizontal ranges are equal. The angle of projection of one is (a) 76 (c) 56

(b) 84 (d) 34

11. A ball is projected upwards from the top of a tower with a velocity 50 ms−1 making an angle 30° with the horizontal. The height of tower is 70 m. After how many seconds from the instant of throwing, will the ball reach the ground. ( g = 10 ms−2 ) (a) 2 s

(b) 5 s

(c) 7 s

(d) 9 s

12. Average velocity of a particle in projectile motion between its starting point and the highest point of its trajectory is (projection speed = u, angle of projection from horizontal = θ) u 1 + 3 cos 2 θ 2 u (d) 1 + cos 2 θ 2

(a) u cos θ (c)

(b)

u 2 + cos 2 θ 2

13. A train is moving on a track at 30 ms−1. A ball is thrown from it perpendicular to the direction of motion with 30 ms−1 at 45° from horizontal. Find the distance of ball from the point of projection on train to the point where it strikes the ground. (a) 90 m

(b) 90 3 m

(c) 60 m

(d) 60 3 m

14. A body is projected at time t = 0 from a certain point on a planet’s surface with a certain velocity at a certain angle with the planet’s surface (assumed horizontal). The horizontal and vertical displacements x and y (in metre) respectively vary with time t in second as, x = (10 3 ) t and y = 10 t − t 2. The maximum height attained by the body is (a) 75 m (c) 50 m

(b) 100 m (d) 25 m

15. A particle is fired horizontally from an inclined plane of inclination 30° with horizontal with speed 50 ms−1. If g = 10 ms−2, the range measured along the incline is (a) 500 m

(b)

1000 m 3

(c) 200 2 m

(d) 100 3 m

16. A fixed mortar fires a bomb at an angle of 53° above the horizontal with a muzzle velocity of

80 ms−1. A tank is advancing directly towards the mortar on level ground at a constant speed of 5 m/s. The initial separation (at the instant mortar is fired) between the mortar and tank, so that the tank would be hit is [ Take g = 10 ms−2 ] (a) 662.4 m (c) 486.6 m

(b) 526.3 m (d) None of these

Subjective Questions 1. At time t = 0, a small ball is projected from point A with a velocity of 60 m/s at 60° angle with horizontal. Neglect atmospheric resistance and determine the two times t1 and t2 when the velocity of the ball makes an angle of 45° with the horizontal x-axis.

2. A particle is projected from ground with velocity 20 2 m/s at 45°. At what time particle is at height 15 m from ground? ( g = 10 m/ s2 )

Chapter 7

Projectile Motion — 249

3. A particle is projected at an angle 60° with horizontal with a speed v = 20 m/s. Taking g = 10 m/ s2. Find the time after which the speed of the particle remains half of its initial speed.

4. Two particles A and B are projected from ground towards each other with speeds 10 m/s and 5 2 m/s at angles 30° and 45° with horizontal from two points separated by a distance of 15 m. Will they collide or not? 5√2 m/s 10 m/s

A

45°

30°

B

15 m

5. Two particles move in a uniform gravitational field with an acceleration g. At the initial moment the particles were located over a tower at one point and moved with velocities v1 = 3 m/ s and v2 = 4 m/ s horizontally in opposite directions. Find the distance between the particles at the moment when their velocity vectors become mutually perpendicular.

6. A ball is thrown from the ground to clear a wall 3 m high at a distance of 6 m and falls 18 m away from the wall. Find the angle of projection of ball.

7. A body is projected up such that its position vector varies with time as r = { 3 t $i + ( 4 t − 5 t 2 )$j} m. Here, t is in seconds. Find the time and x-coordinate of particle when its y-coordinate is zero.

8. A particle is projected along an inclined plane as shown in figure. What is the speed of the particle when it collides at point A ? ( g = 10 m/ s2 ) u = 10 m/s A 30° 30° O

9. In the above problem, what is the component of its velocity perpendicular to the plane when it strikes at A ?

10. Two particles A and B are projected simultaneously from two towers of heights 10 m and 20 m respectively. Particle A is projected with an initial speed of 10 2 m/s at an angle of 45° with horizontal, while particle B is projected horizontally with speed 10 m/s. If they collide in air, what is the distance d between the towers? 10 m/s 10√2 m/s A

B

45° 20 m

10m

d

250 — Mechanics - I 11. A particle is projected from the bottom of an inclined plane of inclination 30° with velocity of 40 m/s at an angle of 60° with horizontal. Find the speed of the particle when its velocity vector is parallel to the plane. Take g = 10 m/ s2.

12. Two particles A and B are projected simultaneously in the directions shown in figure with velocities vA = 20 m/ s and vB = 10 m/ s respectively. They collide in air after

1 s. Find 2

(a) the angle θ (b) the distance x. vA = 20 m/s

A

θ x

vB = 10 m/s

B

13. A ball is shot from the ground into the air. At a height of 9.1 m, its velocity is observed to be v = 7.6i$ + 6.1$j in metre per second ($i is horizontal, $jis upward). Give the approximate answers. (a) (b) (c) (d)

To what maximum height does the ball rise? What total horizontal distance does the ball travel? What are the magnitude and What are the direction of the ball’s velocity just before it hits the ground?

14. A particle is projected with velocity 2 gh, so that it just clears two walls of equal height h which are at a distance of 2h from each other. Show that the time of passing between the walls is 2

h . g

[Hint : First find velocity at height h. Treat it as initial velocity and 2h as the range.]

15. A particle is projected at an angle of elevation α and after t second it appears to have an elevation of β as seen from the point of projection. Find the initial velocity of projection.

16. A projectile aimed at a mark, which is in the horizontal plane through the point of projection, falls a cm short of it when the elevation is α and goes b cm far when the elevation is β.Show that, if the speed of projection is same in all the cases the proper elevation is  b sin 2α + a sin 2β  1 sin−1   2 a+b  

17. Two particles are simultaneously thrown in horizontal direction from two points on a riverbank, which are at certain height above the water surface. The initial velocities of the particles are v1 = 5 m/ s and v2 = 7.5 m/ s respectively. Both particles fall into the water at the same time. First particle enters the water at a point s = 10 m from the bank. Determine (a) the time of flight of the two particles, (b) the height from which they are thrown, (c) the point where the second particle falls in water.

18. A balloon is ascending at the rate v = 12 km/ h and is being carried horizontally by the wind at vw = 20 km/ h.If a ballast bag is dropped from the balloon at the instant h = 50 m, determine the time needed for it to strike the ground. Assume that the bag was released from the balloon with the same velocity as the balloon. Also, find the speed with which the bag strikes the ground?

Chapter 7

Projectile Motion — 251

19. A projectile is fired with a velocity u at right angles to the slope, which is inclined at an angle θ with the horizontal. Derive an expression for the distance R to the point of impact.

u R θ

.

20. An elevator is going up with an upward acceleration of 1 m/ s2. At the instant when its velocity is 2 m/s, a stone is projected upward from its floor with a speed of 2 m/s relative to the elevator, at an elevation of 30°. (a) Calculate the time taken by the stone to return to the floor. (b) Sketch the path of the projectile as observed by an observer outside the elevator. (c) If the elevator was moving with a downward acceleration equal to g, how would the motion be altered?

21. Two particles A and B are projected simultaneously in a vertical plane as shown in figure. They collide at time t in air. Write down two necessary equations for collision to take place. y (m)

u2 θ2

20

B

u1 10

θ1 A

10

30

x (m)

LEVEL 2 Objective Questions Single Correct Option 1. Two bodies were thrown simultaneously from the same point, one straight up, and the other, at an angle of θ = 30° to the horizontal. The initial velocity of each body is 20 ms−1. Neglecting air resistance, the distance between the bodies at t = 1.2 later is (a) 20 m (c) 24 m

(b) 30 m (d) 50 m

2. A particle is dropped from a height h. Another particle which is initially at a horizontal distance d from the first is simultaneously projected with a horizontal velocity u and the two particles just collide on the ground. Then u 2h 2h (c) d = h (a) d 2 =

2u 2h g (d) gd 2 = u 2h (b) d 2 =

252 — Mechanics - I 3. A ball is projected from point A with velocity 10 ms−1perpendicular to the

90° A

inclined plane as shown in figure. Range of the ball on the inclined plane is (a)

40 m 3

(b)

20 m 3

(c)

12 m 3

(d)

60 m 3

30°

4. A heavy particle is projected with a velocity at an angle with the horizontal

(a) O

t

(b) O

x

(c) O

slope

slope

slope

slope

into the uniform gravitational field. The slope of the trajectory of the particle varies as

t

(d) O

x

5. A particle starts from the origin of coordinates at time t = 0 and moves in the xy plane with a constant acceleration α in the y-direction. Its equation of motion is y = βx 2. Its velocity component in the x-direction is (a) variable

(b)

2α β

(c)

α 2β

(d)

α 2β

6. A projectile is projected with speed u at an angle of 60° with horizontal from the foot of an inclined plane. If the projectile hits the inclined plane horizontally, the range on inclined plane will be (a)

u 2 21 2g

(b)

3u 2 4g

(c)

u2 2β

(d)

21 u 2 8g

7. A particle is projected at an angle 60° with speed 10 3 m/s, from the

10√3 m/s 10 3 m/s √

point A, as shown in the figure. At the same time the wedge is made to move with speed 10 3 m/s towards right as shown in the figure. Then the time after which particle will strike with wedge is (a) 2 s

(b) 2 3 s

(c)

4 s 3

60°

30°

(d) None of these

8. A particle moves along the parabolic path x = y 2 + 2 y + 2 in such a way that Y -component of

velocity vector remains 5 ms−1 during the motion. The magnitude of the acceleration of the particle is (a) 50 ms −2

(b) 100 ms −2

(c) 10 2 ms −2

(d) 0.1 ms −2

9. A shell fired from the base of a mountain just clears it. If α is the angle of projection, then the angular elevation of the summit β is (a)

α 2

 tan α  (c) tan −1    2 

 1 (b) tan −1    2

H β

(d) tan −1 (2 tan α )

10. In the figure shown, the two projectiles are fired simultaneously. The

20 √ 3 m/s

minimum distance between them during their flight is (a) 20 m (b) 10 3 m (c) 10 m (d) None of the above

20 m/s 60°

30° 20 √ 3 m

Chapter 7

Projectile Motion — 253

More than One Correct Options 1. Two particles projected from the same point with same speed u at angles of projection α and β strike the horizontal ground at the same point. If h1 and h2 are the maximum heights attained by the projectile, R is the range for both and t1 and t2 are their times of flights, respectively , then (a) α + β =

π 2

(b) R = 4 h1h2

(c)

t1 = tan α t2

(d) tan α =

h1 h2

2. A ball is dropped from a height of 49 m. The wind is blowing horizontally. Due to wind a constant horizontal acceleration is provided to the ball. Choose the correct statement (s). [Take g = 9.8 ms−2] (a) (b) (c) (d)

Path of the ball is a straight line Path of the ball is a curved one The time taken by the ball to reach the ground is 3.16 s Actual distance travelled by the ball is more then 49 m

3. A particle is projected from a point P with a velocity v at an angle θ with horizontal. At a certain point Q it moves at right angles to its initial direction. Then (a) velocity of particle at Q is v sin θ (b) velocity of particle at Q is v cot θ (c) time of flight from P to Q is (v/g ) cosecθ (d) time of flight from P to Q is (v/g ) sec θ

4. At a height of 15 m from ground velocity of a projectile is v = (10 $i + 10$j). Here, $j is vertically upwards and $i is along horizontal direction then ( g = 10 ms−2 )

(a) (b) (c) (d)

particle was projected at an angle of 45° with horizontal time of flight of projectile is 4 s horizontal range of projectile is 100 m maximum height of projectile from ground is 20 m

5. Which of the following quantities remain constant during projectile motion? (a) Average velocity between two points dv (c) dt

(b) Average speed between two points d 2v (d) 2 dt

6. In the projectile motion shown is figure, given tAB = 2 s then ( g = 10 ms−2 )

B

A 15 m O 20 m

(a) (b) (c) (d)

40 m

particle is at point B at 3 s maximum height of projectile is 20 m initial vertical component of velocity is 20 ms −1 horizontal component of velocity is 20 ms −1

B

254 — Mechanics - I Comprehension Based Questions Passage (Q. Nos. 1 to 2) Two inclined planes OA and OB intersect in a horizontal plane having their inclinations α and β with the horizontal as shown in A figure. A particle is projected from point P with velocity u along a direction perpendicular to plane OA. The particle strikes plane OB perpendicularly at Q.

u B P

Q

a α

β O

1. If α = 30° , β = 30°, the time of flight from P to Q is (a)

u g

(b)

3u g

(c)

2u g

(d)

2. If α = 30° , β = 30° and a = 4.9 m, the initial velocity of projection is (a) 9.8 ms −1

(b) 4.9 ms −1

(c) 4.9 2 ms −1

2u g

(d) 19.6 ms −1

Match the Columns 1. Particle-1 is just dropped from a tower. 1 s later particle-2 is thrown from the same tower horizontally with velocity 10 ms−1. Taking g = 10 ms−2, match the following two columns at t = 2 s. Column I

Column II

(a)

Horizontal displacement between two

(p) 10 SI units

(b)

Vertical displacement between two

(q) 20 SI units

(c) (d)

Magnitude of relative horizontal component of velocity Magnitude of relative vertical component of velocity

(r) 10 2 SI units (s) None of the above

R = 20 m. Here, H is maximum height and R the horizontal 2 range. For the given condition match the following two columns.

2. In a projectile motion, given H =

Column I (a) Time of flight

Column II (p) 1

(b) Ratio of vertical component of velocity and horizontal (q) 2 component of velocity (c) Horizontal component of velocity (in m/s) (r) 10 (d) Vertical component of velocity (in m/s)

(s) None of the above

3. A particle can be thrown at a constant speed at different angles. When it is thrown at 15° with horizontal, it falls at a distance of 10 m from point of projection. For this speed of particle match following two columns. Column I

Column II

(a) Maximum horizontal range which can be taken with this speed (b) Maximum height which can be taken with this speed (c) Range at 75°

(p) 10 m

(d) Height at 30°

(s) None of the above

(q) 20 m (r) 15 m

Chapter 7

Projectile Motion — 255

4. In projectile motion, if vertical component of velocity is increased to two times, keeping horizontal component unchanged, then Column I

Column II

(a) Time of flight

(p) will remain same

(b) Maximum height

(q) will become two times

(c) Horizontal range

(r) will become four times

(d) Angle of projection with horizontal

(s) None of the above

5. In projectile motion shown in figure. A u θ O

B

Column I

Column II

(a) Change in velocity between O and A

(p) u cos θ

(b) Average velocity between O and A

(q) u sin θ

(c) Change in velocity between O and B

(r) 2 u sin θ

(d) Average velocity between O and B

(s)

None of the above

6. Particle-1 is projected from ground (take it origin) at time t = 0, with velocity ( 30$i + 30$j) ms−1. Particle-2 is projected from (130 m, 75 m) at time t = 1 s with velocity ( −20 $i + 20 $j) ms−1. Assuming $j to be vertically upward and $i to be in horizontal direction, match the following two columns at t = 2 s. Column I

Column II

(a) horizontal distance between two

(p) 30 SI units

(b) vertical distance between two

(q) 40 SI units

(c) relative horizontal component of velocity between two

(r)

50 SI units

(d) relative vertical component of velocity between two

(s)

None of the above

7. The trajectories of the motion of three particles are shown in the figure. Match the entries of Column I with the entries of Column II. Neglect air resistance. y

A

B

C

x

256 — Mechanics - I Column I

Column II

(a) Time of flight is least for

(p) A (b) Vertical component of velocity is greatest for (q) B (c) Horizontal component of velocity is greatest for (r) C (d) Launch speed is least for (s) same for all

Subjective Questions 1. Determine the horizontal velocity v0 with which a stone must be projected horizontally from a point P, so that it may hit the inclined plane perpendicularly. The inclination of the plane with the horizontal is θ and point P is at a height h above the foot of the incline, as shown in the figure. P

v0

h θ

2. A particle is dropped from point P at time t = 0. At the same time another particle is thrown from point O as shown in the figure and it collides with the particle P. Acceleration due to gravity is along the negative y-axis. If the two particles collide 2 s after they start, find the initial velocity v0 of the particle which was projected from O. Point O is not necessarily on ground. 2m

y

P

10 m v0 O

θ

x

3. Two particles are simultaneously projected in the same vertical plane from the same point with velocities u and v at angles α and β with horizontal. Find the time that elapses when their velocities are parallel.

4. A projectile takes off with an initial velocity of 10 m/s at an angle of elevation of 45°. It is just able to clear two hurdles of height 2 m each, separated from each other by a distance d. Calculate d. At what distance from the point of projection is the first hurdle placed? Take g = 10 m/ s2.

5. A stone is projected from the ground in such a direction so as to hit a bird on the top of a telegraph post of height h and attains the maximum height of 2h above the ground. If at the instant of projection, the bird were to fly away horizontally with a uniform speed, find the ratio between the horizontal velocity of bird and the horizontal component of velocity of stone, if the stone hits the bird while descending.

Chapter 7

Projectile Motion — 257

6. A particle is released from a certain height H = 400 m. Due to the wind, the particle gathers the

horizontal velocity component vx = ay where a = 5 s−1 and y is the vertical displacement of the particle from the point of release, then find

(a) the horizontal drift of the particle when it strikes the ground, (b) the speed with which particle strikes the ground. (Take g = 10 m/s 2)

7. A train is moving with a constant speed of 10 m/s in a circle of radius

y

16 m. The plane of the circle lies in horizontal x-y plane. At time t = 0, π train is at point P and moving in counter-clockwise direction. At this instant, a stone is thrown from the train with speed 10 m/s relative to train towards negative x-axis at an angle of 37° with vertical z-axis. Find

x

P

(a) the velocity of particle relative to train at the highest point of its trajectory. (b) the co-ordinates of points on the ground where it finally falls and that of the highest point of its trajectory. 3 Take g = 10 m/s 2, sin 37° = 5

8. A particle is projected from an inclined plane OP1 from A with velocity v1 = 8 ms−1 at an angle 60° with horizontal. An another particle is projected at the same instant from B with velocity v2 = 16 ms−1 and perpendicular to the plane OP2 as shown in figure. After time 10 3 s there separation was minimum and found to be 70 m. Then find distance AB. P1

v1

v2 P2

60° A

90° 45°

B

30° O

9. A particle is projected from point O on the ground with velocity

u = 5 5 m/s at angle α = tan−1 ( 0.5 ). It strikes at a point C on a

Y

fixed smooth plane AB having inclination of 37° with horizontal as shown in figure. If the particle does not rebound, calculate

5√5 m/s

(a) coordinates of point C in reference to coordinate system as shown O in the figure. (b) maximum height from the ground to which the particle rises. ( g = 10 m/s 2).

α

C

B

y (10/3) m

37° A

D

X

10. A plank fitted with a gun is moving on a horizontal surface with speed of 4 m/s along the positive x-axis. The z-axis is in vertically upward direction. The mass of the plank including the mass of the gun is 50 kg. When the plank reaches the origin, a shell of mass 10 kg is fired at an angle of 60° with the positive x-axis with a speed of v = 20 m/s with respect to the gun in x-z plane. Find the position vector of the shell at t = 2 s after firing it. Take g = 9.8 m/ s2.

Answers Introductory Exercise 7.1 1.

2s

2. False 3. True $ $ 4. v = (40$i + 10 j ) m/s , s = (80$i + 40 j ) m $ 5. t = 5 s , d = 100 m, v = (20$i − 30 j ) ms −1

Introductory Exercise 7.2 1 1. (a) 20 2 m/s at angle tan−1   with horizontal, (b) 100 m.  2 2. Between two points lying on the same horizontal line 3. u cos α 4. 2u sin α , downwards $ $ 5. (a) 80 m, 20 m, 4s (b) (20 $i + 10 j ) ms −1 (c) (20 $i − 20 j ) ms −1 6. (a) 30 ms −1 (vertically downwards) (b) 20.62 ms −1 7.

5 ms −1 2

8. (a)

20 s (b) 20 20 m (c) 49 m/s, θ = tan−1 ( 5 ) with horizontal

9. No 10.

a (1 + b2 ) 2c

Introductory Exercise 7.3 1. 1.69 s, 39 m 2. 6.31 s, 145.71 m 3. 4. 5. 6.

2.31 s, 53.33 mm (a) A vertical straight line (b) A parabola (a) zero (b) 20 ms −1 in horizontal direction (c) 40 m 60°

Exercises LEVEL 1 Assertion and Reason 1. (d)

2. (a)

3. (a)

4. (b)

5. (d)

6. (a)

7. (c)

8. (a or b)

9. (d)

10. (b)

Single Correct Option 1. (a)

2. (c)

3. (a)

4. (b)

5. (b)

6. (b)

11. (c)

12. (b)

13. (a)

14. (d)

15. (b)

16. (d)

7. (d)

8. (c)

9. (d)

10. (d)

Chapter 7

Projectile Motion — 259

Subjective Questions 1 t1 = 2.19s, t 2 = 8.20 s 4. No

2. 3 s and 1 s −1

5. 2.5 m

6. tan

7. time = zero, 0.8 s,x-coordinate = 0, 2.4 m 10. 20 m

11.

8.

40 m/s 3

3. 3 s

 2    3

10 m/s 3

9. 5 m/s

12. (a) 30° (b) 5 3 m

13. (a) 11 m, (b) 23 m (c) 16.6 m/s (d) tan−1 (2), below horizontal 15. u =

gt cos β

17. (a) 2s (b) 19.6 m (c) 15 m

sin (α − β )

19. R =

18. 3.55 s, 32.7 m/s

2 u2 tan θ sec θ g

20. (a) 0.18 s (c) a straight line with respect to elevator and projectile with respect to ground 21. (u1 cos θ1 + u2 cos θ 2 ) t = 20 ...(i) (u1 sin θ1 − u2 sin θ 2 ) t = 10 ...(ii)

LEVEL 2 Single Correct Option 1. (c)

2. (b)

3. (a)

4. (a)

5. (d)

6. (d)

7. (a)

8. (a)

9. (c)

More than One Correct Options 1. (all)

2. (a,c,d)

3. (b,c)

4. (b,d)

5. (c,d)

6. (all)

Comprehension Based Questions 1. (b)

2. (a)

Match the Columns 1. 3. 5. 7.

(a) →(p), (b) →(s), (c) →(p), (d) →(p) (a) →(q), (b) →(p), (c) →(p), (d) →(s) (a) →(q), (b) →(s), (c) →(r), (d) →(p) (a) →(s), (b) →(s), (c) →(r), (d) →(p)

2. (a) →(s), (b) →(q), (c) →(r), (d) →(s) 4. (a) →(q), (b) →(r), (c) →(q), (d) →(s) 6. (a) →(r), (b) →(r), (c) →(r), (d) →(s)

Subjective Questions 1. v0 = 3. t = 5.

2 gh 2 + cot2 θ

2. 26 ms −1 at angle θ = tan−1 (5) with x-axis

uv sin (α − β ) g (v cos β − u cos α )

2 2+1

4. 4.47 m, 2.75 m 6. (a) 2.67 km (b) 0.9 km/s

$ 7. (a) (−6 $i + 10 j ) ms −1 (b) (−4.5 m, 16 m, 0), (0.3 m, 8.0 m, 3.2 m) 8. 250 m $] m 10. [ 24 $i + 15 k

9. (a) (5 m, 1.25 m) (b) 4.45 m

10. (b)

08 Laws of Motion Chapter Contents 8.1

Types of Forces

8.2

Free Body Diagram

8.3

Equilibrium

8.4

Newton's Laws of Motion

8.5

Constraint Equations

8.6

Pseudo Force

8.7

Friction

8.1 Types of Forces There are basically three forces which are commonly encountered in mechanics.

Field Forces These are the forces in which contact between two objects is not necessary. Gravitational force between two bodies and electrostatic force between two charges are two examples of field forces. Weight ( w = mg ) of a body comes in this category.

Contact Forces Two bodies in contact exert equal and opposite forces on each other. If the contact is frictionless, the contact force is perpendicular to the common surface and known as normal reaction. If, however the objects are in rough contact and move (or have a tendency to move) relative to each other without losing contact then frictional force arise which oppose such motion. Again each object exerts a frictional force on the other and the two forces are equal and opposite. This force is perpendicular to normal reaction. Thus, the contact force ( F ) between two objects is made up of two forces. F1

A F2

B

Fig. 8.1

(i) Normal reaction ( N ) (ii) Force of friction ( f ) and since these two forces are mutually perpendicular. F = N2 + f

2

Note In this book normal reaction at most of the places has been represented by N. But at some places, it is also represented by R. This is because N is confused with the SI unit of force newton.

Consider two wooden blocks A and B being rubbed against each other. In Fig. 8.1, A is being moved to the right while B is being moved leftward. In order to see more clearly which forces act on A and which on B, a second diagram is drawn showing a space between the blocks but they are still supposed to be in contact. A N

f F1

F2

f

N B

Fig. 8.2

In Fig. 8.2, the two normal reactions each of magnitude N are perpendicular to the surface of contact between the blocks and the two frictional forces each of magnitude f act along that surface, each in a direction opposing the motion of the block upon which it acts. Note Forces on block B from the ground are not shown in the figure.

Laws of Motion — 263

Chapter 8

Attachment to Another Body Tension (T ) in a string and spring force ( F = kx ) come in this group. Regarding the tension and string, the following three points are important to remember: 1. If a string is inextensible the magnitude of acceleration of any number of masses connected through the string is always same. a

a

M

m

a

F

a

m M

Fig. 8.3

2. If a string is massless, the tension in it is same everywhere. However, if a string has a mass and it is

accelerated, tension at different points will be different. 3. If pulley is massless and frictionless, tension will be same on both sides of the pulley. T1 T1

T T m

T1 T2

m

m T2

T T

T2

M String and pulley are massless and there is no friction between pulley and string

T3 T4

M

M

String is massless but pulley is not massless and frictionless

String and pulley are not massless and there is a friction between pulley and string

Fig. 8.4

Spring force ( F = kx ) has been discussed in detail in the chapter of work, energy and power.

Hinge Force In the figure shown there is a hinge force on the rod (from the hinge). There are two methods of finding a hinge force : Stri n

g

Hinge Rod

Fig. 8.5

(i) either you find its horizontal ( H ) and vertical (V ) components (ii) or you find its magnitude and direction.

264 — Mechanics - I Extra Points to Remember ˜

Normal reaction is perpendicular to the common tangent direction and always acts towards the body. It is just like a pressure force (F = PA) which is also perpendicular to a surface and acts towards it. For example N2 Ladder

Wall

N1

Ground Fig. 8.6

Normal reaction on ladder from ground is N1 and from wall is N2 . ˜ Tension in a string is as shown in Fig. 8.7. In the figure : T1 goes to block A (force applied by string on block A). T2 and T3 to pulley P1 T4 , T5 and T7 to pulley P2 T8 to block B and T6 to roof If string and pullies are massless and there is no friction in the pullies, then T1 = T2 = T3 = T4 = T5 = T6 and T7 = T8 ˜ If a string is attached with a block then it can apply force on the block only in a direction away from the block (in the form of tension).

T1 T2

P1 T6 T5

A T3 T4

P2 T7 T8 B

Fig. 8.7

T String attached with a block

Fig. 8.8

If the block is attached with a rod, then it can apply force on the block in both directions, towards the block (may be called push) or away from the block (called pull) F

or

F

Rod attached with a block Fig. 8.9 ˜

All forces discussed above make a pair of equal and opposite forces acting on two different bodies (Newton's third law).

8.2 Free Body Diagram No system, natural or man made, consists of a single body alone or is complete in itself. A single body or a part of the system can, however be isolated from the rest by appropriately accounting for its effect on the remaining system. A free body diagram (FBD) consists of a diagrammatic representation of a single body or a sub-system of bodies isolated from its surroundings showing all the forces acting on it.

Chapter 8

Laws of Motion — 265

Consider, for example, a book lying on a horizontal surface. A free body diagram of the book alone would consist of its weight ( w = mg ), acting through the centre of gravity and the reaction ( N ) exerted on the book by the surface. N

Mass of book = m

w = mg

Fig. 8.10 V

Example 8.1 A cylinder of weight W is resting on a V-groove as shown in figure. Draw its free body diagram.

Fig. 8.11

The free body diagram of the cylinder is as shown in Fig. 8.12. Here, w = weight of cylinder and N 1 and N 2 are the normal reactions between the cylinder and the two inclined walls. Solution

C N1

w

Fig. 8.12 V

Example 8.2 Three blocks A, B and C are placed one over the other as shown in figure. Draw free body diagrams of all the three blocks. A B C

Fig. 8.13

Solution

Free body diagrams of A, B and C are shown below.

wA N1 FBD of A

Here, and

N1

N2

wB

wC

N2 FBD of B Fig. 8.14

N 1 = normal reaction between A and B N 2 = normal reaction between B and C N 3 = normal reaction between C and ground.

N3 FBD of C

N2

266 — Mechanics - I V

Example 8.3 A block of mass m is attached with two strings as shown in figure. Draw the free body diagram of the block.

θ

Fig. 8.15

Solution

The free body diagram of the block is as shown in Fig. 8.16. T1

θ

T2 mg Fig. 8.16

8.3 Equilibrium Forces which have zero resultant and zero turning effect will not cause any change in the motion of the object to which they are applied. Such forces (and the object) are said to be in equilibrium. For understanding the equilibrium of an object under two or more concurrent or coplanar forces let us first discuss the resolution of force and moment of a force about some point.

Resolution of a Force When a force is replaced by an equivalent set of components, it is said to be resolved. One of the most useful ways in which to resolve a force is to choose only two components (although a force may be resolved in three or more components also) which are at right angles also. The magnitude of these components can be very easily found using trigonometry. F

F2

B

θ F1

A

C

Fig. 8.17

In Fig. 8.17,

F1 = F cos θ = component of F along AC F2 = F sin θ = component of F perpendicular to AC or along AB

Finding such components is referred to as resolving a force in a pair of perpendicular directions. Note that the component of a force in a direction perpendicular to itself is zero. For example, if a force of 10 N is applied on an object in horizontal direction then its component along vertical is zero. Similarly, the component of a force in a direction parallel to the force is equal to the magnitude of the force. For example component of the above force in the direction of force (horizontal) will be 10 N. In the opposite direction the component is −10 N.

Chapter 8 V

Laws of Motion — 267

Example 8.4 Resolve a weight of 10 N in two directions which are parallel and perpendicular to a slope inclined at 30° to the horizontal. Solution

Component perpendicular to the plane

w⊥ = w cos 30°

w| |

60°

3 = (10) =5 3 N 2

30° w⊥ w = 10 N

30°

and component parallel to the plane

Fig. 8.18

 1 w| | = w sin 30° = (10)   = 5 N  2 V

Example 8.5 Resolve horizontally and vertically a force F = 8 N which makes an angle of 45° with the horizontal. Solution

Horizontal component of F is

FV

 1 FH = F cos 45° = ( 8)    2 F

=4 2N and vertical component of F is FV = F sin 45°

45°

 1 = ( 8)   = 4 2 N  2 V

FH

Fig. 8.19

Example 8.6 A body is supported on a rough plane inclined at 30° to the horizontal by a string attached to the body and held at an angle of 30° to the plane. Draw a diagram showing the forces acting on the body and resolve each of these forces (a) horizontally and vertically, (b) parallel and perpendicular to the plane. Solution The forces are The tension in the string T The normal reaction with the plane N The weight of the body w and the friction f

N

f w 30°

Fig. 8.20

(a) Resolving horizontally and vertically f cos 30°

T 30°

N

N sin 60°

T sin 60°

T

30° w f

f sin 30°

60° N cos 60°

Fig. 8.21

60° T cos 60°

268 — Mechanics - I Resolving horizontally and vertically in the senses OX and OY as shown, the components are Force f N T w

Y

Components Parallel to OX (horizontal) Parallel to OY (vertical) − f cos 30° − N cos 60° T cos 60° 0

− f sin 30° N sin 60° T sin 60° −w

X

O

Fig. 8.22

(b) Resolving parallel and perpendicular to the plane T N f

T sin 30°

30°

w sin 30° 30°

T cos 30°

w cos 30° w

Fig. 8.23

Resolving parallel and perpendicular to the plane in the senses OX ′ and OY ′ as shown, the components are : Force

Components Parallel to OX′ (parallel to plane) Parallel to OY′ (perpendicular to plane)

f

−f

0

N

0

N

T

T cos 30°

T sin 30°

w

− w sin 30°

− wcos 30°

X'

Y'

O

Fig. 8.24

Moment of a Force The general name given to any turning effect is torque. The magnitude of torque, also known as the moment of a force F is calculated by multiplying together the magnitude of the force and its perpendicular distance r⊥ from the axis of rotation. This is denoted by C or τ (tau). i.e.

C = Fr⊥

or τ = Fr⊥

Direction of Torque The angular direction of a torque is the sense of the rotation it would cause. Consider a lamina that is free to rotate in its own plane about an axis perpendicular to the lamina and passing through a point A on the lamina. In the diagram the moment about the axis of rotation of the force F1 is F1 r1 anticlock-wise and the moment of the force F2 is F2 r2 clockwise. A convenient way to differentiate between clockwise and anticlock-wise torques is to allocate a positive sign to one sense (usually, but not invariably, this is anticlockwise) and negative sign to the other. With this convention, the moments of F1 and F2 are + F1 r1 and −F2 r2 (when using a sign convention in any problem it is advisable to specify the chosen positive sense).

F2

F1 r2 A

r1

Fig. 8.25

Chapter 8

Laws of Motion — 269

Zero Moment If the line of action of a force passes through the axis of rotation, its perpendicular distance from the axis is zero. Therefore, its moment about that axis is also zero. Note Later in the chapter of rotation we will see that torque is a vector quantity. V

Example 8.7 ABCD is a square of side 2 m and O is its centre. Forces act along the sides as shown in the diagram. Calculate the moment of each force about (a) an axis through A and perpendicular to the plane of square. (b) an axis through O and perpendicular to the plane of square.

Solution

3N

O

A

2N

C 5N B

Fig. 8.26

Magnitude of force

2N

5N

4N

3N

Perpendicular distance from A

0

2m

2m

0

Moment about A

0

−10 N-m

+8 N-m

0

Magnitude of force

2N

5N

4N

3N

Perpendicular distance from O

1m

1m

1m

1m

+2 N-m

–5 N-m

+4 N-m

–3 N-m

Moment about O

Example 8.8 Forces act as indicated on a rod AB which is pivoted at A. Find the anticlockwise moment of each force about the pivot. 3F 2F 30° A

B a

a

F 2a

Fig. 8.27

sin

30

°

Solution

4a

V

4N

Taking anticlockwise moments as positive we have:

(a)

(b)

D

3F 2F 30°

A

B F Fig. 8.28

Magnitude of force

2F

F

3F

Perpendicular distance from A

a

2a

4a sin 30° = 2 a

Anticlockwise moment about A

+2 Fa

–2 Fa

+6 Fa

270 — Mechanics - I Coplanar Forces in Equilibrium When an object is in equilibrium under the action of a set of two or more coplanar forces, each of three factors which comprise the possible movement of the object must be zero, i.e. the object has (i) no linear movement along any two mutually perpendicular directions OX and OY. (ii) no rotation about any axis. The set of forces must, therefore, be such that (a) the algebraic sum of the components parallel to OX is zero or ΣFx = 0 (b) the algebraic sum of the components parallel to OY is zero or ΣF y = 0 (c) the resultant moment about any specified axis is zero or Στ any axis = 0 Thus, for the equilibrium of a set of two or more coplanar forces ΣFx = 0 ΣF y = 0 and

Στ any axis = 0

Using the above three conditions, we get only three set of equations. So, in a problem number of unknowns should not be more than three. V

Y

Example 8.9 A rod AB rests with the end A on rough horizontal ground and the end B against a smooth vertical wall. The rod is uniform and of weight w. If the rod is in equilibrium in the position shown in figure. Find

B

(a) frictional force at A (b) normal reaction at A (c) normal reaction at B.

30°

or

∴ or

∴

NA

NB = fA −w=0 NA =w

Y B

NB

…(i) …(ii)

NA O

w 30° fA

Fig. 8.30

N A ( 2l cos 30° ) − N B ( 2l sin 30° ) − w( l cos 30° ) = 0 3N A − N B −

3 w=0 2

…(iii)

Solving these three equations, we get (a) f A =

X

Fig. 8.29

Solution Let length of the rod be 2l. Using the three conditions of equilibrium. Anticlockwise moment is taken as positive. (i) ΣF X = 0 ⇒ ∴ NB − fA = 0

(ii) ΣFY = 0 ⇒ or (iii) ΣτO = 0

A

O

3 3 w (b) N A = w (c) N B = w 2 2

Exercise : What happens to N A , N B and f A if (a) Angle θ = 30° is slightly increased, (b) A child starts moving on the ladder from A to B without changing the angle θ. Ans (a) Unchanged, decreases, decrease, (b) Increases, increase, increase

A

X

Chapter 8

Laws of Motion — 271

Equilibrium of Concurrent Coplanar Forces If an object is in equilibrium under two or more concurrent coplanar forces the algebraic sum of the components of forces in any two mutually perpendicular directions OX and OY should be zero, i.e. the set of forces must be such that (i) the algebraic sum of the components parallel to OX is zero, i.e. ΣFx = 0. (ii) the algebraic sum of the components parallel to OY is zero, i.e. ΣF y = 0. Thus, for the equilibrium of two or more concurrent coplanar forces ΣFx = 0 ΣF y = 0 The third condition of zero moment about any specified axis is automatically satisfied if the moment is taken about the point of intersection of the forces. So, here we get only two equations. Thus, number of unknown in any problem should not be more than two. V

Example 8.10 An object is in equilibrium under four concurrent forces in the directions shown in figure. Find the magnitudes of F1 and F2 . F1

4N

30° 60° O

30°

8N

F2

Fig. 8.31

Solution

The object is in equilibrium. Hence,

(i) ΣFx = 0 ∴

8 + 4 cos 60° − F2 cos 30° = 0 3 8 + 2 − F2 =0 2 20 F2 = N 3

or or

(ii) ΣF y = 0 ∴ F1 + 4 sin 60° − F2 sin 30° = 0 or or or

F1

F1 +

4 3 F2 − =0 2 2 F 10 F1 = 2 − 2 3 = −2 3 2 3 4 F1 = N 3

4N

Y

30° 60° 30°

O

F2

Fig. 8.32

X 8N

272 — Mechanics - I Lami’s Theorem

If an object O is in equilibrium under three concurrent forces F1 , F2 and F3 as shown in figure. Then, F1 F F = 2 = 3 sin α sin β sin γ F2

γ α

F1

β

F3

Fig. 8.33

This property of three concurrent forces in equilibrium is known as Lami’s theorem and is very useful method of solving problems related to three concurrent forces in equilibrium. V

Example 8.11 One end of a string 0.5 m long is fixed to a point A and the other end is fastened to a small object of weight 8 N. The object is pulled aside by a horizontal force F, until it is 0.3 m from the vertical through A. Find the magnitudes of the tension T in the string and the force F. Solution

∴ and if Then and

T B

C

AC = 0.5 m, BC = 0.3 m

F

8N

AB = 0.4 m ∠BAC = θ. AB cos θ = = AC BC sin θ = = AC

Fig. 8.34

0.4 4 = 0.5 5 0.3 3 = 0.5 5

Here, the object is in equilibrium under three concurrent forces. So, we can apply Lami’s theorem. F 8 T or = = sin (180° − θ ) sin ( 90° + θ ) sin 90° or

A

F 8 = =T sin θ cos θ

∴

T=

8 8 = = 10 N cos θ 4/5

and

F=

8 sin θ ( 8) ( 3/ 5) = = 6N cos θ ( 4/ 5)

A θ

T θ

B

C

F

8N

Fig. 8.35

Ans.

Laws of Motion — 273

Chapter 8 V

Example 8.12 The rod shown in figure has a mass of 2 kg and length 4 m. In equilibrium, find the hinge force (or its two components) acting on the rod and 3 4 tension in the string. Take g = 10 m/ s2 , sin 37° = and cos 37° = . 5 5 String 37°

Hinge

Fig. 8.36

Solution T V 37° H w

Fig. 8.37

In the figure, only those forces which are acting on the rod has been shown. Here H and V are horizontal and vertical components of the hinge force. T sin 37°= 0.6T

V

y H O

2m

2 m T cos 37° = 0.8T

x

20 N

Fig. 8.38

Σ Fx = 0 ⇒ H − 0.8T = 0 Σ F y = 0 ⇒ V + 0.6 T − 20 = 0 ΣτO = 0 ⇒ Clockwise torque of 20 N = anticlock-wise torque of 0.6 T. All other forces ( H , V and 0.8 T pass through O, hence their torques are zero).

...(i) ...(ii)

∴ 20 × 2 = 0.6T × 4 Solving these three equations, we get

...(iii)

T = 16.67 N, H = 13.33 N and

V = 10 N

Ans.

274 — Mechanics - I Hinge force (F) V =10 N

F

θ

H = 13.33 N

Fig. 8.39

F = (13.33) 2 + (10) 2 = 16.67 N tan θ =

Ans.

10 13.33

 10  θ = tan −1   = 37°  13.33

∴

INTRODUCTORY EXERCISE

Ans.

8.1

1. The diagram shows a rough plank resting on a cylinder with one end of the plank on rough ground. Neglect friction between plank and cylinder. Draw diagrams to show (a) the forces acting on the plank, (b) the forces acting on the cylinder.

Fig. 8.40

2. Two spheres A and B are placed between two vertical walls as shown in figure. Friction is absent everywhere. Draw the free body diagrams of both the spheres. B A

Fig. 8.41 C

3. A point A on a sphere of weight w rests in contact with a smooth vertical wall and is

B

supported by a string joining a point B on the sphere to a point C on the wall. Draw free body diagram of the sphere.

A

Fig. 8.42

4. A rod AB of weight w1 is placed over a sphere of weight w 2 as shown in figure. Ground is rough and there is no friction between rod and sphere and sphere and wall. Draw free body diagrams of sphere and rod separately. Wall

A C

AC < CB

B

Note

No friction will act between sphere and ground, think why ?

Ground Fig. 8.43

Chapter 8

Laws of Motion — 275

5. A rod OA is suspended with the help of a massless string AB as shown in Fig. 8.44. Rod is hinged at point O. Draw free body diagram of the rod.

O

A

6. A rod AB is placed inside a rough spherical shell as shown in Fig.8.45.

B

Fig. 8.44

Draw the free body diagram of the rod.

B A

Fig. 8.45

7. Write down the components of four forces F1, F2, F3 and F4 along ox and oy directions as shown in Fig. 8.46.

F2 = 4N 60°

F1 = 4N 30° F4 = 4N

y o

x

F3 = 6N

Fig. 8.46

8. All the strings shown in figure are massless. Tension in the horizontal string is 30 N. Find W. °

45

w

Fig. 8.47

9. The 50 kg homogeneous smooth sphere rests on the 30° incline A and against the smooth vertical wall B. Calculate the contact forces at A and B.

A B 30°

Fig. 8.48

10. In question 3 of the same exercise, the radius of the sphere is a. The length of the string is also a. Find tension in the string.

11. A sphere of weight w = 100 N is kept stationary on a rough inclined plane by a horizontal string AB as shown in figure. Find (a) tension in the string, (b) force of friction on the sphere and (c) normal reaction on the sphere by the plane.

A C 30°

Fig. 8.49

B

276 — Mechanics - I

8.4 Newton's Laws of Motion It is interesting to read Newton’s original version of the laws of motion. Law I Every body continues in its state of rest or in uniform motion in a straight line unless it is compelled to change that state by forces impressed upon it. The change of motion is proportional to the magnitude of force impressed and is made in the direction of the straight line in which that force is impressed.

Law II

To every action there is always an equal and opposite reaction or the mutual actions of two bodies upon each other are always directed to contrary parts. The modern versions of these laws are: 1. A body continues in its initial state of rest or motion with uniform velocity unless acted on by an unbalanced external force. 2. The acceleration of a body is inversely proportional of its mass and directly proportional to the resultant external force acting on it, i.e. F ΣF = F net = m a or a = net m 3. Forces always occur in pairs. If body A exerts a force on body B, an equal but opposite force is exerted by body B on body A. Law III

Working with Newton’s First and Second Laws Normally any problem relating to Newton’s laws is solved in following four steps: 1. First of all we decide the system on which the laws of motion are to be applied. The system may be a single particle, a block or a combination of two or more blocks, two blocks connected by a string, etc. The only restriction is that all parts of the system should have the same acceleration. 2. Once the system is decided, we make the list of all the forces acting on the system. Any force applied by the system on other bodies is not included in the list of the forces. 3. Then we make a free body diagram of the system and indicate the magnitude and directions of all the forces listed in step 2 in this diagram. 4. In the last step we choose any two mutually perpendicular axes say x and y in the plane of the forces in case of coplanar forces. Choose the x-axis along the direction in which the system is known to have or is likely to have the acceleration. A direction perpendicular to it may be chosen as the y-axis. If the system is in equilibrium any mutually perpendicular directions may be chosen. Write the components of all the forces along the x-axis and equate their sum to the product of the mass of the system and its acceleration, i.e. …(i) ΣFx = ma This gives us one equation. Now, we write the components of the forces along the y-axis and equate the sum to zero. This gives us another equation, i.e. …(ii) ΣF y = 0 Note

(i) If the system is in equilibrium we will write the two equations as ΣFx = 0 and ΣFy = 0 (ii) If the forces are collinear, the second equation, i.e. ΣFy = 0 is not needed.

Chapter 8

Laws of Motion — 277

Extra Points to Remember ˜

˜

If a is the acceleration of a body, then ma force does not act on the body but this much force is required to provide a acceleration to the body. The different available forces acting on the body provide this ma force or, we can say that vector sum of all forces acting on the body is equal to ma. The available forces may be weight, tension, normal reaction, friction or any externally applied force etc. If all bodies of a system has a common acceleration then that common acceleration can be given by Net pulling/pusing force NPF a= = TM Total mass Net pulling/pushing force (NPF) is actually the net force. Example Suppose two unequal masses m and 2m are attached to the ends of a light inextensible string which passes over a smooth massless pulley. We have to find the acceleration of the system. We can assume that the mass 2m is pulled downwards by a force equal to its weight, i.e. 2mg. Similarly, the mass m is being a pulled by a force of mg downwards. Therefore, net pulling force on the system is 2mg − mg = mg and total mass being pulled is 2 m + m = 3m. ∴

a m

2m

Acceleration of the system is a=

mg 2mg

Net pulling force mg g = = Total mass to be pulled 3m 3

Fig. 8.50

Note While finding net pulling force, take the forces (or their components) which are in the direction of motion (or opposite to it) and are single (i.e. they are not forming pair of equal and opposite forces). For example weight ( mg ) or some applied force F. Tension makes an equal and opposite pair. So, they are not to be included, unless the system in broken at some place and only one tension is considered on the system under consideration. ˜

V

After finding that common acceleration, we will have to draw free body diagrams of different blocks to find normal reaction or tension etc.

Example 8.13 Two blocks of masses 4 kg and 2 kg are placed side by side on a smooth horizontal surface as shown in the figure. A horizontal force of 20 N is applied on 4 kg block. Find

20 N

4kg

2kg

Fig. 8.51

(a) the acceleration of each block. (b) the normal reaction between two blocks. Solution (a) Both the blocks will move with same acceleration (say a) in horizontal direction. a 20 N

4kg

y

2kg x

Fig. 8.52

Let us take both the blocks as a system. Net external force on the system is 20 N in horizontal direction. Using ΣFx = ma x or

20 = ( 4 + 2)a = 6a 10 m /s 2 a= 3

Ans.

278 — Mechanics - I Alternate Method a=

Net pushing force 20 10 = = m/s 2 Total mass 4+2 3

(b) The free body diagram of both the blocks are as shown in Fig. 8.53. 20

4kg

y

N N

2kg

a

x

a

Fig. 8.53

ΣFx = ma x

Using

10 3 40 20 N = 20 − = newton 3 3 This can also be solved as under 10 20 For 2 kg block, N = 2a = 2 × = newton 3 3 Here, N is the normal reaction between the two blocks. 20 − N = 4a = 4 ×

For 4 kg block,

Ans.

Note In free body diagram of the blocks we have not shown the forces acting on the blocks in vertical direction, because normal reaction between the blocks and acceleration of the system can be obtained without using ΣFy = 0. V

Example 8.14 Three blocks of masses 3 kg, 2 kg and 1 kg are placed side by side on a smooth surface as shown in figure. A horizontal force of 12 N is applied on 3 kg block. Find the net force on 2 kg block. 12N

3kg 2kg 1kg

Fig. 8.54

Solution Since, all the blocks will move with same acceleration (say a) in horizontal direction. Let us take all the blocks as a single system. y 12 N

3kg 2kg 1kg

x

a Fig. 8.55

Net external force on the system is 12 N in horizontal direction. Using ΣFx = ma x , we get, 12 = ( 3 + 2 + 1)a = 6a 12 or a= = 2 m/s 2 6

Chapter 8

Laws of Motion — 279

Alternate Method a=

Net pushing force 12 = = 2 m/s 2 Total mass 3+ 2+1

Now, let F be the net force on 2 kg block in x-direction, then using ΣFx = ma x for 2 kg block, we get F = ( 2)( 2) = 4 N

Ans.

Note Here, net force F on 2 kg block is the resultant of N1 and N2 ( N1 > N2 ) where, N1 = normal reaction between 3 kg and 2 kg block, and N2 = normal reaction between 2 kg and 1 kg block. Thus, F = N1 − N2

V

Example 8.15 In the arrangement shown in figure. The strings are light and inextensible. The surface over which blocks are placed is smooth. Find

4kg

2kg

1kg

F = 14N

Fig. 5.56

(a) the acceleration of each block, (b) the tension in each string.

Solution (a) Let a be the accelera-

y

T2

4kg

tion of each block and T1 and T2 be the tensions, in the two strings as shown in figure. Taking the three blocks and the two strings as the system.

2kg

T1

1kg

F = 14N

x

Fig. 8.57

a

4kg

2kg

F = 14N

1kg

Fig. 8.58

Using

ΣFx = ma x

or 14 = ( 4 + 2 + 1)a or

a=

14 = 2 m/s 2 7

Ans.

Alternate Method a=

Net pulling force 14 = = 2 m/s 2 Total mass 4 + 2+1

(b) Free body diagram (showing the forces in x-direction only) of 4 kg block and 1kg block are shown in Fig. 8.59. a = 2 m/s2

4kg

y

a = 2 m/s2 T2

T1

1kg

Fig. 8.59

F = 14N

x

280 — Mechanics - I ΣFx = ma x F − T1 = (1)( a ) 14 − T1 = (1)( 2) = 2

Using For 1 kg block, or

V

∴

T1 = 14 − 2 = 12 N

For 4 kg block,

T2 = ( 4 )( a )

∴

T2 = ( 4 )( 2) = 8 N

Ans. Ans. F = 120 N

Example 8.16 Two blocks of masses 4 kg and 2 kg are attached by an inextensible light string as shown in figure. Both the blocks are pulled vertically upwards by a force F = 120 N. Find

4kg

(a) the acceleration of the blocks, (b) tension in the string. (Take g = 10 m/ s 2 ).

2kg

Solution (a)

Let a be the acceleration of the blocks and T the tension in the string as shown in figure.

Fig. 8.60

F = 120N y 4 kg x

T 2 kg

Fig. 8.61 F = 120N

Taking the two blocks and the string as the system shown in figure Fig. 8.62. Using ΣF y = ma y , we get or

F – 4g − 2g = ( 4 + 2)a 120 − 40 − 20 = 6a or 60 = 6a

∴

a = 10 m/s 2

a

4g

Ans. 2g

Alternate Method a=

Net pulling force 120 − 60 = = 10 m/s 2 Total mass 4+2 F = 120N 4kg

4g + 2g = 60N

2kg

Fig. 8.63

Fig. 8.62

Chapter 8

Laws of Motion — 281 T

(b) Free body diagram of 2 kg block is as shown in Fig. 8.64. Using ΣF y = ma y we get, T − 2g = 2a or T − 20 = ( 2)(10) ∴

T = 40 N

2kg

a

Ans.

2g

Fig. 8.64 V

Example 8.17 In the system shown in figure pulley is smooth. String is massless and inextensible. Find acceleration of the system a, tensions T1 and T2 . ( g = 10 m/ s2 )

T1 T1 a 2 kg 4 kg

a

T2 6 kg

a

Fig. 8.65

Here, net pulling force will be Weight of 4 kg and 6 kg blocks on one side – weight of 2 kg block on the other side. Therefore, Net pulling force a= Total mass Solution

= =

( 6 × 10) + ( 4 × 10) − ( 2)(10) 6+ 4 + 2 20 m/s 2 3

T1 a

For T1 , let us consider FBD of 2 kg block. Writing equation of motion, we get 20 100 Ans. = N T1 − 20 = 2a or T1 = 20 + 2 × 3 3 For T2 , we may consider FBD of 6 kg block. Writing equation of motion, we get 60 − T2 = 6a ∴

 20 T2 = 60 − 6a = 60 − 6    3 =

60 N 3

2 kg

Ans.

w2 = 20 N T2 a 6 kg

w6 = 60 N

Fig. 8.66

Ans.

Exercise: Draw FBD of 4 kg block. Write down the equation of motion for it and check whether the values calculated above are correct or not.

282 — Mechanics - I V

Example 8.18 In the system shown in figure all surfaces are smooth. String is massless and inextensible. Find acceleration a of the system and tension T in the string. ( g = 10 m/ s2 ) a

2 kg

T

T

4 kg

a

Fig. 8.67

Here, weight of 2 kg is perpendicular to motion (or a). Hence, it will not contribute in net pulling force. Only weight of 4 kg block will be included. Net pulling force a= ∴ Total mass Solution

=

( 4 )(10) 20 = m/s 2 ( 4 + 2) 3

Ans.

T

For T, consider FBD of 4 kg block. Writing equation of motion.

4 kg

a

40 − T = 4a ∴ T = 40 − 4a W4 = 40 N  20 40 Fig. 8.68 Ans. = 40 − 4   = N  3 3 Exercise: Draw FBD of 2 kg block and write down equation of motion for it. Check whether the values calculated above are correct or not.

(a) the acceleration of the system and (b) tensions in the strings. Neglect friction. ( g = 10 m/ s 2 )

B

Example 8.19 In the adjacent figure, masses of A, B and C are 1 kg, 3 kg and 2 kg respectively. Find

C

A

V

60°

30°

Fig. 8.69

Solution (a) In this case net pulling force

= m A g sin 60° + mB g sin 60° − mC g sin 30°  3 3  1 = (1)(10) + ( 3)(10)   − ( 2)(10)    2 2  2 = 24.64 N Total mass being pulled = 1 + 3 + 2= 6 kg 21.17 ∴ Acceleration of the system a = = 4.1 m/s 2 6

Ans.

Chapter 8

Laws of Motion — 283

(b) For the tension in the string between A and B.

T1

FBD of A

a

m A g sin 60° − T1 = ( m A )( a ) ∴ T1 = m A g sin 60° − m A a = m A ( g sin 60° − a )   3 T1 = (1) 10 × − 4.1 ∴ 2  

A

mA g sin 60°

Fig. 8.70

= 4.56 N

Ans.

For the tension in the string between B and C. FBD of C ∴ ∴

T2

a

T2 − mC g sin 30° = mC a T2 = mC ( a + g sin 30° )

C mC g sin 30°

  1  T2 = 2 4.1 + 10     2  

Fig. 8.71

= 18.2 N

INTRODUCTORY EXERCISE

Ans.

8.2

1. Three blocks of masses 1 kg, 4 kg and 2 kg are placed on a smooth horizontal plane as shown in figure. Find (a) the acceleration of the system, (b) the normal force between 1 kg block and 4 kg block, (c) the net force on 2 kg block.

120 N 1kg

4kg

50 N

2kg

Fig. 8.72

2. In the arrangement shown in figure, find the ratio of tensions in the strings attached with 4 kg block and that with 1 kg block. 4kg 3kg 1kg

Fig. 8.73

3. Two unequal masses of 1 kg and 2 kg are connected by an inextensible light string passing over a smooth pulley as shown in figure. A force F = 20 N is applied on 1 kg block. Find the acceleration of either block. ( g = 10 m /s 2 ).

1kg

F

2kg

Fig. 8.74

284 — Mechanics - I 4. In the arrangement shown in figure what should be the mass of block A, so that the system remains at rest? Neglect friction and mass of strings.

2kg A

30° 2kg

Fig. 8.75

5. Two blocks of masses 2 kg and 4 kg are released from rest over a smooth inclined plane of inclination 30° as shown in figure. What is the normal force between the two blocks? g

k g 2

4k

30°

Fig. 8.76

6. What should be the acceleration a of the box shown in Fig. 8.77 so that the block of mass m exerts a force

mg on the floor of the box? 4

a A

Fig. 8.77

7. In the figure shown, find acceleration of the system and tensions T1, T2 and T3. (Take g = 10 m /s 2)

2kg T3 1kg

T2

3kg T1 4 kg Fig. 8.78

8. In the figure shown, all surfaces are smooth. Find (a) acceleration of all the three blocks, (b) net force on 6 kg, 4 kg and 10 kg blocks and

100 N

Fig. 8.79

(c) force acting between 4 kg and 10 kg blocks.

9. Three blocks of masses m1 = 10 kg, m 2 = 20 kg

40 N

6kg 4kg 10kg

m1

T1

m2

and m 3 = 30 kg are on a smooth horizontal table, connected to each other by light Fig. 8.80 horizontal strings. A horizontal force F = 60 N is applied to m 3, towards right. Find (a) tensions T1 and T2 and (b) tension T2 if all of a sudden the string between m1 and m 2 snaps.

T2

m3

F

Chapter 8

Laws of Motion — 285

8.5 Constraint Equations In the above article, we have discussed the cases where different blocks of the system had a common acceleration and that common acceleration was given by Net pulling / pushing force a= Total mass Now, the question is, if different blocks have different accelerations then what? In those cases, we take help of constraint equations. These equations establish the relation between accelerations (or velocities) of different blocks of a system. Depending upon different kinds of problems we have divided the constraint equations in following two types. Most of them are directly explained with the help of some example (s) in their support.

Type 1 V

Example 8.20 Using constraint method find the relation between accelerations of 1 and 2.

x1 x2

11

1 2

Fig. 8.81

2

Fig. 8.82

Solution At any instant of time let x1 and x 2 be the displacements of 1 and 2 from a fixed line (shown dotted). Here x1 and x 2 are variables but,

x1 + x 2 = constant or (length of string) x1 + x 2 = l Differentiating with respect to time, we have v1 + v 2 = 0 or v1 = − v 2 Again differentiating with respect to time, we get a1 + a 2 = 0 or a1 = − a 2 This is the required relation between a1 and a 2 , i.e. accelerations of 1 and 2 are equal but in opposite directions. Note (i) In the equation x1 + x2 = l, we have neglected the length of string over the pulley. But that length is also constant. (ii) In constraint equation if we get a1 = − a2 , then negative sign does not always represent opposite directions of a1 and a2 . The real significance of this sign is, x2 decreases if x1 increases and vice-versa.

286 — Mechanics - I V

Example 8.21 Using constraint equations find the relation between a1 and a2 .

2 1

Fig. 8.83

In Fig. 8.84, points 1, 2, 3 and 4 are movable. Let their displacements from a fixed dotted line be x1 , x 2 , x 3 and x 4 Solution

x1 + x 3 = l1 ( x1 − x 3 ) + ( x 4 − x 3 ) = l2 ( x1 − x 4 ) + ( x 2 − x 4 ) = l3 On double differentiating with respect to time, we will get following three constraint relations …(i) a1 + a 3 = 0 …(ii) a1 + a 4 − 2a 3 = 0 a1 + a 2 − 2a 4 = 0 Solving Eqs. (i), (ii) and (iii), we get

x3 x2

x1 4 2

…(iii)

a 2 = − 7a1 Which is the desired relation between a1 and a 2 . V

x4

3

Fig. 8.84

Example 8.22 In the above example, if two blocks have masses 1 kg and 2 kg respectively then find their accelerations and tensions in different strings. Solution Pulleys 3 and 4 are massless. Hence net force on them should be zero. Therefore, if

we take T tension in the shortest string, then tension in other two strings will be 2T and 4T. 4T 2T

4T

T 4T

T 2T

a

1 kg 2T 1

T

T

2 kg

7a

2 W1=10 N Fig. 8.85

W2 = 20 N

7a

Chapter 8

Laws of Motion — 287

Further, if a is the acceleration of 1 in upward direction, then from the constraint equation a 2 = − 7a1 , acceleration of 2 will be 7a downwards. Writing the equation, F net = ma for the two blocks we have 4T + 2T + T − 10 = 1× a or 7T − 10 = a 20 − T = 2 × ( 7a ) or 20 − T = 14a Solving these two equations we get,

…(i) …(ii)

T = 1.62 N a = 1.31m/s

and

Ans. 2

Ans.

Note In a problem if ‘a’ comes out to be negative after calculations then we will change the initially assumed directions of accelerations.

Type 2 V

Example 8.23 The system shown in figure is released from rest. Find acceleration of different blocks and tension in different strings.

1 kg

2 kg 3 kg

Fig. 8.86

Solution

2T a

1 kg

2T a

P T T

2 kg

ar

Net acceleration = ar – a

3 kg ar Net acceleration = a + ar

Fig. 8.87

288 — Mechanics - I (i) Pulley P and 1 kg mass are attached with the same string. Therefore, if 1 kg mass has an acceleration ‘a’ in upward direction, then pulley P will have an acceleration ‘a’ downwards. (ii) 2 kg and 3 kg blocks are attached with the same string passing over a moveable pulley P. Therefore their relative acceleration, a r (relative to pulley) will be same. Their net accelerations (relative to ground) are as shown in figure. (iii) Pulley P is massless. Hence net force on this pulley should be zero. If T is the tension in the string connecting 2 kg and 3 kg mass, then tension in the upper string will be 2T. Now writing the equation, F net = ma for three blocks, we have: 1 kg block: 2T a 1 kg

w1 = 10 N

Fig. 8.88

…(i)

2T − 10 = 1 × a 2 kg block:

T

ar – a

2 kg

w2 = 20 N

Fig. 8.89

…(ii)

T − 20 = 2( a r − a ) 3 kg block: T

3 kg ar + a w3 = 30 N

Fig. 8.90

30 − T = 3 ( a r + a ) Solving Eqs. (i), (ii) and (iii) we get, T = 8.28 N, a = 6.55 m/s 2

…(iii) and

a r = 0.7 m/s 2 .

Now, acceleration of 3kg block is ( a + a r ) or 7.25 m / s 2 downwards and acceleration of 2kg is ( a r − a ) or − 5.85 m / s 2 upwards. Since, this comes out to be negative, hence acceleration of 2 kg block is 5.85 m / s 2 downwards.

Laws of Motion — 289

Chapter 8 Extra Points to Remember ˜

In some cases, acceleration of a block is inversely proportional to tension force acting on the block (or its component in the direction of motion or acceleration).If tension is double (as compared to other block), then acceleration will be half. In Fig. (a): Tension force on block-1 is double (=2T ) than the tension force on block-2 (=T ). Therefore, acceleration of block -1 will be half. If block-1 has an acceleration ‘a' in downward direction, then block -2 will have an acceleration ‘2a’ towards right. In Fig. (b): Tension force on block-1 is three times (2T + T = 3T ) than the tension force on block-2 (=T ). Therefore acceleration of block-2 will be three times. If block-1 has an acceleration ‘a’ in upwards direction, then acceleration of block-2 will be ‘3a’ downwards.

INTRODUCTORY EXERCISE

2a

T

2 T 2T

T 2T 2T 1

a a

T 2

T

3a

1 (b)

(a)

Fig. 8.91

8.3

1. Make the constraint relation between a1, a 2 and a 3.

1

2

3

Fig. 8.92

2. At certain moment of time, velocities of 1 and 2 both are 1 m/s upwards. Find the velocity of 3 at that moment. 1

2

3

Fig. 8.93

3. Consider the situation shown in figure. Both the pulleys and the string are light and all the surfaces are smooth. (a) Find the acceleration of 1 kg block. (b) Find the tension in the string. ( g = 10 m /s 2 ).

A

2kg

1kg

Fig. 8.94

290 — Mechanics - I 4. Calculate the acceleration of either blocks and tension in the

M

string shown in figure. The pulley and the string are light and all surfaces are smooth.

M

Fig. 8.95

5. Find the mass M so that it remains at rest in the adjoining figure. Both the pulley and string are light and friction is absent everywhere. ( g = 10 m /s 2 ). M

3kg 2kg

Fig. 8.96

6. In Fig. 8.97 assume that there is negligible friction between the blocks and table. Compute the tension in the cord connecting m 2 and the pulley and acceleration of m 2 if m1 = 300 g, m 2 = 200 g and F = 0.40 N.

m2

m1

F

Fig. 8.97

7. In the figure shown,a 3 = 6 m /s (downwards) and a 2 = 4 m /s (upwards). Find acceleration of 1. 2

2

1

2 3 Fig. 8.98

8. Find the acceleration of the block of mass M in the situation shown in the figure. All the surfaces are frictionless.

M 2M

30°

Fig. 8.99

Chapter 8

Laws of Motion — 291

8.6 Pseudo Force Before studying the concept of pseudo force let us first discuss frame of reference. Frame of reference is the way of observation the things.

Inertial Frame of Reference A non-accelerating frame of reference is called an inertial frame of reference. A frame of reference moving with a constant velocity is an inertial frame of reference.

Non-inertial Frame of Reference An accelerating frame of reference is called a non-inertial frame of reference. Note

(i) A rotating frame of reference is a non-inertial frame of reference, because it is also an accelerating one. (ii) Earth is rotating about its axis of rotation and it is revolving around the centre of sun also. So, it is non-intertial frame of reference. But for most of the cases, we consider its as an inertial frame of reference.

Now let us come to the pseudo force. Instead of ground (or inertial frame of reference) when we start watching the objects from a non-inertial (accelerating) frame of reference its motion conditions are felt differently.

For example Suppose a child is standing inside an accelerating lift. From ground frame of reference this child appears to be accelerating but from lift (non-inertial) frame of reference child appears to be at rest. To justify this changed condition of motion, from equations point of view we have to apply a pseudo force. This pseudo force is given by F p = − ma Here, ‘m’ is the mass of that body/object which is being observed from non-inertial frame of reference and a is the acceleration of frame of reference. Negative sign implies that direction of pseudo force F p is opposite to a. Hence whenever you make free body diagram of a body from a non-inertial frame, apply all real forces (actually acting) on the body plus one pseudo force. Magnitude of this pseudo force is ‘ma’ and the direction is opposite to a. Example Suppose a block A of mass m is placed on a lift ascending with an acceleration a 0 . Let N be the normal reaction between the block and the floor of the lift. Free body diagram of A in ground frame of reference (inertial) is shown in Fig. 8.100. N

A

a0

mg

Fig. 8.100

∴ or

N − mg = ma 0 N = m( g + a 0 )

…(i)

292 — Mechanics - I But if we draw the free body diagram of A with respect to the elevator (a non-inertial frame of reference) without applying the pseudo force, as shown in Fig. 8.101, we get N'

(At rest)

A

mg

Fig. 8.101

N ′ − mg = 0 N ′ = mg

or

…(ii)

N'

Since, N ′ ≠ N , either of the equations is wrong. If we apply a pseudo force in non-inertial frame of reference, N ′ becomes equal to N as shown in Fig. 8.102. Acceleration of block with respect to elevator is zero. ∴

(At rest)

A

N ′ − mg − ma 0 = 0 N ′ = m( g + a 0 )

or ∴

…(iii)

N′= N V

Example 8.24 All surfaces are smooth in following figure. Find F, such that block remains stationary with respect to wedge.

mg + FP Here FP = ma0

Fig. 8.102

m F

M θ

Fig. 8.103

Solution

Acceleration of (block + wedge) a =

F ( M + m)

Let us solve the problem by both the methods. Such problems can be solved with or without using the concept of pseudo force.

From Inertial Frame of Reference (Ground)

N cos θ

FBD of block w.r.t. ground (Apply real forces): With respect to ground block is moving with an acceleration a. ∴

ΣF y = 0

⇒

N cos θ = mg

and ⇒

N sin θ

…(i)

ΣFx = ma N sin θ = ma

y

mg

…(ii)

a

Fig. 8.104

x

Chapter 8

Laws of Motion — 293

From Eqs. (i) and (ii), we get a = g tan θ ∴

F = ( M + m) a = ( M + m) g tan θ

From Non-inertial Frame of Reference (Wedge) N cos θ

FBD of block w.r.t. wedge (real forces + pseudo force) w.r.t. wedge, block is stationary ∴

ΣF y = 0 ⇒ N cos θ = mg

…(iii)

ΣFx = 0 ⇒ N sin θ = ma

…(iv)

N sin θ

FP = ma

From Eqs. (iii) and (iv), we will get the same result mg

F = ( M + m) g tan θ.

i.e. V

Fig. 8.105

Example 8.25 A bob of mass m is suspended from the ceiling of a train moving with an acceleration a as shown in figure. Find the angle θ in equilibrium position. θ

a

Fig. 8.106

Solution

This problem can also be solved by both the methods.

Inertial Frame of Reference (Ground) FBD of bob w.r.t. ground (only real forces) T

T cos θ

θ

a T sin θ y x

mg

mg

Fig. 8.107

With respect to ground, bob is also moving with an acceleration a. ΣFx = 0 ⇒ T sin θ = ma ∴ and ΣF y = 0 ⇒ T cos θ = mg From Eqs. (i) and (ii), we get a a tan θ = or θ = tan −1   g g

…(i) …(ii)

294 — Mechanics - I Non-inertial Frame of Reference (Train) FBD of bob w.r.t. train (real forces + pseudo force): T cos θ

T θ ma

FP = ma mg

T sin θ mg

Fig. 8.108

with respect to train, bob is in equilibrium ∴

ΣFx = 0

⇒

T sin θ = ma

∴

ΣF y = 0

⇒

T cos θ = mg

…(iii) …(iv)

From Eqs. (iii) and (iv), we get the same result, i.e.  a θ = tan −1    g

INTRODUCTORY EXERCISE

8.4

1. Two blocks A and B of masses 1kg and 2 kg have accelerations (2$i ) m /s 2 and ( −4$j) m /s 2. Find (a) Pseudo force on block A as applied with respect to the block B. (b) Pseudo force on block B as applied with respect to the block A.

2. Pseudo force with respect to a frame moving with constant velocity is zero. Is this statement true or false?

3. Problems of non-intertial frames can be solved only with the concept of pseudo force. Is this statement true or false?

8.7 Friction Regarding the frictional force (f ) following points are worthnoting : 1. It is the tangential component of net contact force (F) acting between two bodies in contact. 2. It starts acting when there is tendency of relative motion (different velocities) between two bodies in contact or actual relative motion takes place. So, friction has a tendency to stop relative motion between two bodies in contact. 3. If there is only tendency of relative motion then static friction acts and if actual relative motion takes place, then kinetic friction acts.

Chapter 8

Laws of Motion — 295

4. Like any other force of nature friction force also makes a pair of equal and opposite forces acting

on two different bodies. 5. Direction of friction force on a given body is opposite to the direction of relative motion (or its

tendency) of this body. 6. B A

C

(i)

(ii)

Fig. 8.109

In Fig. (i), motion of block A means its relative motion with respect to ground. So, in this case friction between block and ground has a tendency to stop its motion. In Fig. (ii), relative motion between two blocks B and C means their different velocities. So, friction between these two blocks has a tendency to make their velocities same. 7. Static friction is self adjusting in nature. This varies from zero to a limiting value f L . Only that much amount of friction will act which can stop the relative motion. 8. Kinetic friction is constant and it can be denoted by f k . 9. It is found experimentally that limiting value of static friction f L and constant value of kinetic friction f k both are directly proportional to normal reaction N acting between the two bodies. ∴ f L or f k ∝ N ⇒ f L =µ s N and f k =µ k N Here, µ s = coefficient of static friction and µ k = coefficient of kinetic friction. Both µ s and µ k are dimensionless constants which depend on the nature of surfaces in contact. Value of µ k is usually less than the value of µ s i.e. constant value of kinetic friction is less than the limiting value of static friction. Note (i) In problems, if µ s and µ k are not given separately but only µ is given. Then use

fL = fk = µN (ii) If more than two blocks are placed one over the other on a horizontal ground then normal reaction between two blocks will be equal to the weight of the blocks over the common surface. A B C D

Fig. 8.110

For example,

N1 = normal reaction between A and B = mA g N2 = normal reaction between B and C = ( mA + mB ) g and so on.

296 — Mechanics - I Extra Points to Remember Friction force is electromagnetic in nature.

˜

The surfaces in contact, however smooth they may appear, actually have imperfections called asperities. When one surface rests on the other the actual area of contact is very less than the surface area of the face of contact.

Actual contact area

Fig. 8.111

The pressure due to the reaction force between the surfaces is very high as the true contact area is very small. Hence, these contact points deform a little and cold welds are formed at these points. So, in order to start the relative sliding between these surfaces, enough force has to be applied to break these welds. But, once the welds break and the surfaces start sliding over each other, the further formation of these welds is relatively slow and weak and hence a smaller force is enough to keep the block moving with uniform velocity. This is the reason why limiting value of static friction is greater than the kinetic friction.

Note By making the surfaces extra smooth, frictional force increases as actual area of contact increases and the two bodies in contact act like a single body. V

Example 8.26 Suppose a block of mass 1 kg is placed over a rough surface and a horizontal force F is applied on the block as shown in figure. Now, let us see what are the values of force of friction f and acceleration of the block a if the force F is gradually increased. Given that µ s = 0.5, µ k = 0.4 and g = 10 m/ s2 . Solution Free body diagram of the block is a

y mg

F x

N

f

Fig. 8.113

∴ or and

N N fL fk

ΣF y = 0 − mg = 0 = mg = (1)(10) = 10 N = µ s N = ( 0.5)(10) = 5 N = µ k N = ( 0.4 )(10) = 4 N

F

Fig. 8.112

Chapter 8

Laws of Motion — 297

Below is explained in tabular form, how the force of friction f depends on the applied force F . F

f

Fnet = F − f

Static or kinetic friction

Relative motion or tendency of relative motion

Acceleration of F block a = net m

0

0

0

static

Neither tendency nor actual relative motion

0

2N

2N

0

static

Tendency

0

4N

4N

0

static

Tendency

0

5N

5N

0

static

Tendency

0

Diagram

F=2N f=2N

F=4N f=4N

F=5N fL = 5 N

a = 2 m/s2

6N

4N

2N

kinetic

Actual relative motion

2 m/s 2

F=6N fk = 4 N a = 4 m/s2

8N

4N

4N

kinetic

Actual relative motion

4 m/s 2

F=8N fk = 4 N

Graphically, this can be understood as under: Note that f = F till F ≤ f L . Therefore, slope of line OA will be 1 ( y = mx ) or angle of line OA with F-axis is 45°. Here, a = 0 for F ≤ 5 N F − fK F − 4 and a= = = F − 4 for F > 5 N m 1 a-F graph is as shown in Fig. 8.115. When F is slightly increased from 5 N, acceleration of block increases from 0 to 1 m /s 2 . Think why?

f (N) A

fL = 5 N fk = 4 N

45° O

5

F (N)

Fig. 8.114

a (m/s2)

1 5

F (N)

Fig. 8.115

Note Henceforth, we will take coefficient of friction as µ unless and until specially mentioned in the question µ s and µ k separately.

298 — Mechanics - I Angle of Friction (λ) At a point of rough contact, where slipping is about to occur, the two forces acting on each object are the normal reaction N and frictional force µN . The resultant of these two forces is F and it makes an angle λ with the normal reaction, where µN …(i) tan λ = = µ or λ = tan −1 (µ ) N This angle λ is called the angle of friction.

N

F

λ µN

Fig. 8.116

Angle of Repose (α) Suppose a block of mass m is placed on an inclined plane whose inclination θ can be increased or decreased. Let, µ be the coefficient of friction between the block and the plane. At a general angle θ, m

θ

Fig. 8.117

Normal reaction

N = mg cos θ

Limiting friction f L = µN = µ mg cos θ and the driving force (or pulling force) F = mg sin θ

(Down the plane)

From these three equations we see that, when θ is increased from 0° to 90°, normal reaction N and hence, the limiting friction f L is decreased while the driving force F is increased. There is a critical angle called angle of repose (α ) at which these two forces are equal. Now, if θ is further increased, then the driving force F becomes more than the limiting friction f L and the block starts sliding. Thus, f L = F at θ = α or µ mg cos α = mg sin α or tan α = µ or

α = tan −1 (µ )

…(ii)

From Eqs. (i) and (ii), we see that angle of friction ( λ ) is numerically equal to the angle of repose. or

λ =α

From the above discussion we can conclude that If θ < α, F < f L the block is stationary. If θ = α, F = f L the block is on the verge of sliding. and if θ > α, F > f L the block slides down with acceleration F − fL a= = g (sin θ − µ cos θ ) m

Chapter 8 Variation of N , f L and F with θ, is shown graphically in Fig. 8.118.

N, fL, F mg

N = mg cos θ

F

N ∝ cos θ

or

Laws of Motion — 299

µmg

f L = µmg cos θ

N

f L ∝ cos θ

or

fL

F = mg sin θ or

F ∝ sin θ

Normally

µ mg F

F

θ N = mg cos θ

θ N < mg cos θ

θ N > mg cos θ

Solved Examples TYPED PROBLEMS Type 1. Resolution of forces

Concept Different situations of this type can be classified in following two types: (i) Permanent rest, body in equilibrium, net force equal to zero, net acceleration equal to zero or moving with constant velocity. (ii) Accelerated and temporary rest.

How to Solve? l

l

l

In the first situation, forces can be resolved in any direction. Net force (or summation of components of different forces acting on the body) in any direction should be zero. In the second situation forces are normally resolved along acceleration and perpendicular to it. In a direction perpendicular to acceleration net force is zero and along acceleration net force is ma. In temporary rest situation velocity of the body is zero but acceleration is not zero. The direction of acceleration in this case is the direction in which the body is supposed to move just after few seconds. Three situations of temporary rest are shown below. v=0 a≠0 a a u

v=0 a≠0 v=0 a≠0

u θ a

Note In the second situation also, we can resolve the forces in any direction. In that case, net force along this direction = (mass) (component of acceleration in this direction) V

Example 1 A ball of mass 1 kg is at rest in position P by means of two light strings OP and RP. The string RP is now cut and the ball swings to position Q. If θ = 45°. Find the tensions in the strings in positions OP (when RP was not cut) and OQ (when RP was cut). (Take g = 10 m/ s2 ).

O θ θ

P R

Solution In the first case, ball is in equilibrium (permanent rest). Therefore, net force on the ball in any direction should be zero.

Q

Chapter 8 ∴ or or

Laws of Motion — 305

(Σ F) in vertical direction = 0 T1 cos θ = mg mg T1 = cos θ

T1 θ

mg

Substituting m1 = 1 kg, g = 10 m/s 2 and θ = 45° we get, T1 = 10 2 N

Note Here, we deliberately resolved all the forces in vertical direction because component of the tension in RP in vertical direction is zero. In a direction other than vertical we will also have to consider component of tension in RP , which will unnecessarily increase our calculation. O In the second case ball is not in equilibrium (temporary rest). After few seconds it will move in a direction perpendicular to OQ. Therefore, net force on the ball at Q θ is perpendicular to OQ or net force along OQ = 0. T2 ∴ T2 = mg cos θ Q 90° Substituting the values, we get T2 = 5 2 N θ Here, we can see that T1 ≠ T2 a mg

Type 2. To find tension at some point (say at P) if it is variable

How to Solve? l l

l l l l

l l l

Find acceleration (a common acceleration) of the system by using the equation net pulling or pushing force a= total mass In some cases, ‘a’ will be given in the question. Cut the string at P and divide the system in two parts. Make free body diagram of any one part (preferably of the smaller one). In its FBD make one tension at point P in a direction away from the block with which this part of the string is attached. Write the equation, Fnet = ma for this part. You will get tension at P. In this equation m is not the total mass. It is mass of this part only.

306 — Mechanics - I V

Example 2 In the given figure mass of string AB is 2 kg. Find tensions at A, B and C, where C is the mid point of string. F = 100 N

a 2 kg A C B 4 kg

Solution

a= =

F − weight of 2 kg – weight of 4 kg – weight of string mass of 2 kg + mass of 4 kg + mass of string 100 − 20 − 40 − 20 2+4+2

( g = 10 m/s 2)

TA A

20 = = 2.5 m/s 2 8 Refer Fig. (a) or Refer Fig. (b) or or Refer Fig. (c) or or

TC C

TA − mAB g − 40 = (mAB + 4)a TA − 20 − 40 = (2 + 4)(2.5) TA = 75 N TC − mBC g − 40 = (mBC + 4)a TC − 10 − 40 = (1 + 4)(2.5) TC = 62.5 N TB − 40 = 4a TB = 40 + 4 × 2.5 TB = 50 N

B

Ans.

B

a

a

4 kg

4 kg

(a)

(b) TB

Ans.

B

a

4 kg (c)

Ans.

Note Tension at a general point P can be given by : Here

TP = [( Σ mass below P ) × geff ] geff = g + a = 12.5 m/ s2

Type 3. Based on constraint relation between a block (or a plank) and a wedge. These type of problems can be understood by following two examples: V

Example 3 In the figure shown find relation between magnitudes of a A and a B .

A

90° B θ

Chapter 8

Laws of Motion — 307

Solution xA = xB sin θ

...(i)

Here, θ = constant A xB q

xA q

Double differentiating Eq. (i) with respect to time, we get a A = aB sin θ V

Example 4 In the arrangement shown in the figure, the rod of mass m held by two smooth walls, remains always perpendicular to the surface of the wedge of mass M. Assuming all the surfaces to be frictionless, find the acceleration of the rod and that of the wedge. Solution Let acceleration of m be a 1 (absolute) and that of M be a 2 (absolute). Writing equations of motion only in the directions of a1 or a 2. For m mg cos α − N = ma1 For M, N sin α = Ma 2 Here, N = normal reaction between m and M As discussed above, constraint equation can be written as, a1 = a 2 sin α

Ans. Fixed wall

a1

m

a2

α

…(i) …(ii)

...(iii)

Solving above three equations, we get acceleration of rod,

a1 =

and acceleration of wedge

a2 =

mg cos α sin α  M  m sin α +  sin α   mg cos α M m sin α + sin α

Ans.

Ans.

Type 4. Based on constraint relation which keeps on changing.

Concept (i) In the constraint relations discussed so far the relation between different accelerations was fixed. For example: In the two illustrations discussed above a A = a B sinθ or a1 = a 2 sinα but these relations were fixed, as θ or α was constant. (ii) In some cases, θ or α keeps on changing. Therefore, the constraint relation also keeps on changing. (iii) In this case, constraint relation between different accelerations becomes very complex. So, normally constraint relation between velocities is only asked.

308 — Mechanics - I V

Example 5 In the arrangement shown in the figure, the ends P and Q of an unstretchable string move downwards with uniform speed U . Pulleys A and B are fixed. Mass M moves upwards with a speed (JEE 1982) A

B θθ Q

P M

(a) 2U cos θ

(b)

U cos θ

(c)

2U cos θ

(d) U cos θ

Solution In the right angle ∆ PQR

P

c

Q

l2 = c2 + y2

y

Differentiating this equation with respect to time, we get dl dy  dy l  dl  or  −  =  −  = 0 + 2y  dt  y  dt  dt dt dy l 1 and − dl /dt = U − = vM , = dt y cos θ

l

2l

Here,

vM =

Hence,

U cos θ

θ

θ R

U

U

M

c = constant, l and y are variables

∴ The correct option is (b).

Note Here θ is variable. Therefore the constraint relation v M = V

U is also variable. cos θ

Example 6 In the adjoining figure, wire PQ is smooth, ring A has a mass 1 kg and block B, 2 kg. If system is released from rest with θ = 60°, find P A

Q

θ

B

(a) constraint relation between their velocities as a function of θ. (b) constraint relation between their accelerations just after the release at θ = 60°. (c) tension in the string and the values of these accelerations at this instant. v1 , a 1 Solution M and Q are two fixed fixed points. Therefore, MQ = constant = c l = length of string = constant. (a) In triangle MQA, (l − y)2 = x2 + c2 Differentiating w.r.t time, we get  dx  dy 2 (l − y)  −  = 2x  +  + 0  dt   dt 

P A

x θ (l – y)

M

Q

c Q y B

v2, a2

Chapter 8 or ∴

Laws of Motion — 309

 dy  dx (l − y)  +  = x  −   dt   dt  dy  x   dx =  −  dt  l − y  dt 

…(i) …(ii)

y is increasing with time, +

∴

dy = v2 dt

x is decreasing with time ∴

−

dx = v1 dt

x = cos θ l− y

and

Substituting these values in Eq. (ii), we have v2 = v1 cos θ (b) Further differentiating Eq. (i) we have, 2  d 2x  dx 2 d 2y  dy ( l − y ) − = − x ⋅ 2 +       dt   dt   dt 2  dt  dx dy Just after the release, v1 , v2 , and all are zero. Substituting in Eq. (iii), we have, dt dt d 2y  x   d 2x  =  − dt 2  l − y  dt 2 

Ans.

…(iii)

…(iv)

d 2y d 2x = a 2 and − 2 = a1 2 dt dt x 1 = cos θ = cos 60° = l− y 2

Here,

a2 =

Substituting in Eq. (iv), we have

a1 2

…(v)

(c) For A Equation is A

a1

T

60° B

a2

T w = 20 N

T cos 60° = mA a1 For B

or

T = (1)a1 = a1 2

20 − T = mB a 2 20 − T = 2a 2

…(vii)

Solving Eqs. (v), (vi) and (vii), we get 40 20 T= N ⇒ a1 = m/s 2 3 3 10 and a2 = m/s 2 3

Note Eq. (iii) converts into a simple Eq. (iv), just after the release when v 1 , v 2 ,

…(vi)

dx dy and all are zero. dt dt

310 — Mechanics - I Type 5. To find whether the block will move or not under different forces kept over a rough surface

How to Solve? l l l l

l l l l l l l

The rough surface may be horizontal, inclined or vertical. Resolve the forces along the surface and perpendicular to the surface. In most of the cases, acceleration perpendicular to the surface is zero. So net force perpendicular to the surface should be zero. By putting net force perpendicular to the surface equal to zero we will get normal reaction N. After finding, N, calculate µ sN, µ k N or µN. Calculate net force along the plane and call it the driving force F. Now, if F ≤ µ sN (f = Force of friction) Then f =F and Fnet = 0 or a = 0, If F > µ sN F Then, (in the direction of F) f = µ k N and Fnet = F − f or a = net m

V

(a) find the force of friction acting on the block. (b) state whether the block will move or not. If yes then with what acceleration? Solution Resolving the force in horizontal (along the plane) and in vertical (perpendicular to the plane) directions (except friction) Here, R is the normal reaction. Σ Fy = 0 ⇒ R = 26 N µ sR = 0.6 × 26 = 16.6 N µ kR = 0.4 × 26 = 10.4 N Σ Fx = net driving force F = 14 N (a) Since, F ≤ µ sR ∴ Force of friction f = F = 14 N This friction will act in the opposite direction of F. (b) Since, F ≤ µ sR, the block will not move. V

45° 4N

2 kg

µs = 0.6, µk = 0.4

R y 4N

10N F

6N W = 20 N

Example 8 In the figure shown, (a) find the force of friction acting on the block. (b) state whether the block will move or not. If yes then with what acceleration? 4N 2

10√2 N

6N

Example 7 In the figure shown,

kg

37°

µs= 0.4, µk= 0.3

x

Chapter 8

Laws of Motion — 311

Solution Resolving the forces along the plane and perpendicular to

y

the plane. (except friction) Here, R is the normal reaction.

Σ Fy = 0 ⇒ R = 16 N µ sR = 0.4 × 16 = 6.4 N µ kR = 0.3 × 16 = 4.8 N Σ Fx = net driving force F = (12 − 4) N = 8 N (a) Since, F > µ sR, therefore kinetic friction or 4.8 N will act in opposite direction of F. (b) Since, F > µ sR, the block will move in the direction of F (or downwards) with an acceleration, F F − f 8 − 4.8 a = net = = = 1.6 m/s 2 m m 2 This acceleration is in the direction of F (or downwards).

V

R

4N x

F

7°

3 sin 20 12 N = 37°

20 cos 37° = 16 N W = 20 N

Example 9 A block of mass 1 kg is pushed against a rough vertical wall with a 1 force of 20 N, coefficient of static friction being . Another horizontal force of 4 10 N is applied on the block in a direction parallel to the wall. Will the block move? If yes, in which direction? If no, find the frictional force exerted by the wall on the block. ( g = 10 m / s2 ) Solution Normal reaction on the block from the wall will be ( Fnet = 0, perpendicular to the wall) F = 20 N

Therefore, limiting friction

A

R = F = 20 N  1 fL = µR =   (20) = 5 N  4

Weight of the block is w = mg = (1)(10) = 10 N A horizontal force of 10 N is applied to the block. Both weight and this force are along the wall. The resultant of these two forces will be 10 2 N in the direction shown in figure. Since, this resultant is greater than the limiting friction. The block will move in the direction of Fnet with acceleration F −f 10 2 − 5 a = net L = = 9.14 m/s 2 m 1

10 N 45°

10 N

Fnet = 10√2 N

Type 6. To draw acceleration versus time graph. Following three examples will illustrate this type. V

Example 10 In the figure shown, F is in newton and t in seconds. Take g =10 m/ s2 . (a) Plot acceleration of the block versus time graph. (b) Find force of friction at, t = 2 s and t = 8 s.

2 kg µ = 0.6

F = 2t

312 — Mechanics - I Solution (a) Normal reaction, R = mg = 20N. Limiting value of friction, f1 = µR = 0.6 × 20 = 12N The applied force F (= 2t ) crosses this limiting value of friction at 6s. Therefore, upto 6 s block remains stationary and after 6s it starts moving. After 6 s, friction becomes constant at 12 N but the applied force keeps on increasing. Therefore, acceleration keeps on increasing. For t ≤ 6 s ...(i) f = F = 2t ∴

Fnet = F − f = 0 F a = net = 0 m

For t > 6 s F = 2t f = 12 N = f1 Fnet = F − f = 2t − 12 F 2t − 12 a = net = = (t − 6) ∴ m 2 ∴ a-t graph is a straight line with slope =1 and intercept = − 6. Corresponding a-t graph is as shown. (b) At t = 2 s, f = 4 N [from Eq. (i)] At t = 8 s, f = 12 N [from Eq. (ii)] V

…(ii) a(m/s2)

45° 6

t (s)

–6

Example 11 Repeat the above problem, if instead of µ we are given µ s and µ k , where µ s =06 . and µ k =04 .. Solution (a) R = mg = 20 N µ sR = 0.6 × 20 = 12 N µ kR = 0.4 × 20 = 8 N Upto 6 s, situation is same but after 6 s, a constant kinetic friction of 8N will act. At 6 s, friction will suddenly change from 12 N (= µ sR) to 8N (= µ kR) and direction of friction is opposite to its motion. Therefore, at 6 s it will start with an initial acceleration. decrease in friction 12 − 8 ai = = = 2 m/s 2 mass 2 For t ≤ 6 s

∴

f = F = 2t Fnet = F − f = 0 F a = net = 0 m

For t ≥ 6 s

∴

F = 2t f = µ kR = 8N Fnet = F − f = 2t − 8 F 2t − 8 a = net = = (t − 4) m 2

…(i)

a(m/s2)

2

45°

t (s)

6 At t = 6 s, we can see that, ai = 2 m/s 2 Further, a - t graph is a straight line of slope = 1 and intercept = − 4. Corresponding a - t graph is as shown in figure.

Chapter 8 V

Laws of Motion — 313

Example 12 Two blocks A and B of masses 2 kg and A 4 kg are placed one over the other as shown in figure. A time varying horizontal force F = 2t is applied on the B upper block as shown in figure. Here t is in second and F is in newton. Draw a graph showing accelerations of A and B on y-axis and time on x-axis. Coefficient of friction 1 between A and B is µ = 2 and the horizontal surface over which B is placed is

F = 2t

smooth. ( g = 10 m/ s2 )

Concept In the given example, block A will move due to the applied force but block B moves due to friction (between A and B). But there is a limiting value of friction between them. Therefore, there is a limiting value of acceleration (of block B). Up to this acceleration they move as a single block with a common acceleration, but after that acceleration of B will become constant (as friction acting on this block will become constant). But acceleration of A will keep on increasing as a time increasing force is acting on it. Solution Limiting friction between A and B is  1 fL = µmA g =   (2) (10) = 10 N  2 Block B moves due to friction only. Therefore, maximum acceleration of B can be f 10 a max = L = = 2.5 m/s 2 mB 4 Thus, both the blocks move together with same acceleration till the common A acceleration becomes 2.5 m/s 2, after that acceleration of B will become constant fL = 10 N while that of A will go on increasing. To find the time when the acceleration of both the blocks becomes 2.5 m/s 2 (or when slipping will start between A and B) we will write F 2t 2.5 = = (mA + mB ) 6 ∴ Hence, for

F = 2t

B fL = 10 N

For t ≥ 7.5 s

t = 7.5 s t ≤ 7.5 s a A = aB =

F 2t t = = mA + mB 6 3

Thus, a A versus t or aB versus t graph is a straight line passing through origin of slope For, and or

t ≥ 7.5 s aB = 2.5 m/s 2 = constant F − fL aA = mA aA =

2t − 10 2

or

aA = t − 5

1 . 3

314 — Mechanics - I a A or a B aA

45°

2.5 m/s 2

a

=a A

B

aB

tan θ = 1 3

θ

t

7.5 s

Thus, a A versus t graph is a straight line of slope 1 and intercept –5. While aB versus t graph is a straight line parallel to t-axis. The corresponding graph is as shown in above figure.

Type 7. When two blocks in contact are given different velocities and after some time, due to friction their velocities become equal.

Concept In the figure shown, if v1 > v 2 (or v1 ≠ v 2 ) then there is a relative v1 Rough A motion between the two blocks. v2 B As v1 > v 2 , relative motion of A is towards right and relation motion of B is towards left. Since, relative motion is there, so kinetic friction Smooth (or limiting value of friction) will act in the opposite direction of relative motion. This friction (and acceleration due to this force) with decrease the velocity of A and increase the velocity of B. After some time when their velocities become equal, frictional force between them becomes zero and they continue to be moving with that common velocity (as the ground is smooth).

How to Solve? l

l l

Find value of kinetic friction or limiting value of friction (f = µ k N or µN ) between the two blocks and then  f accelerations of these blocks  =  . Then write v1 = v2 , as their velocities become same when relatative  m motion is stopped. or …(i) u1 + at 1 = u 2 + a2t Substituting the proper values of u1, a1, u 2 and a2 in Eq. (i), we can find the time when the velocities become equal.

V

Example 13 Coefficient of friction between two blocks shown in figure is µ = 0.6 . The blocks are given velocities in the directions shown in figure. Find

2 kg 18 m/s

1 kg

(a) the time when relative motion between them is Smooth stopped. (b) the common velocity of the two blocks. (c) the displacements of 1 kg and 2 kg blocks upto that instant. (Take g = 10 m/ s2) Note Assume that lower block is sufficiently long and upper block does not fall from it.

3 m/s

Chapter 8

Laws of Motion — 315

Solution Relative motion of 2 kg block is towards right. Therefore, maximum friction on this block will act towards left 2 kg 12 N

3 m/s

a2

–

+

12 N 1 kg

18 m/s

a1

f = µN = (0.6) (2) (10) = 12 N 12 a2 = − = − 6 m/s 2 2 12 a1 = = 12 m/s 2 1 (a) Relative motion between them will stop when, v1 = v2 or u1 + a1t = u2 + a 2t or −18 + 12t = 3 − 6t 7 Solving we get, t= s 6 (b) Substituting value of ' t ' in Eq. (i) either on RHS or on LHS we have, common velocity = − 4 m/s 1 (c) s1 = u1 t + a1t 2 2  7 1  7 = (−18)   + (12)    6 2  6 = − 12.83 m 1 s2 = u2t + a 2t 2 2  7 1  7 = (3)   + (−6)    6 2  6 = − 0.58 m

...(i) Ans.

Ans.

2

Ans.

2

Ans.

Type 8. Acceleration or retardation of a car

Concept A car accelerates or retards due to friction. On a horizontal road maximum available friction is µN or µ mg (as N = mg). Therefore, maximum acceleration or retardation of a car on a horizontal road is f µ mg a max = max = =µ g m m On an inclined plane maximum value of friction is µN or µ mg cos θ (as N = mg cosθ). Now mg sinθ is a force which is always downwards but the frictional force varying from 0 to µ mg cos θ can be applied in upward or downward direction by the application of brakes or accelerator.

316 — Mechanics - I V

Example 14 On a horizontal rough road, value of coefficient of friction µ =04 .. Find the minimum time in which a distance of 400m can be covered. The car starts from rest and finally comes to rest. Solution Maximum friction on horizontal rough road, fmax = µ mg ∴ Maximum acceleration or retardation of the car may be f µ mg a max or a = max = =µ g m m = 0.4 × 10 = 4 m/s 2 Let, the car accelerates and retards for time ‘t’ with 4 m/s 2. 1 2 1 2 Then, at + at = 400 m 2 2 2 or at = 400 m or 4t 2 = 400 or t = 10 s Therefore, the minimum time is 20 s (10 s of acceleration and 10 s of retardation).

V

Ans.

Example 15 A car is moving up the plane. Angle of inclination is θ and coefficient of friction is µ. (a) What is the condition in which car can be accelerated? If this condition is satisfied then find (b) maximum acceleration of the car. (c) minimum retardation of the car. (d) maximum retardation of the car. Solution (a) mg sin θ in all conditions is downwards but direction of

v

friction may be upwards or downwards. We will have to press accelerator for upward friction and brakes for downward friction. To accelerate the car friction should be upwards. Therefore, car can be accelerated if maximum upward friction > mg sin θ or

µ mg cos θ > mg sin θ or µ > tan θ

g

sin

θ

m

(b) Maximum acceleration maximum upwards force mass µ mg cos θ − mg sin θ = = (µ g cos θ − g sin θ ) m

=

(c) Minimum retardation will be zero, when upward friction = mg sin θ (d) Maximum retardation maximum downward force = m mg sin θ + µ mg cos θ = m = ( g sin θ + µ g cos θ) This is the case, when maximum friction force acts in downward direction.

θ

Miscellaneous Examples V

Example 16 In the adjoining figure, angle of plane θ is increased from 0° to 90°. Plot force of friction ‘f’ versus θ graph.

m

µ s , µk

Solution Normal reaction N = mg cos θ. Limiting value of static friction, fL = µ sN = µ s mg cos θ

θ

Constant value of kinetic friction fK = µ kN = µ kmg cos θ Driving force down the plane, F = mg sin θ Now block remains stationary and f = F until F becomes equal to fL mg sin θ = µ s mg cos θ

or

or tan θ = µ s or θ = tan −1 (µ s ) = θ r (say) After this, block starts moving and constant value of kinetic friction will act. Thus, For θ ≤ tan −1(µ s ) or θr f = F = mg sin θ or f ∝ sin θ At, θ = 0°, f = 0 and at θ = tan −1 (µ s ) or θ r or f = mg sin θ r or µ smg cos θ r −1 For θ > tan (µ s ) or θr f = fk = µ kmg cos θ or f ∝ cos θ At θ = tan −1 (µ s ) or θ r f = µ kmg cos θ r and at θ = 90° f =0 Corresponding f versus θ graph is as shown in figure In the figure, OP is sine graph and MN is cos graph, f1 = mg sin θ r = µ s mg cos θ r f2 = µ k mg cos θ r V

f f1 f2

P M

θr = tan–1(µs)

N O

θr

90°

θ

Example 17 Figure shows two blocks in contact sliding down an inclined surface of inclination 30°. The friction coefficient between the block of mass 2.0 kg and the incline is µ 1 = 0.20 and that between the block of mass 4.0 kg and the incline is µ 2 = 0.30. Find the acceleration of 2.0 kg block. ( g = 10 m/ s2 ). g 4k

g 2k

o

30

Solution Since, µ1 < µ 2, acceleration of 2 kg block down the plane will be more than the acceleration of 4 kg block, if allowed to move separately. But, as the 2.0 kg block is behind the 4.0 kg block both of them will move with same acceleration say a. Taking both the blocks as a single system:

318 — Mechanics - I Force down the plane on the system = (4 + 2) g sin 30°  1 = (6)(10)   = 30 N  2 Force up the plane on the system = µ1 (2)( g ) cos 30° + µ 2(4)( g ) cos 30° = (2 µ1 + 4 µ 2) g cos 30° = (2 × 0.2 + 4 × 0.3) (10) (0.86) ≈ 13.76 N ∴ Net force down the plane is F = 30 − 13.76 = 16.24 N ∴ Acceleration of both the blocks down the plane will be a. F 16.24 a= = = 2.7 m/s 2 4+2 6 V

Ans.

Example 18 Figure shows a man standing stationary with respect to a horizontal conveyor belt that is accelerating with 1 ms −2 . What is the net force on the man? If the coefficient of static friction between the man’s shoes and the belt is 0.2, upto what maximum acceleration of the belt can the man continue to be stationary relative to the belt? Mass of the man = 65 kg. ( g = 98 . m / s2 ) Solution As the man is standing stationary w.r.t. the belt, ∴

Acceleration of the man = Acceleration of the belt

= a = 1 ms −2 Mass of the man, m = 65 kg Net force on the man Ans. = ma = 65 × 1 = 65 N Given coefficient of friction, µ = 0.2 ∴ Limiting friction, fL = µmg If the man remains stationary with respect to the maximum acceleration a 0 of the belt, then ∴ V

ma 0 = fL = µmg a 0 = µg = 0.2 × 9.8 = 1.96 ms −2

Example 19 Two blocks of masses m = 5 kg and M = 10 kg are connected by a string passing over a pulley B as shown. Another string connects the centre of pulley B to the floor and passes over another pulley A as shown. An upward force F is applied at the centre of pulley A. Both the pulleys are massless. Find the acceleration of blocks m and M, if F is (a) 100 N (b) 300 N (c) 500 N (Take g = 10 m / s2)

Ans. F

A

B

m

M

Chapter 8

Laws of Motion — 319

Solution Let T0 = tension in the string passing over A T = tension in the string passing over B 2T0 = F and 2T = T0 ∴ T = F /4 (a) T = F /4 = 25 N weights of blocks are mg = 50 N Mg = 100 N

F

As T < mg and Mg both, the blocks will remain stationary on the floor.

T0

A

(b) T = F /4 = 75 N As T < Mg and T > mg, M will remain stationary on the floor, whereas m will move. T0 Acceleration of m, T − mg 75 − 50 a= = m 5

T0

B

T

= 5 m/s 2

T

Ans.

(c) T = F /4 = 125 N As T > mg and Mg both the blocks will accelerate upwards. Acceleration of m, T − mg 125 − 50 a1 = = = 15 m/s 2 m 5 Acceleration of M, T − Mg 125 − 100 a2 = = = 2.5 m/s 2 M 10 V

Example 20 Consider the situation shown in figure. The block B moves on a frictionless surface, while the coefficient of friction between A and the surface on which it moves is 0.2. Find the acceleration with which the masses move and also the tension in the strings. (Take g = 10 m/ s2 ).

4 kg

8 kg

A

B

c C 20 kg

Solution Let a be the acceleration with which the masses move and T1 and T2 be the tensions

in left and right strings. Friction on mass A is µmg = 8N. Then equations of motion of masses A, B and C are For mass A …(i) T1 − 8 = 4a For mass B …(ii) T2 = 8a For mass C …(iii) 200 − T1 − T2 = 20a Adding the above three equations, we get 32a = 192 or a = 6 m/s 2 From Eqs. (i) and (ii), we have T2 = 48 N and T1 = 32 N

320 — Mechanics - I V

Example 21 Two blocks A and B of masses 1 kg and 2 kg respectively are connected by a string, passing over a light frictionless pulley. Both the blocks are resting on a horizontal floor and the pulley is held such that string remains just taut.At moment t = 0, a force F = 20 t newton starts acting on the pulley along vertically upward direction as shown in figure. Calculate F = 20 t

A

B

(a) velocity of A when B loses contact with the floor. (b) height raised by the pulley upto that instant. (Take, g = 10 m / s2) Solution (a) Let T be the tension in the string. Then, 2T = 20 t or T = 10 t newton Let the block A loses its contact with the floor at time t = t1. This happens when the tension in string becomes equal to the weight of A. Thus, or

T = mg 10 t1 = 1 × 10

or

t1 = 1 s

…(i)

Similarly, for block B, we have 10t2 = 2 × 10 or t2 = 2 s i.e. the block B loses contact at 2 s. For block A, at time t such that t ≥ t1 let a be its acceleration in upward direction. Then, 10t − 1 × 10 = 1 × a = (dv/dt ) or dv = 10(t − 1) dt Integrating this expression, we get v

∫0 or

dv = 10 ∫

t

…(ii)

…(iii)

(t − 1) dt

1

v = 5t 2 − 10t + 5

…(iv)

Substituting t = t2 = 2 s v = 20 − 20 + 5 = 5 m/s (b) From Eq. (iv), dy = (5t 2 − 10t + 5) dt where, y is the vertical displacement of block A at time t (≥ t1 ). Integrating, we have

…(v) …(vi)

Chapter 8 y =h

∫y = 0

dy = ∫

t =2 t =1

Laws of Motion — 321

(5t 2 − 10t + 5) dt 2

2

 t2   t3  2 5 h = 5   − 10   + 5 [t ]1 = m 3  2 1  3 1 h 5 ∴ Height raised by pulley upto that instant = = m 2 6 V

Ans.

Example 22 Find the acceleration of the body of mass m2 in the arrangement shown in figure. If the mass m2 is η times great as the mass m1 ,and the angle that the inclined plane forms with the horizontal is equal to θ. The masses of the pulleys and threads, as well as the friction, are assumed to be negligible.

m1 θ

m2

Solution Here, by constraint relation we can see that the acceleration of m2 is two times that of m1. So, we assume if m1 is moving up the inclined plane with an acceleration a, the acceleration of mass m2 going down is 2a. The tensions in different strings are shown in figure. 2T

2T

a m1

θ

T

T m2

2a

The dynamic equations can be written as For mass m1: 2T − m1 g sin θ = m1a For mass m2 : m2g − T = m2(2a ) Substituting m2 = ηm1 and solving Eqs. (i) and (ii),we get 2 g (2η − sin θ ) Acceleration of m2 = 2a = 4η + 1 V

Example 23 In the arrangement shown in figure the mass of the ball is η times as great as that of the rod. The length of the rod is l, the masses of the pulleys and the threads, as well as the friction, are negligible. The ball is set on the same level as the lower end of the rod and then released. How soon will the ball be opposite the upper end of the rod ?

…(i) …(ii) Ans.

322 — Mechanics - I Solution

From constraint relation we can see that the acceleration of the rod is double than that of the acceleration of the ball. If ball is going up with an acceleration a, rod will be coming down with the acceleration 2a, thus, the relative acceleration of the ball with respect to rod is 3a in upward direction. If it takes time t seconds to reach the upper end of the rod, we have 2T T 2l …(i) t= a 3a Let mass of ball be m and that of rod is M, the dynamic equations of these are For rod Mg − T = M (2a ) For ball 2T − mg = ma Substituting m = ηM and solving Eqs. (ii) and (iii), we get  2 − η a=  g  η + 4 t=

From Eq. (i), we have

V

2l(η + 4) 3 g (2 − η)

2T

T 2a

…(ii) …(iii)

Ans.

Example 24 Figure shows a small block A of mass m kept at the left end of a plank B of mass M = 2m and length l. The system can slide on a horizontal road. The system is started towards right with the initial velocity v. The friction coefficients between the road and the plank is 1/2 and that between the plank and the block is 1/4. Find A B l

(a) the time elapsed before the block separates from the plank. (b) displacement of block and plank relative to ground till that moment. Solution

There will be relative motion between block and plank and plank and road. So at each surface limiting friction will act. The direction of friction forces at different surfaces are as shown in figure. f1

A

B

f2

f1

Here,

 1 f1 =   (mg )  4

and

 1  3 f2 =   (m + 2m) g =   mg  2  2

Retardation of A is

a1 =

f1 g = m 4

Chapter 8

Laws of Motion — 323

f2 − f1 5 = g 2m 8

and retardation of B is

a2 =

Since,

a 2 > a1

Relative acceleration of A with respect to B is 3 g 8 Initial velocity of both A and B is v. So, there is no relative initial velocity. Hence, a r = a 2 − a1 =

s=

(a) Applying

1 3 ar t2 = gt 2 2 16 l t =4 3g l=

or ∴ (b) Displacement of block

1 2 at 2

sA = uAt −

Ans.

1 a t2 2 A

or

sA = 4v

l 1 g  16l − ⋅ ⋅  3g 2 4  3g 

or

sA = 4v

l 2 − l 3g 3

Displacement of plank

sB = uB t −

or

sB = 4v

l 1  5   16l −  g   3g 2 8   3g 

or

sB = 4v

l 5 − l 3g 3

g   a A = a1 =   4

1 a t2 2 B

5    aB = a 2 = g  8  Ans.

Note We can see that sA − sB = l. Which is quite obvious because block A has moved a distance l relative to plank.

Exercises LEVEL 1 Assertion and Reason Directions :

Choose the correct option.

(a) If both Assertion and Reason are true and the Reason is correct explanation of the Assertion.

(b) If both Assertion and Reason are true but Reason is not the correct explanation of Assertion. (c) If Assertion is true, but the Reason is false. (d) If Assertion is false but the Reason is true.

1. Assertion : If net force on a rigid body in zero, it is either at rest or moving with a constant linear velocity. Nothing else can happen. Reason : Constant velocity means linear acceleration is zero.

2. Assertion : Three concurrent forces are F1 , F2 and F3 . Angle between F1 and F2 is 30° and between F1 and F3 is 120°. Under these conditions, forces cannot remain in equilibrium. Reason : At least one angle should be greater than 180°.

3. Assertion : Two identical blocks are placed over a rough inclined plane. One block is given an upward velocity to the block and the other in downward direction. If µ =

1 and θ = 45° the ratio 3

of magnitudes of accelerations of two is 2 : 1. 1+µ Reason : The desired ratio is . 1−µ

4. Assertion : A block A is just placed inside a smooth box B as shown in figure. Now, the box is

given an acceleration a = ( 3$j − 2i$ ) ms−2. Under this acceleration block A cannot remain in the position shown. B y A x

Reason : Block will require ma force for moving with acceleration a.

5. Assertion : A block is kept at rest on a rough ground as shown. Two forces F1 and F2 are acting on it. If we increase either of the two forces F1 or F2, force of friction acting on the block will increase. Reason : By increasing F1, normal reaction from ground will increase. F1 F2

Chapter 8

Laws of Motion — 325

6. Assertion : In the figure shown, force of friction on A from B will always be right wards. Reason : Friction always opposes the relative motion between two bodies in contact. F1

A B

F2

Smooth

7. Assertion : In the figure shown tension in string AB always lies between m1g and m2g. ( m1 ≠ m2 )

A B

m1 m2

Reason : Tension in massless string is uniform throughout.

8. Assertion : Two frames S1 and S 2 are non-inertial. Then frame S 2 when observed from S1 is inertial. Reason : A frame in motion is not necessarily a non-inertial frame.

9. Assertion : Moment of concurrent forces about any point is constant. Reason : If vector sum of all the concurrent forces is zero, then moment of all the forces about any point is also zero.

10. Assertion : Minimum force is needed to move a block on rough surface, if θ = angle of friction. Reason : Angle of friction and angle of repose are numerically same. F θ

Rough

11. Assertion : When a person walks on a rough surface, the frictional force exerted by surface on the person is opposite to the direction of his motion. Reason : It is the force exerted by the road on the person that causes the motion.

326 — Mechanics - I

Objective Questions Single Correct Option 1. Three equal weights A, B and C of mass 2 kg each are hanging on a string passing over a fixed frictionless pulley as shown in the figure. The tension in the string connecting weights B and C is approximately (a) zero (b) 13 N (c) 3.3 N (d) 19.6 N

A

B C

2. Two balls A and B of same size are dropped from the same point under gravity. The mass of A is greater than that of B. If the air resistance acting on each ball is same, then (a) both the balls reach the ground simultaneously (b) the ball A reaches earlier (c) the ball B reaches earlier (d) nothing can be said

3. A block of mass m is placed at rest on an inclined plane of inclination θ to the horizontal. If the coefficient of friction between the block and the plane is µ, then the total force the inclined plane exerts on the block is

(a) mg

(b) µ mg cos θ

(c) mg sin θ

(d) µ mg tan θ

4. In the figure a block of mass 10 kg is in equilibrium. Identify the string in which the tension is zero.

30° B

(a) B (b) C (c) A (d) None of the above

C

A 10 kg

5. At what minimum acceleration should a monkey slide a rope whose breaking strength is

2 rd of 3

its weight ? (a)

2g 3

(b) g

(c)

g 3

(d) zero

6. For the arrangement shown in the figure, the reading of spring balance is (a) 50 N (b) 100 N 5 kg

(c) 150 N

10 kg

(d) None of the above

7. The time taken by a body to slide down a rough 45° inclined plane is twice that required to slide down a smooth 45° inclined plane. The coefficient of kinetic friction between the object and rough plane is given by (a) (c)

1 3

(b) 3 4

(d)

3 4 2 3

Chapter 8

Laws of Motion — 327

8. The force required to just move a body up the inclined plane is double the force required to just prevent the body from sliding down the plane. The coefficient of friction is µ. If θ is the angle of inclination of the plane than tanθ is equal to

(a) µ

(b) 3µ

(d) 0.5 µ

(c) 2µ

9. A force F1 accelerates a particle from rest to a velocity v. Another force F2 decelerates the same particle from v to rest, then (a) F1 is always equal to F2 (b) F2 is greater than F1 (c) F2 may be smaller than, greater than or equal to F1 (d) F2 cannot be equal to F1

10. A particle is placed at rest inside a hollow hemisphere of radius R. The coefficient of friction between the particle and the hemisphere is µ =

1 3

. The maximum height up to which the

particle can remain stationary is (a)

 3 R (b) 1 − 2 

R 2

(c)

3 R 2

(d)

3R 8

11. In the figure shown, the frictional coefficient between table and block is 0.2. Find the ratio of tensions in the right and left strings. T2

T1 5kg

5kg 5kg 15kg

(a) 17 : 24

(b) 34 : 12

(c) 2 : 3

(d) 3 : 2

12. A smooth inclined plane of length L having inclination θ with the horizontal is inside a lift which is moving down with a retardation a. The time taken by a body to slide down the inclined plane from rest will be (a)

2L ( g + a ) sin θ

(b)

2L ( g − a ) sin θ

(c)

2L a sin θ

(d)

2L g sin θ

13. A block rests on a rough inclined plane making an angle of 30° with horizontal. The coefficient of static friction between the block and inclined plane is 0.8. If the frictional force on the block is 10 N, the mass of the block in kg is (g = 10 m/s 2) (a) 2.0 (c) 1.6

(b) 4.0 (d) 2.5

14. In figure, two identical particles each of mass m are tied together with an inextensible string. This is pulled at its centre with a constant force F. If the whole system lies on a smooth horizontal plane, then the acceleration of each particle towards each other is 3 2 2 (c) 3 (a)

F m F m

1 F 2 3 m F (d) 3 m (b)

30° 30°

F

328 — Mechanics - I 15. A block of mass m is placed at rest on a horizontal rough surface with angle of friction φ. The

block is pulled with a force F at an angle θ with the horizontal. The minimum value of F required to move the block is

(a)

mg sin φ cos(θ − φ )

(b)

mg cos φ cos (θ − φ )

(c) mg tan φ

(d) mg sin φ

16. A block of mass 4 kg is placed on a rough horizontal plane. A time dependent horizontal force

F = kt acts on the block. Here k = 2 Ns −1. The frictional force between the block and plane at time t = 2 s is (µ = 0.2)

(a) 4 N (c) 12 N

(b) 8 N (d) 10 N

17. A body takes time t to reach the bottom of a smooth inclined plane of angle θ with the horizontal. If the plane is made rough, time taken now is 2 t. The coefficient of friction of the rough surface is 3 tan θ 4 1 (c) tan θ 4

2 tan θ 3 1 (d) tan θ 2

(b)

(a)

18. A man of mass m slides down along a rope which is connected to the ceiling of an elevator with deceleration a relative to the rope. If the elevator is going upward with an acceleration a relative to the ground, then tension in the rope is (b) m( g + 2a ) (d) zero

(a) mg (c) m( g + a )

19. A 50 kg person stands on a 25 kg platform. He pulls on the rope which is attached to the platform via the frictionless pulleys as shown in the figure. The platform moves upwards at a steady rate if the force with which the person pulls the rope is

(a) 500 N

(b) 250 N

(c) 25 N

(d) None of these

20. A ladder of length 5 m is placed against a smooth wall as shown in figure. The coefficient of friction is µ between ladder and ground. What is the minimum value of µ, if the ladder is not to slip?

A

O

(a) µ =

1 2

(b) µ =

1 4

AB = 5 m AO = 4 m OB = 3 m

B

(c) µ =

3 8

(d) µ =

5 8

Chapter 8

Laws of Motion — 329

21. If a ladder weighing 250 N is placed against a smooth vertical wall having coefficient of friction between it and floor 0.3, then what is the maximum force of friction available at the point of contact between the ladder and the floor? (a) 75 N

(b) 50 N

(c) 35 N

(d) 25 N

22. A rope of length L and mass M is being pulled on a rough horizontal floor by a constant horizontal force F = Mg. The force is acting at one end of the rope in the same direction as the length of the rope. The coefficient of kinetic friction between rope and floor is 1/2. Then, the tension at the midpoint of the rope is

(a)

Mg 4

(b)

2Mg 5

(c)

Mg 8

(d)

 

23. A heavy body of mass 25 kg is to be dragged along a horizontal plane µ =

Mg 2

1   . The least force 3

required is (1 kgf = 9.8 N ) (a) 25 kgf

(b) 2.5 kgf

(c) 12.5 kgf

(d) 6.25 kgf

24. A block A of mass 4 kg is kept on ground. The coefficient of friction between the block and the ground is 0.8. The external force of magnitude 30 N is applied parallel to the ground. The resultant force exerted by the ground on the block is ( g = 10 m/ s2 ) (a) 40 N (c) zero

(b) 30 N (d) 50 N

25. A block A of mass 2 kg rests on another block B of mass 8 kg which rests on a horizontal floor. The coefficient of friction between A and B is 0.2 while that between B and floor is 0.5. When a horizontal force F of 25 N is applied on the block B, the force of friction between A and B is (a) 3 N (c) 2 N

(b) 4 N (d) zero

26. A body of mass 10 kg lies on a rough inclined plane of inclination

 3 θ = sin−1   with the horizontal. When the force of 30 N is applied on  5 the block parallel to and upward the plane, the total force by the plane on the block is nearly along (a) OA (c) OC

(b) OB (d) OD

A

B

30N C O D θ

27. In the figure shown, a person wants to raise a block lying on the ground to a height h. In which case he has to exert more force. Assume pulleys and strings are light

(i)

(a) Fig. (i) (c) Same in both

(ii)

(b) Fig. (ii) (d) Cannot be determined

330 — Mechanics - I 28. A man of mass m stands on a platform of equal mass m and pulls himself by two ropes passing over pulleys as shown in figure. If he pulls each rope with a force equal to half his weight, his upward acceleration would be g 2 g (b) 4 (c) g (d) zero (a)

29. A varying horizontal force F = at acts on a block of mass m kept on a smooth horizontal surface.

An identical block is kept on the first block. The coefficient of friction between the blocks is µ. The time after which the relative sliding between the blocks prevails is

(a)

2 mg a

(b)

2 µmg a

(c)

µmg a

(d) 2 µ mga

30. Two particles start together from a point O and slide down along straight smooth wires inclined at 30° and 60° to the vertical plane and on the same side of vertical through O. The relative acceleration of second with respect to first will be of magnitude (a)

g 2

(b)

3g 2

(c)

g 3

(d) g

Subjective Questions 1. Find the values of the unknown forces if the given set of forces shown in figure are in equilibrium. R F

3N

60°

10N

2. Determine the tensions T1 and T2 in the strings as shown in figure. 60° T1 60° T2

W = 4 × 9.8 N

3. In figure the tension in the diagonal string is 60 N. (a) Find the magnitude of the horizontal forces F1 and F2 that must be applied to hold the system in the position shown. (b) What is the weight of the suspended block ?

F1

45° F2 W

Chapter 8

Laws of Motion — 331

4. A ball of mass 1 kg hangs in equilibrium from two strings OA and OB as shown in figure. What are the tensions in strings OA and OB? (Take g = 10 m/ s2). A

30°

B

60° T1

90°

120°

T2

O 150°

w = 10N

5. A rod OA of mass 4 kg is held in horizontal position by a massless string AB as shown in figure. Length of the rod is 2 m. Find B

60°

A

O

(a) tension in the string, (b) net force exerted by hinge on the rod. ( g = 10 m/s 2)

6. Two beads of equal masses m are attached by a string of length 2a and are free to move in a smooth circular ring lying in a vertical plane as shown in figure. Here, a is the radius of the ring. Find the tension and acceleration of B just after the beads are released to move. A

B

C

7. Two blocks of masses 1 kg and 2 kg are connected by a string AB of mass 1 kg. The blocks are placed on a smooth horizontal surface. Block of mass 1 kg is pulled by a horizontal force F of magnitude 8 N. Find the tension in the string at points A and B. 2kg

A

B

1kg

F = 8N

8. Two blocks of masses 2.9 kg and 1.9 kg are suspended from a rigid support S by two inextensible wires each of length 1 m, as shown in the figure. The upper wire has negligible mass and the lower wire has a uniformly distributed mass of 0.2 kg. The whole system of blocks, wires and support have an upward acceleration of 0.2 m/ s2. Acceleration due to gravity is 9.8 m/ s2. (a) Find the tension at the mid-point of the lower wire. (b) Find the tension at the mid-point of the upper wire.

S

2.9 kg

1.9 kg

332 — Mechanics - I 9. Two blocks shown in figure are connected by a heavy uniform rope of mass 4 kg. An upward force of 200 N is applied as shown.

F = 200 N 5 kg

(a) What is the acceleration of the system ? (b) What is the tension at the top of the rope ? (c) What is the tension at the mid-point of the rope ? (Take g = 9.8 m/s 2)

4 kg 7 kg

10. A 4 m long ladder weighing 25 kg rests with its upper end against a smooth wall and lower end on rough ground. What should be the minimum coefficient of friction between the ground and the ladder for it to be inclined at 60° with the horizontal without slipping? (Take g = 10 m/ s2).

11. A plumb bob of mass 1 kg is hung from the ceiling of a train compartment. The train moves on an inclined plane with constant velocity. If the angle of incline is 30°. Find the angle made by the string with the normal to the ceiling. Also, find the tension in the string. ( g = 10 m/ s2 )

12. Repeat both parts of the above question, if the train moves with an acceleration a = g/ 2 up the plane.

13. Two unequal masses of 1 kg and 2 kg are connected by a string going over a clamped light smooth pulley as shown in figure. The system is released from rest. The larger mass is stopped for a moment 1.0 s after the system is set in motion. Find the time elapsed before the string is tight again. 1kg 2kg

14. In the adjoining figure, a wedge is fixed to an elevator moving upwards

a

with an acceleration a. A block of mass m is placed over the wedge. Find the acceleration of the block with respect to wedge. Neglect friction. m θ

15. In figure m1 = 1 kg and m2 = 4 kg. Find the mass M of the hanging block which will prevent the smaller block from slipping over the triangular block. All the surfaces are frictionless and the strings and the pulleys are light. m1 m2 30° M

Note In exercises 16 to 18 the situations described take place in a box car which has initial velocity v = 0 but acceleration a = (5 m / s 2 ) i$ . (Take g = 10 m / s 2) y x z

a = 5 m/s2 ^i v=0

Chapter 8

Laws of Motion — 333

16. A 2 kg object is slid along the frictionless floor with initial velocity (10 m/s) $i (a) Describe the motion of the object relative to car (b) when does the object reach its original position relative to the box car. $ 17. A 2 kg object is slid along the frictionless floor with initial transverse velocity (10 m/s ) k. Describe the motion (a) in car’s frame and (b) in ground frame.

18. A 2 kg object is slid along a rough floor (coefficient of sliding friction = 0.3) with initial velocity

(10 m/s ) $i. Describe the motion of the object relative to car assuming that the coefficient of static friction is greater than 0.5.

19. A block is placed on an inclined plane as shown in figure. What must be the frictional force between block and incline if the block is not to slide along the incline when the incline is accelerating to the right at 3 m/s 2 3  2  sin 37° =  ? (Take g = 10 m/ s )  5

m a = 3 m/s2 37°

F

20. A 6 kg block is kept on an inclined rough surface as shown in figure. Find the force F required to (a) keep the block stationary, (b) move the block downwards with constant velocity and (c) move the block upwards with an acceleration of 4 m/s 2. (Take g = 10 m/s 2)

6 kg 60°

21. A block of mass 200 kg is set into motion on a frictionless horizontal surface with the help of frictionless pulley and a rope system as shown in figure. What horizontal force F should be applied to produce in the block an acceleration of 1 m/ s2 ?

µs = 0.6 µk = 0.4

a = 1 m/s2 F

22. A cube of mass 2 kg is held stationary against a rough wall by a force F = 40 N passing through centre C. Find perpendicular distance of normal reaction between wall and cube from point C. Side of the cube is 20 cm. Take g = 10 m/ s2.

F

C

23. A 20 kg monkey has a firm hold on a light rope that passes over a frictionless pulley and is attached to a 20 kg bunch of bananas. The monkey looks upward, sees the bananas and starts to climb the rope to get them. (a) As the monkey climbs, do the bananas move up, move down or remain at rest ? (b) As the monkey climbs, does the distance between the monkey and the bananas decrease, increase or remain constant ? (c) The monkey releases her hold on the rope. What happens to the distance between the monkey and the bananas while she is falling ? (d) Before reaching the ground, the monkey grabs the rope to stop her fall. What do the bananas do ?

334 — Mechanics - I 24. In the pulley-block arrangement shown in figure, find the relation between acceleration of blocks A and B.

B A

25. In the pulley-block arrangement shown in figure, find relation between aA , aB and aC .

B C

A

26. In the figure shown, find : (g = 10 m/ s2 ) T2 1 kg

2 kg T1 3 kg

(a) acceleration of 1 kg, 2 kg and 3 kg blocks and (b) tensions T1 and T2.

27. Find the acceleration of the blocks A and B in the situation shown in the figure.

4 kg A

B

5 kg

Chapter 8

Laws of Motion — 335

28. A conveyor belt is moving with constant speed of 6 m/s. A small block is just dropped on it. Coefficient of friction between the two is µ = 0.3. Find

6 m/s

(a) The time when relative motion between them will stop. (b) Displacement of block upto that instant. ( g = 10 m/s 2) .

29. Coefficient of friction between two blocks shown in figure is µ = 0.4. The blocks are given velocities of 2 m/s and 8 m/s in the directions shown in figure. Find 1 kg

2 m/s

2 kg

8 m/s

Smooth

(a) the time when relative motion between them will stop. (b) the common velocities of blocks upto that instant. (c) displacements of 1 kg and 2 kg blocks upto that instant. (g = 10 m/s 2)

30. A 2 kg block is pressed against a rough wall by a force F = 20 N as shown in figure. Find acceleration of the block and force of friction acting on it. (Take g = 10 m/ s2)

20 N

µs = 0.8 µk = 0.6

2 kg

Wall

31. A 2 kg block is kept over a rough ground with coefficient of friction µ = 0.8 as shown in figure. A time varying force F = 2t (F in newton and t in second) is applied on the block. Plot a graph between acceleration of block versus time. (g = 10 m/ s2) 2 kg

F = 2t

µ = 0.8

32. A 6 kg block is kept over a rough surface with coefficients of friction µ s = 0.6 and µ k = 0.4 as shown in figure. A time varying force F = 4t (F in newton and t in second) is applied on the block as shown. Plot a graph between acceleration of block and time. (Take g = 10 m/ s2) 6 kg

µs = 0.6 µk = 0.4

F = 4t

LEVEL 2 Objective Questions Single Correct Option 1. What is the largest mass of C in kg that can be suspended

A

without moving blocks A and B ? The static coefficient of friction for all plane surface of contact is 0.3. Mass of block A is 50kg and block B is 70kg. Neglect friction in the pulleys. (a) 120 kg (b) 92 kg (c) 81 kg (d) None of the above

B

C

2. A sphere of mass 1 kg rests at one corner of a cube. The cube is moved with a

y

velocity v = ( 8t $i − 2t )$j , where t is time in second. The force by sphere on the cube at t = 1 s is (g = 10 ms −2) [Figure shows vertical plane of the cube] 2

(a) 8 N (c) 20 N

(b) 10 N (d) 6 N

x

3. A smooth block of mass m is held stationary on a smooth wedge of mass M

and inclination θ as shown in figure. If the system is released from rest, then the normal reaction between the block and the wedge is

m

M (a) mg cos θ θ (b) less than mg cos θ (c) greater than mg cos θ (d) may be less or greater than mg cos θ depending upon whether M is less or greater than m

4. Two blocks of masses m1 and m2 are placed in contact with each other on a horizontal platform as shown in figure. The coefficient of friction between m1 and platform is 2µ and that between block m2 and platform is µ. The platform moves with an acceleration a. The normal reaction between the blocks is m1

m2 a

(b) zero only if m1 = m2 (d) non zero only if a > µg

(a) zero in all cases (c) non zero only if a > 2 µg

5. A block of mass m is resting on a wedge of angle θ as shown in the figure. With what minimum acceleration a should the wedge move so that the mass m falls freely? a

θ

(a) g (c) g cot θ

(b) g cos θ (d) g tan θ

Chapter 8

Laws of Motion — 337

6. To a ground observer the block C is moving with v0 and the block A with

v1

v1and B is moving with v2 relative to C as shown in the figure. Identify the correct statement. (a) v1 − v2 = v0 (b) v1 = v2 (c) v1 + v0 = v2 (d) None of the above

A

B v2

C v0

7. In each case m1 = 4 kg and m2 = 3 kg. If a1 , a2 and a3 are the respective accelerations of the block m1 in given situations, then a1 m1

a3 m1

a2 m2

m2 30°

m2

m1

(a) a1 > a 2 > a3 (c) a1 = a 2 = a3

(b) a1 > a 2 = a3 (d) a1 > a3 > a 2

8. For the arrangement shown in figure the coefficient of friction between

the two blocks is µ. If both the blocks are identical and moving, then the acceleration of each block is

F − 2 µg 2m F (c) − µg 2m

(b)

(a)

F 2m

m F m

(d) zero

Smooth

9. In the arrangement shown in the figure the rod R is restricted to move in the

R

vertical direction with acceleration a1, and the block B can slide down the fixed wedge with acceleration a2. The correct relation between a1 and a2 is given by

(a) a 2 = a1 sin θ (b) a 2 sin θ = a1 (c) a 2 cos θ = a1 (d) a 2 = a1 cos θ

B

a1

a2 θ Fixed

10. In the figure block moves downwards with velocity v1, the wedge moves rightwards with velocity v2. The correct relation between v1 and v2 is

θ v2

(a) v2 = v1 (c) 2v2 sin θ = v1

v1

(b) v2 = v1 sin θ (d) v2 (1 + sin θ ) = v1

338 — Mechanics - I 11. In the figure, the minimum value of a at which the cylinder starts rising up the inclined surface is

a

θ

(a) g tan θ

(b) g cot θ

(c) g sin θ

(d) g cos θ

12. When the trolley shown in figure is given a horizontal acceleration a, the pendulum bob of mass m gets deflected to a maximum angle θ with the vertical. At the position of maximum deflection, the net acceleration of the bob with respect to trolley is g2 + a 2

(b) a cos θ

(c) g sin θ − a cos θ

(d) a sin θ

(a)

θ l m

a

13. In the arrangement shown in figure the mass M is very heavy compared to m (M >> m). The tension T in the string suspended from the ceiling is (a) 4 mg (c) zero

(b) 2 mg (d) None of these m

14. A block rests on a rough plane whose inclination θ to the horizontal can be varied. Which

M

of the following graphs indicates how the frictional force F between the block and the plane varies as θ is increased ? F

O

F

F

90° (a)

θ

θ

90°

O

F

90°

O

(b)

θ

90°

O

θ

(d)

(c)

15. The minimum value of µ between the two blocks for no slipping is m µ 2m F Smooth

F (a) mg

F (b) 3mg

(c)

2F 3mg

(d)

4F 3mg

16. A block is sliding along an inclined plane as shown in figure. If the a

acceleration of chamber is a as shown in the figure. The time required to cover a distance L along incline is (a)

2L g sin θ − a cos θ

(c)

2L g sin θ + a cos θ

(b)

2L g sin θ + a sin θ

(d)

2L g sin θ

m θ

Chapter 8

Laws of Motion — 339

17. In the figure, the wedge is pushed with an acceleration of 10 3 m/s 2. It is seen that the block starts climbing up on the smooth inclined face of wedge. What will be the time taken by the block to reach the top? 2 s (a) 5

1m 30°

1 (b) s 5 5 (d) s 2

(c) 5 s

a = 10 √3 m/s2

18. Two blocks A and B are separated by some distance and tied by a string as shown in the figure. The force of friction in both the blocks at t = 2 s is m1 = 1 kg

m2 = 2 kg

F' = 2t

F = 15N µ2 = 0.5

µ1 = 0.6

(a) 4 N (→ ), 5 N(←) (c) 0 N(→ ), 10 N(←)

(b) 2 N(→ ), 5 N(←) (d) 1 N(←), 10 N(←)

19. All the surfaces and pulleys are frictionless in the shown arrangement. Pulleys P and Q are massless. The force applied by clamp on pulley P is P

Y

2m

X Q m

α = 30°

mg ^ ^ (− 3 i − 3 j) 6 mg (c) 2 6 (a)

(b)

mg ^ ^ ( 3 i + 3 j) 6

(d) None of these

20. Two blocks of masses 2 kg and 4 kg are connected by a light string and kept on horizontal surface. A force of 16 N is acted on 4 kg block horizontally as shown in figure. Besides, it is given that coefficient of friction between 4 kg and ground is 0.3 and between 2 kg block and ground is 0.6. Then frictional force between 2 kg block and ground is µ = 0.6 2kg

4kg

F = 16N

µ = 0.3

(a) 12 N (c) 4 N

(b) 6 N (d) zero

21. A smooth rod of length l is kept inside a trolley at an angle θ as shown in the figure. What should be the acceleration a of the trolley so that the rod remains in equilibrium with respect to it? (a) g tan θ (c) g sin θ

(b) g cos θ (d) g cot θ

θ

a

340 — Mechanics - I 22. A car begins from rest at time t = 0, and then accelerates along a

(a) 0.2 (c) 0.4

v (m/s)

straight track during the interval 0 < t ≤ 2 s and thereafter with constant velocity as shown in the graph. A coin is initially at rest on the floor of the car. At t = 1 s, the coin begins to slip and its stops slipping at t = 3 s. The coefficient of static friction between the floor and the coin is ( g = 10 m/ s2 )

v = 2t 2

(b) 0.3 (d) 0.5

1

2

3

4 t(s)

23. A horizontal plank is 10.0 m long with uniform density and mass 10 kg. It rests on two supports which are placed 1.0 m from each end as shown in the figure. A man of mass 80 kg can stand upto distance x on the plank without causing it to tip. The value of x is

x 1m

1m

1 (a) m 2

1 (b) m 4

3 (c) m 4

(d)

1 m 8

24. A block is kept on a smooth inclined plane of angle of inclination θ that moves with a constant acceleration so that the block does not slide relative to the inclined plane. If the inclined plane stops, the normal contact force offered by the plane on the block changes by a factor (a) tan θ (c) cos 2 θ

(b) tan 2 θ (d) cot θ

25. A uniform cube of mass m and side a is resting in equilibrium on a rough 45° inclined surface. The distance of the point of application of normal reaction measured from the lower edge of the cube is (a) zero (c)

a 2

a 3 a (d) 4 (b)

mg is applied on the upper surface of a uniform cube of mass m and side a 3 1 which is resting on a rough horizontal surface having µ = . The distance between lines of 2 action of mg and normal reaction is

26. A horizontal force F =

a 2 a (c) 4 (a)

(b)

a 3

(d) None of these

27. Two persons of equal heights are carrying a long uniform wooden plank l l and from nearest end of the rod. The 4 6 ratio of normal reaction at their heads is

of length l. They are at distance (a) 2 : 3 (c) 4 : 3

(b) 1 : 3 (d) 1 : 2

A

B

Chapter 8

Laws of Motion — 341

28. A ball connected with string is released at an angle 45° with the vertical as shown in the figure. Then the acceleration of the box at this instant will be (mass of the box is equal to mass of ball) Smooth

g 4 g (c) 2 (a)

(b)

g 3

45° m

Surface m

(d) g

external agent with acceleration 9 ms−2 vertically downwards. Force exerted on the rod by the wedge will be

(a) 120 N (b) 200 N (c) 160 N (d) 180 N

9m/s2

29. In the system shown in figure all surfaces are smooth. Rod is moved by

10kg 37°

30. A thin rod of length 1 m is fixed in a vertical position inside a train, which is moving

horizontally with constant acceleration 4 ms−2. A bead can slide on the rod and friction coefficient between them is 0.5. If the bead is released from rest at the top of the rod, it will reach the bottom in

(a) 2 s

(b) 1 s

(c) 2 s

(d) 0.5 s

31. Mr. X of mass 80 kg enters a lift and selects the floor he wants. The lift now accelerates upwards

at 2 ms−2 for 2 s and then moves with constant velocity. As the lift approaches his floor, it decelerates at the same rate as it previously accelerates. If the lift cables can safely withstand a tension of 2 × 104 N and the lift itself has a mass of 500 kg, how many Mr. X’s could it safely carry at one time?

(a) 22 (c) 18

(b) 14 (d) 12

32. A particle when projected in vertical plane moves along smooth surface with initial velocity 20 ms−1 at an angle of 60°, so that its normal reaction on the surface remains zero throughout the motion. Then the slope of the surface at height 5m from the point of projection will be

60°

(a) 3 (c) 2

(b) 1 (d) None of these

33. Two blocks A and B, each of same mass are attached by a thin inextensible string through an ideal pulley. Initially block B is held in position as shown in figure. Now, the block B is released. Block A will slide to right and hit the pulley in time tA . Block B will swing and hit the surface in time tB . Assume the surface as frictionless, then l

l

A

(a) tA > tB

(b) tA < tB

B

(c) tA = tB

(d) data insufficient

342 — Mechanics - I 34. Three blocks are kept as shown in figure. Acceleration of 20 kg block with respect to ground is 10kg

100N µ = 0.5

20kg

µ = 0.25

30kg

(a) 5 ms −2

(b) 2 ms −2

µ=0

(c) 1 ms −2

(d) None of these

R from the ground as 5 shown in the figure. Wedge is moving with velocity 20 ms−1 towards left then the velocity of the sphere at this instant will be

35. A sphere of radius R is in contact with a wedge. The point of contact is

20m/s

R/5

(a) 20 ms −1

(b) 15 ms −1

(c) 16 ms −1

(d) 12 ms −1

36. In the figure it is shown that the velocity of lift is 2 ms−1 while string is winding on the motor

2m/s

shaft with velocity 2 ms−1 and shaft A is moving downward with velocity 2 ms−1 with respect to lift, then find out the velocity of block B

A B

(a) 2 ms −1↑

(b) 2 ms −1↓

(c) 4 ms −1↑

(d) None of these

37. A monkey pulls the midpoint of a 10 cm long light inextensible string connecting two identical objects A and B lying on smooth table of masses 0.3 kg continuously along the perpendicular bisector of line joining the masses. The masses are found to approach each other at a relative acceleration of 5 ms −2 when they are 6 cm apart. The constant force applied by monkey is (a) 4 N

(b) 2 N

(c) 3 N

(d) None of these −1

38. In the figure shown the block B moves with velocity 10 ms . The velocity of A in the position shown is

37° A

B

(a) 12.5 ms −1

(b) 25 ms −1

(c) 8 ms −1

(d) 16 ms −1

Chapter 8

Laws of Motion — 343

39. In the figure mA = mB = mc = 60 kg. The coefficient of friction between C and ground is 0.5, B and ground is 0.3, A and B is 0.4. C is pulling the string with the maximum possible force without moving. Then the tension in the string connected to A will be C

A B

(a) 120 N (c) 100 N

(b) 60 N (d) zero

40. In the figure shown the acceleration of A is aA = (15 $i + 15$j ). Then the acceleration of B is ( A remains in contact with B ) A

x

B y 37°

(a) 5 $i

(b) − 15 $i (d) − 5 i$

(c) − 10 $i

A

41. Two blocks A and B each of mass m are placed on a smooth horizontal surface. Two horizontal forces F and 2F are applied on the blocks A and B respectively as shown in figure. The block A does not slide on block B. Then the normal reaction acting between the two blocks is (a) F

(b)

F 2

(c)

F 3

B 2F

F m 30°

(d) 3F

42. Two beads A and B move along a semicircular wire frame as shown in figure. The beads are connected by an inelastic string which always remains tight. B At an instant the speed of A is u , ∠ BAC = 45° and BOC = 75°, where O is the centre of the semicircular arc. The speed of bead B at that instant is (a) 2u u (c) 2 2

m

C

O (centre)

(b) u (d)

A u

2 u 3

D

43. If the coefficient of friction between A and B is µ, the maximum acceleration of the wedge A for which B will remain at rest with respect to the wedge is B

45°

(a) µg

 1 + µ (b) g   1 − µ

1 − µ (c) g    1 + µ

(d)

g µ

344 — Mechanics - I 44. A pivoted beam of negligible mass has a mass suspended from one end and an Atwood’s machine suspended from the other. The frictionless pulley has negligible mass and dimension. Gravity is directed downward and M 2 = 3M3 , l2 = 3l1. Find the ratio M1 / M 2 which will ensure that the beam has no tendency to rotate just after the masses are released. M1 =2 M2 M (c) 1 = 4 M2

(b)

(a)

l1

l2

M1

M1 =3 M2

M2

M3

(d) None of these

45. A block of mass m slides down an inclined right angled trough. If the coefficient of friction between block and the trough is µ k, acceleration of the block down the plane is

(a) g (sin θ + 2 µ k cos θ ) (b) g (sin θ + µ k cos θ ) (c) g (sin θ − 2 µ k cos θ ) (d) g (sin θ − µ k cos θ )

θ

46. If force F is increasing with time and at t = 0, F = 0, where will slipping first start? F

3 kg 2 kg

µ = 0.5 µ = 0.3 µ = 0.1

1 kg

(a) between 3 kg and 2 kg (c) between 1 kg and ground

(b) between 2 kg and 1 kg (d) Both (a) and (b)

47. A plank of mass 2 kg and length 1 m is placed on horizontal floor. A small block of mass 1 kg is placed on top of the plank, at its right extreme end. The coefficient of friction between plank and floor is 0.5 and that between plank and block is 0.2. If a horizontal force = 30 N starts acting on the plank to the right, the time after which the block will fall off the plank is (g = 10 ms−2 )  2 (a)   s  3

(b) 1 .5 s

(c) 0.75 s

 4 (d)   s  3

More than One Correct Options 1. Two blocks each of mass 1 kg are placed as shown. They are connected by a string which passes over a smooth (massless) pulley. F

m2 m1

There is no friction between m1 and the ground. The coefficient of friction between m1 and m2 is 0.2. A force F is applied to m2. Which of the following statements is/are correct?

(a) The system will be in equilibrium if F ≤ 4 N (b) If F > 4 N tension in the string will be 4 N (c) If F > 4 N the frictional force between the blocks will be 2 N (d) If F = 6 N tension in the string will be 3 N

Chapter 8

Laws of Motion — 345

2. Two particles A and B, each of mass m are kept stationary by applying a horizontal force F = mg on particle B as shown in figure. Then O α T1 A β

T2 B F = mg

(a) tan β = 2 tan α (c) 2 T1 = 5T2

(b) 2T1 = 5T2 (d) α = β v (m/s)

3. The velocity-time graph of the figure shows the motion of a wooden

block of mass 1 kg which is given an initial push at t = 0 along a horizontal table. (a) The coefficient of friction between the block and the table is 0.1 (b) The coefficient of friction between the block and the table is 0.2 (c) If the table was half of its present roughness, the time taken by the block to complete the journey is 4 s (d) If the table was half of its present roughness, the time taken by the block to complete the journey is 8 s

4

4

t (s)

4. As shown in the figure, A is a man of mass 60kg standing on a block B of mass 40 kg kept on ground. The coefficient of friction between the feet of the man and the block is 0.3 and that between B and the ground is 0.2. If the person pulls the string with 125 N force, then A

B

(a) B will slide on ground (b) A and B will move with acceleration 0.5 ms −2 (c) the force of friction acting between A and B will be 125 N (d) the force of friction acting between B and ground will be 250 N

5. In the figure shown A and B are free to move. All the surfaces are smooth. Mass of A is m. Then A B θ

(a) the acceleration of A will be more than g sin θ (b) the acceleration of A will be less than g sin θ (c) normal reaction on A due to B will be more than mg cos θ (d) normal reaction on A due to B will be less than mg cos θ

346 — Mechanics - I 6. M A = 3 kg, M B = 4 kg, and MC = 8 kg. Coefficient of friction between any two surfaces is 0.25. Pulley is frictionless and string is massless. A is connected to wall through a massless rigid rod. A B F C

(a) value of F to keep C moving with constant speed is 80 N (b) value of F to keep C moving with constant speed is 120 N (c) if F is 200 N then acceleration of B is 10 ms −2 (d) to slide C towards left, F should be at least 50 N (Take g = 10 ms −2 )

7. A man pulls a block of mass equal to himself with a light string. The coefficient of friction between the man and the floor is greater than that between the block and the floor (a) if the block does not move, then the man also does not move (b) the block can move even when the man is stationary (c) if both move then the acceleration of the block is greater than the acceleration of man (d) if both move then the acceleration of man is greater than the acceleration of block

8. A block of mass 1 kg is at rest relative to a smooth wedge moving

1kg

leftwards with constant acceleration a = 5 ms−2. Let N be the normal reaction between the block and the wedge. Then (g = 10 ms−2 )

(a) N = 5 5 newton 1 (c) tan θ = 2

a

(b) N = 15 newton

θ

(d) tan θ = 2

9. For the given situation shown in figure, choose the correct options (g = 10 ms −2)

(a) At t = 1 s, force of friction between 2 kg and 4 kg is 2 N (b) At t = 1 s, force of friction between 2 kg and 4 kg is zero (c) At t = 4 s, force of friction between 4 kg and ground is 8 N (d) At t = 15 s, acceleration of 2kg is 1 ms −2

2kg

µs = 0.4 µk = 0.2

4kg

F = 2t

µs = 0.6 µk = 0.4

10. In the figure shown, all the strings are massless and friction is absent everywhere. Choose the correct options. (a) T1 > T3 (b) T3 > T1 (c) T2 > T1 (d) T2 > T3

T2 1kg 2kg T1 1kg

T3 2kg

11. Force acting on a block versus time graph is as

F (N)

shown in figure. Choose the correct options. ( g = 10 ms−2 )

10

(a) (b) (c) (d)

At t = 2 s, force of friction is 2 N At t = 8 s, force of friction is 6 N At t = 10 s, acceleration of block is 2 ms −2 At t = 12 s, velocity of block is 8 ms −1

2kg

10

t(s)

µ = 0.3

F

Chapter 8

Laws of Motion — 347

12. For the situation shown in figure, mark the correct options. µ = 0.4 2kg 4kg

F = 2t

Smooth

(a) At t = 3 s, pseudo force on 4 kg block applied from 2 kg is 4 N in forward direction (b) At t = 3 s, pseudo force on 2 kg block applied from 4 kg is 2 N in backward direction (c) Pseudo force does not make an equal and opposite pairs (d) Pseudo force also makes a pair of equal and opposite forces

13. For the situation shown in figure, mark the correct options.

F θ

(a) Angle of friction is tan −1 (µ ) (b) Angle of repose is tan −1 (µ ) (c) At θ = tan −1 (µ ), minimum force will be required to move the block µMg (d) Minimum force required to move the block is . 1 + µ2

M µ

Comprehension Based Questions Passage 1 (Q. Nos. 1 to 5) A man wants to slide down a block of mass m which is kept on a fixed m inclined plane of inclination 30° as shown in the figure. Initially the block is not sliding. To just start sliding the man pushes the block down the incline with a force 30° F. Now, the block starts accelerating. To move it downwards with constant speed the man starts pulling the block with same force. Surfaces are such that ratio of maximum static friction to kinetic friction is 2. Now, answer the following questions.

1. What is the value of F? (a)

mg 4

(b)

mg 6

(c)

mg 3 4

(d)

mg 2 3

(d)

1 2 3

2. What is the value of µ s , the coefficient of static friction? (a)

4 3 3

(b)

2 3 3

(c)

3 3 3

3. If the man continues pushing the block by force F, its acceleration would be (a)

g 6

(b)

g 4

(c)

g 2

(d)

g 3

4. If the man wants to move the block up the incline, what minimum force is required to start the motion? (a)

2 mg 3

(b)

mg 2

(c)

7mg 6

(d)

5mg 6

5. What minimum force is required to move it up the incline with constant speed? (a)

2 mg 3

(b)

mg 2

(c)

7mg 6

(d)

5mg 6

348 — Mechanics - I Passage 2 (Q. Nos. 6 to 7) A lift with a mass 1200 kg is raised from rest by a cable with a tension 1350 kg- f . After some time the tension drops to 1000 kg- f and the lift comes to rest at a height of 25 m above its initial point. (1 kg- f = 9.8 N)

6. What is the height at which the tension changes? (a) 10.8 m (c) 14.3 m

(b) 12.5 m (d) 16 m

7. What is the greatest speed of lift? (a) 9.8 ms −1 (c) 5.92 ms −1

(b) 7.5 ms −1 (d) None of these

Passage 3 (Q. Nos. 8 to 9) Blocks A and B shown in the figure are connected with a bar of negligible weight. A and B each has mass 170 kg, the coefficient of friction between A and the plane is 0.2 and that between B and the plane is 0.4 (g = 10 ms−2)

A 8m

8. What is the total force of friction between the blocks and the plane? (a) 900 N (c) 600 N

B

θ

(b) 700 N (d) 300 N

15 m

9. What is the force acting on the connecting bar? (a) 140 N (c) 75 N

(b) 100 N (d) 125 N

Match the Columns 1. 2kg

F

µs, µk

a (m/s2)

F (N) 2 1

1

t (s)

4

t (s)

Force acting on a block versus time and acceleration versus time graph are as shown in figure. Taking value of g = 10 ms−2, match the following two columns. Column I (a) Coefficient of static friction (b) Coefficient of kinetic friction (c) Force of friction (in N ) at t = 0.1 s a (d) Value of , where a is acceleration of block ( in m/s 2 ) at 10 t =8s

Column II (p) 0.2 (q) 0.3 (r) 0.4 (s) 0.5

Chapter 8

Laws of Motion — 349

2. Angle θ is gradually increased as shown in figure. For the given situation match the following two columns. (g = 10 ms−2)

2kg

µ=1

θ

Column I

Column II

(a) Force of friction when θ = 0° (b) Force of friction when θ = 90° (c) Force of friction when θ = 30°

(p) 10 N (q) 10 3 N 10 (r) N 3 (s) None of the above

(d) Force of friction when θ = 60°

3. Match the following two columns regarding fundamental forces of nature. Column I (a) (b) (c) (d)

Column II

Force of friction Normal reaction Force between two neutrons Force between two protons

(p) (q) (r) (s)

field force contact force electromagnetic force nuclear force

4. In the figure shown, match the following two columns. (g = 10 ms−2) 5 m/s2

10N 2kg

µs = 0.4 µk = 0.3

F

Column I

Column II

(a) Normal reaction (b) Force of friction when F = 15 N (c) Minimum value of F for stopping the block moving down (d) Minimum value of F for stopping the block moving up

(p) 5 N (q) 10 N (r) 15 N (s) None of the above

5. There is no friction between blocks B and C. But ground is rough. Pulleys are smooth and

massless and strings are light. For F = 10 N, whole system remains stationary. Match the following two columns.(mB = mC = 1 kg and g = 10 ms−2) P3 B

P2

Smooth P4 A

C

P1 F

350 — Mechanics - I Column I (a) (b) (c) (d)

Column II

Force of friction between A and ground Force of friction between C and ground Normal reaction on C from ground Tension in string between P3 and P4

(p) (q) (r) (s)

10 N 20 N 5N None of the above

6. Match Column I with Column II. Note Applied force is parallel to plane. Column I

Column II

(a) If friction force is less than applied force then friction may be (b) If friction force is equal to the force applied, then friction may be (c) If a block is moving on ground, then friction is (d) If a block kept on ground is at rest, then friction may be

(p) Static (q) Kinetic (r) Limiting (s) No conclusion can be drawn

7. For the situation shown in figure, in Column I, the statements regarding friction forces are mentioned, while in Column II some information related to friction forces are given. Match the entries of Column I with the entries of Column II (Take g = 10 ms−2 ) µ = 0.2

2kg

µ = 0.1

3kg 5kg

F = 100N

Smooth

Column I

Column II

(a) Total friction force on 3 kg block is (b) Total friction force on 5 kg block is (c) Friction force on 2 kg block due to 3 kg block is (d) Friction force on 3 kg block due to 5 kg block is

(p) Towards right (q) Towards left (r) Zero (s) Non-zero

8. If the system is released from rest, then match the following two columns.

2kg √3 µ= 2

30°

3kg

Chapter 8 Column I (a) (b) (c) (d)

Laws of Motion — 351 Column II

Acceleration of 2 kg mass Acceleration of 3 kg mass Tension in the string connecting 2 kg mass Frictional force on 2 kg mass

(p) (q) (r) (s)

2 SI unit 5 SI unit Zero None of these

Subjective Questions 1. A small marble is projected with a velocity of 10 m/s in a direction 45° from the y-direction on the smooth inclined plane. Calculate the magnitude v of its velocity after 2s. (Take g = 10 m/ s2) 10 m/s °y 45 x 45°

2. Determine the acceleration of the 5 kg block A. Neglect the mass of the pulley and cords. The block B has a mass of 10 kg. The coefficient of kinetic friction between block B and the surface is µ k = 0.1. (Take g = 10 m/ s2 )

B

A

3. A 30 kg mass is initially at rest on the floor of a truck. The coefficient of static friction between the mass and the floor of truck in 0.3 and coefficient of kinetic friction is 0.2. Initially the truck is travelling due east at constant speed. Find the magnitude and direction of the friction force acting on the mass, if : (Take g = 10 m/ s2) (a) The truck accelerates at 1.8 m/s 2 eastward (b) The truck accelerates at 3.8 m/s 2 westward.

4. A 6 kg block B rests as shown on the upper surface of a 15 kg wedge A. Neglecting friction, determine immediately after the system is released from rest (a) the acceleration of A (b) the acceleration of B relative to A. (Take g = 10 m/ s2 ) B

A

30°

352 — Mechanics - I 5. In the arrangement shown in the figure, the rod of mass m held by two smooth walls, remains always perpendicular to the surface of the wedge of mass M. Assuming all the surfaces are frictionless, find the acceleration of the rod and that of the wedge. Fixed wall

m α `

6. At the bottom edge of a smooth vertical wall, an inclined plane is kept at an angle of 45°. A uniform ladder of length l and mass M rests on the inclined plane against the wall such that it is perpendicular to the incline. A l

B

O

45°

(a) If the plane is also smooth, which way will the ladder slide. (b) What is the minimum coefficient of friction necessary so that the ladder does not slip on the incline?

7. A plank of mass M is placed on a rough horizontal surface and a constant horizontal force F is applied on it. A man of mass m runs on the plank. Find the range of acceleration of the man so that the plank does not move on the surface. Coefficient of friction between the plank and the surface is µ. Assume that the man does not slip on the plank.

m

M

F

8. Find the acceleration of two masses as shown in figure. The pulleys are light and frictionless and strings are light and inextensible.

M

m

9. The upper portion of an inclined plane of inclination α is smooth and the lower portion is rough. A particle slides down from rest from the top and just comes to rest at the foot. If the ratio of smooth length to rough length is m : n , find the coefficient of friction.

Chapter 8

Laws of Motion — 353

10. Block B rests on a smooth surface. If the coefficient of static friction between A and B is µ = 0.4. Determine the acceleration of each, if

(a) F = 30 N and (b) F = 250 N ( g = 10 m/s 2)

100 N A B

F 250 N

11. Block B has a mass m and is released from rest when it is on top of wedge A, which has a mass 3m. Determine the tension in cord CD while B is sliding down A. Neglect friction. B D

C

A

θ

12. Coefficients of friction between the flat bed of the truck and crate are µ s = 0.8 and µ k = 0.7. The coefficient of kinetic friction between the truck tires and the road surface is 0.9. If the truck stops from an initial speed of 15 m/s with maximum braking (wheels skidding). Determine where on the bed the crate finally comes to rest. (Take g = 10 m/s 2) 3.2 m

13. The 10 kg block is moving to the left with a speed of 1.2 m/s at time t = 0. A force F is applied as shown in the graph. After 0.2 s, the force continues at the 10 N level. If the coefficient of kinetic friction is µ k = 0.2. Determine the time t at which the block comes to a stop. (g = 10 m/ s2) F (N) 20 v0 =1.2 m/s

10 F 0 0

10kg

t (s)

0.2

14. The 10 kg block is resting on the horizontal surface when the force F is applied to it for 7 s. The variation of F with time is shown. Calculate the maximum velocity reached by the block and the total time t during which the block is in motion. The coefficients of static and kinetic friction are both 0.50. ( g = 9.8 m/ s2 ) F(N) 100

40 10 kg

F 4

7

t(s)

354 — Mechanics - I 15. If block A of the pulley system is moving downward with a speed of 1 m/s while block C is moving up at 0.5 m/s, determine the speed of block B.

A C

B

16. The collar A is free to slide along the smooth shaft B mounted in the frame. The plane of the frame is vertical. Determine the horizontal acceleration a of the frame necessary to maintain the collar in a fixed position on the shaft. ( g = 9.8 m/s 2 ) A B

a

30°

17. In the adjoining figure all surfaces are frictionless. What force F must by applied to M1 to keep M3 free from rising or falling?

M2

M3

F

M1

18. The conveyor belt is designed to transport packages of various weights. Each 10 kg package has

a coefficient of kinetic friction µ k = 0.15. If the speed of the conveyor belt is 5 m/s, and then it suddenly stops, determine the distance the package will slide before coming to rest. ( g = 9.8 m/ s2 ) B

19. In figure, a crate slides down an inclined right-angled trough. The coefficient of kinetic friction

between the crate and the trough is µ k. What is the acceleration of the crate in terms of µ k ,θ and g?

90°

θ

20. A heavy chain with a mass per unit length ρ is pulled by the constant force F along a horizontal surface consisting of a smooth section and a rough section. The chain is initially at rest on the rough surface with x = 0. If the coefficient of kinetic friction between the chain and the rough

Chapter 8

Laws of Motion — 355

surface is µ k , determine the velocity v of the chain when x = L. The force F is greater than µ kρgL in order to initiate the motion. L x Rough

x=0

F

Smooth

21. A package is at rest on a conveyor belt which is initially at rest. The belt is started and moves to the right for 1.3 s with a constant acceleration of 2 m/s2. The belt then moves with a constant deceleration a2 and comes to a stop after a total displacement of 2.2 m. Knowing that the coefficients of friction between the package and the belt are µ s = 0.35 and µ k = 0.25, determine (a) the deceleration a2 of the belt, (b) the displacement of the package relative to the belt as the belt comes to a stop. ( g = 9.8 m/s 2 )

22. Determine the normal force the 10 kg crate A exerts on the smooth cart B, if the cart is given an acceleration of a = 2 m/s 2 down the plane. Also, find the acceleration of the crate. Set θ = 30° . ( g = 10 m/ s2). A B

a θ

23. A small block of mass m is projected on a larger block of mass 10 m and length l with a velocity v

as shown in the figure. The coefficient of friction between the two blocks is µ 2 while that between the lower block and the ground is µ 1. Given that µ 2 > 11 µ 1. v

m

10 m l

(a) Find the minimum value of v, such that the mass m falls off the block of mass 10 m. (b) If v has this minimum value, find the time taken by block m to do so.

24. A particle of mass m and velocity v1 in positive y direction is projected on to a belt that is moving with uniform velocity v2 in x-direction as shown in figure. Coefficient of friction between particle and belt is µ. Assuming that the particle first touches the belt at the origin of fixed x-y coordinate system and remains on the belt, find the co-ordinates (x, y) of the point where sliding stops. Y

v2

Belt O v1

356 — Mechanics - I 25. In the shown arrangement, both pulleys and the string are massless and all the surfaces are frictionless. Find the acceleration of the wedge.

m2

m1

m3

26. Neglect friction. Find accelerations of m, 2m and 3m as shown in the figure. The wedge is fixed. m 3m

2m

30°

27. The figure shows an L shaped body of mass M placed on smooth horizontal surface. The block A is connected to the body by means of an inextensible string, which is passing over a smooth pulley of negligible mass. Another block B of mass m is placed against a vertical wall of the body. Find the minimum value of the mass of block A so that block B remains stationary relative to the wall. Coefficient of friction between the block B and the vertical wall is µ. m B

M

A

Answers Introductory Exercise 8.1 1. See the hints

2. See the hints

3. See the hints

4. See the hints

5. See the hints

6. See the hints

7. F1x = 2 3 N, F2 x = − 2 N, F3 x = 0, F4 x = 4 N, F1y = 2 N,F2y = 2 3 N, F3y = − 6 N, F4y = 0 9. NA =

8. 30 N

10.

2 W 3

1000 500 N, NB = N 3 3

11. (a) 26.8 N

(b) 26.8 N

(c) 100 N

Introductory Exercise 8.2 1. (a) 10 ms −2

2. 4

(b) 110 N (c) 20 N

5. zero

6.

3.

3g 4

10 ms −2 3

4. 3 kg

7. 4 ms −2 , 24 N, 42 N, 14 N

8. (a) 3 ms −2 (b) 18 N, 12 N, 30 N, (c) 70 N

9. (a) 10 N, 30 N (b) 24 N

Introductory Exercise 8.3 1. 2a1 + a2 + a3 = 0

2. 3 m/s downwards

5. 4.8 kg

6.

3. (a)

12 2 N, ms −2 35 7

2g 10 (b) N 3 3

4. (a)

g Mg , 2 2

7. 1 ms −2 (upwards)

8.

g (up the plane) 3

Introductory Exercise 8.4 $ 1. (a) (4 j ) N (b) (−4$i ) N

2. True

3. False

Introductory Exercise 8.5 1. (a) zero, 20 N (b) 6 ms −2 , 8 N

2. (a)

mg mg mg , 0, 0 (b) , 0, 0 (c) 2 2 2

 3 − 1  3 − 1  g  mg ,  ,  2   2     

Exercises LEVEL 1 Assertion and Reason 1. (d)

2. (a)

3. (a)

4. (b)

5. (d)

6. (a)

7. (b)

8. (d)

9. (d)

10. (b)

11. (d)

Single Correct Option (c)

6. (d)

(c)

10. (b)

11. (a)

12. (a)

13. (a)

14. (b)

15. (a)

16. (a)

17. (a)

18. (b)

19. (b)

20. (c)

21. (a)

22. (d)

23. (c)

24. (d)

25. (d)

26. (a)

27. (a)

28. (d)

29. (b)

30. (a)

1.

(b)

2.

(b)

3.

(a)

4.

(d)

5.

7.

(b)

8.

(b)

9.

358 — Mechanics - I Subjective Questions 1. F = 1016 . N, R = 2.4 N

2. 45.26 N, 22.63 N

3. F1 = F2 = W = 30 2 N

4. 5N, 5 3 N

40 40 5. (a) N (b) N 3 3

6.

mg g , 2 2

7. 4N, 6N

9. (a) 2.7 ms −2 (b) 136.5 N (c) 112.5 N  2  11. 30° , 10 N 12. tan−1  , 5 7 N  3

8. (a) 20 N (b) 50 N 10. 0.288

1 14. ( g + a) sin θ, down the plane s 3 15. 6.83 kg 16. (a) x = x0 + 10t − 2.5 t 2 vx = 10 − 5t (b) t = 4 s 13.

17. (a) x = x0 − 2.5 t 2 , z = z0 + 10t , vx = − 5t , vz = 10 ms −1 (b) x = x0 , z = z0 + 10t , vx = 0, vz = 10 ms −1 18. For t ≤ 1. 25 s :

After 1.25 s : 9 19. mg 25 21. F = 100 N

x = x0 + 10t − 4t 2 vx = 10 − 8t Block remains stationary 20. (a) 34 N (up) (b) 40 N (up) (c) 88 N (up) 22. 5 cm

23. (a) move up (b) constant (c) constant (d) stop 24. aB = − 3aA 25. aA + 2aB + aC = 0 120 50 70 120 ms −2 , a2 = ms −2 (downwards ) a3 = N ms −2 (downwards ) (b) T1 = T 2 = 11 11 11 11 2 g 27. 28. (a) 2 s (b) 6 m g (downwards ), (upwards ) 7 7 29. (a) 1 s (b) 6 ms −1 (c) 4m, 7m (both towards right) 26. (a) a1 =

30. 4 ms −2 (downwards), 12 N (upwards ) 31. a = 0 for t ≤ 8 s, a = t − 8 for t ≥ 8 s 2

a (m/s )

45° 8

32. a = 0 for t ≤ 9s,

t (s)

2 a =  t − 4 for t ≥ 9 s 3  2

a (m/s )

2 9

t (s)

Laws of Motion — 359

Chapter 8

LEVEL 2 Single Correct Option 1. (c)

2. (b)

3. (b)

4. (d)

5. (c)

6. (a)

7. (b)

8. (c)

9. (b)

10. (d)

11. (a)

12. (c)

13. (b)

14. (b)

15. (c)

16. (c)

17. (b)

18. (d)

19. (b)

20. (c)

21. (d)

22. (c)

23. (a)

24. (c)

25. (a)

26. (b)

27. (c)

28. (b)

29. (b)

30. (d)

31. (b)

32. (d)

33. (b)

34. (c)

35. (b)

36. (d)

37. (b)

38. (d)

39. (d)

40. (d)

41. (d)

42. (a)

43. (b)

44. (b)

45. (c)

46. (c)

47. (a)

More than One Correct Options 1. (a,c,d)

2. (a,c)

3. (a,d)

8. (a,c)

9. (b,c)

10. (b,c, d)

4. (c,d) 11. (all)

5. (a,d)

6. (a,c)

12. (b, c)

13. (all)

7. (a,b, c)

Comprehension Based Questions 1. (b)

2. (a)

3. (d)

4. (c)

5. (d)

6. (c)

7. (c)

8. (a)

9. (a)

Match the Columns 1. (a) → (r)

(b) → (q)

(c) → (p)

(d) → (s)

2. (a) → (s)

(b) → (s)

(c) → (p)

(d) → (p)

3. (a) → (q,r) (b) → (q,r)

(c) → (s)

(d) → (p, s)

4. (a) → (s)

(b) → (p)

(c) → (s)

(d) → (s)

5. (a) → (p)

(b) → (s)

(c) → (q)

(d) → (p)

6. (a) → (q)

(b) → (p,r) (c) → (q)

7. (a) → (q, s) (b) → (p,s) 8. (a) → (r)

(b) → (r)

(c) → (p,s) (c) → (s)

(d) → (p,r) (d) → (q,s) (d) → (q)

Subjective Questions 2 3. (a) 54 N (due east) (b) 60 N (due west) 4. (a) 6.36 ms −2 (b) 5.5 ms −2 ms −2 33 mg cos α sin α mg cos α µ (M + m) g µ (M + m) g 1 F F 6. (a) Clockwise (b) 5. 7. , − ≤ a≤ + M M 3 m m m m m sin α + m sin α + sin α sin α  5m − M   m + n  tan α 8. aM =   g. (upwards) am = 5aM 9. µ =    m   25m + M  mg 10. (a) aA = aB = 0.857 m/s 2 (b) aA = 21 m/s 2 , aB = 16 11. 12. 2.77 m sin 2θ . m/s 2 2 M 13. t = 0.33 s 14. 5.2 m/s , 5.55 s 15. zero 16. 5.66 m/s 2 17. 3 (M1 + M 2 + M3 ) g 18. 8.5 m M2 1. 10 ms −1

19. g (sin θ − 23. (a) vmin =

2.

2F − µ k gL ρ

2 µ k cos θ) 20. 22( µ 2 − µ 1) gl 10

(b) t =

2mm 1 3g 25. (m2 + m3 ) (m1 + m2 ) + m2 m3

21. (a) 6.63 m/s 2

20l 11g (µ 2 − µ 1)

13 26.am = g, a2m = 34

24. x = v2

(b) 0.33 m 22. 90 N, 1 ms −2 v12 + v22 2 µg

,y=

397 3 g, a3m = g 34 17

v1

v12 + v22 2 µg

27. mA =

M + m µ −1

but µ > 1

09 Work, Enengy and Power Chapter Contents 9.1

Introduction to Work

9.2

Work Done

9.3

Conservative and Non-conservative Forces

9.4

Kinetic Energy

9.5

Work-Energy Theorem

9.6

Potential Energy

9.7 9.8 9.9

Three Types of Equilibrium Power of a Force Law of Conservation of Mechanical Energy

9.1 Introduction to Work In our daily life ‘work’ has many different meanings. For example, Ram is working in a factory. The machine is in working order. Let us work out a plan for the next year, etc. In physics however, the term ‘work’ has a special meaning. In physics, work is always associated with a force and a displacement. We note that for work to be done, the force must act through a distance. Consider a person holding a weight at a distance ‘h’ off the floor as shown in figure.

T

h

T

h

No work is done by the man holding the weight at a fixed position. The same task could be accomplished by tying the rope to a fixed point.

Fig. 9.1

In everyday usage, we might say that the man is doing a work, but in our scientific definition, no work is done by a force acting on a stationary object. We could eliminate the effort of holding the weight by merely tying the string to some object and the weight could be supported with no help from us. Let us now see what does ‘work’ mean in the language of physics.

9.2 Work Done There are mainly three methods of finding work done by a force: (i) Work done by a constant force. (ii) Work done by a variable force. (iii) Work done by the area under force and displacement graph.

Work done by a Constant Force Work done by a constant force is given by (F = force, S = displacement) W = F⋅S = FS cos θ = (magnitude of force) (component of displacement in the direction of force) = (magnitude of displacement) (component of force in the direction of displacement) Here, θ is the angle between F and S. Thus, work done is the dot product of F and S.

Special Cases (i) If θ = 0°, W = FS cos 0° = FS (ii) If θ = 90°, W = FS cos 90° = 0 (iii) If θ =180°, W = FS cos180° = − FS

Chapter 9

Work, Energy and Power — 363

Extra Points to Remember Work done by a force may be positive, negative or even zero also, depending on the angle (θ) between the force vector F and displacement vector S. Work done by a force is zero when θ = 90°, it is positive when 0° ≤ θ < 90° and negative when 90°< θ ≤ 180°. For example, when a person lifts a body, the work done by the lifting force is positive (as θ = 0° ) but work done by the force of gravity is negative (as θ = 180° ).

˜

Work depends on frame of reference. With change of frame of reference, inertial force does not change while displacement may change. So, the work done by a force will be different in different frames. For example, if a person is pushing a box inside a moving train, then work done as seen from the frame of reference of train is F ⋅ S while as seen from the ground it is F⋅ ( S + S 0 ) . Here S 0 , is the displacement of train relative to ground.

˜

Suppose a body is displaced from point A to point B, then S = rB − rA = ( xB − xA ) $i + ( yB − yA ) $j + ( zB − zA ) k$

˜

Here, ( xA , yA , zA ) and ( xB , yB , zB ) are the co-ordinates of points A and B. V

Example 9.1 A body is displaced from A = (2 m, 4 m, −6 m) to rB = ( 6$i − 4$j + 2k$ ) m under a constant force F = ( 2$i + 3$j − k$ ) N . Find the work done. r A = ( 2$i + 4$j − 6k$ ) m

Solution

∴

S = rB − r A = ( 6$i − 4$j + 2k$ ) − ( 2$i + 4$j − 6k$ ) = 4i$ − 8$j + 8k$

W = F ⋅ S = (2$i + 3$j − k$ ) ⋅ ( 4$i − 8$j + 8k$ ) = 8 − 24 − 8 = − 24 J Note

Ans.

Work done is negative. Therefore angle between F and S is obtuse. V

Example 9.2 A block of mass m = 2 kg is pulled by a force F = 40 N upwards through a height h = 2 m. Find the work done on the block by the applied force F and its weight mg. ( g = 10 m/s2 ) F

Fig. 9.2

Solution

Weight

mg = ( 2) (10) = 20 N

Work done by the applied force WF = F h cos 0°. As the angle between force and displacement is 0° or

WF = ( 40) ( 2) (1) = 80 J

Ans.

Similarly, work done by its weight Wmg = ( mg ) ( h ) cos 180° or

Wmg = ( 20) ( 2) ( −1) = − 40 J

Ans.

364 — Mechanics - I V

Example 9.3 Two unequal masses of 1 kg and 2 kg are attached at the two ends of a light inextensible string passing over a smooth pulley as shown in Fig. 9.3. If the system is released from rest, find the work done by string on both the blocks in 1 s. (Take g = 10 m/s2 ). Solution

1kg

Net pulling force on the system is

2kg

F net = 2g − 1g = 20 − 10 = 10 N Total mass being pulled m = (1 + 2) = 3 kg Therefore, acceleration of the system will be F 10 a = net = m/s 2 m 3 Displacement of both the blocks in 1 s is 1 1 10 5 S = at 2 =   (1) 2 = m 2 2  3 3 Free body diagram of 2 kg block is shown in Fig. 9.4 (b). Using ΣF = ma, we get 10 20 40 20 − T = 2a = 2   or T = 20 − = N  3 3 3 ∴ Work done by string (tension) on 1 kg block in 1 s is W1 = (T ) ( S ) cos 0° 40 =    3

Fig. 9.3

a

1kg

T a

1g

2kg

2kg

2g

20 N (b)

(a)

 5 (1) = 200 J    3 9

Ans.

a

Fig. 9.4

Similarly, work done by string on 2 kg block in 1 s will be W2 = (T )( S ) (cos 180° ) 40 5 200 =     ( −1) = − J  3   3 9

Ans.

Work Done by a Variable Force So far we have considered the work done by a force which is constant both in magnitude and direction. Let us now consider a force which acts always in one direction but whose magnitude may keep on varying. We can choose the direction of the force as x-axis. Further, let us assume that the magnitude of the force is also a function of x or say F ( x ) is known to us. Now, we are interested in finding the work done by this force in moving a body from x1 to x 2 . Work done in a small displacement from x to x + dx will be dW = F ⋅ dx Now, the total work can be obtained by integration of the above elemental work from x1 to x 2 or x2

x2

x1

x1

W = ∫ dW = ∫ F ⋅ dx

Chapter 9

Work, Energy and Power — 365

Note In this method of finding work done, you need not to worry for the sign of work done. If we put proper limits in integration then sign of work done automatically comes. x2

It is important to note that ∫ F dx is also the area under F - x graph between x = x1 to x = x 2 . x1

F

F

dW

x1

dx

W = Area X

x2

x2

x1

X

Fig. 9.5

Spring Force An important example of the above idea is a spring that obeys Hooke’s law. Consider the situation shown in figure. One end of a spring is attached to a fixed vertical support and the other end to a block which can move on a horizontal table. Let x = 0 denote the position of the block when the spring is in its natural length. When the block is displaced by an amount x (either compressed or elongated) a restoring force (F) is applied by the spring on the block. The direction of this force F is always towards its mean position ( x = 0) and the magnitude is directly proportional to x or F ∝x ∴ F = − kx Here, k is a constant called force constant of spring and depends on the nature of spring. From Eq. (i) we see that F is a variable force and F - x graph is a straight line passing through origin with slope = − k. Negative sign in Eq. (i) implies that the spring force F is directed in a direction opposite to the displacement x of the block. Let us now find the work done by this force F when the block is displaced from x = 0 to x = x. This can be obtained either by integration or the area under F - x graph. Thus,

x

W = ∫ dW = ∫ Fdx = 0

x

∫0

− kx dx = −

x=0 F

x

Fig. 9.6

(Hooke’s law) …(i) F

x=x

X

Fig. 9.7

1 2

kx 2

Here, work done is negative because force is in opposite direction of displacement. Similarly, if the block moves from x = x1 to x = x 2 . The limits of integration are x1 and x 2 and the work done is x2

W = ∫ − kx dx = x1

1 2

k ( x12 − x 22 )

366 — Mechanics - I V

Example 9.4 A force F = (2 + x ) acts on a particle in x-direction where F is in newton and x in metre. Find the work done by this force during a displacement from x = 1.0 m to x = 2.0 m. Solution As the force is variable, we shall find the work done in a small displacement from x to x + dx and then integrate it to find the total work. The work done in this small displacement is dW = F dx = ( 2 + x ) dx W =∫

Thus,

2. 0

1. 0

dW =

2. 0

∫1. 0 ( 2 + x ) dx 2. 0

 x2  = 2x +  = 3.5 J 2  1. 0  V

k

( x ≠ 0) acts on a particle in x-direction. Find the x2 work done by this force in displacing the particle from. x = + a to x = + 2a. Here, k is a positive constant.

Example 9.5

Solution

A force F = −

Ans.

W = ∫ F dx = ∫

+2 a +a

= −

+2 a

 −k  k  2  dx =   x   x + a

k 2a

Ans.

Note It is important to note that work comes out to be negative which is quite obvious as the force acting on the k  particle is in negative x-direction  F = − 2  while displacement is along positive x-direction. (from x = a to  x  x = 2a)

Work Done by Area Under F-S or F -x Graph This method is normally used when force and displacement are either parallel or antiparallel (or one dimensional). As we have discussed above W = ∫ Fdx = area under F - x graph So, work done by a force can be obtained from the area under F - x graph. Unlike the integration method of finding work done in which sign of work done automatically comes after integration, in this method area of the graph will only give us the magnitude of work done. If force and displacement have same sign, work done will be positive and if both have opposite signs, work done is negative. Let us take an example. F(N)

V

Example 9.6 A force F acting on a particle varies with the position x as shown in figure. Find the work done by this force in displacing the particle from (a) x = − 2 m to x = 0 (b) x = 0 to x = 2 m.

10 –2 2 –10

Fig. 9.8

x(m)

Chapter 9

Work, Energy and Power — 367

Solution (a) From x = − 2 m to x = 0, displacement of the particle is along positive x-direction

while force acting on the particle is along negative x-direction. Therefore, work done is negative and given by the area under F-x graph with projection along x-axis. 1 Ans. ∴ W = − ( 2) (10) = − 10 J 2 (b) From x = 0 to x = 2 m, displacement of particle and force acting on the particle both are along positive x-direction. Therefore, work done is positive and given by the area under F -x graph, 1 Ans. or W = ( 2) (10) = 10 J 2

INTRODUCTORY EXERCISE

9.1

1. A block is displaced from (1m, 4m, 6m) to (2$i + 3$j − 4k$ ) m under a constant force F = (6$i − 2$j + k$ ) N. Find the work done by this force.

2. A block of mass 2.5 kg is pushed 2.20 m along a frictionless horizontal table by a constant force 16 N directed 45° above the horizontal. Determine the work done by (a) the applied force, (b) the normal force exerted by the table, (c) the force of gravity and (d) determine the total work done on the block

3. A block is pulled a distance x along a rough horizontal table by a horizontal string. If the tension in the string is T, the weight of the block is w, the normal reaction is N and frictional force is F. Write down expressions for the work done by each of these forces.

4. A bucket tied to a string is lowered at a constant acceleration of g/4. If mass of the bucket is m and it is lowered by a distance l then find the work done by the string on the bucket. 1 4 is pulled by a force F acting at 45° with horizontal as shown in Fig. 9.9. The block is displaced by 2 m. Find the work done on the block by (a) the force F (b) friction (c) gravity.

5. A 1.8 kg block is moved at constant speed over a surface for which coefficient of friction µ = . It

F 45°

Fig. 9.9

6. A block is constrained to move along x-axis under a force F = − 2x . Here, F is in newton and x in metre. Find the work done by this force when the block is displaced from x = 2 m to x = − 4 m. 4 7. A block is constrained to move along x-axis under a force F = 2 ( x ≠ 0). Here, F is in newton x and x in metre. Find the work done by this force when the block is displaced from x = 4 m to x = 2 m.

368 — Mechanics - I 8. Force acting on a particle varies with displacement as shown in Fig. 9.10. Find the work done by this force on the particle from x = − 4 m to x = + 4 m. F(N) 10 –4

–2 2

x(m)

4

–10

Fig. 9.10 Fx (N)

9. A particle is subjected to a force Fx that varies with position as shown in figure. Find the work done by the force on the body as it moves (a) from x = 10.0 m to x = 5.0 m, (b) from x = 5.0 m to x = 10.0 m, (c) from x = 10.0 m to x = 15.0 m, (d) what is the total work done by the force over the distance x = 0 to x = 15.0 m ?

4 3 2 1 0

x (m) 2 4 6 8 10 12 14 16 Fig. 9.11

10. A child applies a force F parallel to the x-axis to a block moving on a horizontal surface. As the child controls the speed of the block, the x-component of the force varies with the x-coordinate of the block as shown in figure. Calculate the work done by the forceF when the block moves Fx (N)

2 1

2

3

x (m)

6

4

–1

Fig. 9.12

(a) from x = 0 to x = 3.0 m (c) from x = 4.0 m to x = 7.0 m

(b) from x = 3.0 m to x = 4.0 m (d) from x = 0 to x = 7.0 m

9.3 Conservative and Non-Conservative Forces In the above article, we considered the forces which were although variable but always directed in one direction. However, the most general expression for work done is dW = F ⋅ dr rf

W =∫

Here,

dr = dx$i + dy$j + dzk$

ri

dW = ∫

rf

and

ri

F ⋅ dr

ri = initial position vector and r f = final position vector Conservative and non-conservative forces can be better understood after going through the following two examples.

Chapter 9 V

Work, Energy and Power — 369

Example 9.7 An object is displaced from point A(2 m, 3 m, 4 m) to a point B ( 1 m, 2 m, 3 m) under a constant force F = ( 2$i + 3$j + 4k$ ) N . Find the work done by this force in this process. Solution

W =∫

rf

F ⋅ dr = ∫

ri

(1 m, 2 m, 3 m) (2 m, 3 m, 4 m)

( 2i$ + 3$j + 4k$ ) ⋅ ( dxi$ + dy$j + dzk$ )

= [ 2x + 3 y + 4 z ] (1 m, 2 m, 3 m) = −9 J (2 m, 3 m, 4 m)

Ans.

Alternate Solution Since, F = constant, we can also use. W = F⋅S S = r f − r i = ( i$ + 2$j + 3k$ ) − ( 2$i + 3$j + 4k$ )

Here,

= ( − i$ − $j − k$ ) ∴

W = ( 2$i + 3$j + 4k$ ) ⋅ ( − $i − $j − k$ ) = − 2 − 3 − 4 = − 9J

V

Ans.

Example 9.8 An object is displaced from position vector r1 = (2$i + 3$j) m to r2 = ( 4$i + 6$j) m under a force F = ( 3x 2 $i + 2 y$j) N . Find the work done by this force. Solution

W =∫ =∫

r2 r1 r2

r1

F ⋅ dr = ∫

r2 r1

( 3x 2 i$ + 2 y$j ) ⋅ ( dx$i + dy$j + dzk$ )

( 3x 2 dx + 2 y dy ) = [ x 3 + y 2 ] ((42,, 63))

= 83 J

Ans.

In the above two examples, we saw that while calculating the work done we did not mention the path along which the object was displaced. Only initial and final coordinates were required. It shows that in both the examples, the work done is path independent or work done will be same along all paths. The forces in which work is path independent are known as conservative forces. 1 Thus, if a particle or an object is displaced from position A to B position B through three different paths under a conservative force 2 field. Then, W1 = W2 = W3 Further, it can be shown that work done in a closed path is zero A 3 under a conservative force field. (WAB = − WBA or Fig. 9.13 WAB + WBA = 0). Gravitational force, Coulomb’s force and spring force are few examples of conservative forces. On the other hand, if the work is path dependent or W1 ≠ W2 ≠ W3 , the force is called a non-conservative. Frictional forces, viscous forces are non-conservative in nature. Work done in a closed path is not zero in a non-conservative force field.

Note The word potential energy is defined only for conservative forces like gravitational force, electrostatic force and spring force etc.

370 — Mechanics - I We can differentiate the conservative and non-conservative forces in a better way by making a table as given below. S.No

Conservative Forces

Non-conservative Forces

1.

Work done is path independent

Work done is path dependent

2.

Work done in a closed path is zero

Work done in a closed path is not zero

3.

The word potential energy is defined for conservative forces Examples are: gravitational force, electrostatic force, spring force etc.

The word potential energy is not defined for non-conservative forces. Examples are: frictional force, viscous force etc.

4.

Extra Points to Remember ˜

Gravitational force is a conservative force. Its work done is path independent. For small heights it only depends on the height difference ‘h’ between two points. Work done by gravitational force is (± mgh), in moving the mass ‘m’ from one point to another point. This is (+ mgh) if the mass is moving downwards (as the force mg and displacement both are downwards, in the same direction) and (− mgh) if the mass is moving upwards. B WAB = –mgh

h

WBA = +mgh A

Fig. 9.14 ˜ ˜

The magnetic field (and therefore the magnetic force) is neither conservative nor non-conservative. Electric field (and therefore electric force) is produced either by static charge or by time varying magnetic field. First is conservative and the other non-conservative.

9.4 Kinetic Energy Kinetic energy (KE) is the capacity of a body to do work by virtue of its motion. If a body of mass m has a velocity v, its kinetic energy is equivalent to the work which an external force would have to do to bring the body from rest upto its velocity v. The numerical value of the kinetic energy can be calculated from the formula. 1 KE = mv 2 2 This can be derived as follows: Consider a constant force F which acting on a mass m initially at rest. This force provides the mass m a velocity v. If in reaching this velocity, the particle has been moving with an acceleration a and has been given a displacement s, then (Newton’s law) F = ma v 2 = 2as Work done by the constant force = Fs or

 v2  1 W = ( ma )   = mv 2  2a  2

Chapter 9

Work, Energy and Power — 371

But the kinetic energy of the body is equivalent to the work done in giving the body this velocity. 1 Hence, KE = mv 2 2 Regarding the kinetic energy the following two points are important to note: 1. Since, both m and v 2 are always positive. KE is always positive and does not depend on the direction of motion of the body. 2. Kinetic energy depends on the frame of reference. For example, the kinetic energy of a person of 1 mass m sitting in a train moving with speed v is zero in the frame of train but 2 mv 2 in the frame of earth.

9.5 Work Energy Theorem This theorem is a very important tool that relates the works to kinetic energy. According to this theorem: Work done by all the forces (conservative or non-conservative, external or internal) acting on a particle or an object is equal to the change in kinetic energy of it. ∴

Wnet = ∆KE = K f − K i

Let, F1 , F2 ...be the individual forces acting on a particle. The resultant force is F = F1 + F2 + ...and the work done by the resultant force is W = ∫ F ⋅ dr = ∫ ( F1 + F2 + ... ) ⋅ dr = ∫ F1 ⋅ dr + ∫ F2 ⋅ dr + ...

where, ∫ F1 ⋅ dr is the work done on the particle by F1 and so on. Thus, work energy theorem can also be written as: work done by the resultant force which is also equal to the sum of the work done by the individual forces is equal to change in kinetic energy. Regarding the work-energy theorem it is worthnoting that : (i) If Wnet is positive then K f − K i = positive, i.e. K f > K i or kinetic energy will increase and if Wnet is negative then kinetic energy will decrease. (ii) This theorem can be applied to non-inertial frames also. In a non-inertial frame it can be written as work done by all the forces (including the pseudo forces) = change in kinetic energy in non-inertial frame. Let us take an example. m m

a

(a)

m m

f = ma (b)

m

fp = ma

f = ma (c)

Fig. 9.15

372 — Mechanics - I Refer Figure (a) A block of mass m is kept on a rough plank moving with an acceleration a. There is no relative motion between block and plank. Hence, force of friction on block is f = ma in forward direction.

Refer Figure (b) Horizontal force on the block has been shown from ground (inertial) frame of reference. If the plank moves a distance s on the ground, the block will also move the same distance s (as there is no slipping between the two). Hence, work done by friction on the block (w.r.t. ground) is W f = fs = mas From work-energy principle if v is the speed of block (w.r.t. ground), then 1 KE = W f or mv 2 = mas or v = 2as 2 Thus, velocity of block relative to ground is 2as.

Refer Figure (c) Free body diagram of the block has been shown from accelerating frame (plank). Here, f p = pseudo force = ma Work done by all the forces, W = W f + Wp = mas − mas = 0 From work-energy theorem, 1 or mv r2 = W = 0 vr = 0 2 Thus, velocity of block relative to plank is zero. Note Work-energy theorem is very useful in finding the work-done by a force whose exact nature is not known to us or to find the work done of a variable force whose exact variation is not known to us. V

Example 9.9 An object of mass 5 kg falls from rest through a vertical distance of 20 m and attains a velocity of 10 m/s. How much work is done by the resistance of the air on the object ? (g = 10 m/s 2 ) Solution Applying work-energy theorem,

or ∴

work done by all the forces = change in kinetic energy 1 Wmg + Wair = mv 2 2 1 2 1 Wair = mv − Wmg = mv 2 − mgh 2 2 1 2 = × 5 × (10) − ( 5) × (10) × ( 20) 2 = − 750 J

Ans.

Chapter 9 V

Work, Energy and Power — 373

Example 9.10 An object of mass m is tied to a string of length l and a variable force F is applied on it which brings the string gradually at angle θ with the vertical. Find the work done by the force F .

θ

In this case, three forces are acting on the object: 1. tension (T ) 2. weight ( mg ) and 3. applied force ( F ) Using work-energy theorem

l

Solution

m

F

Fig. 9.16

h = l (1 – cosθ) θ

l T F

h

mg

Fig. 9.17

W net = ∆KE or WT + Wmg + WF = 0 as ∆KE = 0 because Ki = K f = 0 Further, WT = 0, as tension is always perpendicular to displacement. Wmg = − mgh or Wmg = − mgl (1 − cos θ )

…(i)

Substituting these values in Eq. (i), we get WF = mgl (1 − cos θ )

Ans.

Note Here, the applied force F is variable. So, if we do not apply the work energy theorem we will first find the magnitude of F at different locations and then integrate dW ( = F ⋅ dr ) with proper limits. V

Example 9.11 A body of mass m was slowly hauled up the hill as shown in the Fig. 9.18 by a force F which at each point was directed along a tangent to the trajectory. Find the work performed by this force, if the height of the hill is h, the length of its base is l and the coefficient of friction is µ. Four forces are acting on the body: weight ( mg ) normal reaction ( N ) friction ( f ) and the applied force ( F )

Solution

1. 2. 3. 4.

m

F

l

Fig. 9.18

h

374 — Mechanics - I Using work-energy theorem or Wmg + W N Here, ∆KE = 0, because K i = 0 = K f

W net = ∆KE + W f + WF = 0

…(i) F

Wmg = − mgh WN = 0 (as normal reaction is perpendicular to displacement at all points) W f can be calculated as under f = µ mg cos θ ∴ ( dW AB ) f = − f ds = − (µ mg cos θ ) ds = − µ mg ( dl ) ∴ f = − µ mg Σ dl = − µ mgl Substituting these values in Eq. (i), we get

WF = mgh + µmgl Note

B

ds

A f

θ dl

Fig. 9.19

(as ds cos θ = dl )

Ans.

Here again, if we want to solve this problem without using work-energy theorem we will first find magnitude of applied force F at different locations and then integrate dW ( = F ⋅ dr ) with proper limits. V

Example 9.12 The displacement x of a particle moving in one dimension, under the action of a constant force is related to time t by the equation t= x +3 where, x is in metre and t in second. Calculate: (a) the displacement of the particle when its velocity is zero, (b) the work done by the force in the first 6 s. Solution As t =

x+3

i.e. So, (a) v will be zero when

x = ( t − 3) 2

…(i)

v = ( dx / dt ) = 2( t − 3)

…(ii)

2( t − 3) = 0 i.e. Substituting this value of t in Eq. (i),

t=3

x = ( 3 − 3) 2 = 0 i.e. when velocity is zero, displacement is also zero. (b) From Eq. (ii), ( v ) t = 0 = 2( 0 − 3) = − 6 m/s and ( v ) t = 6 = 2( 6 − 3) = 6 m/s So, from work-energy theorem 1 1 w = ∆KE = m[ v 2f − v i2 ] = m[ 62 − ( −6) 2 ] = 0 2 2 i.e. work done by the force in the first 6 s is zero.

Ans.

Ans.

Chapter 9

INTRODUCTORY EXERCISE

Work, Energy and Power — 375

9.2

1. A ball of mass 100 gm is projected upwards with velocity 10 m/s. It returns back with 6 m/s. Find work done by air resistance.

2. Velocity-time graph of a particle of mass 2 kg moving in a straight line is as shown in Fig. 9.20. Find the work done by all the forces acting on the particle. v (m/s) 20

2

t (s)

Fig. 9.20

3. Is work-energy theorem valid in a non-inertial frame? 4. A particle of mass m moves on a straight line with its velocity varying with the distance travelled according to the equation v = α x , where α is a constant. Find the total work done by all the forces during a displacement from x = 0 to x = b .

5. A 5 kg mass is raised a distance of 4 m by a vertical force of 80 N. Find the final kinetic energy of the mass if it was originally at rest. g = 10 m /s 2.

6. An object of mass m has a speed v 0 as it passes through the origin. It is subjected to a retarding force given by Fx = − Ax . Here, A is a positive constant. Find its x-coordinate when it stops.

7. A block of mass M is hanging over a smooth and light pulley through a light string. The other end of the string is pulled by a constant force F. The kinetic energy of the block increases by 40J in 1s. State whether the following statements are true or false: (a) The tension in the string is Mg (b) The work done by the tension on the block is 40 J (c) The tension in the string is F (d) The work done by the force of gravity is 40J in the above 1s

8. Displacement of a particle of mass 2 kg varies with time as s = (2t 2 − 2t + 10) m. Find total work done on the particle in a time interval from t = 0 to t = 2 s.

9. A block of mass 30 kg is being brought down by a chain. If the block acquires a speed of 40 cm/s in dropping down 2 m. Find the work done by the chain during the process. (g = 10 m /s 2 )

9.6 Potential Energy The energy possessed by a body or system by virtue of its position or configuration is known as the potential energy. For example, a block attached to a compressed or elongated spring possesses some energy called elastic potential energy. This block has a capacity to do work. Similarly, a stone when released from a certain height also has energy in the form of gravitational potential energy. Two charged particles kept at certain distance has electric potential energy.

376 — Mechanics - I Regarding the potential energy it is important to note that it is defined for a conservative force field only. For non-conservative forces it has no meaning. The change in potential energy ( dU ) of a system corresponding to a conservative force is given by dU = − F ⋅ dr = − dW

∫i

or

f

dU = − ∫

F ⋅ dr

ri

∆U = U

or

rf

f

−U i = − ∫

rf ri

F ⋅ dr

We generally choose the reference point at infinity and assume potential energy to be zero there, i.e. if we take ri = ∞ (infinite) and U i = 0 then we can write r

U = − ∫ F ⋅ dr = − W ∝

or potential energy of a body or system is the negative of work done by the conservative forces in bringing it from infinity to the present position. Regarding the potential energy it is worth noting that: 1. Potential energy can be defined only for conservative forces and it should be considered to be a property of the entire system rather than assigning it to any specific particle. 2. Potential energy depends on frame of reference. 3. If conservative force F and potential energy associated with this force U are functions of single variable r or x then : dU dU or − F =− dr dx Now, let us discuss three types of potential energies which we usually come across.

Elastic Potential Energy In Article 9.2, we have discussed the spring forces. We have seen there that the work done by the 1 spring force (of course conservative for an ideal spring) is − 2 kx 2 when the spring is stretched or compressed by an amount x from its unstretched position. Thus,  1  U = − W = −  − kx 2   2 

or

U =

1 2 kx 2

(k = spring constant)

Note that elastic potential energy is always positive.

Gravitational Potential Energy The gravitational potential energy of two particles of masses m1 and m2 separated by a distance r is given by mm U = −G 1 2 r Here, G = universal gravitation constant = 6.67 × 10 −11

N-m 2 kg 2

Chapter 9

Work, Energy and Power — 377

If a body of mass m is raised to a height h from the surface of earth, the change in potential energy of the system (earth + body) comes out to be mgh (R = radius of earth) ∆U = h  1 +   R ∆U ≈ mgh

or

if

h 2 Ω ⇒ ∴

1. ρ =

G

x + 20

Applying We have

−3 2

πd V (3.14 )(2.00 × 10 ) (100.0) = 4 lI (4 )(31.4 )(10.0) × 10−2

∆ρ ∆l ∆I   ∆d ∆V 2. × 100 = 2 + + + × 100 ρ d V l I     0.01  0.1   0.1   0.1   = 2 ×    +  +  +  2.00  100.0  31.4   10.0  

 0.1 0.1 = + × 100 = 1.1% 100 10 

2Ω

G

1.00 × 10−4 Ω - m (to three significant figures)

⇒

100 – x

R

3.6

= 2.41 % V R= I ∆R   ∆V   ∆I   × 100 =   +    × 100 R  V   I  

R

x

 336 × 100  =   − 0.6 × 2 cm 4 × 512     = 15.2 cm

2

100 − x > x

2Ω

  v =n   4 (l + 0.6 r)  Because, e = 0.6 r, where r is radius of pipe. For first resonance, n = 1 v f = ∴ 4 (l + 0.6 r) v or l= − 0.6r 4f

INTRODUCTORY EXERCISE

3.7

× 100

80 – x

P R = Q S 2 x = R 100 − x R x + 20 = 2 80 − x

…(i) …(ii)

Solving Eqs. (i) and (ii), we get, R = 3 Ω ∴ Correct option is (a). 2. Using the concept of balanced, Wheatstone bridge, we have, P R = Q S X 10 ∴ = (52 + 1) (48 + 2) 10 × 53 ∴ X = = 10.6 Ω 50 ∴ Correct option is (b). 3. Slide wire bridge is most sensitive when the resistance of all the four arms of bridge is same. Hence, B is the most accurate answer.

Chapter 3 INTRODUCTORY EXERCISE

1.

3.8

Therefore, the unknown resistance X lies between 14.2 Ω and 14.3 Ω. 2. Experiment can be done in similar manner but now K 2 should be pressed first then K 1. 3. BC, CD and BA are known resistances. The unknown resistance is connected between A and D.

P R = Q X  Q  1 X =  R=  R  P  10

⇒

Experiments — 483

R lies between 142 Ω and 143 Ω.

Exercises Objective Questions R 20 7. = 80 80 ∴

∴ ∴

R = 20 Ω

Ans. ∴

11. Thermal capacity = ms −1 o −1

= (0.04 kg) (4.2 × 10 J kg 2

C )

= 16.8 J/ °C 1 12. Intercept = f Therefore,

f =

Ans.

1 1 = Intercept 0.5

=2m 13. Deflection is zero for R = 324 Ω Now,

1 MSD 50 1 MSD = 50 (VC) = 50 (0.001 cm ) VC =

17. 100 VCD = 99 MSD

 Q  1  X =  R=  (324 )  P  100

= 3.24 Ω 1 14. 1 MSD = (1 cm ) = 1 mm 10 10 VSD = 8 MSD 8 1 VSD = MSD 10 Least count = 1 MSD − 1 VSD 8 LC = 1 mm − mm ∴ 10 2 ∴ LC = mm 10 2 ∴ LC = cm = 0.02 cm 100 15. 50 VSD = 49 MSD 49 1 VSD = MSD 50 ∴ VC = 1 MSD − 1 VSD 49 ∴ VC = 1 MSD − MSD 50

Ans. 1 MSD = 0.5 mm 1 16. LC = MSD n Here, n = number of vernier scale divisions 1 1  0.005 cm =  cm  n  10 01 . 1000 n= = 0.005 50 Ans. ∴ n = 20 ∴

∴ ∴

99 MSD 100 LC = 1 MSD − 1 VSD 99 LC = 1 MSD − MSD 100 LC = 0.01 MSD

1 VSD =

∴

Ans.

LC = 0.01(1 mm ) = 0.01 mm 0.8 cm 18. 1 VSD = = 0.08 cm 10 1 MSD = 0.1 cm ∴ LC = 1 MSD − 1 VSD = 0.1 cm − 0.08 cm = 0.02 cm 19. A = l × b = 10 × 1.0 = 10 cm 2

∴

∆A ∆l ∆b = + A l b  ∆l ∆b ∆A = ±  +  ×A  l b

Ans.

484 — Mechanics - I  0.1 0.01 =± −  × 10  10.0 1.00 = ± 0.2 cm 2

21. Distance moved in one rotation = 0.5 mm 0.5 mm = 0.01 mm 50 divisions Screw gauge has negative zero error. This error is (50 − 20) 0.01 mm or (30) (0.01) mm. Thickness of plate = (2 × 0.5 mm ) + (30 + 20) (0.01 mm) Ans. = 1.5 mm

Least count, LC =

22.

e l1

e

l2

λ 4 3λ l2 + e = 4 Solving these two equations, we get l − 3l1 e= 2 2 l2 − 3l1 23. e = 2 1 ∴ e = 1 cm = m 100 1 l − 3 (015 . ) = 2 ∴ 100 2 ∴ l2 = 0.47 m ∴ l = 47 cm v 340 24. λ = = = 1 m = 100 cm f 340 l1 + e =

Length of air columns may be, λ 3λ 5λ , , ... or 25 cm, 75 cm 125 cm.... 4 4 4 Minimum height of water column = 120 − maximum height of air column = 120 − 75 = 45 cm v 25. f = λ λ 3λ Now, l1 = , l2 = 4 4 λ l2 − l1 = ⇒ 2

λ = 2 (l2 − l1 )

or

Substituting in Eq. (i), we have v f = 2 (l2 − l1 ) 1 ⇒ f ∝ l2 − l1 f1 l′2 − l1′ 90 − 30 3 = = = f2 l2 − l1 30 − 10 1

∴

26. Heat lost by aluminium = 500 × s × (100 − 46.8) cal Heat lost = 26600 s ∴ Heat gained by water and calorimeter = 300 × 1 × (46.8 − 30) + 500 × 0.093 × (46.8 − 30) ∴ Heat gained = 5040 + 781.2 = 5821.2 Now, heat lost = Heat gained ∴ 26600 s = 5821.2 ∴ s ≈ 0.22 cal g−1 (o C)−1

27. Heat lost = Heat gained ∴ ∴

m1s1∆T1 = m2s2∆T2 m s ∆T s1 = 2 2 2 m1∆T1 =

∴

0.5 × 4.2 × 103 × 3 J kg−1 oC−1 0.2 × 77

s1 = 0.41 × 103 J kg−1 oC−1

Ans.

28. Heat lost = Heat gained 0.20 × 103 × s(150 − 40) = 150 × 1 × (40 − 27) + 0.025 × 103 × (40 − 27) {Q swater = 1 cal g−1 (o C)−1} Ans.

...(i)

∴ 22000 s = 1950 + 325 ∴ 22000 s = 2275 Ans. ∴ s = 010 . cal g−1 (o C)−1 R1 50 29. = =1 R2 50 ...(i) ∴ R1 = R2 = R When 24 Ω is connected in parallel with R2, then the balance point is 70 cm, so  24 R    RP  24 + R  30 = = R R 70 (Q RP < R ) 24 3 = ∴ 24 + R 7

Chapter 3 ∴ 168 = 72 + 3R ∴ 96 = 3R ∴ R = 32 Ω R1 + 10 50 30. = =1 R2 50 ∴ Again,

R1 + 10 = R2 R1 40 2 = = R2 60 3

∴ ∴ Ans.

...(i)

∴ ∴ ∴ X 20 1 32. = = Y 80 4 ∴ Since, ∴ ∴ ∴

R2 = 4 R1 R1 + 15 40 2 = = R2 60 3 R1 + 15 2 = 4 R1 3 R1 = 9 Ω

and

3. 4. 5.

6. Ans.

Y = 4X 4X l = Y 100 − l 4X l = 4 X 100 − l Ans.

33. For meter bridge to be balanced P 40 2 = = Q 60 3 2 ∴ P= Q 3 When Q is shunted, i.e. a resistance of 10 Ω is connected in parallel across Q, the net resistance 10 Q becomes . 10 + Q Now, the balance point shifts to 50 cm, i.e. P =1  10Q     10 + Q  2 10 = ∴ 3 10 + Q

Ans.

Subjective Questions

7.

2l = 100 l = 50 cm

20 + 2Q = 30 Q =5Ω 10 P= Ω 3

2. The bridge method is better because it is the null

∴ 3R1 = 2R2 Substituting the value of R2 from Eq. (i), we get 3R1 = 2(R1 + 10) Ans. ∴ R1 = 20 Ω R1 20 1 31. = = R2 80 4 ∴

Experiments — 485

8.

point method which is superior to all other methods. Because the graph in this case is a straight line. In the case of second resonance, energy gets distributed over a larger region and as such second resonance becomes feebler. The bridge becomes insensitive for too high or too low values and the readings become undependable. When determining low resistance, the end resistance of the meter bridge wire and resistance of connecting wires contribute towards the major part of error. No, the resistance of the connecting wires is itself of the order of the resistance to be measured. It would create uncertainty in the measurement of low resistance. 20 VSD = 19 MSD 19 ∴ 1 VSD = MSD 20 LC = 1 MSD − 1 VSD 19 1 = 1 MSD − MSD = MSD 20 20 1 cm = = 0.05 cm 20 1 The value of one main scale division = cm 20 Number of divisions on vernier scale = 20 Value of one main scale division LC = Number of divisions on vernier scale 1 1 20 Ans. = = = 0.0025 cm 20 400 Value of one main scale division Number of divisions on vernier scale 1 mm 01 . cm = = 10 10 Ans. = 0.01 cm

9. Least count =

Reading (diameter) = MS reading + (coinciding VS reading × Least Count)

486 — Mechanics - I = 4.3 cm + (7 × 0.01) = 4.3 + 0.07 = 4.37 cm Diameter = 4.37 cm 4.37 = 2.185 cm ∴ Radius = 2 = 2.18 cm To the required number of significant figures. 1mm 10. Pitch of the screw = 2 = 0.5 mm 0.5 Least count = 50 Ans. = 0.01 mm Reading = Linear scale reading + (coinciding circular scale × least count) = 3.0 mm + (32 × 0.01) = 3.0 + 0.32 Ans. = 3.32 mm 0.5 mm 11. LC = = 0.01 mm 50 Thickness = 5 × 0.5 mm + 34 × 0.01 mm = 2.84 mm 1 mm 12. LC = = 0.02 mm 50 Negative zero error = (50 − 44 ) × 0.02 = 0.12 mm Thickness = (3 × 1) mm + (26 × 0.02) mm + 0.12 mm = 3.64 mm 1 mm 13. LC = = 0.01 mm 100 The instrument has a positive zero error, e = + n (LC) = + (6 × 0.01) = + 0.06 mm

Linear scale reading = 2 × (1 mm ) = 2 mm Circular scale reading = 62 × (0.01 mm ) = 0.62 mm ∴ Measured reading = 2 + 0.62 = 2.62 mm or True reading = 2.62 − 0.06 = 2.56 mm 14. The instrument has a negative error, e = (−5 × 0.01) cm or e = – 0.05 cm Measured reading = (2.4 + 6 × 0.01) = 2.46 cm True reading = Measured reading – e = 2.46 − (−0.05) = 2.51 cm Therefore, diameter of the sphere is 2.51 cm. 15. We have, Least count of vernier callipers 1mm = = 0.1 mm = 0.01 cm 10 Side of cube = (10) (1mm ) + (1) (LC) or a = 10 mm + 0.1 mm or a = 10.1 mm or a = 1.01 cm mass m ρ= = volume a3 2.736 = (1.01)3 = 2.65553 g/cm 3 = 2.66 g/cm 3

Ans.

4. Units and Dimensions Exercises Single Correct Option 1. 2. 3. 4.

nh L = mvR = 2π ∴ [ L ] = [ h ] = [ mvR ] Velocity gradient is change in velocity per unit depth. Coefficient of friction is unitless and dimensionless. Dipole moment = (charge) × (distance)

Electric flux = (electric field) × (area) Hence, the correct option is (d).  MLT −2  F 5. [ η ] =   =    av   LLT −1  F 6. a = t F b= 2 t l l 7. R = ⇒ σ= σA RA From

H = I 2Rt

we have

R=

∴

12. Q ω k is dimensionless

13. Let E a v b F c = km Then, [ ML2T −2 ]a [ LT −1 ]b [ MLT −2 ]c = [ M ] a = 1, b = − 2 and c = 0 c

[ MLT −2 ] a [ L ] b [ T ] c = [ M ]

Equating the powers we get, a = 1, b = − 1 and c=2 nh 15. L = Iω = 2π h ∴ = [ω ] I 2πx 16. Q is dimensionless. λ ∴ [λ ] = [x ] = [L ] = [A ]

 lI 2t  [σ ] =    HA 

Ans.

17. In option (b), all three are related to each other.  Capacitance X  =   Z 2   (Magnetic induction )2 

18. [Y ] = 

 M −1L−2Q2T 2  = 2 −2 −2   M Q T 

−2 2 

 MLT L F ⋅s =   IL  AL 

= [M −3L−2T 4Q4 ]

[ g ] = LT ] -2

10. If unit of length and time is double, then value of g will be halved. µ i 11. B = 0 2π r

∴

b

∴

,

19. C = or

F il F µ0 i = il 2π r

Ans.

14. [ F ] [ L ] [ T ] = [ M ] a

H I 2t

B=

1 [k ] =   = [T ] ω 

∴

= [M −1L−3T 3A2 ]

But

Ans.

Equating the powers, we get

 LA2T  = 2 -2 2   ML T L 

8. φ = Bs =

F  [µ 0 ] =  2  i 

∴

(F = ilB )

or

∆q ε 0 A = ∆V d

A ∆q = L ∆V (∆q) L ε0 = A.(∆V ) ∆V X = ε0 L ∆t ε0

488 — Mechanics - I = but ∴

(∆q)L ∆V L A (∆V ) ∆t

[ A ] = [ L2 ] ∆q X = = current ∆t

αZ  20.   = [ M 0L0T 0 ] kθ

Further

 kθ  [α ] =   Z α  [ p] =   β 

α   kθ  [β ] =   =    p   Zp  Dimensions of k θ are that to energy. Hence,  ML2T −2  [β ] =  −1 −2   LML T 

∴

= [ M 0L2T 0 ]

F=

4.

[ ε0 ] =

1 q q . 1 2 4 πε 0 r 2 [ q1 ][ q2 ] [ IT ]2 = 2 [ F ][ r ] [ MLT −2 ][ L2 ]

= [ M −1 L−3 T 4I2 ] 1 Speed of light, c = ε0 µ 0 ∴ [µ 0 ] =

1 1 = 2 −1 −3 4 2 [ ε 0 ][ c ] [ M L T I ][ LT −1 ]2 = [ MLT −2I−2 ]

L both are time constants. Their units is R

5. CR and

second. 1 R and have the SI unit (second )−1. Further, ∴ CR L 1 resonance frequency ω = LC

Match the Columns More than One Correct Options 1. (a) Torque and work both have the dimensions [ML2T −2 ].

(d) Light year and wavelength both have the dimensions of length i.e. [L]. 2. Reynold’s number and coefficient of friction are dimensionless quantities. Curie is the number of atoms decaying per unit time and frequency is the number of oscillations per unit time. Latent heat and gravitational potential both have the same dimension corresponding to energy per unit mass. φ weber 3. (a) L = or henry = i ampere e  di  (b) e = − L   ⇒ ∴ L = −  dt  (di / dt ) volt- second or henry = ampere 2U 1 2 (c) U = Li ⇒ ∴ L = 2 2 i joule or henry = (ampere)2 1 (d) U = Li 2 = i 2 Rt 2 ∴ L = Rt or henry = ohm-second

1 kT 2 [ML2T −2 ] = [ k ] [K]

1. (a) U = ⇒

[ K ] = [ ML2T −2K −1 ] dv (b) F = ηA dx [MLT −2 ] ⇒ [ η ] = 2 −1 −1 [L LT L ]

⇒

= [ ML−1 T −1 ] (c)

E = hν ⇒

[ML2T 2 ] = [ h ][ T −1 ] [ h ] = [ ML2T −1 ]

⇒ (d)

dQ k A∆θ = dt l ⇒ [k ] =

[ML2T −3L] = [ MLT −3K −1 ] [L2K]

2. Angular momentum L = I ω [ L ] = [ Iω ] = [ ML2 ][ T −1 ] = [ ML2T −1 ] Q Latent heat, L = (as Q = mL) m 2 −2  Q   ML T  2 −2 [L ] =   =  ⇒  = [L T ] m  M 

∴

Units and Dimensions — 489

Chapter 4 τ = F × r⊥ [ τ ] = [ F × r⊥ ] = [ MLT −2 ][ L ]

Torque ∴

e (dt ) ≡ volt-second/ampere (di ) F = ilB F B ≡ ≡ newton/ampere-metre il

∴

L≡

= [ ML2T −2 ] Capacitance C =

1 q2 2U

 1 q2   as U =  2 C 

 q 2   Q2  ∴ [C] =   =  = [ M −1L−2T 2Q2 ] 2 −2   U   ML T  2U 1 2  Inductance L = 2  as U = Li   2  i 2 Q   U   Ut  [L ] =  2  =  2  ∴  as i =   t i Q      ML2T −2T 2  2 −2 =  = [ ML Q ] Q2   RA l  Resistivity ρ =  as R = ρ   l A  H  A = 2   (as H = i 2Rt ) i t  l   Ht   A  Q  =  2    as i =   t Q   l  2 2 2 −  ML T TL  3 −1 −2 =  = [ ML T Q ] 2 Q L   The correct table is as under Column I Angular momentum

[ML2T –1]

Latent heat

[L2T –2]

Torque

[M L2T –2]

Capacitance

[M –1L–2T 2Q2]

Inductance

[M L2Q–2]

Resistivity

[M L3T –1Q–2]

3. t ≡

∴

∴

L R

Column II

⇒ ∴ L ≡ tR ≡ ohm-second q2 U ≡ 2C q2 C ≡ ≡ coulomb2/joule U q ≡ CV q C ≡ ≡ coulomb /volt V −e L≡ di / dt

∴

Column I

Column II

Capacitance

coulomb/( volt )−1 coulomb2 joule−1

Inductance

ohm-second, volt second/ampere−1

Magnetic induction

newton (ampere-metre)−1

GM eM s r2 = Gravitational force between sun and earth ⇒ GM eM s = Fr2

4. (a) F =

∴

[ GM eM s ] = [ Fr 2 ] = [ MLT −2 ][ L2 ] = [ ML3T −2 ] 3RT = rms speed of gas molecules M 3RT 2 = vrms M  3 RT  −1 2 2 2 −2  M  = [ v rms ] = [ LT ] = [ L T ] Bqv = magnetic force on a charged particle F =v Bq

(b) vrms = ∴ or (c) F = ∴ or (d) vo = ∴

 F2  2 2 −2  2 2  = [ v ] = [L T ] B q  GM e = orbital velocity of earth's satellite Re GM e = vo2 Re

 GM e  2 2 −2 or   = [ vo ] = [L T ]  Re  (p) W = qV ⇒ (Coulomb) (Volt) = Joule or [(Volt) (Coulomb) (Metre)] = [(Joule) (Metre)] = [ ML2T −2 ][ L ] = [ ML3T −2 ] (q) [(kilogram) (metre)3 (second )−2] = [ ML3T −2 ] (r) [(metre)2 (second )−2 ] = [ L2T −2 ]

490 — Mechanics - I 1 (s)U = CV 2 2 (farad) ( volt )2 = Joule ⇒ or [(farad ) ( volt )2 (kg)−1 ] = [(Joule)(kg)−1 ] = [ ML2T −2 ][ M −1 ] = [ L2T −2 ]

Subjective Questions 1. [ a ] = [ y ] = [L] ⇒

[ωt ] = [M 0L0T 0 ] 1 ∴ [ω ] = = [T -1 ] [t ] θ is angle, which is dimension less. 2. 1N = 105 dyne

7. [T ] = [ p a d b E c ] = [ML−1T −2 ] a [ML−3 ] b [ML2T −2 ] c Equating the powers of both sides, we have a+ b+ c=0 − a − 3b + 2c = 0 −2a − 2c = 1 Solving these three equations, we have 5 1 a=− ,b= 6 2 1 and c= 3 −1 −2 8. [ Y ] = [ML T ] −2  MgL   MLT L  =    πr2l  2  LL 

= [ML−1T −2 ]

1 m 2 = 104 cm 2 ∴

2.0 × 1011

N m2

2.0 × 1011 × 105 dyne cm 2 104 12 = 2.0 × 10 dyne/cm 2 =

3. 1 dyne = 10−5 N 1 cm = 10−2 m ∴ 72 dyne/cm

72 × 10−5 N 10−2 m N = 0.072 m =

4. h =

Dimensions of RHS and LHS are same. Therefore, the given equation is dimensionally correct. 9. [ E ] = k [ m ] x [ n ] y [ a ] z where, k is a dimensionless constant. ∴ [ ML2T −2 ] = k [ M ] x [ T −1 ] y [ L ] z Solving we get,

10. Since dimension of Fv = [ Fv ] = [ MLT −2 ][ LT −1 ] = [ ML2T −3 ] So,

[ h ] = [ML2T -2 ][T] 1 2 at  + 2   1   =  u + at − a  2 

5. S t = ut +

1  2 u (t − 1) + a (t − 1)  2  

Equation is dimensionally correct.

6. (a) Young's modulus =

F/A ∆l / l

Hence, the MKS units are N/m 2. (c) Power of a lens (in dioptre) 1 = f (in metre)

x = 1, y = 2 z=2 E = k m n2a2

and ∴

E J = = J-s γ s −1 = [ML2T -1 ]

... (i) ... (ii) ... (iii)

β  2 −3  x 2  should also be [ ML T ] [β] = [ ML2T −3 ] [ x2 ] [β ] = [ ML4T −3 ]

Ans.

β  and Fv + 2  will also have dimension [ ML2T −3 ], x   so LHS should also have the same dimension [ ML2 T −3 ]. [α ] So = [ ML2T −3 ] [ t2 ] [α ] = [ ML2T −1 ]

11. [ b ] = [ V] = [ L ] 3

[a] = [P] = [ML−1T −2 ] [V ] 2 ⇒

[a] = [ML5T −2 ]

Ans.

Chapter 4 a  12.  = [M 0L0T 0 ]  RTV ⇒

dx

  is dimensionless.  a2 − x 2    1 −1  a  −1  a sin  x   has the dimension of [ L ].   

dx

14. 

[a] = [RTV] = [ML2T −2 ][L3 ] = [ML5T −2 ]

  is dimensionless.  2ax − x     x  Therefore, an sin −1  − 1  should also be a  

15. 

[b] = [V] = [L3 ] F  (s)  qv 

13. (a) Magnetic flux, φ = Bs = 

[as F = Bqv ] ∴



 Fs   MLT −2L2  [φ ] =   =  −1   qv   QLT  = [ ML2T −1Q−1 ]

−2  F   MLT  (b) [Rigidity Modulus] =   =    A   L2  −1 −2

= [ ML T ]

Units and Dimensions — 491

2

dimensionless. Hence, n = 0

16. (a) [ density] = [ F ]x [ L ] y [ T ]z ∴

[ ML−3 ] = [ MLT −2 ]x [ L ] y [ T ]z

Equating the powers we get, x = 1, y = − 4 and z = 2 ∴

[ density ] = [ FL−4 T 2 ]

In the similar manner, other parts can be solved.

5. Vectors INTRODUCTORY EXERCISE

5.1

3. A − 2B + 3C = (2$i + 3$j) − 2($i + $j) + 3k$ = ($j + 3k$ )

3. Apply R = A 2 + B 2 + 2 AB cos θ ∴

4. Apply S = A 2 + B 2 − 2 AB cos θ

= 10 units

5. R = S ⇒ A 2 + B 2 + 2 AB cos θ =

4. (a) Antiparallel vectors

A 2 + B 2 − 2 AB cos θ

(b) Perpendicular vectors (c) A lies in x - y plane and B is along positive z- direction. So, they are mutually perpendicular vectors (d)

Solving we get, cos θ = 0 or θ = 90°

INTRODUCTORY EXERCISE

5.2

1. A or | A| = (3) + (−4 ) + (5) = 5 2 units 2

2

2

B

Directions of cosines are, A 3 cos α = x = A 5 2 Ay −4 and cos β = = A 5 2 A 1 cos γ = z = A 2

2.

| A − 2B + 3C| = (1)2 + (3)2

3 ^j θ 45°

−3 ^i

O

4 ^i

A

θ = 135°

y

INTRODUCTORY EXERCISE Fx

3. A ⋅ B = AB cos θ

O x

60°

5.3

4. 2A = 4 i$ − 2$j − 3B = − 3$j − 3k$

Fy F = 10 N

Fx = 10 cos 60° = 5 N (along negative x-direction) Fy = 10 sin 60°= 5 3 N (along negative y-direction)

$i $j k$ ⇒ (2A) × (−3B) = 4 −2 0 0 −3 −3 = $i (6 − 0) + $j(0 + 12) + k$ (−12 − 0) = 6$i + 12$j − 12k$

Exercises 11. τ = r × F

Single Correct Option 6. A ⋅ B = AB cos θ

$i $j k$ = 3 2 3 2 −3 4

= (3) (5) cos 60° = 7.5

9. A × B = AB sinθ

0 ≤ sinθ ≤ 1 0 ≤ A × B ≤ AB ∴ 10. W = F ⋅ s = 9 + 16 = 25 J

= $i (8 + 9) + $j (6 − 12) + k$ (−9 − 4 ) Ans.

= (17$i − 6$j − 13k$ ) N - m

Ans.

Chapter 5 12.

(0.5)2 + (0.8)2 + c2 = 1

Ans.

26. | A × B | = 3 A ⋅ B ⇒ AB sin θ = 3 AB cos θ

13. R = A 2 + A 2 + 2 AA cos θ R = A atθ = 120° 14. (2i$ + 3$j + 8k$ ) ⋅ (4 $i − 4 $j + α k$ ) = 0 8 − 12 + 8α = 0 1 α= ∴ 2 15. A ⋅ B = 9 + 16 − 25 = 0

Ans.

or

tan θ = 3

⇒

θ = 60° |A + B | =

∴

∴ Angle between A and B is 90°. 16. A + B = 7 and ∴

A−B=3 A = 5 and B = 2

A 2 + B 2 + AB

27. (B × A) is ⊥ to both A and B Ans.

∴

(B × A) ⋅ A = 0

28. R = P + Q + 2PQ cos θ 2

2

Substituting the values of P , Q and R we get, cos θ = 0 ⇒ θ = 90°

Ans.

29.

18. s = rB − rA = (−2i$ + 6$j + 4 k$ ) − (3$j − k$ ) = − 2i$ + 3$j + 5k$

A 2 + B 2 + 2 AB cos 60°

=

Ans.

17. P ⊥ Q = P ⋅ Q = 0

Q

2

135°

Ans.

C = − (A + B) |C| = − (A + B) A ~ B ≤C ≤ A + B

larger magnitude.

22. A = | A | = (3) + (6) + (2) 2

P

| P | = |Q | = x |R | = 2 x

30. (P + Q ) ⋅ (P − Q ) = P 2 + PQ cos θ − PQ cos θ − Q 2

2

= P2 − Q 2

=7 A  α = cos−1  x  = cos−1  A

90°

R

21. Resultant is always inclined towards a vector of 2

x√

135°

2 x√

20. A + B + C = 0 ⇒ or or

Vectors — 493

Since dot product may be positive (if P > Q ), negative (if Q > P ) or zero (if P = Q ). Therefore angle between (P + Q ) and (P − Q ) may be acute, obtuse or 90°. 31. Scalar triple product of three vectors = volume of parallel applied = 0

 3    7

= angle of A with positive x-axis. Similarly, β and γ angles. 23. R = A + B = 12$i + 5$j R = | R | = (12)2 + (5)2 = 13 $R = R R 24. Component of A along B = A cos θ A⋅B 2+ 3 5 = = = B 1+ 1 2

25. C 2 = A 2 + B 2 + 2 AB cos θ At θ = 90°; C 2 = A 2 + B 2

∴

2 3 −2 5 n 1 =0 −1 2 3

Solving this equation, we get n = 18 32. (a + b) × (a − b) Ans.

= a × a − a × b + b × a − b× b = 2 (b × a) As a × a and b × b are two null vectors and − a × b = b × a

494 — Mechanics - I 33.

R = (3P )2 + (2P )2 + 2 (3P ) (2P ) cos θ

...(i)

2R = (6P )2 + (2P )2 + 2 (6P ) (2P ) cos θ

...(ii)

(b) If A and B are parallel or antiparallel to each other then their cross product is a null vector and they may be said to be equal otherwise for any other angle, A × B= −B× A

Solving these two equations we get, 1 or θ = 120° cos θ = − 2

34.

R 2x

θ

(c)

Ans.

35. A ⋅ B = 0 ∴

A⊥B A⋅C = 0 ∴ A ⊥C A is perpendicular to both B and C and B × C is also perpendicular to both B and C. Therefore, A is parallel to B × C. R = 8N

B

a ⋅ b   ab 

1. θ = cos−1 

2. If the angle between A and B is θ, the cross product

5. A

Ans.

Subjective Questions

B cos θ = 8N

θ

cos θ = 0 θ = 90°

(d) A + B = C only at θ = 0°

2x sin θ = x θ = 30° α = 120°

36.

A 2 + B 2 − 2 AB cos θ

∴ or

x

2x sin θ

x

∴ or

=

R = 2x cos θ

α

A 2 + B 2 + 2 AB cos θ

B sin θ

A

...(i) A + B = 16 B sin θ = A and B cos θ = 8 Squaring and adding these two equations, we get ...(ii) B 2 = A 2 + 64 Solving Eqs. (i) and (ii), we get A = 6 N and B = 10 N

6.

Ans.

37. (A + B) ⊥ (A − B)

7.

∴ ∴

(A + B) ⋅ (A − B) = 0 A 2 + BA cos θ − AB cos θ − B 2 = 0

or

A=B

Ans.

38. A ~ B ≤ C ≤ A + B

8.

will have a magnitude | A × B| = AB sin θ or 15 = 5 × 6 sin θ 1 or sin θ = 2 Thus, θ = 30° or 150° The angle between the force F and the displacement r is 180°. Thus, the work done is W =F⋅r = Fr cos θ = (12 N) (2.0 m ) (cos 180° ) = − 24 N -m = − 24 J A × B is perpendicular to both A and B. So this is parallel or antiparallel to C. Now, cross product of two parallel or antiparallel vectors is zero. Hence, C × (A × B) = 0. A × B is perpendicular to A. Now, dot product of two perpendicular vectors is zero. Hence, A ⋅ (A × B) = 0 S = S1 + S2 + S3 = [(5 cos 37° )$i + (5 sin 37° )$j ] + 3$i + 2$j = (4 $i + 3$j) + 3$i + 2$j = (7i$ + 5$j) m

39. A ~ B ≤ C ≤ A + B Match the Columns

5m

S

1. (a) AB sin θ = | AB cos θ | ∴ ⇒ or

tan θ = ± 1 v = 45° 135°

7m

Ans.

S = (7)2 + (5)2 = 74 m

Chapter 5 tanθ =

5 7

=3 5m tan θ =

 5 θ = tan −1    7

or

9. Suppose α is the angle between A and B and β the angle between C and B. But α =/ β. Given that, A⋅ B = C⋅ B ⇒ AB cosα = CB cosβ ⇒ A cosα = C cosβ or A =/ C as α =/ β 10. A + B = R = 3$i + $j , A − B = S = i$ + 5$j

= 20 cos 45° cos15° $i + 20 cos 45° cos15° $j = R $i + R $j x

y

Here,

Rx = R y = 10 2 cos15°

∴

| R| = Rx2 + R y2 = 20 cos15°

Since,

Rx = R y , therefore, θ = 45°

18. Their x components should be equal and opposite.

2

= (8) + (10) + (4 )

 2 θ = cos−1  −   3

or

19. Component of A along another vector (say B ) is

Area of parallelogram = | A × B | 2

6 cosθ = − 4

∴

= $i (8 − 0) + $j (−6 − 4 ) + k$ (0 − 4 ) = 8$i − 10$j − 4 k$

Ans.

13. See the hint of Q-23 of objective type problems. 2 −3 7 0 2 0 1 −1

14. A ⋅ (B × C) = 1

A⋅B B 20. Find A × B and then prove that (A × B) ⋅ A = 0 and (A × B) ⋅ B = 0. It means (A × B) ⊥ to both A and B. given by, A cos θ =

2

= 13.4 units

21.

B

= 2 (0 − 2) − 3 (0 + 1) + 7 (1 − 0) = −4 − 3+ 7= 0 15. (a) A + B = R Rx = Ax + Bx ∴ Bx = Rx − Ax = 10 − 4 = 6 m Similarly, By = Ry − Ay = 9 − 6 = 3 m (b) B =

3 1 = 6 2

= 10(cos 30° + cos 60° )$i + 10(cos 60°+ cos 30° )$j = 10(2 cos 45° cos15° )$i + 10(cos 45° cos15° )$j

12. A × B = 2 4 −6

2

=

 P ⋅ Q (c) Angle between P and Q is, θ = cos−1    PQ  17. R = [(10 cos 30° )i$ + (10 cos 60° )$j ] + [(10 cos 60° )$i + (10 cos 30° )$j ]

2 3 −4 = = 3 −a b a = − 4.5 b= −6 $i $j k$

1 0

Bx

 1 θ = tan −1    2 $ 16. (a) a + b + c = P = i − 2$j (b) a + b − c = $i + 4 $j = Q

11. Ratio of their coefficients should be same. ∴ and

By

∴

Now, angle between R and S is  R ⋅ S θ = cos−1    RS  ∴

Vectors — 495

Bx2 + B y2 = (6)2 + (3)2

x

or ∴ or ∴

R θ x α 2

R = x cos θ A

R = x cos θ x = x cos θ 2 1 cos θ = 2 θ = 60° α = 150°

Ans.

496 — Mechanics - I 22.

AB + BC + CA = 0 ∴ c$i + (a sin B$j − a cos B$i )

y

+ (− b cos A$i − b sin A$j) = 0

3P

∴ (c − a cos B − b cos A ) i$ 4P

2P

+ (a sin B − b sin A ) $j = 0

O

x

y

P

R = (Pi$ + 2P$j − 3Pi$ − 4 P$j) = (−2P$i − 2P$j) | R | = (2P ) + (2P ) 2

C b

=2 2P

A

Ans.

23. R 2 = P 2 + Q 2 + 2PQ cos θ

....(i)

S 2 = P 2 + Q 2 − 2PQ cos θ

...(ii)

Adding these two equations we get, R 2 + S 2 = 2 (P 2 + Q 2 )

24. From polygon law of vector addition we can see that,

a

C

2

A

B c

B

x

Putting coefficient of $j = 0, we find that a b = sin A sin B Now taking B as origin and BC as the x-axis, we can prove other relation.

6. Kinematics INTRODUCTORY EXERCISE

2. (a) v ⋅ a = 6 − 4 − 4 = −2 m /s 2

6.1

5. (a) On a curvilinear path (a path which is not straight

3

(b) Since dot product is negative. So angle between v and a is obtuse. (c) As angle between vand a at this instant is obtuse, speed is decreasing. INTRODUCTORY EXERCISE

6.2

1. v and a both are constant vectors. Further, these two vectors are antiparallel. 2. a is function of time and v and a are neither parallel nor antiparallel. INTRODUCTORY EXERCISE

6.3

1. Distance may be greater than or equal to magnitude of displacement.

2. Constant velocity means constant speed in same direction. Further if any physical quantity has a constant value (here, the velocity) then its average value in any interval of time is equal to that constant value 3. Stone comes under gravity ∴ F = mg F or Ans. a= =g m 1 4. In 15 s, it will rotate 90° or th circle. 4 d R

line), even if speed is constant, velocity will change due to change in direction. ∴ a ≠0 (b) (i) On a curved path velocity will definitely change (at least due to change in direction). ∴ a ≠0 (ii) In projectile motion, path is curved, but acceleration is constant (= g ). (iii) Variable acceleration on curved path is definitely possible. 2πR 6. (a) T = Ans. = 2513 . s v (b) (i) Since speed is constant. Therefore, average speed = constant speed Ans. = 1.0 cm/s (ii) Average velocity =

d R

(4 )( 2 )(4.0) = 0.9 cm/s 2513 . (iii) Velocity vector will rotate 90° =

Ans.

∴ | ∆v| = v 2 + v 2 − 2vv cos 90°

S 90°

= 2v ∴ | a av | =

Total distance d ∴ Average speed = = Total time t (2πR / 4 ) πR π (2.0) = = = 15 30 30  π Ans. =   cm/s  15 Total displacement s Average velocity = = Total time t 2 R ( 2 )(2.0) 2 2 = = cm/s t 15 15

S 90° R

R

=

s 2 R 4 2R = = t (T / 4 ) T

Ans.

=

2v | ∆v| 4 2v = = ∆t (T / 4 ) T (4 2)(1.0) = 0.23 cm/s2 25.13

INTRODUCTORY EXERCISE d d + d2 1. Average speed = = 1 t t1 + t2 v1t1 + v2t2 = t1 + t2 (4 × 2) + (6 × 3) = 2 +3

= 5.2 m/s

Ans.

6.4

Ans.

498 — Mechanics - I 2.

2 m/s A d1

C

3d/4 6 m/s

d/4 D 4 m/s

t

B

t

2t = d1 (d / 4 ) (3d / 4 ) + =t 4 6 3d =t ⇒ 16 total distance Average speed = total time d +d = 1 = 2t 11 = m/s 3

...(i)

...(ii)

16t 3 2t

2t +

INTRODUCTORY EXERCISE

6.5

1. S t = (displacement upto t second) − [ displacement upto (t − 1) sec ] 1 2 1 = (ut + at ) − [ u (t − 1) + a (t − 1)2 ] 2 2 1 = u + at − a 2 1 2  1  2. st = ut + at  − u (t − 1) + a (t − 1)2  2 2     1 1 2 = (u)(1) + (a)(2t ) − (a)(1) 2 2 The first term is (u) (1), which we are writing only u. Dimensions of u are [ LT −1 ] and dimensions of 1 (which is actually 1 second) are [T] [(u)(1)] = [ LT −1 ][ T ]

∴

= [L ] = st Therefore dimension of (u) (1) are same as the dimensions of s t . Same logic can be applied with other terms too. 5. In 4 s, it reaches upto the highest point and then changes its direction of motion. 1 1 s = ut + at 2 = (40) (6) + (−10) (6)2 2 2 Ans. = 60 m u  1 d = | s | 0− 4 + | s | 4−6 =   +  g (t − t0 )2  2g 2 2

=

ut +

1 2 at 1 2 = u + at t 2

Ans.

i.e. v -t function is linear. In linear function, (final value + initial value) average value = 2 v f + vi v1 + v2 ∴ vav = = 2 2 2h 2 × 125 8. t = = =5s g 10 s 125 (downwards) vav = = = 25 m/s t 5 1 9. (a) S = ut + at 2 2 1 = (2.5 × 2) + (0.5) (2)2 2 = 6 m = distance also (b) v = u + at 7.5 = 2.5 + 0.5 × t ⇒ t = 10 s (c) v 2 = u2 + 2as ∴

Ans.

Ans.

Ans.

(7.5)2 = (2.5)2 + (2) (0.5) s

⇒

s = 50 m = distance also 2

u (50)2 = = 125 m 2g 2 × 10 u 50 (b) Time of ascent = = =5s g 10

10. (a) h =

 125 (c) v 2 = u2 + 2as = (50)2 + 2(−10)    2  Solving, we get speed v ≈ 35 m/s INTRODUCTORY EXERCISE dv 1. (a) a = = 5 − 2t dt At t=2s a = 1 m/s2 3

3

0

0

6.6

Ans.

(b) x = ∫ vd t = ∫ (10 + 5t − t 2 ) dt 3

 t3  = 10 t + 2.5 t 2 −  3 0  = (10 × 3) + (2.5) (3)2 −

2

(40) 1 + × 10 × (6 − 4 )2 = 100 m 2 × 10 2

s 6. uav = = t 7. v = u + at

= 43.5 m

(3)3 3 Ans.

Chapter 6 2. (a) Acceleration of particle,

Kinematics — 499

INTRODUCTORY EXERCISE 1. u = 2$i

dv a= = (6 + 18t ) cm/s 2 dt

6.7

^j 2 m/s2

At t = 3 s, a = (6 + 18 × 3) cm/s 2 = 60cm/s 2

8

s

∫0 ds = ∫ 5 (3 + 6t + 9t

∴

2

2 m/s

a = (2 cos 60° ) i$ + (2 sin 60° ) $j = ($i + 3 $j) t=2s (i) v = u + at = (2$i ) + ($i + 3 $j) (2) = 4 $i + 2 3 $j

) dt 8

∴

s = [ 3t + 3t 2 + 3t 3 ]5

or

s = 1287 cm

∴

3. (a) Position, x = (2t − 3)2 v=

Velocity,

dv = 8 m/s 2 dt

∴

At t = 2 s,

(i) a =

v = 4 (2 × 2 − 3) = 4 m/s

∴

a = 8m/s 2

5. v ∝ t

3/ 4

dv dt s = ∫ vdt a=

⇒ a ∝ t −1/ 4 ⇒ s ∝ t7/ 4

Ans.

dv = (2$j) m/s2 = constant dt v = u + a t can be applied. 1

1

0

0

(ii) s = ∫ vdt = ∫ (2i$ + 2t$j) dt = [ 2t $i + t 2$j ] 10

(b) At origin, x = 0

At t = 0, v = 0 dv (c) a = = 2 + 12t dt At t=2s a = 26 m/s2

| s | = (6)2 + (2 3 )2

2. v = (2$i + 2t $j)

= 1.0 m

or (2t − 3) = 0 ∴ v=4 ×0=0 4. (a) At t = 0, x = 2.0 m dx (b) v = = 2t + 6t 2 dt

Ans. 3 $j) (2)2

=4 3m

x = (2 × 2 − 3)2

and

| v | = (4 )2 + (2 3 )2

= 2 7 m/s 1 1 (ii) s = u t + a t 2 = (2$i ) (2) + ($i + 2 2 = (6i$ + 2 3 $j)

dx = 4 (2t − 3)m/s dt

and acceleration, a =

60°

^i

(b) Given, v = (3 + 6t + 9t 2 ) cm/s ds or = (3 + 6t + 9t 2 ) dt or ds = (3 + 6t + 9t 2 ) dt

= (2i$ + $j) m t

v

Ans.

Ans.

3.

∫$ dv = ∫ adt = ∫ (2t$i + 3t $j) dt 2

0

2i

Solving we get v = 2i$ + t 2i$ + t 3$j At t = 1 s, v = (3i$ + $j ) m/s t

t

t

0

0

0

3 2 ∫ dr = ∫ vdt = ∫ [(2 + t )i$ + t $j ] dt

Ans.

 t3 t4 r =  2t +  $i + $j ⇒ 3 4  7 1 At t = 1 s, r = $i + $j 3 4  7 1 Therefore, the co-ordinates are  ,  m  3 4

Ans.

500 — Mechanics - I INTRODUCTORY EXERCISE

6.8

4. Applying sine law in ∆ ABC , we have

1. (b) v = Slope of x - t graph

N

2. Distance travelled = displacement

B m/s 150 °-q 150 C q Net velocity = v 20 m/s 30°

= area under v - t graph Acceleration = Slope of v - t graph 3. Acceleration = Slope of v - t graph Distance travelled = displacement = area under v - t graph s x −x 4. (a) Average velocity = = f i t t x10 sec − x0 sec 100 = = = 10 m/s 10 10 (b) Instantaneous velocity = Slope of x - t graph 5. From 0 to 20 s, displacement s1 = area under v - t graph (as v is negative) = + 50 m From 20 to 40 s, displacement s2 = area under v - t graph (as v is negative) = − 50m Total distance travelled = s1 + | s2 | = 100 m s s +s Average velocity = = 1 2 t t 50 − 50 = =0 40

6.9

INTRODUCTORY EXERCISE 2 1. vA = Slope of A = = 0.4 m/s 5 12 vB = Slope of B = = 2.4 m/s 5 ∴ vAB = vA − vB = − 2 m/s 2. aA = aB = g (downwards)

∴

aAB = aA − aB = 0

Ans.

400 3. (a) t = = 40 s 10

E

A

150 v 20 = = sin 30° sin θ sin (150° − θ ) From first and third we have, 20 sin 30° 1 sin (150° − θ ) = = 150 15 −1  1  ∴ 150° − θ = sin    15 -q 0° 15 150 m/s

N B

E

A

Hence, we can see that direction of 150 m/s is   1 150° − θ or sin −1    , east of the line AB  15    1 (b) 150° − θ = sin −1   = 3.8°  15 ∴

θ = 146.2°

From first and second relation,

Ans. B

Ans.

v=

C

150 sin θ sin 30°

Substituting the values we get, 400 m

v = 167 m/s

10 m/s

∴ A

2 m/s

(b) BC = (2 m/s) (t ) = 2 × 40 = 80 m

t= =

Ans.

AB v 500 × 1000 min = 49.9 min 167 × 60

≈ 50 min

Ans.

Chapter 6 5. Let vr = velocity of river

2 vb = vbr − vr2

vbr = velocity of river in still water and ω = width of river Given, tmin = 10 min or

ω = 10 vbr

K (i)

ω

v br

12.5 =

Now,

ω = vb

ω 2 vbr

K (iii)

6. 2√2 m/s

ω

A For minimum time

45°

Drift in this case will be, x = vr t ∴ 120 = 10 vr K (ii) Shortest path is taken when vb is along AB. In this case,

3 m/s 2 m/s

B

vbr

− vr2

Solving these three equations, we get vbr = 20 m/min, vr = 12 m/min and ω = 200 m

A

vr

Kinematics — 501

5 m/s Net velocity of boatman

vb

ω 20 = = 10 s 2 2 (b) Drift = 5t = 50 m (a) t =

A Shortest path

Exercises LEVEL 1 Assertion and Reason 1. a = − 2 ($i − $j) = − 2 v i,e. a and v are constant vectors and antiparallel to each other. So motion is one dimensional. First retarded then accelerated in opposite direction. 2. In the given situation, v -t graph is a straight line. But s -t graph is also a straight line. v

s

v0

Þ t

v = constant = v0

s t area of v -t graph = t  1  v0t0  v0 2 = = t0 2

3. Average velocity =

t s = v 0t

4. It is not necessary that if v = 0 then acceleration is also zero. If a particle is thrown upwards, then at highest point v = 0 but a =/ 0. 5. If acceleration is in opposite direction of velocity then speed will be decreasing even if magnitude of acceleration is increasing. 6. a = 2t ≠ constant Therefore, velocity will not increase at a constant rate.

7. If a particle is projected upwards then s = 0, when it returns back to its initial position.

502 — Mechanics - I s =0 t But its acceleration is constant = g At point A, sign of slope of s -t graph is not changing. Therefore, sign of velocity is not changing. ds v1 = 1 = (2 − 8t ) dt ds v2 = 2 = (−2 + 8t ) dt ∴ v12 = v1 − v2 = (4 − 16t ) v12 does not keep on increasing. dv a= dt So, direction of a and d v is same. 2 (h/ 2) h t1 = = g g

v

∴ Average velocity =

8. 9.

10.

11.

Velocity of second particle at height

h is, 2

a1 F1

Downward journey

In 15 s, velocity vector (of same magnitude) will rotate 90°. ∴

| ∆v | = | v f − vi | = v 2 + v 2 − 2vv cos 90o π 2 cm/s 30

u2 or u ∝ h 2g d d + d2 5. vav = = 1 t t1 + t2 (18) (11) + (42) (v ) 21 = 60 ⇒ v = 25.3 m/s dx 6. v = = 32 − 8t 2 dt v = 0 at t = 2 s dv a= = −16 t dt at t = 2s, a = −32 m/s2

Ans.

4. h =

or

h g

t1 = t2

mg − F 12. a= m F = g − = depends on m m Since m1 ≠ m2 ∴ a1 ≠ a2 or t1 ≠ t2

F a

mg

Single Correct Option

7. a = bt dv = bt dt

∴

1. Packet comes under gravity. Therefore only force is mg. F mg (downwards) = =g m m 2. Air resistance (let F ) is always opposite to motion (or velocity) Retardation in upward journey F + mg F a1 = 1 =g+ 1 m m a=

v

Acceleration in downward journey mg − F2 F a2 = =g− 2 m m Since, a1 > a2 ⇒ T1 < T2 2πR (2π ) (1)  π  3. v = = =   cm/s  30 T 60

v = u − 2g (h/ 2)

⇒

mg

mg

= 2v=

= gh − gh = 0

a2

Upward journey

2

h is its highest point. 2 gh u t2 = = = ∴ g g

F2

or ∴

v

t

v0

0

∫ dv = ∫ (bt ) dt v = v0 +

bt 2 2

t t  bt 2   dt s = ∫ vdt = ∫  v0 + 2 0 0

= v0t +

bt 3 6

Ans.

Ans.

Kinematics — 503

Chapter 6 8. t =

2h = g

2×5 =1s 10

11. vx =

dx =5 dt dy dt = (4 t + 1)

vy =

3 t0 h, t

2 t0 1

Let t0 is the interval between two drops. Then 2 t0 = t ∴ t0 = 0.5 s nd 2 drop has taken t0 time to fall. Therefore distance fallen, 1  1 d = gt02 =   (10) (0.5)2  2 2 = 1.25 m ∴ Height from ground = h − d = 5 − 1.25 = 3.75 m

9.

At 45°, vx = v y ⇒ t =1s 1 2 12. h = gT 2 T At second, distance fallen 3 1 T  h d = g  = 2  3 9 ∴ Height from ground = h − d =

Now,

10. vx = ∴

t

14

2

Ans. Q P

PQR = 5 a ∴ Ans.

∫ dx = ∫ (8t − 2) dt

x − 14 = (4 t 2 − 2t ) − (12) ∴

x = 4 t 2 − 2t + 2 dy vy = =2 dt y

∴ ∴ or

...(i)

t

∫ dy = ∫ 2dt 4

Ans.

R

dx = (8t − 2) dt x

8h 9

13. It should follow the path PQR.

1 2 1 gt = 20 (t − 1) + g (t − 1)2 2 2 Solving this equation we get, ∴ t = 1.5 s 1 d = 20 (t − 1) + g (t − 1)2 2 = 11.25 cm

Ans.

2

y − 4 = 2t − 4 y = 2t y ∴ t= 2 Substituting this value of t in Eq. (i) we have, Ans. x = y2 − y + 2

t=

5a u

Ans.

14. At 4 s u = at = 8 m/s 1 1 s1 = at 2 = × 2 × 4 2 = 16 m 2 2 From 4 s to 8 s a = 0, v = constant = 8 m/s ∴ s2 = vt = (8)(4 ) = 32 m From 8 s to 12 s s3 = s1 = 16 m Ans. ∴ sTotal = s1 + s2 + s3 = 64 m

15. Retardation is double. Therefore retardation time will be half. Let t0 = acceleration time t0 Then = retardation time 2 t t0 + 0 = t 2 2t t0 = ∴ 3

504 — Mechanics - I Now,

s = s1 + s2  1  1 t  =   (a)(t0 )2 +   (2a) 0   2  2  2 =

Now applying v = u + at , we have v = (+ 20) + (−10) (3) = − 10 m/s ∴ Velocity is 10 m/s, downwards. dv 20. a = = 0.2 v 2 dt

2

3 2 at0 4

 3   2t  =   (a)   4  3 

2

v

at 2 = 3 16. At the time of overtaking, s1 = s2 1 2 1 2ut + at = ut + (2a)t 2 ∴ 2 2 2u ∴ t= a ∴

 2u 1  2u s1 (or s2 ) = (2u)  + (a)   a 2  a =

17. t =

6u2 a

Ans.

= 15 +

2

= ∫ dt 0

 5  v  = 2 10

∴

5 5 − =2 v 10 v = 2.0 m/s

∴ or

Ans.

=

v12 v22 + 2a1 2a2

=

(10)2 (20)2 + 2 × 2 (2 × 1) Ans.

22. s1 = s2 1 × 10 × t 2 = (40) (t − 2) 2 1 − × 10 × (t − 2)2 2 Solving this equation, we get t=5s 1 Then s1 = (40) (5) − × 10 × (5)2 2 Ans. = 75 m 1 23. d = g (T − t )2 2 ∴

(40) t −

Free fall

u2 2g

T–t

(10)2 2 × 10

d

H t

h = 20 m

h

Initial velocity u = 2gh = 2 × 10 × 20 = 20 m/s

Ans.

21. d = d1 + d2

2

= 225 m

At 2 s, net area = 0 ∴ s=0 and the particle crosses its initial position.

or

10

80 m 8 = s. 30 m/s 3

Total height = 15 +

−2

v

Now in vertical direction 2u t= r ar tar or ur = (ar = g = 10 m/s2 ) 2 (8/ 3) (10) = 2 40 Ans. = m/s 3 18. s = net area of v -t graph

19.

∫ − 5v

∴

∴

h=H −d =H −

1 g (T − t )2 2

Kinematics — 505

Chapter 6 29. F = 3t 2 − 32

t2 2

24. x =

2

a=

4

dx x t =t y= = dt 2 8 dy t 3 vy = = dt 2 vx =

∴ ∴

∴

vx = 2 m/s v y = 4 m/s v = vx $i + v y $j = (2$i + 4 $j) m/s

∴

x = (t + 3)

− 3.2) dt

Ans.

(10)2 = u2 − gh

∴

u = (100 + gh)

Now,

h=

u2 100 + gh = 2g 2g h h=5+ 2 h = 10 m

(as 2g = 20 m/s2 ) Ans.

dv is the magnitude of total acceleration. dt d | v| represents the time rate of change of While dt speed (called the tangential acceleration, a component of total acceleration) as | v | = v. (b) These two are equal in case of one dimensional motion. x 2. (a) x = 2t ⇒ t = 2

1. (a)

Ans.

27. v = 25 + 25 s v 2 = (5)2 + 2 (12.5) s

Now compare with v 2 = u2 + 2as

28. Retardation during upward motion a1 = 10 + 2 = 12 m/s2

y = t 2 or

2 m/s2

 x y=   2

2

∴

a1

x 2 = 4 y is the trajectory (b) r = x i$ + y$j = (2t $i ) + (t 2$j) dr v= = (2i$ + 2t $j) units dt dv (c) a = = (2$j) units dt

a2 v

10 m/s2 2 m/s2

10 m/s2

Acceleration during downward motion, a2 = 10 − 2 = 8 m/s2

∴

2

Subjective Questions

2

or

0

2

∴

Ans.

vi = 0 ν f = area = 40 + 50 = 90 m/s

v

0

∴

26. ∆v = v f − vi = area under a-t graph

or

10

 h

2

dx dt = 2 (t + 3) ∴ v -t equation is linear

⇒

5

30. v 2 = u2 − 2g    2

v=

or

5

v − 10 = − 3.5 v = 6.5 m/s

∴

x =t+3

25.

v

∫ dv = ∫ adt = ∫ (0.3 t

At t = 2 s and

F = (0.3 t 2 − 3.2) m

2s t= a 1 t∝ a t1 a 8 = 2 = = t2 a1 12

3. B

2 3

Ans.

Ans. Ans.

N

C

D

Ans.

W

E

A S

506 — Mechanics - I (a) Distance = AB + BC + CD = (500 + 400 + 200) = 1100 m (b) Displacement = AD = ( AB − CD )2 + BC 2 = (500 − 200)2 + (400)2 = 500 m Total distance (c) Average speed = Total time 1100 = = 55m/min 20 AD (d) Average velocity = t 500 = = 25 m/min (along AD) 20 4. (a) The distance travelled by the rocket in 1 min (= 60 s) in which resultant acceleration is vertically upwards and 10 m/s2 will be h1 = (1/ 2) × 10 × 602 = 18000 m = 18 km …(i) and velocity acquired by it will be …(ii) v = 10 × 60 = 600 m/s Now, after 1 min the rocket moves vertically up with velocity of 600 m/s and acceleration due to gravity opposes its motion. So, it will go to a height h2 till its velocity becomes zero such that 0 = (600)2 − 2gh2 or

h2 = 18000 m [as

g = 10 m /s2 ] …(iii)

= 18 km So, from Eqs. (i) and (iii) the maximum height reached by the rocket from the ground Ans. h = h1 + h2 = 18 + 18 = 36 km (b) As after burning of fuel the initial velocity from Eq. (ii) is 600 m/s and gravity opposes the motion of rocket, so the time taken by it to reach the maximum height (for which v = 0), 0 = 600 − gt or t = 60 s i.e. after finishing fuel the rocket further goes up for 60 s, or 1 min. Ans. 2

5. h =

u2 (20) = = 20 m 2g 2 × 10 20 m/s h

60 m

Let t be the time when particle collides with ground. 1 Then using the equation s = ut + at 2 2 1 we have, − 60 = (20) t + (−10) t 2 2 Solving this equation, we get t=6s Total distance (a) Average speed = Total time 20 + 20 + 60 = 6 Ans. = 16.67 m/s s 60 (b) Average velocity = = t 6 = 10 m/s (downwards) Ans. Total displacement 6. Average velocity = Total time vt0 + 2vt0 + 3vT 2.5 v = ∴ t0 + t0 + T Solving, we get T = 4 t0

Ans.

7. Retardation time is 8 s (double). Therefore, retardation should be half or 2 m/s2. s1 = acceleration displacement 1 1 = a1 t12= × 4 × (4 )2= 32 m 2 2 s2 = retardation displacement 1 = a2 t22 2 1 = × 2 × (8)2 = 64 m 2 v f − vi 0−0 (a) aav = = =0 t 12 (b) and (c) d = s = s1 + s2 = 96 m ∴ Average speed = Average velocity d = t s 96 or = = 8 m/s t 12  22  21 (2)      7   22 2πR 8. T = =6s = 1 v T t=2s= 3 ∴ Particle will rotate by 120°.

Ans.

Ans.

Chapter 6

(a) vav

s = = t

2R sin

θ 2

(θ = 120° )

t

Kinematics — 507

Solving this equation we get, v t t= 0 + 0 g 2

Ans.

12. (a) Velocity = slope of s -t graph R

S 120 ° R

=

3 R 21 3 = m/s t 44

Ans.

v 2 + v 2 − 2vv cos 120° t 3v 3 Ans. = = m/s2 t 2 9. At minimum distance, their velocities are same, ∴ vA = vB or uA + aA t = uB + aB t (b) aav =

or

| ∆v | = t

(3 + t ) = (1 + 2t )

t=2s

or

Minimum distance, dmin = Initial distance − extra displacement of A upto this instant due to its greater speed = 10 − (sA − sB ) = 10 + sB − sA 1 1     = 10 +  u B t + aB t 2 −  uA t + aA t 2     2 2 1 1     = 10 + (1) (2) + (2) (2)2  − (3) (2) + (1) (2)2  2 2     =8m

Ans.

10. Given h1 − h2 = 10 m h2 h1 2

∴ Sign of velocity = sign of slope of s -t graph (b) Let us discuss any one of them, let the portion cd, slope of s -t graph (= velocity ) of this region is negative but increasing in magnitude. Therefore velocity is negative but increasing. Therefore sign of velocity and acceleration both are negative. 13. Displacement s = net area of v -t graph = 40 + 40 + 40 − 20 = 100 m Distance d = | Total area | = 40 + 40 + 40 + 20 = 140 m s 100 50 (a) Average velocity = = Ans. = m/s t 14 7 d 140 (b) Average speed = = Ans. = 10 m /s t 14 14. v1 = speed of person v2 = speed of escalator l l and v2 = v1 = t1 t2 l l t= ∴ = v1 + v2 l l + t1 t2 t1 t2 90 × 60 Ans. = = = 36 s t1 + t2 90 + 60

15. At time t = t1, slope of s -t graph (= velocity ) is positive and increasing. Therefore velocity is positive and acceleration is also positive. At time t = t2, slope is negative but decreasing (in magnitude). Therefore velocity is negative but decreasing in magnitude. Hence, acceleration is positive. 16. Comparing with v = u + at, we have, u = 40 m/s and a = − 10 m/s2 a

1

1 2 1 gt − g (t − 1)2 = 10 2 2 Solving equation, we get t = 15 . s

X = – 60m

∴

11. At the time of collision, s1 = s2 ∴

v0 t −

1 2 1 gt = v0 (t − t0 ) − g (t − t0 )2 2 2

u X=0 t=0

X = + 60m t1

Ans. t3

t2

Distance of 60 m from origin may be at x = + 60 m and x = − 60 m.

508 — Mechanics - I From the figure, we can see that at these two points particle is at three times, t1 , t2 and t3. For X = + 60 m or t1 and t2 1 s = ut + at 2 2 1 ⇒ 60 = (+40) t + (−10) t 2 2 Solving this equation, we get Ans. t1 = 2 s and t2 = 6 s For X = − 60 m or t3 1 2 at 2 1 ∴ − 60 = (+ 40) t + (−10) t 2 2 Solving this equation, we get positive value of t as Ans. t3 = 2 (2 + 7) s s = ut +

17. Displacement = Area under velocity-time graph Hence,

or

sOA =

1 × 2 × 10 = 10 m 2

sAB = 2 × 10 = 20 m sOAB = 10 + 20 = 30 m sBC =

or and or

1 × 2(10 + 20) = 30 m 2

sOABC = 30 + 30 = 60 m sCD =

1 × 2 × 20 = 20 m 2

sOABCD = 60 + 20 = 80 m

v f − vi

v12 − v6 12 − 6 (−10) − (20) = = − 5 m/s2 Ans. 6 (b) ∆v = v f − vi = net area of a -t graph ∴ v14 − v0 = 40 + 30 + 40 − 20 = 90 But v0 = 0 Ans. ∴ v14 = 90 m/s r − r s 19. (a) vav = = f i t t (6$i + 4 $j) − ($i + 2$j) = 4 $ Ans. = (125 . i + 0.5 $j) m/s ∆v v f − vi (b) a av = = t t (2$i + 10$j) − (4 $i + 6$j) = 4 Ans. = (− 0.5$i + $j ) m/s2

18. (a) aav =

t

=

(c) We cannot calculate the distance travelled from the given data. 20. In falling 5 m, first stone will take 1 s (from 1 h = gt 2 ) 2 2h ...(i) First stone, t= g Second stone, (t − 1) =

s (m)

2 (h − 2s) g

Solving these equations, we get h = 45 m

80

...(ii)

Ans.

21. If we start time calculations from point P and apply

60

the equation, a

a

30 10

u=0

2

4

6

8

t (s)

Between 0 to 2 s and 4 to 6 s motion is accelerated, hence displacement-time graph is a parabola. Between 2 to 4 s motion is uniform, so displacement-time graph will be a straight line. Between 6 to 8 s motion is decelerated hence displacement-time graph is again a parabola but inverted in shape. At the end of 8 s velocity is zero, therefore, slope of displacement-time graph should be zero. The corresponding graph is shown in above figure.

1 at 2 2 0

s = ut +

P

v = at0

+ve

1 2 at 2

1  1 2 2  − at0  = (+ at0 ) t + (− a) t  2  2 Solving we get, t = ( 2 + 1) t0 = 2.414 t0 ∴ Total time, T = t + t0 = (3.414 ) t0

Ans.

Chapter 6 22. (a) h = 15 +

u2 2g (5)2 2 × 10 = 16.25 m

After next 2 s (a) v2 = v1 + a 2 t2 = (4 $i ) + (2$i − 4 $j) (2) = (8$i − 8$j) m/s

= 15 +

(b) t =

2h = g

2 × 16.25 = 1.8 s 10

Ans.

1 a 2 t22 2 1 = (6$i + 4 $j) + (4 $i ) (2) + (2$i − 4 $j) (2)2 2 = (18$i − 4 $j) m

(b) r2 = r1 + v1t2 +

Ans.

1 ...(i) × a × (6)2 2 ...(ii) 15 = u + (a) (6) Solving these two equations, we get u = 5 m/s 5 and Ans. a = m/s2 3  5 (c) (5)2 = (2)   s (v 2 = 2 as)  3

23. (a) and (b) 60 = u (6) +

∴ s = 7.5 m 1 24. s = s0 + ut + at 2 2 at

t = 0, s = s0 = 2 m

at

t = 10 s,

Ans.

...(i)

1 s = 0 = s0 + ut + at 2 2 or or

∴

Co-ordinates are, x = 18 m and y= −4 m 1 26. x = uxt + ax t 2 2 1 ∴ 29 = (0) (t ) + × (4.0) t 2 2 ∴ t 2 = 14.5 s2 or t = 3.8 s 1 (a) y = uy t + ay t 2 2 1 = (8) (3.8) + × 2 × 14.5 2 = 44.9 m ≈ 45 m = (8j$ ) + (4.0$i + 2.0$j) (3.8) = (15.2$i + 15.6$j)

...(ii)

∴ Speed = | v | = (15.2)2 + (15.6)2

v = u + at ∴

0 = u + 6a

∴ u + 6a = 0 Solving Eqs. (ii) and (iii) we get, u = − 1.2 m/s and a = 0.2 m/s2 Now, ∴

v = u + at v = (−1.2) + (0.2) (10) = 0.8 m/s

25. After 2 s v1 = u + a 1 t1 = 0 + (2$i ) (2) = (4 $i ) m/s 1 r1 = ri + a 1 t12 2 1 = (2i$ + 4 $j) + (2$i ) (2)2 2 $ $ = (6i + 4 j) m

Ans.

(b) v = u + at

0 = 2 + 10u + 50a 10u + 50a = −2

Kinematics — 509

(at t = 6 s) ...(iii)

= 22 m/s

dv ds = (10) (−3) = − 30 m/s2

a = v⋅

Ans.

Ans.

Ans.

dv 3m/s 27. =− ds m

Ans.

28. v = 3t 2 − 6t ⇒ v = 0 at t = 2 s For t < 2 s, velocity is negative. At t = 2 s, velocity is zero and for t > 2 s velocity is positive. 3. 5

∴

s1 =

3. 5

∫ vdt = ∫ (3t 0

2

− 6t ) dt

0

= 6.125 m = displacement upto 3.5 s 2

2

0

0

s2 = ∫ vdt = ∫ (3t 2 − 6t ) dt

510 — Mechanics - I =−4m = displacement upto 2 s O

–4m

33. Kinetic energy will first increase and then decrease.

6.125 m

∴

d = distance travelled in 3.5 s = 4 + 4 + 6.125 Ans. = 14.125 m d 14.125 Average speed = = t 3.5 Ans. = 4.03 m/s s1 6.125 Average velocity = = t 3.5 Ans. = 1.75 m/s

29. a =

4 v

or

∴ ∴ ∴

∫ vdv = ∫ 4 dt

E

C

t

–

2

v2 − 18 = 4 t − 8 2 v = 8t + 20 4 dv = a= dt 8t + 20 t = 3s a = 0.603 m/s2

At

B +

O

t

6

a A

dv 4 = dt v v

(in downward journey) v = gt 1 1 2 KE = mv = mg 2 t 2 ∴ 2 2 Hence, KE versus t graph is a parabola. 34. Speed first increases. Then just after collision, it becomes half and now it decreases. Pattern of velocity is same (with sign). In the answer, downward direction is taken as positive. Further, v = gt Hence, v -t graph is straight line. 35. Area of a-t graph gives change in velocity ∴ ∆v = v f − vi = area of a-t graph Since, v f = vi

30. v -s equation is

D

∴ Net area of a-t graph should be zero Area OABC = Area CDE ∴ Substituting the values we get, tE = (2 + 3 ) s Ans.

v (m/s)

20 2 v = − s + 20 3 dv 2 ∴ = − per second ds 3 30 s (m) At s = 15 m, v = 10 m/s dv 20 Ans. a=v⋅ =− m/s2 ds 3 31. a = slope of v -t graph and s = area under v -t graph.

Further, for t ≤ 2s, velocity (i.e. slope of s -t graph) is positive but increasing. Therefore, s -t graph is as under. s

t

Same logic can be applied with other portions too.

32. Same as Q.No. 31.

Ans.

36. (a) a = slope of v -t graph (b) s = rf − ri = net area of v -t graph ∴ rf = ri + net area = 10 + 10 + 20 + 10 − 10 − 10 = 30 m (c) (i) For 0 ≤ t ≤ 2 s u = initial velocity = 0 s0 = initial displacement = 10 m a = slope of v -t graph = + 5 m/s2 1 s = s0 + ut + at 2 = 10 + 2.5 t 2 ∴ 2 (ii) For 4 s ≤ t ≤ 8 s u = initial velocity = 10 m/s = velocity at 4 s s0 = (10 m) + area of v -t graph upto 4 s = 10 + 10 + 20 = 40 m a = slope of v-t graph = −5 m/s2 1 ∴ s = s0 + u (t − 4 ) + a (t − 4 )2 2 Ans. = 40 + 10 (t − 4 ) − 2.5 (t − 4 )2

Kinematics — 511

Chapter 6 37. (a) a1 = a2 = −g ∴

(b) For automobile

a12 = a 1 − a 2 = 0

Ans.

(b) u21 = u2 − u1 = 20 − (− s) = + 25 m/s

Ans.

(c) u12 = − u21 = − 25 m/s Since, a12 = 0 ∴ u12 = constant = − 25 m/s

Ans.

(d) a12 = 0. Therefore, relative motion between them is uniform with constant velocity 25 m/s. d 20 Ans. t= = = 0.8 s ∴ v 25 38. (a) When the two meet,

1 1 a 2 t 2 = × 3.5 × (7.39)2 2 2

s2 =

= 95.5 m ∴ Initial distance between them, = s2 − s1 = 35.5 m

B

Net

4 cos q

18 m/s g

1

45° A

(4 sin q + 2)

2 m/s = constant

4 cosθ = 4 sin θ + 2 Solving this equation we get,

12 m 2

θ ≈ 24.3°

5m

or

N Q

Ans.

s2 = 2 × 3.65 = 7.3 m Height = 5 + 7.3 = 12.30 m

Net velocity = v

s1 =

∴

indicates

the

Ans.

P

50

0

E

km /h

(b)

v = (500)2 − (200)2 = 100 21 km/h PQ 1000 10 ∴ t= = = h v 100 21 21

downward

1 a1 t2 2

200 km/h

θ = sin −1 (0.4 ), west of north

∴

42.

1 60 = × 2.2 × t 2 2 t = 7.39 s

q

q

(b) vball = u − gt = 18 − 9.8 × 3.65 = − 17.77 m/s ∴ Velocity of ball with respect to elevator = velocity of ball − velocity of elevator = (−17.77) − (2) = − 19.77 m/s Ans. ≈ − 19.8 m/s Negative sign direction. 39. (a) For truck

Ans.

200 41. (a) sinθ = = 0.4 500

s2 = s1 + 7 (2 t ) = (18 t ) − (4.9 t 2 ) + 7

Solving we get, t = 3.65 s ∴

Ans.

(c) v1 = a1 t = (2.2) (7.39) = 16.2 m/s and v2 = a 2 t = (3.5) (7.39) = 25.9 m/s 40. Net velocity is along AB or at 45° if,

Ans.

v

v0

Ans.

t1

t2

t

512 — Mechanics - I v0 = x ⇒ t1 = t1 v0 = y ⇒ t2 = t2

v0 x v0 y

2.

 1 1 t1 + t2 = v0  +  = 4 y x

...(i)

Further, area = displacement 1 v0 × t = s ∴ 2 But numerically, t = s = 4 units ∴ v0 = 2 units Substituting in Eq. (i) we get, 1 1 + = 2. x y

dv = a = − 4v + 8 dt da dv ∴ = −4 dt dt = − 4 (−4 v + 8) = 16v − 32  da ∴   = 16 vi − 32  dt  i = (16) (0) − 32 = − 32 m/s2 Further, v

dv

t

∫ 8 − 4 v = ∫ dt 0

0

Solving this equation we get, v = 2 (1 − e−4t )

LEVEL 2

v (m/s)

Single Correct Option 1. Let velocity of rain is

2

5 m/s ^j t (s)

37°

37° ^i

37° 5 m/s (i)

In first case,

(ii)

(iii)

vR = a$i − b$j v = (−4 i$ − 3$j) M

∴ vRM = vR − vM = ( a + 4 ) i$ + (− b + 3) $j It appears vertical ∴ a+ 4 =0 or a= −4 In second case, vM = (4 $i + 3$j) ∴ vRM = vR − vM = (a − 4 ) $i + (− b − 3$j) = − 8$i + (− b − 3$j)  7 It appears at θ = tan −1    8 ∴ ∴

−b − 3 = − 7 b=4

and speed of rain = a2 + b2 = 32 m/s

Ans.

Hence, v -t graph is exponentially increasing graph, terminating at 2 m/s. 3. For collision, (at same instant) rA = rB ∴ (ri + vt )A = (ri + vt )B ⇒ (5$i + 10$j + 5k$ ) t = 30$i + (10$i + 5$j + 5k$ ) t Equating the coefficients of x 5 t = 30 + 10 t ⇒ t = − ve So, they will never collide. d vy 4. = 2t dt dy 2 ∴ v y = t 2 or =t dt t3 or y= 3 x and x = v0t ⇒ t = v Substituting in Eq. (i) we have, x3 y= 3 3v0

...(i)

Ans.

Kinematics — 513

Chapter 6 5.

dt = (2αx + β ) dx  dx 1  ∴ =v=  dt  2αx + β

10.

B

v 30°

2

a=

 dv 1  dx = −2α   ⋅ dt  2αx + β dt

= − 2α (v 2 ) (v ) = − 2αv 3

6. f = v ⋅

A

Ans.

dv = a − bx dx v

x

0

0

∫ vdv = ∫ (a − bx ) dx

or ∴

v = 2ax − bx 2

...(i)

At other station, v = 0 2a Ans. x= ⇒ b Further acceleration will change its direction when, a f = 0 or a − bx = 0 or x = b At this x, velocity is maximum. Using Eq. (i), 2

a  a  a vmax = 2a   − b   =  b  b b

7. a =

v⋅ v

or

∴

dv kx 2 =− dx m

kx 2 ∫ v dv = ∫ − m dx 0 a v 2 ka3 = 2 3m 2ka3 v= 3m

Ans.

dv = (4 ) (− tan 60° ) ds = − 4 3 m/s2 1 9. x1 = gt 2 = 0.5 gt 2 2 1 x1 + x2 = g (2t )2 = 2 gt 2 2 x2 = 2gt 2 − x1 = 1.5 gt 2 x2 − x1 = gt 2 or

t=

v cos 60° = 2 m/s ∴

v = 4 m/s

Ans.

5 11. 90 km/h = 90 × = 25 m/s 18 5 180 km/h = 108 × = 30 m/s 18 At maximum separation their velocities are same. ∴ Velocity of motorcycle = 25 m/s or at = 25 or t=5s But thief has travelled up to 7s. s1 = displacement of thief = v1 t1 = 25 × 7 = 175 m s2 = displacement of motorcycle 1 = × a2 t22 2 1 = × 5 × (5)2 2 = 62.5 m ∴ Maximum separation Ans. = s1 − s2 = 112.5 m

12. Relative velocity of A with respect to B should be

0

8. a = v ⋅

∴

Ans.

F − kx 2 = m m

∴

2 m/s

60°

x2 − x1 g

along AB or absolute velocity components perpendicular to AB should be same. 2u sin θ = u sin 30° ∴ 3  3 Ans. ∴ θ = sin −1    4

13. Deceleration is four times. Therefore, deceleration time should be

1 th. 4 v

vmax = 4t a1 4t

a2 t

t

vmax = (a1 ) (4 t ) = (1) (4 t ) = 4 t

514 — Mechanics - I Area of v -t graph = displacement 1 ∴ 200 = (5t ) (4 t ) 2 or t = 20 s

More than One Correct Options 1. (a) a = − α v

Total journey time = 5t = 22.4 s

Ans. or

14. Area of v -t graph = displacement ∴

1032 =

1 (56 + t0 ) (24 ) or 2

0

t0 = 30 s

t1

t (s)

24 =6s 4 ∴ Acceleration time t2 = 56 − t0 − t1 = 20 s 24 = 1.2 m/s2 ∴ Acceleration = 20 15. vQ = − 20i$

v⋅

∴

0

Deceleration time t1 =

2v 3/ 2 s= 0 3α

∴ Ans.

y

2. a = − 0.5t =

v

∴

4m

∴

Q 20 cm/s

vP = − 20 cos 60$i − 20 sin 60° $j = − 10$i − 10 3 $j Assuming P to be at rest, vQP = vQ − vP = − 10$i + 10 3 $j 10 3 = 3 or θ = 60° 10 where, θ is the angle of vQP from x-axis towards positive y -axis. ∴ tan θ =

vQP

ON = OQ = 4 m OP = 3 m

N 60° M P 60°

\ PN = 1 m = 100 cm

t

∫ dv = ∫ −0.5t dt 0

v = 16 − 0.25 t 2

OP = 3 m OQ = 4 m

60° O

Ans.

dv dt

16

P 20 cm/s

3m

O

dv = −α v ds s 0 1 1/ 2 ∫ ds = − α ∫ v dv 0 v

∴

56 s

x

Ans.

(d) a = − α v

t0

t2

Now,

2 v0 t= α

∴

v (m/s) 24

dv = −α v dt t 0 1 −1/ 2 dt v = − ⋅ dv ∫ α v∫ 0

∴

0.25t 3 3 v = 0 when 16 − 0.25t 2 = 0

s = ∫ vdt = 16t − t=8s

or

Ans.

So direction of velocity changes at 8 s. Up to 8 s distance = displacement ∴ At 4 s 0.25 × (4 )3 Ans. d = 16 × 4 − = 58.67 m 3 (0.25) (8)3 S 8 s = 16 × 8 − 3 = 85.33 m (0.25) (10)3 S 10s = 16 × 10 − 3 = 76.67 cm 85.33 m 8s 0s

60°

10 s

Q

Shortest distance = PM = PN sin 60° 3 = (100) = 50 3 cm 2

76.67 m

Distance travelled in 10 s, d = (85.33) + (85.33 − 76.67) Ans. = 94 m

Kinematics — 515

Chapter 6

s1 1 = a1 t1 t1 + t2 2 1 s2 = (t1 + t3 ) vmax 2 1 = (t1 + 2t2 ) (2a1 t1 ) 2 s2 v2 = = a1 t1 (t1 + t3 )

At 10 s

v1 =

v = 16 − 0.25 (10)2 = − 9 m/s ∴

Speed = 9 m/s

Ans.

3. If a = constant Then | a | is also constant or

dv = constant dt

4. rB = 0 at t = 0

v

Ù N (j) A

vmax

B

E (^i )

3 km

t1

rA = 3$i + 4 $j

vB = (40 cos 37° )i$ + (40 sin 37° )$j = (32$i + 24 $j) Ans. vAB = (−32i$ − 44 $j) km/h vB

37°

At time t = 0, rAB = rA − rB = (3$i + 4 $j) ∴ At time t = t , rAB = (rAB at t = 0) + vAB t = (3 − 32t )$i + (4 − 44 t )$j

5. a1 t1 = a2 t2 vmax = a1 t1 = a2 t2 s1 = Area of v -t graph 1 = (t1 + t2 ) (a1 t1 ) 2 v

Ans.

a2

or t2

t

Ans.

(downwards) s = 0 and a = constant = g F α ...(i) 7. a = = t m m or a∝t i,e. a-t graph is a straight line passing through origin. If u = 0, then integration of Eq. (i) gives, αt 2 or v ∝ t 2 v= 2m Hence in this situation (when u = 0 ) v -t graph is a parabola passing through origin. 8. a-s equation corresponding to given graph is, s a=6− 5 dv  s ∴ v⋅ = 6 −  5 ds 

vmax

t1

t

t3

6. In the complete journey,

or a1

a2

From the four relations we can see that v2 = 2v1 and 2s1 < s2 < 4 s1

t=0 vA = (−20$j)

at

2a1

4 km

t3 = 2t2 tmax = 2a1t1 = 2a2t2

At

v

s

0

0



s

∫ vdv = ∫  6 − 5 ds s2 5 s = 10 m , v = 10 m/s

v = 125 −

Ans.

516 — Mechanics - I Maximum values of v is obtained when dv = 0 which gives s = 30 m ds ∴

vmax

s 9. vav = t Now, ∴

and

(30)2 = 12 × 30 − 5 = 180 m/s d vav = t d ≥ |s | vav ≥ | vav |

Ans.

1 10. v = at and x = at 2 (u = 0) i.e. v -t graph is a straight 2 line passing through origin and x -t graph a parabola passing through origin. 11. For minimum time b tmin = ∴ v

u

l=

= vt −

v2 v2 − 2α 2β

∴

∴

∴

(but v > u )

v = − 2$j and

For t > T , v = + ve At t =T, v = 0 ∴ Particle changes direction of velocity at t = T S = Net area of v -t graph = 0 a = Slope of v -t graph = constant v 13. v = αt1 ⇒ t1 = α v v = βt2 ⇒ t2 = β v

t2

a = 2i$

v⊥a 2 , At t = v y = 0, vx ≠ 0`. 3 Hence, the particle is moving parallel to x-axis. 15. (14 )2 = (2)2 + 2 as ∴

2 as = 192 units  s At mid point, v 2 = (2)2 + 2a    2 192 =4+ = 100 2 ∴ v = 10 m/s ∴

v

t0

⇒ vx =

or

12. For t < T , v = −ve

t1

2lαβ α+β

∴

⇒ Net velocity = v 2 − u2 b t= net velocity

v=

At t = 0,

⇒ Net velocity

v

l v  1 1 +  +  v 2  α β

dx = 2t dt dv ax = x = 2 dt y = t 3 − 2t dy vy = = 3t 2 − 2 dt dv y ay = = 6t dt vx = 0, v y = − 2, ax = 2 and ay = 0

⇒

u

t=

For t to be minimum its first derivation with respect to velocity be zero or, l α+β 0=− 2 + 2αβ v

⇒

For reaching a point exactly opposite

b

1 2 1 αt1 + vt0 + β t 22 2 2 2 2   v 1 v v 1  v = (α )   + v  t − −  + ( β )   α   β 2 α β 2 Now,

14. x = t 2

v

b

 v v t0 = t − t1 − t2 =  t − −   α β

∴

t

XA : AY = 1 : 3 1 XA = s and 4

AY =

 s v12 = (2)2 + 2a    4

Ans. 3 s 4

Chapter 6

3. S = 25 × 2.13 − 5 × (2.13)2

192 = 52 4 v1 = 52 ≠ 5 m/s =4+

∴

Ans.

10 = 2 + at1 (v = u + at ) 8 t1 = 14 = 10 + at2 ∴ a 4 or t1 = 2 t2 Ans. t2 = ∴ a 1 S 1 = (2t ) + a (t 2 ) = distance travelled in first half 2 1 S 2 = 2 (2t ) + a (2t )2 2 S 3 = S 2 − S 1 = distance travelled in second half We can see that, S S1 ≠ 3 2

Comprehension Based Questions 1. Velocity of ball with respect to elevator is 15 m/s (up) and elevator has a velocity of 10 m/s (up). Therefore, absolute velocity of ball is 25 m/s (upwards). Ball strikes the floor of elevator if,

1

= 9.5 m

Ans.

5. uA + aAT = uB − aBT T =4s

Putting we get Now, ∴ ∴

4 (aA + aB ) = uB − uA SA = SB 1 1 uA t + aA t 2 = uB t − aB t 2 2 2 (uB − uA ) t=2 = 2 × 4= 8 s (aA + aB )

....(i)

Ans.

6. S A = S B 1 1 aAt 2 = 15t − aB t 2 2 2 10 + aA t = 30 − aB t

5t +

Ans.

aA = 0.5 m/s

Ans.

Match the Columns

S1 = S2 + 2 10t + 2.5t 2 = 25t − 5t 2 + 2

1. In (a) and (b), if velocity is in the direction of Ans.

2. If the ball does not collides, then it will reach its maximum height in time, u 25 t0 = = = 2.5 s g 10

acceleration (or of the same sign as that of acceleration) then speed increases. And if velocity is in opposite direction, then speed decreases. In (c) slope of s -t graph (velocity) is increasing. Therefore speed is increasing. In (d) slope of s -t graph is decreasing. Therefore, speed is decreasing.

2. If v ⋅ a = 0, speed is constant because angle between

Since, t < t0, therefore as per the question ball is at its maximum height at 2.13 s. hmax = 50 + 2 + 25 × 2.13 − 5 × (2.13)2 = 82.56 m

= 2 + [ 25 × 1 − 5 × (1)2 ] − [10 × 1 + 2.5 (1)2 ]

vA = uA + aA t = (6) + (0.5) (10) = 11 m/s

10 m/s, 5 m/s2

Solving this equation we get, t = 2.13 s

∴ 25 − 10t = 10 + 5t or t =1s Maximum separation = 2 + S 2 − S 1

∴ At 10 s,

2m

∴

4. At maximum separation, their velocities are same

7. 8 = 6 + aAT = 6 + 4 aA

–

50 m

Ans.

∴ (5 + aA t ) − (15 − aB t ) = 10 or vA − vB = 10 m/s

10 m/s2

2

= 30.56 m

or

+

25 m/s

Kinematics — 517

Ans.

v and a in this case is 90°. If v ⋅ a = positive then speed is increasing because angle between v and a in this case is acute. If v ⋅ a = negative then speed is increasing because angle between v and a in this case is obtuse.

518 — Mechanics - I 3. In portion AB, we can see that velocity is positive and increasing. Similarly for other parts we can draw the conclusions. S Area of v -t graph 4. (a) Average velocity = = t Time 20 = = 5 m/s 4 v f − vi (b) Average acceleration = time v 4 s − v1 s 0 − 5 5 = = = − m/s2 4 −1 3 3 d | Total area | (c) Average speed = = t Time 20 + 10 = = 5 m/s 6 (d) Rate of change of speed at 4 s = | a | = | slope of v -t graph | = 5 m/s2

Subjective Questions 1. aav =

∴

Speed = vx2 + v 2y = 2 5 m/s

(d) At t = 0, ax = 2 m/s2 and ∴

ay = 2 m/s2 a = ax2 + a2y = 2 2 m/s

+ve

–ve

= 126 . × 103 m/s2

Ans.

Note v f is upwards (+ve) and v i is downwards ( − ve).

2. v dv = ads

∴

12 m

v

∫0vdv = ∫0

∴

5. (a) x = 0 at t = 2s dx dv (b) v = = 10 t and a = = 10 dt dt v = a at 1s (c) v = 10 t ∴ Velocity is positive all the time. (d) v = 0 at t = 0 s dx 6. vx = = 2t − 2 dt dv ax = x = 2 dt dy vy = = 2t − 4 dt dv y ay = =2 dt (a) It crosses y -axis when, x = 0 ⇒ t =1s At this instant v y is −2 m/s. (b) It crosses x-axis when, y = 0 ⇒ t=2s At this instant vx is + 2 m/s (c) At t = 0, vx = − 2 m/s and v y = − 4 m/s

∆ v v f − vi = ∆t ∆t 2g hf + 2g hi = ∆t 2 × 9.8 × 2 + 2 × 9.8 × 4 = 12 × 10−3

a ds

2

v = area under a -s graph from s = 0to s = 12 m. 2 = 2 + 12 + 6 + 4 = 24 m 2/s2

or

v = 48 m/s = 4 3 m/s

Ans.

3. Let AB = BC = d BD = x and BB′ = s = displacement of point B. AN

vt

BN s A

C B

D

1 at 2 2

CN

From similar triangles we can write, 1 2 at vt s 2 = = d+x x d−x From first two equations we have, d vt 1+ = x s d vt or = −1 x s From last two equations we have, 1 2 at d −1= 2 x s 1 2 at d 2 or = +1 x s

…(i)

…(ii)

Kinematics — 519

Chapter 6

1 (3t )(0.2 t ) = 14 2 Solving this equation, we get 3t = 20.5 s

Equating Eqs. (i) and (ii) we have, 1 2 at vt −1= 2 +1 s s 1 vt − at 2 2 or =2 s

∴

Note Maximum speed 0.2 t is less than 2.5 m/s.

6. Let t0 be the breaking time and a the magnitude of

1  a  v s =   t −   t2  2 2  2

∴

Ans.

1 2 at we have, 2 v a Initial velocity of B is + and acceleration − . 2 2

Comparing with s = ut +

4. Let us draw v-t graph of the given situation, area of which will give the displacement and slope the acceleration.

deceleration. 80.5 km/h = 22.36 m/s, 48.3 km/h = 13.42 m/s. In the first case, (22.36)2 …(i) 56.7 = (22.36 × t0 ) + 2a (13.42)2 and …(ii) 24.4 = (13.42 t0 ) + 2a Solving these two equations, we get t0 = 0.74 s and Ans. a = 6.2 m/s2

7. Absolute velocity of ball = 30 m/s

v y

10 m/s 30 m/s

y d d x

x

2

x

2m

1

t1

t2

s2 − s1 = x d +

28 m

t

t3

1 yd 2

…(i)

1 …(ii) yd 2 Subtracting Eq. (i) from Eq. (ii), we have s3 + s1 − 2s2 = yd or s3 + s1 − 2 s1s3 = yd (s2 = s1s3 ) s3 − s2 = xd + yd +

Dividing by d 2 both sides we have, ( s1 − s3 )2 y = d d2 = slope of v-t graph = a. Hence proved.

(a) Maximum height of ball from ground (30)2 = 28 + 2 + = 76 m 2 × 9.8 (b) Ball will return to the elevator floor when, s1 = s2 + 2 or 10t = (30t − 4.9 t 2 ) + 2 Solving, we get t = 4.2 s

8. (a) Average velocity 5

Displacement = = Time

5. Area of v-t graph = displacement

5

=

v (m/s)

0.2t

t

2t

t (s)

Ans.

∫0 (3t − t

2

∫0v dt 5

) dt

5 = − 0.833 m/s (b) Velocity of particle = 0 at t = 3 s i.e. at 3 s, particle changes its direction of motion. Total distance Average speed = Total time

520 — Mechanics - I =

∴ Total time taken t = t1 + t2 = (200 sec θ − 120 tan θ ) dt For t to be minimum, =0 dθ or 200 sec θ tan θ − 120 sec2 θ = 0

(Distance from 0 to 3 s) + (Distance from 3 s to 5 s) Time 3

d0-3 = ∫ (3t − t 2 )dt = 4.5 m 0

5

d3-5 = ∫ (t 2 − 3t ) dt = 8.67 m 3

4.5 + 8.67 5 = 2.63 m/s

Ans.

9. (a) a = 2t − 2 (from the graph) Now,

v

t

t

∫0 dv = ∫0a dt = ∫0 (2t − 2) dt

∴

v = t 2 − 2t 4

4

2

2

θ = sin −1 (3/5)

or

Average speed =

∴

(b) s = ∫ v dt = ∫ (t 2 − 2t ) dt = 6.67 m

10. (a)

3 (b) tmin = 200 sec θ − 120 tan θ (where, sin θ = ) 5 5 3 = 200 × − 120 × 4 4 Ans. = 250 − 90 = 160 s = 2 min 40 s dy …(i) 11. Given that | vbr | = v y = = u dt dx  2v0  …(ii) | vr | = vx = =   y dt  c  From Eqs. (i) and (ii) we have,

3 m/s

120 m

θ

dy uc = dx 2v0 y

vbr vr

c

y

y

4 m/s

x

or

3 cos θ (4 – 3 sin θ)

Time to cross the river 120 40 t1 = = = 40 sec θ 3 cos θ cos θ  40  Drift along the river x = (4 − 3 sin θ )    cos θ  = (160 sec θ − 120 tan θ ) To reach directly opposite, this drift will be covered by walking speed. Time taken in this,

y

∫0

uc y dy = 2v0

At or

3

∫0 dx

or

ucx y2 = v0

c cv ,x = 0 2 4u cv0 = 2x = 2u

xnet

dv = v (slope of v-s graph) ds At s = 50 m dv 40 v = 20 m/s and = = 0.4 per sec ds 100 ∴ a = 20 × 0.4 = 8 m/s2 At s = 150 m

∴

v = (40 + 5) = 45 m/s dv 10 = = 01 . per sec ds 100 a = 45 × 01 . = 4.5 m/s2

a- s graph θ

From s = 0 to s = 100 m v = 0.4 s 4

160 sec θ − 120 tan θ 1 = 160 sec θ − 120 tan θ

t2 =

and ∴

Ans.

y=

12. a = v

and 5

x

dv = 0.4 ds dv a = v. = 016 . s ds

Ans.

Kinematics — 521

Chapter 6 i.e a - s graph is a straight line passing through origin of slope 0.16 per (sec)2. At s = 100 m, a = 016 . × 100 = 16 m/s2 From s = 100 m to s = 200 m v = 01 . s + 30 dv = 01 . ds dv = (01 . s + 30)(01 . ) = (0.01 s + 3) ds i.e. a-s graph is straight line of slope 0.01 (sec)-2 and intercept 3 m/s 2. At s = 100 m, a = 4 m/s2

∴ a=v

and at s = 200 m, a = 5 m/s2 Corresponding a-s graph is as shown in figure. a (m/s2)

16 5 4 100

s (m)

200

− x) dx a2 x2 x 3 …(iii) or − y = 2a 3a2 This is the desired equation of trajectory. (b) Time taken to cross the river is a a t= = vx v y

∫0

or

dy = ∫

x x (a

0

(c) When the boatman reaches the opposite side, x = a or v y = 0 [from Eq. (i)] Hence, resultant velocity of boatman is v along positive x-axis or due east. (d) From Eq. (iii) a2 a3 a y= − 2= 2a 3a 6 At x = a (at opposite bank) Hence, displacement of boatman will be a s = x $i + y $j or s = a $i + $j 6 14. (a) Since, the resultant velocity is always perpendicular to the line joining boat and R, the boat is moving in a circle of radius 2ω and centre at R. R

13. (a) Let vbr be the velocity of boatman relative to

ω

river, vr the velocity of river and vb is the absolute velocity of boatman. Then,

Q

2ω S

Y P

j

N vr X

O

i

A

vbr

(b) Drifting = QS = 4ω 2 − ω 2 = 3ω.

E

W

S

(c) Suppose at any arbitrary time, the boat is at point B. R

60°

a

vb = vbr + vr Given, | vbr | = v and | vr | = u dy v Now u = vy = = x (a − x ) 2 dt a dx and v = vx = =v dt Dividing Eq. (i) by Eq. (ii), we get dy x (a − x ) x (a − x ) or dy = = dx 2 dx a a2

θ

S Q

v θ

…(i)

vnet θ

B

…(ii)

P

Vnet = 2v cos θ dθ Vnet v cos θ = = dt 2ω ω

v

522 — Mechanics - I ω sec θ dθ = dt v t ω 60° ∴ ∫0dt = v ∫0sec θ dθ ω ° t = [ ln (sec θ + tan θ )]60 ∴ 0 v 1.317ω or Ans. t= v 15. For 0 < s ≤ 60 m 12 s v= s+ 3=3+ 60 5 dv  1 ds = ⋅ dt  5 dt 1 1 s 3 s …(i) = (v ) =  3 +  = + 5 5 5 5 25 3 s or a= + 5 25 i.e. a-s graph is a straight line. 3 At s = 0, a = m/s2 = 0.6 m/s2 5 and at s = 60 m, a = 3.0 m/s2 or

s > 60 m v = constant ∴ a=0 Therefore, the corresponding a-s graph is shown in figure. For

16. From the graph, 22.5 ⋅ s or 150 60  22.5  v ⋅ dv = ∫  22.5 − × s ds 0   150 a = 22.5 −

v

∫0

22.5 (60)2 v2 = 22.5 × 60 − × 2 150 2 v = 46.47 m/s

∴ ∴

17. (a) ux = 3m/s ax = − 1.0 m/s2 Maximum x-coordinate is attained after time u  t = x  = 3 s  ax  At this instant vx = 0 and v y = uy + ayt = 0 − 0.5 × 3 = − 15 . m/s Ans. ∴ v = (− 1.5 $j) m/s 1 2 (b) x = uxt + axt 2 1 = 3 × 3 + (− 10 . ) (3)2 = 4.5 m 2 1 1 y = uy t + ay t 2 = 0 − (0.5) (3)2 = – 2.25 m 2 2 Ans. ∴ r = (4.5 $i – 2.25 $j) m $ $ $ $ 18. (b) v = vx i + v y j and a = ax i + ay j ` v ⋅ a = vxax + v yay

a (m/s2)

v = vx 2 + v y 2

Further

60

From Eq. (i), or

∴

v⋅ a vx ax + v y ay = v vx 2 + v y 2

∴

3.0 0.6 s (m)

120

dv v = dt 5 v dv 1 t ∫3 v = 5 ∫0 dt  v t ln   =  3 5

v = 3 et / 5 or

60

Ans.

dv = at dt or component of a parallel to v = tangential acceleration. 19. (a) vbr = 4 m/s, vr = 2 m/s BC |v | 2 1 tan θ = = r = = AB | vbr | 4 2 =

D t1

∫0 ds = 3 ∫0

θ

et/ 5 dt

60 = 15 (et1 / 5 − 1) or t1 = 8.0 s Time taken to travel next 60 m with speed 15m/s 60 will be =4s 15 Ans. ∴ Total time = 12.0 s

vb

B 100 m

vbr

C 200 m θ

A

vb vr

In this case, vb should be along CD.

√5 θ 2

1

Kinematics — 523

Chapter 6 ∴

vr cos θ = vbr sin α

21. In order that the moving launch is always on the straight line AB, the components of velocity of the current and of the launch in the direction perpendicular to AB should be equal, i.e.

D θ vb vbr

B

α

θ C

u β

vr

α

 2 2   = 4 sin α  5 1 sin α = 5

or

v A B

 1 α = θ = tan −1    2 200 200 (b) t1 = = = 50 s | vbr | 4 ∴

α u

5 = 50 5 m 2 | vb | = | vbr | cos α − | vr | sin θ 6  2  1 =4   −2  = m /s  5  5 5

DC = DB sec θ = (100)

t1 DC 50 5 200 + = 25 + = s 6 2 | vb | 3 5 t2 4 Ans. = t1 3 t2 =

∴

or

20. vb = velocity of boatman = vbr + vr vc = velocity of child = vr vbc = vb − vc = vbr vbc should be along BC. i.e. vbr should be along BC, 0.6 3 where, tan α = = 0.8 4 or α = 37°

and ∴

A

…(i) u sin β = v sin α …(ii) S = AB = (u cos β + v cos α )t1 Further BA = (u cos β − v cos α )t2 …(iii) …(iv) t1 + t2 = t Solving these equations after proper substitution, we get Ans. u = 8m/s and β = 12°

22. Here, absolute velocity of hail stones v before colliding with wind screens is vertically downwards and velocity of hail stones with respect to cars after collision v′HC is vertically upwards. Collision is elastic, hence, velocity of hail stones with respect to cars before collisionvHC and after collision v′HC will make equal angles with the normal to the wind screen. (v ′HC)1 A

Ans.

2β Child C α

v

A

0.8 km

B

1.0 km

0.6 km α

Further

t=

Boat

BC 1 = h | vbr | 20

= 3 min

(vHC)1

β 90° – 2β α1 –v1

β C

α1

vbr B

v

β

Ans.

( vHC )1 = velocity of hail stones – velocity of car 1 = v − v1 From the figure, we can see that β + 90° − 2β + α 1 = 90° or α 1 = β or 2β = 2α 1

524 — Mechanics - I v1 v

In ∆ ABC, tan 2β = tan 2α 1 =

or

Similarly, we can show that v tan 2α 2 = 2 v From Eqs. (i) and (ii), we get v1 tan 2α 1 tan 60° 3 = = = =3 v2 tan 2α 2 tan 30° 1/ 3 v1 ∴ =3 v2 dv kv cos θ k 23. ax = x = − = − vx dt m m ∴

dvx k = − dt or vx m

∫v

0 cos θ 0

−

vx = v0 cos θ 0 e

or

dvx k =− vx m

vx

 k  v y + g  m

…(i)

k t m

…(ii)

 k  v0 sin θ 0 + g  m

or

x

∫0

m k

dx = v0 cos θ 0

t

∫0

−

e

k t m

…(ii)

dt k

t

∫0 dt …(i)

− t mv0 cos θ 0 [1 − e m ] k mv0 cos θ 0 xm = k t = ∞.

x=

or

at

24. In the first case, BC = vt1 and

Y

k t m

k  k   − t  v0 sin θ 0 + g e m − g    m   (b) Eq. (i) can be written as k − t dx = v0 cos θ 0 e m dt

or v y =

Ans.

−

=e

B

Ans.

w = ut1

C

θ kv

u

u

mg

α

X A

Similarly, dv y kv sin θ k  ay = =− − g = −  v y + g m  dt m dv y vy t or = − ∫ dt ∫v0 sin θ 0 k 0 vy + g m or

m k

v

y  k  ln  m v y + g   v0 sin 

=−t θ0

v

In the second case, u sin α = v and w = (u cos α ) t2 Solving these four equations with substitution, we get w = 200 m, u = 20m/min, v =12m/min and α = 36°50′

v

proper

Ans.

7. Projectile Motion INTRODUCTORY EXERCISE 7.1 y

1. v1 ⋅ v2 = 0 ⇒ ⇒

(u 1 + a 1t ) ⋅ (u 2 + a 2t ) = 0 (10i$ − 10t$j) ⋅ (−20$i − 10t$j) = 0

⇒

−200 + 100t = 0

u a O

2

t = 2 sec

⇒

R

4. ∆v = v f − vi

2. It is two dimensional motion. 3. The uniform acceleration is g.

y

^ j u

4. u = (40$i + 30$j) m/s, a = (−10$j) m/s2 , t = 2 sec 1 2 at 2 5. ux = uy = 20 m/s, ay = −10 m/s2 1 2 ayt 2 1 ⇒ −25 = 20t − × 10 × t 2 2 Solving this equation, we get the positive value of, t = 5 sec Now apply, v = u + at and sx = uxt

INTRODUCTORY EXERCISE 7.2

1. u = 40$i + 40$j

^ i

α

Now, v = u + at and s = ut +

s y = uyt +

T x

S

α u

O

x

= (u cos α $i − u sin α $j) − (u cos α $i + u sin α $j) = (−2u sin α ) $j Therefore, change in velocity is 2u sin α in downward direction. u2 sin 2θ u2 sin 2 θ 5. (a) R = , H = g 2g 2u sinθ and T = g Here, u = 20 2 m/s and θ = 45° (b) v = u + at

a = − 10$j t=2s 40 m/s 40Ö2 m/s

a = g = 10 m/s2

where,

u = (20$i + 20$j) m/s a = (−10$j) m/s2

and (c) Horizontal component remains unchanged (= 20i$ ) and vertical component is reversed in direction (= − 20j$ ) Ans. v = (20$i − 20$j) m/s

6. (a) Since, acceleration is constant. Therefore,

45° 40 m/s

(a) Apply v = u + at as a = constant 1 (b) Apply s = ut + at 2 2 3. Average velocity s R (u T ) = = = x = ux t T T = u cos α

∴

∆v a av = a = g = (−10$j) = ∆t ∆v = (−10$j) (∆t ) = (−10$j) (3) = (−10$j) (3) = (−30j$ ) m/s

∴Change in velocity is 30 m/s, vertically downwards. Ans. 1 2 ut + at s 1 2 (b) vav = = = u + at t t 2

526 — Mechanics - I 1 = (20i$ + 20$j) + (−10$j) (3) 2 $ $ = (20i + 5 j) m/s

2u sin θ 7. T = = g

2 × 20 × 10

Ans. 1 2 =2 2s

10 10 5 = = m/s T 2 2 2

Ans.

8. (a) Initial velocity is horizontal. So, in vertical direction it is a case of free fall. 2h 2 × 100 t= = = 20 sec g 10 (c) vx = ux = 20 m/s v y = uy + ayt

θ

vx

v=

+

(2) (20 2 ) sin (45° − 30° ) = 1.69 s (10) (cos 30° ) u2 [sin (2α − β ) − sin β ] R= g cos2 β =

vy

= 49 m/s

(20 2 )2 [sin (2 × 45° − 30° ) − sin 30° ] (10) cos2 30°

= 39m 2u sin (α + β ) 2. T = g cos β 2 × 20 2 sin (45° + 30° ) ≈ 6.31 s (10) cos 30° u2 [sin (2α + β ) + sin β ] R= g cos2 β =

= 0 − 10 20 v 2y

2u sin (α − β ) g cos β

=

(b) x = uxt = 20 20 m

vx2

v

(20 2 )2 [sin (2 × 45° + 30° ) + sin 30° ] (10) cos2 30°

= 145.71 m

3. Using the above equations with α = 0°

 vy θ = tan −1   = tan −1 ( 5 )  vx  u2 sin 60° 9. R = = 3 km g ⇒

a (1 + b2 ) 2c

=

Now the remaining horizontal distance is (50 − 40) m = 10 m. Let v is the speed of player, then

= −10 20 m/s

u=

1. T =

= 40 m

v=

∴

INTRODUCTORY EXERCISE 7.3

u2 sin 2 θ (20)2 sin 90° R= = g 10

vT = 10 or

a (1 + b2 ) =c 2u2

∴

∴ | vav | = (20)2 + (5)2 = 20.62 m/s

g (1 + tan 2 θ ) = c 2u2

and

u2 3 = Rmax = g sin 60° = 2 3 km

Since, Rmax < 5 km So, it can't hit the target at 5 km. 10. Comparing with, gx 2 y = x tan θ − 2 (1 + tan 2 θ ) 2u tan θ = b, g = a

T =

2 (20) sin 30° = 2.31 s (10) cos 30°

R=

(20)2 [sin 30° + sin 30° ] (10) cos2 30°

= 53.33 m 4. (a) Horizontal component of velocities of passenger and stone are same. Therefore relative velocity in horizontal direction is zero. Hence the relative motion is only in vertical direction. (b) With respect to the man, stone has both velocity components, horizontal and vertical. Therefore path of the stone is a projectile. 5. (a) (downwards) a1 = a2 = g ∴

Relative acceleration = 0

Projectile Motion — 527

Chapter 7 u 12 = u 1 − u 2 = (20$j) − (20$i + 20$j) = (−20i$ ) m/s

(b)

∴

d = | u 12 | t = 20 × 2 = 40m 6. The range is maximum at, π β (given in the theory) α= + 4 2 30° Ans. = 45° + = 60° 2

or 20 m/s in horizontal direction (c) u 12 = (−20i$ ) m/s is constant. Therefore relative motion is uniform.

Exercises LEVEL 1

T =

Assertion and Reason if a is constant ∴

a

gT ⇒ u sin θ = 2

H =

∴ dv 9.   = 10 m/s2  dt 

u sin θ (gT / 2) = 2g 2g 2

2

2

H ∝T2

or

R = ux T R1 (ux )1 T1  1  2 = ⋅ =     =1 R2 (ux )2 T2  2  1

u

a

2u sinθ 3. T = g

10. vx = ux

u

vy = ±

90°

v

5. In projectile motion along an inclined plane, normally we take x and y-axis along the plane and perpendicular to it. In that case, ax and ay both are non-zero. 2u sinθ 7. T = =4 g ∴

(u sin θ ) = 20 m/s u2 sin 2 θ (20)2 H = = = 20 m 2g 20

∴

8. H = or

u2 sin 2 θ 2g

Ans.

u sin θ ∝ H 4H =2 H

u2y − 2gh

and

v = vx2 + v 2y

and

u2 = ux2 + u2y

Single Correct Option 1. a = g = constant for small heights. 2. H θ =

u2 sin 2 θ 2g u2 sin 2 (90 − θ ) u2 cos2 θ = 2g 2g 2 sin θ = cos2 θ

H 90 − θ = ∴

3. Rθ =

(u sin θ )1 = (u sin θ )2

R1 = R2

d| v| ≠ 10 m/s2 dt

But

4.

⇒ T ∝ u sin θ

T1 (u sin θ )1 2 = = T2 (u sin θ )2 1

1. In the cases shown below path is straight line, even u

2u sinθ g

∴ ∴

Hθ H 90 − θ

R45° 2 u2 sin 2 θ 1 = g 2

 u2     g

or sin 2 θ =

2θ = 30° or θ = 15°

1 2

528 — Mechanics - I 4. At 45°, range is maximum. At highest point, it has only horizontal component of velocity or 20 cos 45° ≈ 14 m/s. 5. F ⋅ v = 0 ⇒ F ⊥ v Hence path is parabola. 6. vx = ux ∴ v cos 30° = u cos 60° u or v= 3 1 Velocity has become times. Therefore, kinetic 3 energy will become 1/3 times.

7. T1 =

8. Rmax

2u sin θ 2u cos θ , T2 = g g 2 (u sin θ ) (u cosθ ) R= g  gT   gT  2  1  2  2   2  = g 1 = gT1T2 2 v02 at θ = 45° = g

Other stone should be projected at 90° − θ or 30° from horizontal. u2 sin 2 30° ∴ H2 = 2g (52.2)2 (1/ 4 ) = 20 Ans. = 34 m 1 2 11. Using s = ut + at in vertical direction 2 1 −70 = (50 sin 30° ) t + (−10) t 2 ∴ 2 On solving this equation, we get Ans. t=7s

12. S = H 2 + R 2 / 4

H

S

Ans.

R/2

Average velocity =

S S 2S = = t T /2 T

13. Velocity of train in the direction of train is also 30 m/s. So there is no relative motion in this direction. In perpendicular direction, u2 sin 2θ d=R= g

R

∴

(30)2 sin 90° = 90 m 10 14. For maximum value of y dy = 10 − 2t = 0 dt ⇒ t=5s ymax = (10) (5) − (5)2

2 Amax = π Rmax

=

9. Maximum range is obtained at 45° u2 = 1.6 or g T =

u = 4 m/s

2u sin 45° 2 × 4 × (1/ 2 ) = g 10

= 0.4 2 s Number of jumps in given time, t 10 2 n= = = 25 T 0.4 2 ∴ Total distance travelled = 1.6 × 25 = 40 m

∴

15. R = Ans.

(u2 ) sin 2 60° 20 u = 52.2 m/s

102 =

Ans.

2

u [sin (2α + β ) + sin β ] g cos2 β

u = 50 m/s, g = 10 m/s2 , α = 0°, R=

u2 sin 2 θ 10. H 1 = 2g ∴

= 25 m

Ans.

(50)2 [sin (2 × 0 + 30° ) + sin 30° ] 10 cos2 30° (2500)  1 1 =  +  10 × (3/ 4 )  2 2 1000 Ans. = m 3

Chapter 7 16. T =

2u sin θ (2) (8) sin 53° = = 1.28 s g 10

R=

and vertical component of B v2 = 5 2 cos 45° = 5 m/s

u2 sin 2θ (80)2 sin (106° ) = = 615.2 m g 10

Distance travelled by tank, d = (5) T = (5) (1.28) = 6.4 m Total distance = (615.2 + 6.4 ) m ∴ = 621.6 m

Ans.

Subjective Questions y

1. At 45° v y = ± vx = ± ux = ± 60 cos 60° = ± 30 m/s Now, v y = uy + ay t vy − vy ∴ t= ay = ∴

t2 = 8.20 s

Since, v1 = v2 , so they may collide Now the second condition is, R1 + R2 ≥ d (d = 15 m ) (10)2 sin 60° R1 = = 8.66 m ∴ 10 (5 2 )2 sin 90° R2 = = 5m 10 Since, R1 + R2 < d, so they will not collide. 5. v1 ⋅ v2 = 0 when v1 ⊥ v2 ∴

x

(± 30) − 60 sin 60° − 10

t1 = 2.19 s and

Projectile Motion — 529

Ans.

∴

(u 1 + a 1t ) ⋅ (u 2 + a 2t ) = 0 (3$i − 10t $j) ⋅ (4 $i − 10t $j) = 0

or

t = 0.12 s

Now in vertical direction they have no relative motion and in horizontal direction their velocities are opposite. ∴ d = 3t + 4 t = 7t = (7) ( 0.12 ) m ≈ 2.5 m

2. Vertical component of initial velocity u = 20 2 sin 45° = 20 m/s 1 Now, apply s = ut + at 2 (to find t ) in vertical 2 direction, with s = 15 m , u = 20 m/s and a = − 10 m/s2 Ans.

y P = (6, 3) P

3. vx = ux = 20 cos 60° = 10 m/s Given, ∴ or ∴

u 2 2 4 v = u2

3m

v=

x 6m

4 (vx2 + v 2y ) = u2

∴

4 [(10)2 + v 2y ] = (20)2 or vy = 0 Hence, it is the highest point T u sin θ t= = ∴ 2 g (20) sin 60° 10 = 3s

Ans.

x  6. y = x 1 −  tanα  R

∴

18 m

yR x (R − x ) (3) (24 ) 2 = = (6) (18) 3

tan α =

 2 α = tan −1    3

Ans.

7. y -coordinate of particle is zero

=

when, Ans.

4. For collision to take place, relative velocity of A with respect to B should be along AB or their vertical components should be same. Vertical components of A v1 = 10 sin 30° = 5 m/s

4 t − 5t 2 = 0 ∴ at and at

t = 0 and 0.8 s x = 3t t = 0, x = 0 t = 0.8 s, x = 2.4 m

Ans.

530 — Mechanics - I 2u sin (α − β ) (α = 60° , β = 30° ) g cos β y 2 × 10 × sin 30° = (10) cos 30° 2 = s 3 x vx = ux = 10 cos 60° = 5 m/s v y = uy + ay t T =

8.

11. Horizontal component of velocity always remains constant ∴ 40 cos 60° = v cos 30° 40 or v= m/s 3

12. (a) If they collide in air then relative velocity of A with respect to B should be along AB or their vertical components should be same. ∴ 20 sin θ = 10 or Ans. θ = 30°

= (10 sin 60° ) + (−10) (2/ 3 ) =5 3−

20 5 =− m/s 3 3

(b) x = | S A | + | S B | = (20 cos 30° ) (t ) + 0 3 1 = 20 × × 2 2 =5 3m

= v = vx2 + v 2y

∴

= 25 + =

25 3

v y = u2y − 2gh ∴

= (10 sin 30° ) + (− g cos 30° ) T

uy = v 2y + 2gh = (6.1)2 + 2 × 10 × 9.1

10 m/s

= 14.8 m/s

x

y

(a) H =

30° 30°

(b) R = 3 2 = 5 − 10 × × 2 3 = − 5m/s 10. In vertical direction,

SA 10 m

S A = S B + 10 1 1 (10) t − gt 2 = − gt 2 + 10 2 2 t =1s SA

In horizontal direction, d = |S A | + |S B | = (10) t + 10 t = 20 t (as t = 1 s) = 20 m

u2y

=

2g 2uxuy g

(14.8)2 ≈ 11 m 2 × 10 =

2 × 7.6 × 14.8 ≈ 23 m 10

Ans.

(c) u = ux2 + u2y = (7.6)2 + (14.8)2 = 16.6 m/s

Ans.

 uy  (d) θ = tan −1    ux 

SB

∴

Ans.

13. vx = ux = 7.6 m/s

10 m/s 3

9. v y = uy + ay t

∴

Ans.

 14.8 −1 = tan −1   ≈ tan (2),  7.6  below horizontal.

Ans. ...(i)

14. ux2 + u2y = (2 gh )2 = 4 gh B A

SB

C h

O

h D 2h

Chapter 7 v y = u2y − 2gh

...(ii)

...(iii) vx = ux Now for the projectile ABC, vx and v y are the initial components of velocity. 2vxv y 2uxv y 2h = range = = ∴ g g gh or ...(iv) ux = vy Using Eqs. (ii) and (iv) for rewriting Eq. (i) we have, 2

 gh   + (v 2y + 2gh) = 4 gh v   y ∴

v 4y − (2gh) v 2y + g 2h2 = 0 2gh ±

∴

v 2y =

∴

= gh v y = gh

Now,

4 g 2h2 − 4 g 2h2 2

tAC = time of projectile ABC 2v y h = =2 g g

Ans.

15. Let v is the velocity at time t v y = uy + ay t v sin β = u sin α − gt vx = ux ∴ v cos β = u cos α From Eqs. (i) and (ii), we get  u cos α    sin β = u sin α − gt  cos β 

Using ∴

...(i) ...(ii)

16. R − a =

Adding Eqs. (i) and (ii) and by putting R=

=2s (b) Vertical components of velocities are zero. 1 ∴ h = gt 2 2 1 = × 9.8 × (2)2 2 = 19.6 m (c) S x = uxt = 7.5 × 2= 15 m 5 18. ux = 20 km/h = 20 × = 5.6 m/s 18 uy = 12 km/h 5 = 12 × = 3.3 m/s 18 1 Using S = ut + at 2 in vertical direction, we have 2 1 −50 = (3.3) t + (−10) t 2 2 Solving this equation we get, t ≈ 3.55 s At the time of striking with ground, vx = ux = 5.6 m/s v y = uy + ay t = (3.3) + (−10) (3.55) = 32.2 m/s Speed = (32.2)2 + (5.6)2 ≈ 32.7 m/s

19. x = (u sin θ ) T

u2 sin 2α g

1  y =  gT 2 − (u cos θ ) T 2 

Multiplying with b we have, bu2 sin 2 α bR − ab = g R+ b=

u2 sin 2θ , g

we get the result. S 10 17. (a) t = x = ux 5

∴

On solving we get, gt cosβ u= sin (α − β )

Projectile Motion — 531

...(i)

u2 sin 2 β g

Multiplying with a we have, au2 sin 2β aR + ab = g

1 2 gT − (u cos θ ) T y = tan θ = 2 x (u sin θ ) T ∴

...(ii)

(2u sin θ ) (tan θ ) = gT − (2u cos θ ) 2u [sin θ tan θ + cosθ ] T = g 2u sec θ = g

532 — Mechanics - I (ii) yA = yB

u cos q u

q

∴ 10 + (u1 sin θ 1 ) t − u sin q

+ (u2 sin θ 2 ) t −

R

y

q

Single Correct Option 1. Relative acceleration between two is zero. Therefore, relative motion is uniform. u 12 = u 1 + u 2 = (20$j) − (20 cos 30° $i + 20 sin 30° $j) = (10$j − 10 3 $i )

2u2 sinθ sec2 θ g

20. (a) Acceleration of stone is g or 10 m/s2 in downward direction. Acceleration of elevator is 1 m/s2 upwards. Therefore, relative acceleration of stone (with respect to elevator) is 11 m/s2 downwards. Initial velocity is already given relative . 2u sin θ r (2) (2) sin 30° (b) T = r = ar 11 . Ans. = 0.18 s (c) vS = vSE + vE

| u 12 | = (10)2 + (10 3 )2 = 20 m/s ∴

2. T =

∴

2 m/s

Ans.

d2 =

2h g

2hu2 g

Ans.

u2 [sin (2α + β ) + sin β ] g cos2 β

u = 10 m/s, g = 10 m/s2 , β = 30° and α = 60° 30°

Ö3 m/s

4. y = x tan θ −

Stone relative to elevator

3 m/s

∴

2Ö3 m/s

∴ 60° Ö3 m/s Absolute velocity of stone

(d) In that case relative acceleration between stone and elevator will be zero. So, with respect to elevator path is a straight line (uniform) with constant velocity of 20 m/s at 30°. But path, with respect to man on ground will remain unchanged. 21. (i) xA = xB ∴ ∴

d = | u 12 | t = 24 m 2h g d = (u) T = u

3. R =

1 m/s

Elevator

1 2 gt 2

LEVEL 2

Now, R = (x ) sec θ = (u sin θ T ) sec θ  2u sec θ  = (u sin θ )   (sec θ )  g 

2 m/s

(u1 sin θ 1 − u2 sin θ 2 ) t = 10

or

q x

=

1 2 gt = 20 2

10 + (u1 cos θ 1 ) t = 30 − (u2 cos θ 2 ) t (u1 cos θ 1 + u2 cos θ 2 ) t = 20

5.

gx 2 2u2 cos2 θ

  dy g = (tan θ ) −  2  x 2 dx  u cos θ  dy versus x graph is a straight line with negative dx slope and positive intercept.

 d 2x  dx  2  dy dx d2 y and = (2 βx ) ⋅ = 2β  2 +    2  dt   dt dt dx  dt  d2 y = α = ay dt 2 dx d 2x = ax = 0 and = vx dt dt 2 α = 2 β ⋅ vx2

∴ or

vx =

α 2β

Ans.

Chapter 7 6.

R u2 sin 2 (60° ) = 2 2g

8. ay = 0 = 3 u2 4g

= H =

∴

2

=

d2 y dt 2 dx dy = (2 y + 2) ⋅ dt dt

d2 y d 2x  dy = ( 2 + 2 ) ⋅ + 2  y 2 2  dt  dt dt

u2 sin 2 60° 3u2 = 2g 8g

 R AB =   + H 2  2

∴

Projectile Motion — 533

21 u2 8g

ax = a = (2 y + 2) (0) + 2 (5)2 = 50 m/s2 H 2H tan β = = R/2 R

9.

= B

Ans.

(2u2 sin 2 α )/ 2g tanα = 2 (2u2 sin α cos α )/ g

 tan α  β = tan −1    2 

∴

H

Ans.

10. a1 = a2 = g

A

R/2

2

(downwards)

30 m/s

10 m/s

Ans.

7.

1

10Ö3 m/s

2 10Ö3 m/s

10Ö3 m/s

Wedge

20Ö3 m

10Ö3 sin 60° = 15 m/s

20Ö3 m/s

1

2

30° 10Ö3 cos 60° = 5Ö3 m/s Particle

dmin 20 m/s

10Ö3 m/s

15 m/s

60° 5Ö3 m/s Particle with respect to wedge

∴ a12 = 0 ∴ Relative motion between them is uniform. Relative velocity v21 dmin = 20 3 sin 30°

2u sin (α − β ) T = g cos β =

= 10 3 m

2 (10 3 ) ⋅ sin (60° − 30° ) =2s 10 cos 30°

More than One Correct Options 1. α + β = 90° or β = 90° − α u2 sin 2 α 2g 2u sinα t1 = g

h1 = 10Ö3 m/s At Rest

60° 30°

R1 = R2 =

and and

u2 cos2 α 2g 2u cos α t2 = g h2 =

2u2 sin α cosα =R g

Ans.

534 — Mechanics - I 2. Since u = 0, motion of particle is a straight line in the

6. Horizontal component of velocity remains unchanged

direction of anet .

X OA = 20 m =

aH

tOA =

∴

X AB 2

tAB =1s 2

For AB projectile g

T =2s=

anet

t=

2h = g

= 3.16 s

3. Horizontal

component

∴

2 × 49 9.8 Ans. of

velocity

remains

unchanged ∴ v cos θ = v′ cos (90 − θ ) or v′ = v cot θ In vertical ( y) direction, v y = uy + ay t v y − uy ∴ t= ay − v 1 sin (90 − θ ) − v sin θ −g (v cot θ ) ⋅ cos θ + v sin θ = g v cosec θ = g

∴

Maximum height of total projectile, = 15 + 5 = 20 m tO B = tOA + tAB = 1 + 2 = 3 s For complete projectile T = 2 (tOA ) + tAB 2uy =4s= g

∴

uy = 20 m/s AB 40 ux = = = 20 m/s tAB 2

Comprehension Based Questions 1. At Q, component parallel to OB becomes zero y

Ans.

x

4. ux = vx = 10 m/s uy =

u

+ 2gh

90°

= (10)2 + (2) (10) (15) = 20 m/s Angle of projection,  uy  θ = tan −1   = tan −1 (2)  ux  2uy (2) (20) T = = =4s g 10

30°

Ans. ∴

dv = a = constant = g dt d 2 v da = = 0 = constant dt dt 2

Q

30° 30°

vx = ux + ax t 0 = (u cos 30° ) + (− g sin 30° ) t u 3u t = cot 30° = g g

Ans.

2. PQ = range = 2 (PM ) = 2a cos 30°  3 = (2) (4.9)    2

R = uxT = (10) (4 ) = 40 m u2y (20)2 H = = = 20 m 2g 2 × 10

5.

g

uy = 10 m/s u2y (10)2 H = = =5m 2g 2 × 10

=

v 2y

2uy

= 4.9 3 m u2 sin 2 (60° ) 9.8 u2 ( 3 / 2) 4.9 3 = 9.8 u = 9.8 m/s =

∴ ∴

Ans.

Projectile Motion — 535

Chapter 7

(c) R75° = R15° = 10 m

u 60° P

Q

M

30°

60°

(d) H 30° =

30° 90°

a

120° 30°

30°

v

O

=

4. T =

Note Velocity at P is making an angle of 60° with horizontal and velocity at Q is making an angle of 60° with horizontal. That is the reason PQ = range. Because under this condition, points P and Q lie on same horizontal line.

Match the Columns ∆x = (ux )2 (t − 1)

1. (a)

= (10) (2 − 1) = 10 m (b) y1 =

∴

1 2 1 gt = × 10 × (2)2 = 20 m 2 2 1 y2 = g (t − 1)2 = 5 m 2 ∆y = y1 − y2 = 15 m

2uy g

u2 sin 2 30° 2g u2 20 = = 2.5 m 8g 8

⇒ T ∝ uy H =

⇒

∴

R = uxT R ∝T R ∝ ux uy tan θ = ux

⇒ ⇒ ⇒

⇒ tan θ ∝ uy By doubling uy, tan θ will become two times, notθ. ∆v 5. (a) a av = a = (−g$j) = ∆t ∴ ∆v = (− g$j) ∆ t T  = (− g$j)    2

vx 2 − vx 1 = 10 m/s

R = ux T = 40m 40 ux = = 10 m/s ∴ T u2 sin 2 (15° ) 3. 10 = g 2 u ⇒ = 20 m g u2 (a) Rmax = = 20 m g u2 (b) H max = when thrown vertically 2g = 10 m

2g

H ∝ u2y

(c) vx 1 = 0 , vx 2 = 10 m/s (d) v y1 = gt = 10 × 2 = 20 m/s v y2 = g (t − 1) = 10 × 1 = 10 m/s ∴ v y1 − v y2 = 10 m/s u2y 2. H = = 20 m 2g ∴ uy = 20 m/s 2uy T = =4s g

u2y

 u sin θ  = (− g$j)    g  = (− u sin θ ) $j (b) vav =

s t A

S O

H

R/2

(R / 2)2 + H 2 T /2 $ ∆v = (− gj) (∆t ) = (− g$j) T =

(c)

 2u sin θ  = (− g$j)   g   = (−2u sin θ ) $j

536 — Mechanics - I (d) vav =

S R = t T

R ∝ ux

or

(asT → same)

R for C is maximum. Therefore ux is greatest for C. u = u2y + ux2

(uy → same)

R is least for A. Therefore ux and hence u is least for A. T O

B S=R

uT = x T = ux = u cos θ 6. (a) x1 = (ux1 ) t = (30) (2) = 60 m x2 = (130) + (ux2 ) (t − 1) = 130 + (−20) (1)

Subjective Questions 1. ux = v0 cos θ , uy = v0 sin θ , ax = − g sin θ , At Q , vx = 0 ∴

ay = g cos θ

ux + axt = 0

x

θ

v0

P

= 110 m

θ

∴

∆x = 50 m 1 (b) y1 = uy1t − gt 2 2 = (30) (2) − = 40 m

y

h Q

1 (10) (2)2 2

1 y2 = 75 + uy2 (t − 1) − g (t − 1)2 2 1 = 75 + 20 × 1 − × 10 × (1)2 2 = 90 m ∴ ∆y = 50 m (c) vx1 = 30 m/s vx2 = − 20 m/s ∴ vx1 − vx2 = 50 m/s (d) v y1 = uy1 + ayt = (30) + (−10) (2) = 10 m/s v y2 = uy2 + ay (t − 1) = 20 + (−10) (1) = 10 m/s − v y2 = 0

∴ v y1 u2 7. H = y or 2g

Sy

H ∝ u2y

θ

or

2

 v cos θ  1 (g cos θ )  0  = h cos θ 2  g sin θ  Solving this equation we get, 2gh Ans. v0 = 2 + cot 2 θ +

2. Let vx and v y be the components of v0 along x and y directions. ∴ or

R = uxT

…(i)

s y = h cos θ 1 ∴ uyt + ayt 2 = h cos θ 2  v cos θ  ∴ (v0 sin θ )  0   g sin θ 

Since H is same. Therefore, uy is same for all three. 2uy or t ∝ uy ∴ T = g Since uy is same. Therefore T is same for all.

O

v cos θ t= 0 g sin θ

(vx )(2) = 2 vx = 1 m/s v y (2) = 10 v y = 5 m/s v0 = vx2 + v 2y = 26 m/s

∴

tan θ = v y/vx = 51 / θ = tan −1 (5)

Ans.

Projectile Motion — 537

Chapter 7 Note We have seen relative motion between two particles. Relative acceleration between them is zero. 3. v1 = (u cos α ) $i + (u sin α − gt ) $j v = (v cos β ) i$ + (v sin β − gt ) $j 2

vx t2 − t1 = ux t2 1 2 h = uyt − gt 2 gt 2 − 2uyt + h = 0

(t2 − t1 )ux = t2vx or

Now Further or

gt 2 − 4 gh t + 2h = 0

or

These two velocity vectors will be parallel when, the ratio of coefficients of $i and $j are equal. u cos α u sin α − gt ∴ = v cos β v sin β − gt

t1 =

4 gh − 16gh − 8gh h = (2 − 2 ) 2g g

Solving we get,

and

t2 = (2 +

t=

uv sin (α − β ) g (v cos β − u cos α )

Ans.

4. At height 2 m, projectile will be at two times, which are obtained from the equation, 2 = (10 sin 45° )t + or

2 = 5 2t − 5t 2

or

5t − 5 2t + 2 = 0

1 (− 10)t 2 2

or

and Now

or ∴

=5− 5 = 2.75 m

Ans.

u2y 2g

t1

t2

vx h

2h

h x

ux

or

2 × 400 10

dx 1  = 5  gt 2 = 5 5t 2 2  dt x

∫0dx = 5

5

t 2

∫0 t

dt

v y = gt = 89.4 m/s ∴ Speed = vx2 + v 2y = 899 m/s ≈ 0.9 km/s

Ans.

7. At t = 0, vT = (10$j) m/s vST = 10 cos 37° k$ − 10 sin 37° i$ = (8k$ − 6i$ ) m/s ∴ vS = vST + vT = (−6$i + 10$j + 8 k$ ) m/s (a) At highest point vertical component (k$ ) of v

S

will become zero. Hence, velocity of particle at highest point will become (– 6$i + 10$j) m/s . 2v 2×8 (b) Time of flight, T = Z = = 16 . s g 10

y

uy

2H = g

Ans.

or horizontal drift 5 5 x= (8.94 )3 = 2663 m ≈ 2.67 km. 3 (b) When particle strikes the ground vx = 5 y = ( 5 )(400) = 400 5 m/s

Distance of point of projection from first hurdle = (10 cos 45° )t1 10  5 2 − 10    = 10 2  

5. 2h =

h g

= 8.94 s vx = ay = 5 y

Now

5 2 − 50 − 40 t1 = 10 5 2 − 10 = 10 5 2 + 10 t2 = 10 d = (10 cos 45° )(t2 − t1 ) 10  2 10    = 4.47 m = 2  10 

2)

Substituting in Eq. (i) we have, vx 2 = ux 2+1

6. (a) Time of descent t =

2

…(i)

uy = 2 gh

x = xi + vxT 16 = − 6 × 1.6 = − 4.5 m π y = (10) (16 . ) = 16 mand z = 0 Therefore, coordinates of particle where it finally lands on the ground are (− 4.5 m , 16 m , 0).

538 — Mechanics - I T = 0.8 s 2

At highest point, t =

B

16 − (6) (0.8) = 0.3 m π y = (10) (0.8) = 8.0 m v 2 (8)2 z= Z = = 3.2 m 2g 20

∴

x=

and

C 37°

A

Therefore, coordinates at highest point are, (0.3 m, 8.0 m, 3.2 m) 8. | v21 x | = (v1 + v2 ) cos 60° = 12 m/s

Thus, at C, the particle has only horizontal component of velocity vx = u cos α = 5 5 × (2 / 5 ) = 10m/s

∴ v21 = (12)2 + (4 3 )2 = 192 m/s C

Given, that the particle does not rebound after collision. So, the normal component of velocity (normal to the plane AB) becomes zero. Now, the particle slides up the plane due to tangential  4 component vx cos 37° = (10)   = 8 m/s  5

y

A

45°

105°

30° 30°

x

B

Let h be the further height raised by the particle. Then 1 mgh = m (8)2 or h = 3.2 m 2 Height of the particle from the ground = y + h Ans. ∴ H = 1.25 + 3.2 = 4.45 m

O

BC = (v21 ) t = 240 m AC = 70 m

(Given)

Hence, AB = (240) + (70) = 250 m 2

37°

Using v y = uy + ayt , we have v y = u sin α − g (x / u cos α ) (Q t = x /u cos α ) 5 5 10 × 5 = − =0 5 (5 5 × 2 / 5 )

Ans.

| v21 y | = (v2 − v1 ) sin 60° = 4 3 m/s

v21

vx cos 37°

2

Ans.

9. (a) Let, (x , y) be the coordinates of point C

10. For shell uz = 20 sin 60° = 17.32 m/s 50 kg

√5

4 m/s

1

α 2

x = OD = OA + AD 10 10 + 4 y …(i) ∴ x= + y cot 37° = 3 3 As point C lies on the trajectory of a parabola, we have gx 2 …(ii) y = x tan α − 2 (1+ tan 2 α) 2u 1 Given that, tan α = 0.5 = 2 Solving Eqs. (i) and (ii), we get x = 5 m and y = 1.25 m. Hence, the coordinates of point C are (5 m, 1.25 m). Ans. (b) Let v y be the vertical component of velocity of the particle just before collision at C.

40 kg v

∴

z = uzt −

10 kg

(20 cos 60° + v)

1 2 1  gt = (17.32 × 2) –  × 9.8 × 4 2  2

or z = 15 m ⇒ uy = 0 ∴ y=0 For ux conservation of linear momentum gives, 50 × 4 = (40) (v ) + 10(20 cos 60° + v ) or ∴ ∴ ∴

v = 2 m /s ux = (20 cos 60° ) + 2 = 12 m /s x = uxt = (12)(2) = 24 m r = (24 i$ + 15k$ ) m

Ans.

8. Laws of Motion INTRODUCTORY EXERCISE

8.1

No friction will act between sphere and ground because horizontal component of normal reaction from rod (on sphere) will be balanced by the horizontal normal reaction from the wall.

1. w1 = weight of cylinder w2 = weight of plank N 1 = normal reaction between cylinder and plank N 2 = normal reaction on cylinder from ground N 3 = normal reaction on plank from ground f1 = force of friction on cylinder from ground f2 = force of friction on plank from ground

5.

FV O

FH

A

N1

T

w FBD of rod w1

N3

N1

In the figure T = tension in the string, W = weight of the rod, FV = vertical force exerted by hinge on the rod FH = horizontal force exerted by hinge on the rod

w2

N2 f1

f2

2. N = normal reactions

6. In the figure

N3 N3

N1 N2

f1

N 1 = Normal reaction at B, f1 = force of friction at B, N 2= C normal reaction at A, f2 = force of friction at A w = weight of the rod.

w = weights N4 w2

N1

N2

w1

f2 FBD of the rod

T

N

7.

Force

A

W2 FBD of sphere (i )

4 sin 30° = 2 N

=2 3N F2

−4 cos 60° =−2N

4 sin 60° = 2 3 N

F3

0

−6 N

F4

4N

0

N3

N1

Fy

Fx 4 cos 30°

F1 w

N2

N3

w

A

3.

4.

B

N4

C f

B

W1 FBD of rod (ii )

In the figure N 1 = normal reaction between sphere and wall, N 2 = normal reaction between sphere and ground N 3 = normal reaction between sphere and rod and N 4 = normal reaction between rod and ground f = force of friction between rod and ground

8. T1 cos 45° = w T1 sin 45° = 30 N

and T1

45° T2 = 30 N w

∴

w = 30 N

Ans.

540 — Mechanics - I 9. N A cos 30° = 500 N

...(i)

N A sin 30° = N B

INTRODUCTORY EXERCISE

...(ii)

Net pushing force Total mass 120 − 50 = 1+ 4 + 2

a=

1. (a)

NA 30° NB

= 10 m/s2

30° w = 50 kg = 500 N

On solving these two equations, we get 1000 NA = N 3 500 N and NB = 3

120 N

(b) Ans.

10. Net force in vertical direction = 0

1 kg

R a

120 − R = 1 × a = 10 ∴ R = 110 N (c) Fnet = ma = (2) (10) = 20 N

a 30° T

2. T4 = 4 g and T1 = (1) g

30° a N

a

w

∴ T cos 30° = w

or T =

2w 3

Ans.

11. w = 100 N ∴

8.2

∑ (forces in horizontal direction) = 0 T + f cos 30° = N sin 30° N 30°

∴ ...(i)

T C

f 30°

30° w = 100 newton

∑(Forces in vertical direction) = 0 ∴ N cos 30° + f sin 30° = 100 ∑ (moment of all forces aboutC) = 0 ∴ (T ) (R ) = f (R ) or T = f On solving these three equations, we get T = f = 26.8N and N = 100 newton

T4 =4 ∴ T1 Net pulling force F + 10 − 20 3. a = = Total mass 1+ 2 10 = m/s2 3 4. mAg = (2) (g ) + (2)g sin 30°

...(ii)

...(iii)

Ans.

as a = 0 Ans. (F = 20 N) Ans. as a = 0

mA = 3 kg

Ans.

5. Since surface is smooth, acceleration of both is

g sin θ = 10 sin 30° = 5 m/s2 ,down the plane.

The component of mg down the plane (= mg sin θ ) provides this acceleration. So, normal reaction will be zero. mg 6. N = is given N 4 a mg mg − = ma 4 3g ∴ a= mg 4 Net pulling force 7. a = Total mass (3 + 4 − 2 − 1) g Ans. = = 4 m/s2 3+ 4 + 2+1 40 − T1 = 4 a ∴ T1 = 24 N

Ans.

3 kg T1 + 30 − T2 = 3a ∴ T2 = 42 N

Ans.

T3 − 10 = (1) (a) ∴ T3 = 14 N

Ans.

4 kg 1 kg

Laws of Motion — 541

Chapter 8 8. (a) a =

Net pushing force Total mass 100 − 40 = = 3 m/s2 6 + 4 + 10

(b) Net force on any block = ma (c) N − 40 = 10 a = 30 10 kg

40 Newton

N

a

∴

N = 70 newton

Ans.

9. (a) T1 = 10 a, T2 − T1 = 20 a, 60 − T2 = 30 a 10 kg

T1 a

20 kg

T1

T2 a

30 kg

T2

(x2 − x4 ) + (x3 − x4 ) = l2 (length of second string) or …(ii) x2 + x3 − 2x4 = l2 On double differentiating with respect to time, we get …(iii) a1 + a4 = 0 and …(iv) a2 + a3 − 2a4 = 0 But [From Eq. (iii)] a4 = − a1 We have, a2 + a3 + 2a1 = 0 This is the required constraint relation between a1 , a2 and a3. 2. In above solution, we have found that a2 + a3 + 2a1 = 0 Similarly, we can find v 2 + v 3 + 2 v1 = 0 Taking, upward direction as positive we are given v1 = v2 = 1 m/s ∴ v3 = − 3 m/s i.e. velocity of block 3 is 3 m/s (downwards). and

60 N a

T

2 kg

2a

mg = 10 N

8.3

20 m/s2 3 = acceleration of 1 kg block

2a = 2T =

and

displacements from the fixed dotted line be x1 , x2 , x3 and x4 (ignoring the length over the pulley)

4.

T T

M a

x4 x3

x2 2

1 kg

a

1. Points 1, 2, 3 and 4 are movable. Let their

4

2T

Ans.

INTRODUCTORY EXERCISE

1

...(ii)

On solving these two equations, we get 10 T = N 3

(b) T1 = 0, T2 = 20a, 60 − T2 = 30a Net pulling force 60 a= = Total mass 20 + 30 = 1.2 m/s2

x1

...(i)

10 − T = 1 (2a)

1 kg

On solving, we get T1 = 10 N, T2 = 30 N Ans.

On solving, we get T2 = 24 N

2T = 2 (a)

3. 2 kg

Net pulling force 60 a= = Total mass 10 + 20 + 30 = 1 m/s2

3

We have, x1 + x4 = l1 (length of first string)

…(i)

M mg

T = Ma Mg − T = Ma On solving these two equations, we get g a= 2 Mg and T = 2

...(i) ...(ii)

Ans.

542 — Mechanics - I 5. In the figure shown,

Net pulling force 30 − 20 a= = Total mass 3+ 2 = 2 m/s2

8. T − Mg sin 30° = Ma

...(i)

 a 2Mg − 2T = 2M    2

...(ii)

Solving these equations, we get g a= 3

P1

2T

P2 ® Stationary

2T

T

2T

M

P2

a

Mg

T a 2 kg

T

0 n3

°

M

a

2 Mg

INTRODUCTORY EXERCISE

2. Constant velocity means acceleration of frame is Ans.

6. T = m1 (2a) = (0.3) (2a) = 0.6a

m2

2T

2a

INTRODUCTORY EXERCISE

...(ii) Ans. F

a

12 2T = N 35

and

Ans.

7. a3 = a + ar = 6

...(i)

a2 = ar − a = 4 On solving these two equations, we get a = 1 m/s2

a 1

a

ar

2

3

zero.

...(i)

F − 2T = m2 a ∴ 2T = 0.4 − 0.2a On solving these two equations, we get 2 a = m/s2 7 T

8.4

1. (a) FA = − mAa B = (4 $j) N (b) FB = − mB a A = (−4 $i ) N

= 4.8 kg

m1

a 2

30°

T − 20 = 2 × a = 4 T = 24 N Mg = 2T 2T 48 M = = g 10

∴

2M

si

3 kg

2 kg ∴ Now,

Ans.

ar

...(ii)

1. Figure (a)

8.5

N = mg = 40 N

µ s N = 24 N Since, F < µ sN Block will remain stationary and f = F = 20 N Figure (b) N = mg = 20 N µ S N = 12 N and µ KN = 8 N Since, F > µ S N , block will slide and kinetic friction (= 8 N) will act. F − f 20 − 8 a`= = = 6 m /s2 m 2 2. If θ ≤ angle of repose, the block is stationary, a = 0, Fnet = 0 and f = mg sin θ. If θ > angle of repose, the block will move, f = µmg cos θ mg sin θ − µmg cos θ a= m = g sin θ − µg cos θ and Fnet = ma Further, µ = tan α (α = angle of repose = 45°) = tan 45° = 1

Exercises LEVEL 1

 2m2  = m1 g   = m2 g  m1 + m2   m + m2  = m1 g  2  = m2 g  m1 + m2 

Assertion and Reason 1. If net force is zero but net torque ≠ 0, then it can rotate.

2. If three forces are above a straight line (say AB ), then there should be some force below the line AB also to make their resultant = 0.

A

B

3. a1 = g sin θ + µ g cos θ

inertial.

v

9. If vector sum of concurrent forces is zero, then all

a1

g mm

co

Since m1 > m2 , T < m1 g but T > m2 g Similarly, we can prove that T < m2 g and T > m1 g if m2 > m1. Therefore under all conditions T lies between m1g and m2 g.

8. A frame moving with constant velocity (a = 0) is

a2 = g sin θ − µ g cos θ q sin

mg

forces can be assumed a pair of two equal and opposite forces acting at one point. F1

q sq mg

sin

q m

s co mg

q

a2 v

q

a1 1 + µ for θ = 45° = a2 1 − µ

∴

F2

horizontal force in − ve x- direction is required which can be provided only by the right vertical wall of the box. 5. By increasing F1 limiting value of friction will increase. But it is not necessary that actual value of friction acting on block will increase. 7. If m1 > m2 Net pulling force (m1 − m2 ) g a= = Total mass m1 + m2 FBD of m1 m1 g − T = m1 a = T =

2 m1 m2 g (m1 + m2 )

m1 (m1 − m2 ) g m1 + m2

F1 = F2

P

Now, if we take moment about any general point P, then moment of F1 is clockwise and moment of F2 (of same magnitude) is anti-clockwise. Therefore net moment is zero.

10. N = mg − F sinθ F θ

4. For (−2i$ ) m/s2 component of acceleration, a

∴

 2m1  m + m   1 2  m1 + m1  m + m   1 2

∴

F cosθ = µN = µ (mg − F sin θ ) µmg F= = f (θ ) cosθ + µ sin θ

…(i)

For F to be minimum, denominator should be maximum. d or (cosθ + µ sin θ ) = 0 dθ From here we get θ = tan −1 (µ ) = angle of friction. With this value of θ we can also find that minimum value of force required from Eq. (i). 11. Friction (or you can say net force on man from ground) is in the direction of motion.

544 — Mechanics - I Single Correct Option Net pulling force 1. a = Total mass (2 + 2 − 2) g g = = 2+ 2+ 2 3 FBD of C mg mg − T = ma = 3 2 ∴ T = mg 3 2 = (20) = 13.3 N 3 mg − F F 2. a = =g− m m F

7. t = T

∴ a C

2 = 1 mg

Ans.

F = air resistance

a mg

mA > mB ∴ aA > aB and A will reach earlier 3. Only two forces are acting, mg and net contact force (resultant of friction and normal reaction) from the inclined plane. Since the body is at rest. Therefore these two forces should be equal and opposite. (upwards) ∴ Net contact force = mg 4. TA = 10 g = 100 N TB cos 30° = TA 200 TB = N ∴ 3 TB sin 30° = TC 100 ∴ TC = N 3 mg − T T 5. a = =g− T m m T a amin = g − max m 2 mg mg g =g− 3 = m 3 Net pulling force 10 × 10 − 5 × 10 10 6. a = = = m/s2 Total mass 10 + 5 3 50 5 kg T − 5 × 10 = 5 × a = 3 200 ∴ T = N 3 This is also the reading of spring balance.

2S 1 ∝ a a t1 a2 = t2 a1 g sin θ 1 = g sin θ − µ K g cos θ 1 − µK

as, sin θ = cos θ at 45° On solving the above equation, we get 3 µK = 4 8. F1 = mg sin θ + µ mg cos θ F2 = mg sin θ − µ mg cos θ Given that F1 = 2 F2 ∴(mg sin θ + µmg cosθ ) = 2(mg sin θ − µmg cos θ ) On solving, we get tan θ = 3µ 9. During acceleration, F v = a1 t1 = 1 t1 m mv ...(i) ∴ F1 = t1 During retardation, F 0 = v − a2 t2 = v − 2 t2 m mv ...(ii) ∴ F2 = t2 If t1 = t2 then F1 = F2 t1 < t2 then F1 > F2 and t1 > t2 then F1 < F2 10. Angle of repose θ = tan −1 (µ )  1 = tan −1   = 30°  3 R 30° h 30°

So, particle may be placed maximum upto 30°, as shown in figure  3 h = R − R cos 30° = 1 −  R 2 

11. Net pulling force F1 = 15g − 5g F1 = 10g = 100 N Net stopping force F2 = 0.2 × 5 × 10 = 10 N Since F1 > F2, therefore the system will move (anticlockwise) with an acceleration,

Chapter 8 a=

F1 − F2 90 = 15 + 5 + 5 25

Acceleration of particle towards each other component of T towards each other particle m T cos 60° (F / 3 ) (1/ 2) F = = = m m 2 3m

=

= 3.6 m/s2 15 kg block 15 × 10 − T2 = 15 × a = 15 × 3.6 ∴ T2 = 96 N 5 kg block T1 − 5 × 10 = 5 × a = 5 × 3.6 ∴ T1 = 68 N T1 68 17 = = ∴ T2 96 24

15. µ = tan φ N = mg − F sin θ F q m

12. Acceleration of block with respect to lift

F cos θ = µN = (tan φ ) (mg − F sin θ )  sin φ  =  (mg − F sin θ )  cos φ 

Pseudo force = (downward)

ar q

mg+ma

m (g + a) sin θ = (g + a) sin θ m 1 S = ar t 2 2 2S 2L t= = ar (g + a) sin θ ar =

∴

13. Since the block is resting (not moving) ∴ or

f = mg sin θ ≠ µ S mg cos θ f m= g sin θ 10 = 10 × sin 30° = 2 kg

Ans.

14. At point P,

On solving this equation, we get mg sin φ F= cos (θ − φ )

16. N = mg = 40 N At Since ∴ 1 17. S = at 2 2

µN = 0.2 × 40 = 8 N t = 2 s, F = 4 N F < µN f =F =4N

∴

t=

∴

t1 = t2

2S a a2 a1

or

t = 2t

g sin θ − µg cos θ g sin θ

or

t∝

1 a

On solving this equation, we get 3 µ = tan θ 4 18. amr = acceleration of man relative to rope

60° T 30° 30°

F

∴

P

T

∴

Laws of Motion — 545

F = 2T cos 30° F T = 3

Now, ∴

= am − ar am = amr + ar = (a) + (a) = 2a T − mg = m (am ) = m (2a) T = m (g + 2a)

Ans. +

– T

mg

Ans.

Note Man slides down with a deceleration a relative to the rope. So, amr = + a not −a.

546 — Mechanics - I 19. Steady rate means, net force = 0

Now at the mid point a

T

T

T F f

(50 +25) g

∴ or

3T = 75 g = 750 T = 250 N

Ans.

20. N 1 = µN 2

...(i) N2 = w Σ MB = 0  3 w   = N 1 (4 )  2

∴ A

...(ii) ...(iii)

M M T −µ   g =  2 2  M   g a=     2   2 1 Putting µ = , we get 2 Mg T = 2 23. Angle of friction θ = tan −1 (µ ) or

N1

Ans.

 1 θ = tan −1   = 30°  3 F a

N2 B

O

mN2 w

On solving these equations, we get 3 Ans. µ= 8 21. In the figure, N 2 is always equal to w or 250 N A

N1

N2 B O

mN2 w = 250 N

∴ Maximum value of friction available at B is µ N 2 or 75 N.

22. a =

F− f M

F is minimum when denominator is maximum or d (cosα + µ sin α ) = 0 dα or − sin α + µ cos α = 0 1 or tanα = µ = 3 ∴ α = 30° ,the angle of friction θ ∴ At α = 30°, force needed is minimum. Substituting the values in Eq. (i) we have,  1   (25) (g )  3 Fmin = ( 3 / 2) + (1/ 3 ) (1/ 2) = 12.5 g = 12.5 kg f

Mg − µ mg g = = M 2

Ans.

Note From this example we can draw a conclusion T

m Mg 2

Suppose the body is dragged by a force F acting at an angle α with horizontal. Then, N = mg − F sinα and F cosα = µN = µ (mg − F sin α ) µ mg ...(i) ∴ F= cosα + µ sin α

a

that force needed to move the block is minimum when pulled at an angle equal to angle of friction.

Laws of Motion — 547

Chapter 8 24. N = mg = 40 N

29. Maximum friction between two,

µN = 32 N Applied force F = 30 N < µN Therefore friction force, f = F = 30 N ∴ Net contact force from ground on block = N 2 + f 2 = 50 N

Ans.

25. Maximum value of friction between B and ground, = µN = µ (mA + mB ) g = (0.5) (2 + 8) (10) = 50 N Since applied force F = 25 N is less than 50 N. Therefore system will not move and force of friction between A and B is zero. 26. mg sin θ = (10) (10) (3/ 5) = 60 N

fmax = µ N = µ mg upper block moves only due to friction. Therefore its maximum acceleration may be, f amax = max = µ g m Relative motion between them will start when common acceleration becomes µg. Net force at ∴ µg = = Total mass 2m 2µ mg Ans. ∴ t= a 30. a1 = g sin 30° = g / 2 a2 = g sin 60° = 3 g /2

This 60 N > 30 N. Therefore friction force f will act in upward direction. F

ar = | a 1 − a 2 | Angle between a 1 and a 2 is 30°

N

∴

ar = a12 + a22 − 2 a1 a2 cos 30° =

f

g 2

Subjective Questions 1.

.

2

F

R

27. If force applied by man is F. Then in first figure

3N

force transferred to the block is F, while in second figure force transferred to the block is 2F.

60°

F

F

F

F

∴ ∴ ∴ ∴

(i)

F

F

∑ Fx = 0 F − 3 cos 60° − 10 sin 60° = 0 F = 10.16 N ∑ Fy = 0 R + 3 sin 60° − 10 cos 60° = 0 R = 2.4 N

Ans.

Ans.

2. Resolving the tension T1 along horizontal and

(ii)

mg 28. Total upward force = 4   = 2 mg  2 F

60° 60° 10 N

60°

F

x

y

F = N + f = net force by plane on the block. 2

vertical directions. As the body is in equilibrium, 60°

F

T1

Total downward force = (m + m) g = 2 mg Net force = 0 ∴ mg F= 2

T2 4 kg wt.

T1 sin 60° = 4 × 9.8 N

...(i)

548 — Mechanics - I T1 cos 60° = T2 4 × 9.8 4 × 9.8 × 2 T1 = = sin 60° 3 = 45.26 N

...(ii) ∴

∑ M0 = 0 40  l 40   = T (l sin 60° ) or T = N  2 3 T

Ans.

T2 = T1 cos 60° = 45.26 × 0.5 = 22.63 N

V 60°

Ans.

O

H

3. (a) At P

mg = 40 N

H =

Q

F1

T

45° F2

P

∴

20 N 3

and V = 20 N

Net hinge force = H 2 + V 2 =

w

T ...(i) 2 T ...(ii) w = T cos 45° = 2 T AtQ, ...(iii) F1 = 2 From these three equations, we can see that 60 T Ans. = = 30 2 N F1 = F2 = w = 2 2 F2 = T cos 45° =

At P,

4. Various forces acting on the ball are as shown in figure. The three concurrent forces are in equilibrium. Using Lami’s theorem, A

B 30°

60° T1

90° 120°

T2

∴

and

T2 = 10 sin 60° = 10 ×

A T

T

B a mg

For A T cos 45° = ma or T = 2 ma For B mg − T cos 45° = ma

...(i)

g 2 mg Substituting in Eq. (i), we get T = 2 Net pulling force 8 7. a = = Total mass 1+ 2 + 1 ∴

TB

T1 = 10 sin 30° = 10 × 0.5 =5N

Ans.

mg − ma = ma or

a=

= 2 m/s2 ∴ TA

T1 T2 10 = = sin 150° sin 120° sin 90° T1 T2 10 = = sin 30° sin 60° 1

=5 3N T 5. H = T cos 60° = 2 V + T sin 60° = 40

a

O 150°

w = 10N

or

6.

40 N 3

8 − TB = m1a = (1) (2) TB = 6 N TA = m2 a = (2) (2)

∴

=4N

Ans.

8. (a) T1 − 2g = 2a T1

Ans. 0.1 kg

3 2

0.2 m/s2

...(i) ...(ii)

1.9 kg 2g

Ans. ∴

T1 = 2 (g + a) = 2 (9.8 + 0.2) = 20 N

Laws of Motion — 549

Chapter 8 (b) T2 − 5g = 5a

For vertical equilibrium, …(ii) N2 = w Taking moments about B, we get for equilibrium, …(iii) N 1 (4 cos 30° ) − w(2 cos 60° ) = 0 Here, w = 250N Solving these three equations, we get f = 72.17 N and N 2 = 250 N f 72.17 ∴ µ= = N2 250

T2

2.9 kg 0.2 kg

0.2 m/s2

5g

1.9 kg

∴

T2 = 5 (g + a) = 5 (9.8 + 0.2) = 50 N Net pulling force 9. (a) a = Total mass 200 − 16 × 9.8 = 16

= 0.288

11. Constant velocity means net acceleration = 0. Therefore, net force should be zero. Only two forces T and mg are acting on the bob. So they should be equal and opposite. Asked angle θ = 30° T = mg = (1) (10) = 10 N

= 2.7 m/s2 (b) 200 − 50 − T1 = 5a ∴

T1 = 200 − 5 × 2.7 − 50 = 136.5 N (c) T2 − 9g = 9a

Ans.

200 N

a

T

5 kg 50 N T1

q mg

T2

30° a

2 kg 7 kg

12. T sin θ − mg sin 30° = ma = 9g

T2 = 9 (g + a) = 9 (9.8 + 2.7) = 112.5 N 10. In figure, AB is a ladder N1 of weight w which acts at its centre of gravity G. ∠ ABC = 60° G ∴ ∠ BAC = 30° Let N 1 be the reaction of the wall, and N 2 the W N2 reaction of the ground. Force of friction f 60° f between the ladder and B the ground acts along BC. For horizontal equilibrium, f = N1

T

∴

mg 2

a

q 30° 30° mg

A

∴

T sin θ = mg T cos θ = mg cos 30°

3 mg 2 Solving Eqs. (i) and (ii), we get  2 θ = tan −1    3

or

C

and ...(i)

T cos θ =

T =

...(i)

...(ii)

Ans.

7 mg 2

=5 7N

Ans.

550 — Mechanics - I Net pulling force Total mass 2g − (1) g = 2+1 g 10 = = m/s2 3 3 After 1 s, 10 v = at = m/s 3 At this moment string slacks (T `= 0)

13. a =

10 m 3 3

1 kg

2 kg

s1

g

s2

15. a =

Net pulling force Total mass a m1 m2

30°

M

a= u=0 g

String is again tight when, s1 = s2 10 1 2 1 2 ∴ t − gt = gt (g = 10 m/s2 ) 3 2 2 On solving we get, 1 Ans. t= s 3 14. Since, acceleration of block w.r.t. wedge (an accelerating or non-inertial frame of reference) is to be find out. N

m1

a

Mg m1 + m2 + M

...(i)

N cos 30° = m1g

...(ii)

N sin 30° = m1 a

...(iii)

30° N a 30° m1

From Eqs. (ii) and (iii), we get a = g tan 30° =

g 3

Substituting this value in Eq. (i) we get, M = 6.83 kg

Ans.

16. (a) With respect to box (Non-inertial) a = 5 m/s2 (due to pseudo force)

θ mg + FP = mg + ma

( t=

a ne

g+

a

nθ ) si

θ

FBD of ‘block’ w.r.t. ‘wedge’ is shown in figure. The acceleration would had been g sin θ (down the plane) if the lift were stationary or when only weight (i.e. mg) acts downwards. Here, downward force is m(g + a). ∴ Acceleration of the block (of course w.r.t. wedge) will be (g + a) sin θ down the plane.

u = 10 m/s

x = x0 + uxt +

x

1 2 axt 2

= x0 + 10t − 2.5t 2

Ans.

vx = ux + axt = 10 − 5t

Ans.

(b) vx = 0 at 2s = t0 (say) ∴ To return to the original position, time taken = 2t0 = 4 s

Ans.

17. (a) In car's frame (non-inertial) ax = − 5 m/s2 (due to pseudo force) ux = 0 az = 0 uz = 10 m/s

Laws of Motion — 551

Chapter 8 Now apply, v = u + at and s = s0 + ut +

 3 = (6) (10)    2

1 2 at 2

in x and z-directions. (b) In ground frame (inertial) ax = 0, ux = 0, az = 0 and uz = 10 m/s 18. Relative to car (non-inertial) a1 = 5 m/s2

= 52 N (a) Force needed to keep the block stationary is (upwards) F1 = F − µ S N = 52 − 18 (upwards) Ans. = 34 N

ux = 10 m/s

F1 m N s

a2 = µx = 3 m/s2 ax

F

a1 is due to pseudo force ax = − (5 + 3) = −8 m / s

60°

2

(b) If the block moves downwards with constant velocity (a = 0, Fnet = 0), then kinetic friction will act in upward direction. ∴ Force needed, (upwards) F2 = F − µ K N = 52 − 12 (upwards) Ans. = 40 N

Block will stop when vx = 0 = ux + axt = 10 − 8t or at t = 1.25 s So, for t ≤ 1.25 s x = x0 + uxt +

1 2 axt 2

= x0 + 10t − 4 t 2

F2

vx = ux + axt = 10 − 8t

m KN

After this µ s g > 5 m/s2 as µ s > 0.5 Therefore, now the block remains stationary with respect to car.

19. 37° f 37°

52

N

N

60°

(c) Kinetic friction will act in downward direction F3 − 52 − 12 = ma = (6) (4 ) Ans. ∴ F3 = 88 N (upwards)

q

a 37°

F3 a

mg

N cos 37° + f sin 37° = mg N sin 37° − f cos 37° = ma On solving these two equations, we get f = 3.6 m 9 = mg 25 20. N = mg cos 60°

...(i) ...(ii)

N mK

Ans.

21. As shown in Fig. when force F is applied at the end of the string, the tension in the lower part of the

 1 = (6) (10)   = 30 N  2 µ S N = 18 N µ K N = 12 N Driving force F = mg sin 60° = force responsible to move the body downwards

N 52 60° N 2 =1

a = 1 m/s2 F F

T

T

552 — Mechanics - I string is also F. If T is the tension in string connecting the pulley and the block, then, T = 2F But T = ma = (200)(1) = 200 N ∴ 2F = 200 N or F = 100 N 22. N = F = 40 N

25. X A + 2 X B + X C Fix line

XB

T

XC

B

XA A

f

C

F x

= constant on double differentiating with respect to time, we get, Ans. aA + 2aB + aC = 0

C

N

26. P2 2T1 = 2T2

mg = 20 N

Net moment about C = 0 ∴ Anti-clockwise moment of f = clockwise moment of N  20 ∴ (20)   = (40) ⋅ x  2 x = 5 cm

or

∴ 1 kg

T1 = T2 T2 = (1) (a) a  T1 − 20 = 2  ar −   2 a  30 − T1 = 3  ar +   2

2 kg Ans.

3 kg

23. Force diagram on both sides is always similar. Therefore motion of both sides is always similar. For example, if monkey accelerates upwards, then T > 20 g. But same T is on RHS also. T

T

Monkey

Bananas

...(iii) ...(iv)

T2

1 kg

T2 a

T2 P1

a 2

2T2 a 2

P2 ar

20 g

...(i) ...(ii)

T1

20 g 2 kg

Therefore, bananas also accelerate upwards. 24. TA = 3T , TB = T 2T 2T

T T B

A

In such type of problems 1 a∝ T ∴ | aB | = 3 | aA | or aB = − 3aA , as aA and aB are in opposite directions.

T1

ar

3 kg

On solving these equations, we get 120 T1 = T2 = N 11 120 a1 = a = m/s2 11 a 50 a2 = ar − = − m/s2 2 11 50 or a2 = m/s2 (downwards) 11 a 70 a3 = ar + = m/s2 (downwards) Ans. 2 11

Laws of Motion — 553

Chapter 8 27. 2T − 50 = 5a

...(i)

40 − T = 4 (2a) On solving these equations, we get T

2T a 5 kg

4 kg

50 N

and

28. a =

...(ii)

2a

40 N

10 a= m/s2 or 7 20 2a = m/s2 or 7

g 7 2g 7

Ans. Ans.

f = µg = 3 m/s2 m a f = mmg

(a) Relative motion will stop when velocity of block also becomes 6 m/s by the above acceleration. v = at v 6 Ans. t= = =2s ∴ a 3 1 1 (b) S = at 2 = (3) (2)2 = 6 m Ans. 2 2 29. 2 kg block has relative motion towards right. Therefore, maximum friction will acts on it towards left. f = µN = (0.4) (1) (10) = 4 N 4 a1 = = 4 m/s2 1 4 a2 = − = − 2 m/s2 2

(b) Substituting value of ' t ' in Eq. (ii), either on RHS or on LHS, common velocity = 6 m/s Ans. 1 (c) s1 = u1t + a1t 2 2 1 Ans. = (2) (1) + (4 ) (1)2 = 4 m 2 1 s2 = u2 t + a2 t 2 2 1 Ans. = (8) (1) + (−2) (1)2 = 7 m 2 30. N = 20 N = applied force µ S N = 16 N and µ K N = 12 N Driving force (downwards) F = mg Since, F > µS N , block will slide downwards and kinetic friction of 12 N will act in upward direction F − µ KN (downwards) ∴ a= m 20 − 12 (downwards) Ans. = = 4 m/s2 2 31. fmax = µ N = µ mg = 0.8 × 2 × 10 = 16 N Block will not move till driving force F = 2t becomes 16 N. This force becomes 16 N in 8 s For t ≤ 8 s a=0 For t ≥ 8 s F − fmax 2t − 16 a= = ⇒ a=t−8 m 2 i.e. a - t graph is a straight line of slope +1 and intercept −8. Corresponding a - t graph is shown below a (m/s2)

2 m/s 4N

1 kg a1

45° –

8

+

t (s)

32. N = mg = 60 N

4N 2 kg

8 m/s

a2

(a) Relative motion between them will stop when, v 1 = v2 ...(i) ∴ u1 + a1t = u2 + a2t ...(ii) ∴ 2 + 4 t = 8 − 2t or Ans. t =1s

µ S N = 36 N µ K N = 24 N Driving force F = 4 t Block will move when, F = µ S N or 4 t = 36 ⇒ t = 9 s For t ≤ 9 s a=0

554 — Mechanics - I For t ≥ 9 s At 9 s block will start moving. Therefore kinetic friction will act F − µK N a= ∴ m 4 t − 24 = 6 2  =  t − 4 3  ∴ a - t graph is a straight line At t = 9 s , a = 2 m/s2

where, F= force on cube ∴ F = (8$i − 4 $j) − w = (8$i − 4 $j) − (−10$j) = (8$i + 6$j) or

| F | = (8)2 + (6)2 = 10 N

Ans.

3. During motion of block, a component of its acceleration comes in the direction of mg cos θ. Therefore,

The corresponding a - t graph is as shown in figure. a (m/s2) Motion relative to wedge

mg cos q

mg cos θ > N 4. Maximum friction available to m2 is

2 t (s)

( fmax ) = µ m2 g Therefore maximum acceleration which can be provided to m2 by friction, (without the help of normal reaction from m1) is f amax = max = µ g m2

9

LEVEL 2 Single Correct Option 1. Maximum value of friction between A and B ( f1 )max = µN 1 = µmAg = 0.3 × 50 × 10 = 150 N Maximum value of friction between B and ground ( f2 )max = µN 2 = µ (mA + mB ) g = (0.3) (120) (10) = 360 N Force diagram is as shown below

If a > µ g, normal reaction from m1 (on m2 ) is non zero. a 5. = cot θ g a q g

T1 q

∴

C T1

2T2

B f2

f1

A

T2

v1 A v0

mcg

∴

x y B v2

Ans.

dv = (8$i − 4 t$j) dt At 1 s Fnet = ma = (1) (8$i − 4 $j) = (8$i − 4 $j) =W+F

2. a =

Ans.

6. x + y = constant

f1

T2 = ( f1 )max = 150 N T1 = 2T2 + ( f1 )max + ( f2 )max = 300 + 150 + 360 = 810 N T1 = mc g T 810 mc = 1 = = 81 kg g 10

a = g cot θ

∴ or ∴ or

dx dy + =0 dt dt  dx   dy −  =    dt   dt  v 1 − v 0 = v2 v 1 − v 2 = v0

Ans.

Laws of Motion — 555

Chapter 8 7. a1 =

∴

m2 g 30 = m/s2 m1 + m2 7 (m − m2 ) g a2 = 1 m1 + m2 10 = m/s2 7 m g − m1 g sin 30° a3 = 2 m1 + m2 10 = m/s2 7 a1 > a2 = a3

 dx  dy −  =  dt  dt

dw dx =− = v2 dt dt dy = v1 dt x and = sinθ z Substituting these values in Eq. (i) we have v2 (1 + sin θ ) = v1

...(i)

−

Ans.

Ans.

11. On the cylinder

8. T − µ mg = ma

If N = normal reaction between cylinder and inclined plane N sin θ = horizontal component of N ....(i) = ma N cos θ = vertical component of N ...(ii) = mg Dividing Eq. (i) by (ii) we get, a tan θ = g Ans. ∴ a = g tan θ

a T

m mmg

mmg m

F

T

a

∴

9.

 dw x −  +  dt  z

or

T = µ mg + ma F − T − µ mg = ma ∴ F − µ mg − ma − µ mg = ma F or a= − µg 2m a1 = sinθ a2

Ans.

12. With respect to trolley means, assume trolley at rest and apply a pseudo force (= ma, towards left ) on the bob. q

a2

T

a1

ma

q

q

q

∴

anet

a1 = a2 sin θ

mg

10. z = x 2 + c2 x y z c q

13.

w

w+ y+ z=l

Now, or ∴

w+ y+ dw dy + + dt dt

x 2 + c2 = l x

x 2 + c2

⋅

mg sin θ − ma cos θ m = g sin θ − a cos θ Net pulling force (M − m) g a= = Total mass (M + m) anet =

dx =0 dt

Ans. ...(i)

Since, M >> m ∴ M −m≈M + m=M Substituting in Eq. (i), we have a≈g FBD of m T − mg = ma = mg ∴

T = 2 mg

Ans.

556 — Mechanics - I  3  1 = (10 3 )   − (10)    2 2  

14. Let α = angle of repose For θ ≤ α Block is stationary and force of friction,

= 10 m/s2

f = mg sin θ or f ∝ sin θ i.e. it is sine graph For θ ≥ α Block slides downwards

2s = ar

t=

2×1 1 = s 10 5

a2

∴ f = µ mg cos θ or f ∝ cos θ i.e. now it is cosine graph The correct alternative is therefore (b).

N q (Pseudo force) = ma

15. If they do not slip, then net force on system = F F Acceleration of system a = ∴ 3m T =F F FBD of m F − µ mg = ma = 3

q mg

18. ( f1 )max = µ 1m1g = 6 N ( f2 )max = µ 2m2g = 10 N At t = 2 s, F ′ = 4 N Net pulling force, F1 = F `− F ′ = 11 N

a T=F

T = 5N

mmg

∴

µ mg =

2F 3

µ=

or

2F 3 mg

f1 = 1N

Ans.

T = 5N

Relative acceleration along the inclined plane ma cos θ + mg sin θ ar = m = (a cos θ + g sin θ ) t=

2s = ar

2L a cos θ + g sin θ

N ma (pseudo force)

15N

f2 = 10N

16. FBD of m w.r.t. chamber

Ans.

19.

Total maximum resisting force, F2 = ( f1 )max + ( f2 )max = 16 N Since F1 < F2, system will not move and free body diagrams of the two block are as shown in figure. Net pulling force a= Total mass 2mg sin 30° g = = 2m + m 3 m

T q

a

FBD of m a2

Ans.

T = ma = mg /3

q

30°

mg

17. FBD of m w.r.t. wedge Relative acceleration along the inclined plane ma cos θ − mg sin θ ar = m = a cos θ − g sin θ

P

T T

Resultant of tensions R = − (T cos 30° ) $i − (T sin 30° + T ) $j

Chapter 8 Putting T = mg /3 3 mg $ R=− mg $i − j 6 2 Since, pulley P is in equilibrium. Therefore, F+ R=0 where, F = force applied by clamp on pulley mg Ans. F=−R= ( 3 $i + 3 $j) ∴ 6

20. ( f2 kg )max = µ 2m2 g = 0.6 × 2 × 10 = 12 N ( f4 kg )max = µ 4m4 g = 0.3 × 4 × 10 = 12 N Net pulling force F = 16 N and Net resistive force F ′ = ( f2 kg )max + ( f4 kg )max = 24 N Since, F < F ′, system will not move and free body diagrams of two blocks are as shown below. T

2 kg f2

T

4 kg

4 kg

T + 12 = 16 ⇒ T = 4 N

2 kg

f2 = T = 4 N

∴ Maximum value of acceleration of coin which can be provided by the friction, f ...(ii) amax = L = µ S g m Equating Eqs. (i) and (ii), we get 4 = µ S (10) Ans. ∴ µ S = 0.4

23. Just at the time of tipping, normal reaction at 1 will become zero. Σ (moments of all forces about point 2) = 0 N2 1

2 w1

∴

w1 (4 ) = w2 (x ) 4 w1 4 (10 g ) 1 x= = = m w2 80 g 2

or Ans.

w2 x

4m

F = 16N

f4 = 12N

Laws of Motion — 557

24. In vertical direction, net force = 0

21. FBD of rod w.r. t. trolley

N1

q

N1 Pseudo force = ma

a

C

N2

q

q

mg

mg

N 1 = ma N 2 = mg Net torque about point C = 0 l  l  ∴ N 1  sin θ = N 2  cos θ 2  2  or ⇒

(ma) (sin θ ) = (mg ) (cos θ ) a = g cot θ

∴ ...(i) ...(ii)

At

N2 = cos2 θ N1

∴

...(i)

Ans.

25. Σ (moments of all forces about point C ) = 0 Ans.

g =m

a=

Limiting value of static friction, fL = µ S mg

mg cos θ

Under normal condition, normal reaction is, ...(ii) N 2 = mg cos θ

N

dv = 4t dt t = 1 s, a = 4 m/s2

or N 1 =

N 1 cos θ = mg

22. v = 2t 2 ∴

Ans.

cos

f = mg sin q

C

...(i)

q in

s mg

O q q = 45°

q

x

mg cosq

558 — Mechanics - I  a f   = N (x )  2

∴

Net force in OP direction is zero ma mg T + = ∴ 2 2 FBD of box w.r. t. ground

 a (mg sin θ )   = (mg cos θ ) x  2

or

At 45° , sin θ = cos θ a ∴ x= 2 Hence, the normal reaction passes through O. N = mg F=

a 45°

T

mg 3

C

f=F mg = 3

mg

Σ (moments of all forces about point C ) = 0 a a Nx = F ⋅ + f ⋅ ∴ 2 2 mg  a mg  a or (mg ) x =   +   3  2 3  2 x=

or l 2

27. x = −

a 3

29.

NA

a

NB C

∴

28. Let acceleration of box at this instant is ' a' (towards O

P

45°

anet

/s

9

w

T

2

y

right). FBD of ball w.r.t. box

45° mg

Ans.

Let N = normal reaction between rod and wedge. Then N sin 37° will provide the necessary ma force to the wedge

l l l y= − = 2 6 3 Σ (moments of all forces about point C ) = 0 ∴ NAx = NB y NA y 4 Ans. or = = NB x 3

pseudo force = ma

Substituting in Eq. (i), we get g a= 3 9 3 = tan 37° = a 4 ∴ a = 12 m/s2

Ans.

l l = 4 4

x

T cos 45° = ma T = 2 ma

or

x

26.

...(i)

∴

a 37°

m

N sin 37° = ma = (10) (12) = 120 120 120 N = = sin 37° 0.6 = 200 N

30. N = ma

Ans. f

L fL = µN = 0.5 ma Vertical acceleration, mg − fL av av = = g − 0.5 a m mg = 10 − 0.5 × 4 2 = 8 m/s 1 Applying s = at 2 2 2s in vertical direction, we have or t = a 2×1 t= = 0.5 s 8

N

a

Ans.

Chapter 8 31. T − (nm + M ) g = (nm + M ) a

Laws of Motion — 559

34. Maximum value of friction between 10 kg and 20 kg is

∴ n (mg + ma) = T − Mg − Ma

a

T

10 kg

100 N

f1

f1

20 kg a

30 Kg a

(nm + M)

( f1 )max = 0.5 × 10 × 10 = 50 N Maximum value of friction between 20 kg and 30 kg is ( f2 )max = (0.25) (10 + 20) (10) = 75 N Now let us first assume that 20 kg and 30 kg move as a single block with 10 kg block. So, let us first calculate the requirement of f 1 for this 100 − f1 = 10 a f1 = 50 a On solving these two equations, we get f1 = 83.33 N Since, it is greater than ( f1 )max , so there is slip between 10 kg and other two blocks and 50 N will act here. 50 N Now let us check a 20 kg whether there is slip between 20 kg and f2 f2 30 kg or not. For this a 30 kg we will have to calculate requirement of f2 for no slip condition. 50 − f2 = 20 a and f2 = 30 a On solving these two equations, we get f2 = 30 N and a = 1 m/s2

T − M (g + a) m (g + a) T − M (g + a) = max m (g + a)

n=

or

nmax

=

2 × 104 − 500 (10 + 2) 80 (10 + 2)

= 14.58 But answer will be 14.

Ans.

32. It implies that the given surface is the path of the given projectile y = x tan θ −

gx 2 2u cos2 θ

= x tan 60° −

Slope, At or

2

(10) x 2 (2) (20)2 cos2 60°

y = 3 x − 0.05 x 2 dy = 3 − 0.1 x dx y = 5m 5 = 3 x − 0.05 x 2

...(i) ...(ii)

0.05 x 2 − 3 x + 5 = 0

3 ± 3−1 3± 2 = 0.1 0.1 From Eq. (ii) slope at these two points are, − 2 and 2. x=

33. Horizontal displacement of both is same (= l ).

35.

Since, f2 is less than ( f2 )max , so there is no slip between 20 kg and 30 kg and both move together with same acceleration of 1 m/s2. 4R/5 cos θ = = 0.8 R

Horizontal force on A is complete T . But horizontal force on B is not complete T . It is component of T . So, horizontal acceleration of B will be less. T 4R 5

T

q

∴ ∴

tB > tA

q R

θ = 37° v1 3 = tan θ = tan 37° = v2 4

R/5

560 — Mechanics - I 2R

∴

At point P 2T cos θ = F

v1

T =

∴

q q

v2

3 3 v2 = × 20 4 4 = 15 m/s

v1 =

∴

ar = 2a a a = r = 2.5 m/s2 2

36. vAL = vA − vL

Ans.

∴

vA = vAL + vL = (−2) + 2 = 0 Let z = length of string at some instant. Then,

At mass m Component of T along the other mass m is T sin θ T sin θ  F   sin θ  ∴ a= =     2 cos θ   m  m ∴

+ L

F 2 cos θ

F=

2 ma 2 × 0.3 × 2.5 = tan θ (3/ 4 )

=2N

Lift

Ans.

38. x + x + y + c = l = length of string 2

–

2

Differentiating w.r.t. time, we get

2 m/s Fix

x

Ö y2 + c2

x c

x

q B

y

A

y A

Now, ∴

B

Fixed

2

dz − = 2 m/s dt y = x − (z − x ) = 2x − z dy dx  dz =2 + −  dt dt  dt 

(Given)

2 vB =

or ∴

= 2 (2 m/s) + 2 m/s = 6 m/s = vB

Ans.

37. Relative acceleration

 dy −  y + c  dt 

dx = dt

y

2

2

1 vA cos θ

vA = 2 vB cos θ = (2) (10) (0.8) = 16 m/s

39. Maximum force of friction between c and ground is

( fc )max = (0.5) (60) (10) = 300 N Since it is pulling the blocks by the maximum force (without moving). Therefore the applied force is F = 300 N

a m q 5 cm

3 cm

4 cm T a

m

Ans.

q q

A

P P

F

F = 300 N

B

( fAB )max = 0.4 × 60 × 120 × 10 = 240 N ( fBG )max = 0.3 × 120 × 10 = 360 N

Laws of Motion — 561

Chapter 8 Since ( fBG )max is greater than 300 N, blocks will not move. Free body diagrams of block are as shown below.

43. Let a = maximum acceleration of A. Under no slip condition acceleration of B is also a FBD of A w.r. t. ground

T=0 f=0

A

N

y

45°

f=0 f BG = 300 N

40.

a

45°

T=0

B

F = 300 N

x

45° mN

Let a B = ai$

mg

a AB = a A − a B = (15 − a) $i + 15$j

Then,

37° 37°

aAB

Since, a AB is along the plane as shown in figure. 3 15 ∴ tan 37° = = 4 15 − a Solving this equation, we get a = − 5 or a B = (−5i$ )

41. Acceleration, a=

2F − F F = m + m 2m

N = 3F

Ans.

T

T

M2 M3

a

M 2g − M 3g 3M 3g − M 3g = M2 + M3 3M 3 + M 3 g = 2 Now FBD of M 2 gives the equation, M g M 2g − T = M 2 ⋅ a = 2 2 M 2g ∴ T = 2 or 2T = M 2 g Now taking moments of forces about support point M 1g (l1 ) = (2T ) l2 = (M 2g ) (3l1 ) M1 Ans. ∴ =3 M2 a=

Ans.

B

v

...(ii)

2T

a

42. Distance AB = constant

60°

...(i)

44. For M2 and M3

(towards left)

Horizontal forces on B gives the equation, 2F − N sin 30° = m ⋅ a N  F or 2F − =m   2m 2 ∴

Σ Fy = 0 N µN ∴ = mg + 2 2 Σ Fx = ma N µN + = ma ∴ 2 2 Solving these two equations, we get  1 + µ a=g   1 − µ

75° 45° A u

∴ Component of v along BA = component of u along BA or v cos 60° = u ⋅ cos 45° Ans. or v= 2u

45. Resultant of N and N (= 2 N ) is equal to mg cos θ ∴

2 N = mg cos θ

562 — Mechanics - I ∴

N =

mg cos θ 2

N

47. ( f1 )max = between 1 kg and 2 kg 1 kg

N

m = 0.2

2 kg

30 N

m = 0.5 1m

Now kinetic friction will act from two sides mg sin θ − 2µ k N a= ∴ m Substituting the value of N , we get a = g (sin θ − 2 µ k cos θ )

= 0.2 × 1 × 10 = 2 N ( f2 )max = between 2 kg and ground a1 1 kg

Ans.

46. f1 → force of friction between 2 kg and 3 kg ( f1 )max = 0.5 × 3 × 10 = 15 N f2 → force of friction between 2 kg and 1 kg ( f2 )max = 0.3 × 5 × 10 = 15 N f3 → force of friction between 1 kg and ground ( f3 )max = 0.1 × 6 × 10 = 6 N When F > 6 N system will start moving with a common acceleration F −6 F  a= =  − 1 m/s2 3+ 2+1 6 

2N 1 kg

= 0.5 × 3 × 10 = 15 N 2 a1 = = 2 m/s2 1 30 − 15 − 2 a2 = 2 = 6.5 m/s2 ar = a2 − a1 = 4.5 m/s2

2 kg

t=

1 kg 6N

=

F f1 − 6 = (3) a = − 3 2 F   f1 =  6 + − 3   2 F  =  + 3 2 

∴

2×1 2 = s 4.5 3

Ans.

1. Maximum value of friction between two blocks F

m2

T 2N

2N

a 1 kg 6N

f2 − 6 = (1) (a) =

2S r ar

More than One Correct Options

Since F is slightly greater than 6 N ∴ f1 < 15 N or < ( f1 )max ∴ No slipping will occur here f2

30 N

15 N

a

f1

2N a2

F −1 6

F ∴ f2 = + 5 6 Again f2 < ( f2 )max . So no slip will take place here also.

m1

T

fmax = 0.2 × 1 × 10 = 2 N In critical case, T =2N F =T + 2= 4 N ∴ System is in equilibrium if f ≤ 4 N

Ans.

For F > 4 N F − (T + 2) = m2 a = (1) (a) T − 2 = m1 a = (1) (a)

...(i) ...(ii)

Laws of Motion — 563

Chapter 8 a

3. a = slope of v - t graph

m2

= − 1 m/s2

T

F

2N

2N m1

∴

Retardation = 1 m/s2 =

or

µ=

T a

On solving these two equations, we get F T = 2 When, F = 6 N, T = 3 N

Ans.

2. Resultant of mg and mg is 2 mg.

1 1 = = 0.1 g 10

If µ is half, then retardation a is also half. So using v = u − at or 0 = u − at u 1 or t ∝ or t= a a we can see that t will be two times.

4. Maximum force of friction between A and B

T2

( f1 )max = 0.3 × 60 × 10 = 180 N Maximum force of friction between B and ground ( f2 )max = 0.3 × (60 + 40) g = 300 N

b mg B

A

T = 125 N

B

f1 T = 125 N

f1 mg

Therefore T2 should be equal and opposite of this. or ...(i) T2 = 2 mg Further, T2 cos β = mg and T2 sin β = mg or sin β = cos β ⇒ β = 45° T1 cos α = mg + T2 cos β  1 = mg + 2 mg    2 or T1 cos α = 2 mg  1 T1 sin α = T2 sin β = 2 mg    2 ∴

µ mg = µg m

T1 sin α = mg a mg b

Both are stationary f1 = T = 125 N f2 = T + f1 = 250 N

5. ax =

mg sin θ = g sin θ m N

mg

sin

q mg cos q

It is also moving in y -direction ∴ mg cos θ > N mg cos θ − N ay = m

ax

mg

From Eqs. (iv) and (v), we get 1 and T1 = 5 mg tan α = 2 tan β = tan 45° = 1 and T2 = 2 mg ∴ tan β = 2 tan α and

x

...(iv)

...(v)

T1

B

...(ii) ...(iii)

f2

2 T1 = 5 T2

a

ay

Now,

a = ax2 + a2y > g sin θ

y

564 — Mechanics - I 6. Maximum value of friction between A and B is

( f1 )max = 0.25 × 3 × 10 = 7.5 N Maximum value of friction between B and C ( f2 )max = 0.25 × 7 × 10 = 17.5 N and maximum value of friction between C and ground, ( f3 )max = 0.25 × 15 × 10 = 37.5 N F0 = force on A from rod F0

A

7.5 N

7.5 N

F

T

B

17.5 N

(f2)max = 17.5 N

C

T

(f3)max = 37.5 N

If C is moving with constant velocity, then B will also move with constant velocity For B, T = 17.5 + 7.5 = 25 N

9. f1 → force of friction between 2 kg and 4 kg f2 → force of friction between 4 kg and ground ( fS 1 )max = 0.4 × 2 × 10 = 8 N FK1 = 0.2 × 2 × 10 = 4 N ( fS2 )max = 0.6 × 6 × 10 = 36 N FK2 = 0.4 × 6 × 10 = 24 N At t = 1 s, F = 2 N < 36 N, therefore system remains stationary and force of friction between 2 kg and 4 kg is zero. At t = 4 s, F = 8 N < 36 N. Therefore system is again stationary and force of friction on 4 kg from ground is 8 N. At t = 15 s, F = 30 N < 36 N and system is stationary.

10. Net pulling force = 0 ⇒

a=0 T1 = 1 × g = 10 N T3 = 2 × g = 20 N T2 = 20 + T1 = 30 N

For C, F = 17.5 + 25 + 37.5 = 80 N For F = 200 N Acceleration of B towards right = acceleration ofC towards left = a (say ) Then ...(i) T − 7.5 − 17.5 = 4 a ...(ii) 200 − 17.5 − 37.5 − T = 8 a On solving these two equations, we get a = 10 m/s2

7. Since, µ 1 > µ 2 ∴

( f1 )max > ( f2 )max

Further if both move, T − µ mg a= m µ of block is less. Therefore, its acceleration is more.

8. N cos θ = mg = 10

11. fmax = 0.3 × 2 × 10 = 6 N At t = 2 s, F = 2 N < fmax ∴ f =F =2N At t = 8 s, F = 8 N > fmax ∴

At t = 10 s, F = 10 N > fmax ∴

f F− a= m F = fmax For 6 s ≤ t ≤ 10 s F− a= m

f

=

t−6 = 0.5t − 3 2

10

∫ dv = ∫ adt = ∫ (0.5t − 3) dt 0

...(ii)

6

v = 4 m/s After 10 s

N

F − f 10 − 6 = = 2 m/s2 m 2 = constant v′ = v + at = 4 + 2 (12 − 10) = 8 m/s a=

q a

∴ mg

N = 5 5 N and

=6N f 10 − 6 = = 2 m/s2 2 = 6 N at 6 s

v

...(i)

N sin θ = ma = 5 On solving these two equations, we get

f =6N

tan θ =

1 2

12. Maximum force of friction between 2 kg and 4 kg = 0.4 × 2 × 10 = 8 N

Chapter 8 2 kg moves due to friction. Therefore its maximum acceleration may be 8 amax = = 4 m/s2 2 Slip will start when their combined acceleration becomes 4 m/s2 F 2t or 4 = or t = 12 s a= ∴ m 6 At t = 3s F 2t 2 × 3 a2 = a4 = = = m 6 6 = 1 m/s2 Both a2 and a4 are towards right. Therefore pseudo forces F1 (on 2 kg from 4 kg) and F2 (on 4 kg from 2 kg) are towards left F1 = (2) (1) = 2 N F2 = (4 ) (1) = 4 N From here we can see that F1 and F2 do not make a pair of equal and opposite forces.

13. See the hint of of Q.No-10 of Assertion and Reason type questions of Level-1.

Comprehension Based Questions

1. Let µ K = µ, then µ S = 2µ According to first condition, F + mg sin θ = µ S mg cos θ = 2 µ mg cos θ ...(i) According to second condition, mg sin θ = F + µ K mg cos θ ...(ii) = F + µ mg cos θ Putting θ = 30°, we get  3 F + mg /2 = 2µ mg    2 or

3 µ mg = F + 0.5 mg

...(iii)

 3 mg = F + µ mg   2  2 or

0.5 3 µ mg = 0.5 mg − F

Dividing Eq. (iii) and (iv), we get mg F= 6

2. Substituting value of F in Eq. (iii), we have 2 = µK 3 3 4 µ S = 2µ = 3 3 µ=

∴

3. a =

Laws of Motion — 565

F + mg sin θ − µ K mg cos θ m  3  2  (mg / 6) + (mg / 2) −   mg    3 3  2 = m g = 3

Ans.

4. F ′ = mg sin θ + µ S mg cos θ =

 3 mg 4 + mg   2 3 3  2

=

7mg 6

Ans.

5. F ′′ = mg sin θ + µ K mg cos θ

6.

 3  2  = (mg / 2) +   mg    3 3  2 5mg Ans. = 6 1350 × 9.8 − 1200 × 9.8 Acceleration a1 = 1200 vf = 0 = 1.225 m/s2 h 2 1200 g − 1000 g a2 Retardation, a2 = 1200 v = 1.63 m/s2 h a h1 + h2 = 25

1

1

...(i)

v = 2a1 h1

or

2a2 h2

2a1 h1 = 2a2 h2 h1 a2 ∴ = h2 a1 1.63 = = 1.33 1.225 Solving these equations, we get h1 = 14.3 m

u=0

or

...(ii)

Ans.

7. v = 2a1h1 = 2 × 1.225 × 14.3 = 5.92 m/s

...(iv)

Ans.

8. tanθ =

8 ∴ θ = tan −1 15

Ans.  8   = 28°  15

( fA )max = 0.2 × 170 × 10 × cos 28° = 300.2 N ≈ 300 N ( fB )max = 0.4 × 170 × 10 × cos 28° = 600.4 N ≈ 600 N Now, (mA + mB ) g sin θ = (340) (10) sin 28° = 1596 N

566 — Mechanics - I

9.

Since this is greater than ( fA )max + ( fB )max , therefore blocks slides downward and maximum force of friction will act on both surfaces ∴ ftotal = ( fA )max + ( fB )max Ans. = 900 N (mA + mB ) g sin θ − ftotal a= mA + mB 1596 − 900 = = 2.1 m/s2 340 a

F

A

mA

gs

2. (a) At θ = 0° , driving force F = 0 ∴ friction = 0 (b) At θ = 90° , N = 0 ∴ Maximum friction = µN = 0or friction = 0 (c) Angle of repose, θ r = tan −1 (µ ) = 45° Since θ < θ r , block is at rest and f = mg sin θ = 2 × 10 sin 30° = 10 N (d) θ > θ r . Therefore block will be moving f = µ mg cos θ = (1) (2) (10) cos 60° = 10 N

) max (f A

q in

4. (a) N − 10 = ma = 5 × 2 10 N

mA g sin θ − F − ( fA )max = mA a F = mA g sin θ − ( fA )max − mA a

∴

F w

Ans. ∴

N = 20 N µ sN = 8 N µ kN = 6 N W = mg = 20 N (b) When F = 15 N w−F =5N This is less than µ s N ∴ f =5N (c) F = w − µ s N = 20 − 8 = 12 N

Match the Columns

1. F = 2 t µ s mg = 20 µ s µ k mg = 20 µ k (a) Motion starts at 4 s ∴ F = µ s mg ⇒ (2) (4 ) = 20 µ s ∴ µ s = 0.4

force at C = 10 N (b) fc = 0

(c) N c = (mB + mC ) g = 20 N (d) T = F = 10 N

(everywhere)

7. Ground is smooth. So all blocks will move towards

f = F = 2 × 0.1 = 0.2 N

right 2 kg and 5 kg blocks due to friction.

F − µ k mg m 2 × 8 − 0.3 × 20 = 2 2 = 5 m/s

2 kg Kg

a=

a = 0.5 m /s2 10

(upwards)

5. (a) Net pulling force F = net resisting frictional

(c) At t = 0.1 s, when motion has not started,

∴

(downwards)

(d) F = w + µSN = 20 + 8 = 28 N

(b) At 4 s, when motion starts, F − µ k mg a= m 8 − 20 µ k ∴ 1= 2 Solving we get, µ k = 0.3

(d) At 8 s

ms = 0.4

2 kg

mk = 0.3

= 170 × 10 × sin 28° − 300 − 170 × 2.1 = 141 N

5 m/s2

N

F = force on connecting bar

f1

f1 3 kg

F

f2 f2

Ans.

5 kg Kg

Chapter 8

Laws of Motion — 567

8. µ mg cos θ = 15 N

4T a

4a 2T

T 5N

a 2

1 0N

30 N

1

T − 10 − 15 = 2 a  a 30 − 2T = 3    2

...(i) ...(ii)

Solving these two equations we get, a = − 3.63 m/s2 So, if we take the other figure, 2T

T a

15 10

N

a 2

N

This figure is not feasible. Because for ' a' to be down the plane, 10 N > T + 15 which is not possible ∴ a=0 and free body diagrams are as shown below. 15

N

2T = 30 N

a=0

a=0

10

N f=5N

30 N

Subjective Questions

1. It is just like a projectile motion with g to be replaced by g sin 45°. After 2 s, g   × 2 + (10 cos 45° )2 10 sin 45° −   2

= 10 m/s

a

A mAg

Similarly, if a be the acceleration of block A (downwards), then acceleration of block B towards right will be 4a. Equations of motion are For block A, mAg − 4T = mAa or …(i) 50 − 4T = 5a For block B, T − f = 10(4 a) or T − (0.1)(10)(10) = 40a or …(ii) T − 10 = 40a Solving Eqs. (i) and (ii), we get 2 Ans. a= m/s2 33 friction on mass is eastwards. frequired = mass × acceleration = (30 × 1.8) = 54 N Since it is less than µ s mg (eastwards) ∴ f = 54 N (b) When the truck accelerates westwards, force of friction is westwards. frequired = mass × acceleration = 30 × 3.8 = 114 N Since it is greater than µ s mg. Hence Ans. f = fk = µ k mg = 60 N (westwards)

4. Block B will fall vertically downwards and A along the plane. Writing the equations of motion. For block B, mB g − N = mB aB or 60 − N = 6aB (N + mAg ) sin 30° = mAaA or (N + 150) = 30 aA

Ans.

2. Suppose T be the tension in the string attached to block B. Then tension in the string connected to block A would be 4T .

…(i) …(ii)

aB = aA sin 30°

Further

v = vx2 + v 2y =

T

3. (a) When the truck accelerates eastward force of

30 N

T=

f

B

or aA = 2aB Solving these three equations, we get (a) aA = 6.36 m/s2 (b)

aBA = aA cos 30° = 5.5 m/s

…(iii) Ans.

2

Ans.

5. Let acceleration of m be a 1 (absolute) and that of M be a2 (absolute).

568 — Mechanics - I Writing equations of motion.

N′

N f1

f1

N a1

N′ f2 + F

f1 + f2

or

F

mg M

For m For M,

N + mg

N sin a

mg cos a

FBD of man

a2

mg cos α − N = ma1

…(i)

N sin α = Ma2

…(ii)

For the plank not to move F − ( f2 )max ≤ f1 ≤ F + ( f2 )max F − µ (M + m)g ≤ ma ≤ F + µ (M + m)g F µ(M + m)g or a should lie between − m m F µ(M + m)g Ans. and + m m or

Note In the FBD only those forces which are along a1 and a2 have been shown. Constraint equation can be written as, a1 = a2 sin α Solving above three equations, we get acceleration of rod, mg cos α sin α a1 =  M   m sin α +   sin α  and acceleration of wedge mg cos α a2 = M m sin α + sin α

...(iii)

N + mg FBD of plank

8. Writing equations of motion T

5T

Ans.

a1

Ans. For M For m,

6. (a) N 2 and mg pass through G. N 1 has clockwise moment about G, so the ladder has a tendency to slip by rotating clockwise and the force of friction ( f ) at B is then up the plane. (b) ΣM A = 0 l  …(i) ∴ fl = mg  sin 45° 2  ΣFV = 0 …(ii) ∴ mg = N 2 cos 45° + f sin 45° From Eqs. (i) and (ii), 3 N2 = mg N1 A 2 2 mg G and f = 2 2 N2 f f or µ min = B mg N2 1 Ans. = 45° 3

7. Here f1 = force of friction between man and plank

and f2 = force of friction between plank and surface.

a2

Mg

mg

FBD of M

FBD of m

5T − Mg = Ma1 mg − T = ma2

From constraint equation, a2 = 5a1 Solving these equations, we get acceleration of M ,  5m − M  a1 =   g  25m + M 

…(i) …(ii) …(iii)

 5m − M  a2 = 5  g  25m + M  a s n 2a1s1 = 2a2s2 or 1 = 2 = a2 s1 m g sin α m or = µg cos α − g sin α n and of m,

9.

Solving it, we get  m + n µ=  tan α  m 

10. Limiting friction between A and B

Ans.

fL = µN = 0.4 × 100 = 40 N (a) Both the blocks will have a tendency to move together with same acceleration (say a).

Chapter 8 So, the force diagram is as shown.

12. Assuming that mass of truck >> mass of crate.

a A

Retardation of truck a1 = (0.9) g = 9 m/s2

F = 30 N

f

Retardation of crate a2 = (0.7) g = 7 m/s2

f B

Equations of motion are, 30 − f = 10 × a f = 25 × a Solving these two equations, we get a = 0.857 m/s2

or relative acceleration of crate ar = 2 m/s2 . …(i) …(ii)

Truck will stop after time 15 t1 = = 1.67 s 9 and crate will strike the wall at 2s 2 × 3.2 t2 = = = 178 . s ar 2

and f = 2142 . N As this force is less than fL, both the blocks will move together with same acceleration,

As t2 > t1 , crate will come to rest after travelling a distance

aA = aB = 0.857 m/s2 (b)

Laws of Motion — 569

250 − f = 10a f = 25a Solving Eqs. (iii) and (iv), we get f = 178.6 N

…(iii) …(iv)

1 1  15 ar t12 = × 2.0 ×    9 2 2

s=

2

= 2.77 m

Ans.

13. µ kmg = 0.2 × 10 × 10 = 20 N

a

For t ≤ 0.2 s

F + µ kmg m 20 + 20 = = 4 m/s2 10 At the end of 0.2 s,

F = 250 N

A

Retardation

f f B

As f > fL, slipping will take place between two blocks and f = fL = 40 N 250 − 40 aA = 10

v = u − a1t v = 1.2 − 4 × 0.2 = 0.4 m/s For t > 0.2 s 10 + 20 Retardation a2 = = 3 m/s2 10 Block will come to rest after time v 0.4 t0 = = = 013 . s a2 3 ∴ Total time = 0.2 + 013 . = 0.33 s

= 21.0 m/s2 A

F = 250 N

40 N 40 N B

40 aB = = 1.6 m/s2 25

a1 =

Ans.

14. Block will start moving at, F = µ mg Ans.

11. Normal reaction between A and B would be

N = mg cos θ. Its horizontal component is N sin θ. Therefore, tension in cord CD is equal to this horizontal component. Hence, T = N sin θ = (mg cosθ ) (sin θ ) mg Ans. = sin 2 θ 2

or 25t = (0.5) (10) (9.8) = 49 N ∴ t = 1.96 s Velocity is maximum at the end of 4 second. dv 25t − 49 ∴ = = 2.5t − 4.9 dt 10 vmax

∫0

∴ ∴

dv = ∫

4 1.96

(2.5 t − 4.9) dt

vmax = 5.2 m/s

Ans.

570 — Mechanics - I For 4 s < t < 7 s 49 − 40 = 0.9 m/s2 10 − a1 t1 = 5.2 − 0.9 × 3 = 2.5 m/s a1 =

Net retardation ∴ v = vmax For t > 7 s

a=

∴

F = (M 1 + M 2 + M 3 ) a M = (M 1 + M 2 + M 3 ) 3 g M2

Distance travelled before sliding stops is, v2 s= 2a (5)2 = ≈ 8.5 m 2 × 1.47

∴ Total time = (4 − 1.96) + (7 − 4 ) + (0.51) Ans. = 5.55 s

15. Let B and C both move upwards (alongwith their pulleys) with speeds vB and vC then we can see that, A will move downward with speed, 2vB + 2vC . So, with sign we can write, v vB = A − vc ∴ 2 Substituting the values we have, vB = 0 Ans.

19.

A is in equilibrium under three concurrent forces shown in figure, so applying Lami's theorem N 30°

A

mg cos θ 2 mg sin θ – 2µ k N a= m

∴ N =

∴

2 µ k g cos θ

= g (sin θ –

2 µ k cos θ )

17. FBD of M 2 and M 3 in accelerated frame of reference is shown in figure.

Note Only the necessary forces have been shown. Mass M 3 will neither rise nor fall if net pulling force is zero. M 2a = M 3 g

M3 M3g

a

N

N

Ans.

dv Net force F − µ k ρ (L − x ) g = = dx mass ρL v L F − µ ρ (L − x ) g k dx ∴ ∫0 v dv = ∫0 ρL v2 F µ gL = − µ k gL + k 2 ρ 2 v=

∴

2F − µ k gL ρ

Ans.

21. (a) v = a1 t1 = 2.6 m/s 1 2 1 a1t1 = × 2 × (1.3)2 = 1.69 m 2 2 s2 = (2.2 − 1.69) = 0.51 m v2 Now, s2 = 2a2 s1 =

∴ and

M2 Pseudo force = M2a

√2 N

20. v ⋅

∴

30°

ma mg = sin (90 + 60) sin (90 + 30) g cos 60° a= = 5.66 m/s2 cos 30°

i.e.

= g sin θ –

60° mg

Ans.

2 N = mg cos θ

16. FBD of A with respect to frame is shown in figure.

Pseudo force = ma

Ans.

18. Retardation a = µ k g = 0.15 × 9.8 = 1.47 m/s2

49 = 4.9 m/s2 10 v 2.5 t= = = 0.51 s a2 4.9

Retardation a2 = ∴

M3 g M2

or

v2 (2.6)2 = = 6.63 m/s2 2s1 2 × 0.51 v t2 = = 0.4 s a2

a2 =

(b) Acceleration of package will be 2 m/s2 while retardation will be µ k g or 2.5 m/s2 not 6.63 m/s2. For the package,

Laws of Motion — 571

Chapter 8 1 a1 t12 = 1.69 m 2 1 1 s2 = vt2 − a′2 t22 = 2.6 × 0.4 − × 2.5 × (0.4 )2 2 2 = 0.84 m ∴ Displacement of package w.r.t. belt Ans. = (0.84 − 0.51) m = 0.33 m v = a1t1 = 2.6 m/s ⇒ s1 =

Alternate Solution For last 0.4 s | ar | = 6.63 − 2.5 = 4.13m/s 2 1 1 ∴ sr = | ar | t 2 = × 4.13 × (0.4 )2 2 2 = 0.33 m

Relative retardation of upper block 11 ar = a1 + a2 or ar = (µ 2 − µ 1 )g 10 2 Now, 0 = vmin − 2ar l ∴

vmin = 2ar l =

(b) 0 = vmin − ar t v 20l or t = min = ar 11(µ 2 − µ 1 )g

Ans.

Retardation a = µg ∴ Time when slipping will stop is t =

in figure.

aA

sr =

…(i)

vr 2 v12 + v22 = 2a 2 µg

 v 2 + v22  xr = − sr cos θ = −  1   2µg 

30°

=

…(i)

Substituting in Eq. (i), we get

=

23. (a) Force of friction at different contacts are shown in figure.

  v2    v 2 + v 2  1 2 

− v2 v12 + v22 2 µg

 v 2 + v22  yr = sr sin θ =  1   2 µg 

N = 90 N.

vr a

v12 + v22 µg

t=

or

N

Equation of motion is 100 − N = 10aA  1 aA = a sin 30° = (2)  or aA = 1m /s2  2

Ans.

24. vr = v12 + v22

22. Free body diagram of crate A w.r.t ground is shown

mAg = 100 N

22(µ 2 − µ 1 )gl 10

  v1    v 2 + v 2  1 2 

v1 v12 + v22 2 µg vr

v1 = v1

m f1

f2

Here, and

10m

f1 = µ 2mg f2 = µ 1 (11 mg )

Given that µ 2 > 11 µ 1 ∴ f1 > f2 Retardation of upper block f a1 = 1 = µ 2g m Acceleration of lower block f − f2 (µ 2 − 11µ 1 ) g a2 = 1 = m 10

θ

f1

–v2 = v2

In time t, belt will move a distance s = v2t or

v2 v12 + v22 in x-direction. µg

Hence, coordinate of particle, x = xr + s = and

y = yr =

v2 v12 + v22 2µg

v1 v12 + v22 2 µg

Ans.

572 — Mechanics - I mg …(iv) = ma3 2 From constraint equation, …(v) a1 = a2 − a3 We have five unknowns. Solving the above five equations, we get

a1

25. FBD of m1 (showing only

N

T

T − N = m1a1

…(i)

the horizontal forces) Equation of motion for m1 is

T −

T

T

T

N N

a3

a1

N

a2

m2g FBD of m2

FBD of 3m

and

N = m2a1

…(ii)

m2g − T = m2a2

…(iii)

Equation of motion for m3 are …(iv) m3g − T = m3a3 Further from constraint equation we can find the relation, …(v) a1 = a2 + a3 We have five unknowns a1 , a2 , a3, T and N solving, we get 2m1m3 g Ans. a1 = (m2 + m3 ) (m1 + m2 ) + m2m3

26. Writing equations of motion, T − N = 3ma1 N = 2ma1 2mg − T = 2ma2

…(i) …(ii) …(iii)

27.

3 g, 17

2mg FBD of 2m

a3 mg sin 30° =mg 2 FBD of m

19 13 g and a3 = g 34 34 13 Acceleration of m = a3 = g, 34 397 Acceleration of 2m = a12 + a22 = g 34 3 and acceleration of 3m = a1 = g 17 mAg a= mA + M + m a1 =

Equations of motion for m2 are

T

a2

a1

m3g

a1

T

a2 =

Ans.

For the equilibrium of B, mg = µN = µ(ma) = ∴

mA =

(M + m)m (µ − 1)m

mA =

(M + m) µ −1

Note mA > 0 ∴ µ > 1

µmmAg mA + M + m

Ans.

9. Work, Energy and Power INTRODUCTORY EXERCISE

9.1

−4

−4

6. W = ∫ Fdx = ∫ (−2x ) dx

1. W = F ⋅ S = F ⋅ (rf − ri )

2

2

= [ − x 2 ]−2 4 = − [16 − 4 ] = − 12 J

= (6$i − 2$j + k$ ) ⋅ [(2$i + 3$j − 4 k$ ) − ($i + 4 $j + 6k$ )] = − 2J

2

4

2

 1 1 1 = − 4   = − 4  −  = −1 J x 4 2 4 

Ans. Ans.

(c) W mg = (mg ) (S ) cos 90° = 0

Ans.

(d) Only three forces are acting. So, total work done is summation of all above work done. 3. WT = (T ) (x ) cos 0° = Tx WW = (W ) (x ) cos 90° = 0 W N = (N ) (x ) cos 90° = 0 W F = (F ) (x ) cos 180° = − Fx mg 4. mg − T = ma = T 4 3mg a ⇒ T = 4 ∴ WT = (T ) (l ) (cos 180° ) mg 3 = − mgl 4 F 5. N = mg − F sin 45° = 18 − 2 Moving with constant speed means net force = 0 1 F F cos 45° = µN = 18 −  4  2 4F F ∴ = 18 − 2 2 18 2 ∴ F= N 5 (a) W F = FS cos 45°  18 2   1 =  (2)   = 7.2 J  2  5 

From X = − 4 to X = − 2 F = − ve

F  18 −  (2) (−1)  2 = − 7.2 J = (mg ) (S ) cos 90° = 0

(from graph)

S = + ve 1 W 1 = − × 2 × 10 = − 10 J 2

∴

From −2 to 4 F = + ve S = + ve 1 W 2 = + (6 + 2) (10) 2

∴

= + 40 J WT = W 1 + W 2 = 30 J

∴

Ans.

9. (a) From x = 10 m to x = 5 m S = − ve F = + ve W 1 = − Area = −5× 3 = − 15 J

∴

(from graph)

Ans.

(b) From x = 5m to x = 10 m and ∴

S = + ve F = + ve W 2 = + Area = 5 × 3 = 15 J

Ans.

(c) From x = 10m to x = 15 m

(b) W f = (µN ) (S ) cos 180°  1 =   4

Ans.

8. W = area under F -x graph

(b) W N = N S cos 90° = 0

(c) W mg

4 dx x2 4

7. W = ∫ Fdx = ∫

2. (a) W F = FS cos 45°  1 = (16) (2.2)   = 24.9 J  2

Ans.

2

∴

S = + ve F = + ve W 3 = Area =3J

(d) From x = 0 to

x = 15 m

S = + ve and F = + ve

Ans.

574 — Mechanics - I ∴

W 4 = + Area 1 = × 3 × (12 + 6) = 27 J 2 10. (a) From x = 0 to x = 3.0 m

7. (a) T = F , (b)W All forces = 40 J Ans.

S = + ve F = + ve ∴ W 1 = + Area = + 4J (b) From x = 3 m to x = 4 m

8.

ds = (4 t − 2) dt W all = ∆K = K f − K i = K 2 s − K 0 s 1 = m (v 2f − vi2 ) 2 1 = × 2 [(4 × 2 − 2)2 − (4 × 0 − 2)2 ] 2 v=

= 32 J

F = 0 ⇒ W2 = 0 (c) From x = 4 m to x = 7m S = + ve and F = − ve ∴ W 3 = + Area = − 1 J (d) From x = 0 to x = 7 m

9. W all = ∆K = K f − K i = K f ∴

W mg + W chain = K f

or

W chain = K f − W mg =

⇒

2. W All

9.2

INTRODUCTORY EXERCISE 1 = m (v 2f − vi2 ) 2

1 1 m (v 2f − vi2 ) = × 2 × (0 − 202 ) 2 2 = − 400 J

Ans.

∴ W All = ∆ KE = K f − K i 1 1 = m (v 2f − vi2 ) = × m [(α b )2 − 0 ] 2 2 1 2 Ans. = mα b 2 5. W F = F S cos 0° = 80 × 4 × 1 = 320 J

Ans.

x

1 mv02 = ∫ − Axd x 2 0

∴

1 Ax 2 mv02 = 2 2

or

x = v0

m A

U B = − 20 J U B − U A = 40 J

∴

INTRODUCTORY EXERCISE

1. U =

Ans.

9.4

3

x − 4x + 6 3 dU = − x2 + 4 dx F = 0 at x = ± 2m F =−

S

K f − K i = W F = ∫ Fdx 0−

9.3

So, if work done by conservative force is positive then ∆U is negative or potential energy will decrease. But there is no straight forward rule regarding the kinetic energy. 2. U A = − 60 J

4. vx = 0 = 0 : vx = b = α b

W mg = (mg ) (S ) cos 180° = (50) (4 ) (−1) = − 200 J K f = W All = 120 J

Ans.

1. ∆U = − W

1 0 + W air = × 0.1[(6)2 − (10)2 ]= − 3.2 J 2 = ∆ KE = K f − K i =

6.

1 mv 2f − mgh 2

1 × 30 × (0.4 )2 − 30 × 10 × 2 2 = − 597.6 J

1. From work energy theorem, W net = W mg + W air

(as K i = 0)

=

W = W1 + W2 + W3 = 3 J

INTRODUCTORY EXERCISE

Ans.

F x = –2 F=0

S

0

x=2 F=0

F

For x > 2 m , F = − ve i.e. displacement is in positive direction and force is negative. Therefore x = 2 is stable equilibrium position. For x < − 2 m, F = − ve i.e. force and displacement are in negative directions. Therefore, x = − 2 m is unstable equilibrium position.

Chapter 9

Work, Energy and Power — 575

2. At A , x = 0 and F = 0

F

For x > 0, F = + ve. i.e. force is in the direction of displacement. Hence A is unstable equilibrium position. Same concept can be applied with E also. At point C , F = 0. For x > xC , F = − ve Displacement is positive and force is negative (in opposite direction of displacement). Therefore, C point is stable equilibrium point. 3. (a) At x = 0, F = 0. At P, attraction on −q or F1 > attraction F2 F2

x = -a +q

x=0

P

Fnet F1 x = +a +q

–q

∴ Net force is in the direction of displacement. So, equilibrium is unstable. (b) Fnet is in opposite direction of S. Therefore equilibrium is stable. –q

x = -a

+q

Fnet

F

S

x = +a x=0

x>4m F = +ve S

Therefore, equilibrium is unstable.

INTRODUCTORY EXERCISE

+q

1 1 mv 2 m (at )2 W 2 2 1. (a) Pav = = = t t t 1 1 2 = ma t = × 1 × (4 )2 (2) 2 2 = 16 W

at x = 2 m ∴ x = 2 m is stable equilibrium position

= 64 W 1 2. (i) W = Pt = mv 2 2 2 Pt (ii) v = m (iii) Integrating the displacement

Ans.

S =

F = 0 at x = 4 m When x > 4 m, F = + ve When displaced from x = 4 m (towards positive direction) force also acts in the same direction.

Ans.

velocity,

2P  t 3/ 2   = m  3 / 2

8P 3/ 2 t 9m

will

get

Ans.

1 mv 2 = t 2 2 ∴ (b) Pav =

v=

2 t m

W t2 = =t t t

Exercises LEVEL 1

F t m F2 P = F ⋅v = t m P ≠ constant P∝t v = at =

Assertion and Reason 1. F = constant F = constant m

we

3. (a) K E = W = ∫ Pdt = ∫ 2t ⋅ dt = t 2

5. F = (x − 4 )

a=

Ans.

= ma2t = (1) (4 )2 (4 )

∴

4. U = minimum = − 20 J

∴

9.5

(b) Pi = Fv = (ma) (at )

S

F

x=4m F=0

∴ But

Ans. Ans.

576 — Mechanics - I 3. In the figure, work done by conservative force (gravity force) is positive, potential energy is decreasing. But kinetic energy may increase, decrease or remain constant, depending on the value of F. F

External force

Single Correct Option 3. (W f )A = + ve (W f )B = − ve

4. In non-uniform circular motion (when speed ≠

W 7. P = = t =

S q2 is obtuse

q1 is acute

t

(800 × 10 × 10) + 60

Ans. h′ = 0.8 h = 8m

or

∴

1 × 1 × v2 + 2 2 v = 2 2 m/s

∴

10. Maximum range is obtained at 45°. E=

(W f 1 )on B = + f1 S ∴ Net work done by f1 = 0 B A

s

∴

...(i)

1 mv 2 − 0 2

1 mv 2 2 Decrease in mechanical energy in second case 1 = Ei − E f = mv 2 − mgh 2 Further, µ does not depend on angle of inclination.

u 2

1 1  u mv 2 = m   2 2  2 1 2  mu  2  E = = 2 2

2

E′ =

12. Decrease in mechanical energy in first case = Ei − E f =

1 mu2 2

At highest point, v = u cos 45° = s f1

Ans.

0+ 6=

11. (W f 1 )on A = − f1 S

f1

1 × 800 × (20)2 2

9. K i + U i = K f + U f Sr

N

1 mv 2 2

mgh +

8. E f = 80 % of Ei ⇒ mgh′ = (0.8) mgh

q2

S2

Ans.

= 4000 W

N q1

0

= 135 J

Therefore, speed of the particle is increasing. 9. W All = ∆K = K f − K i

S2

s

5

0

6. Pi = F ⋅ v

10.

B

f

F Þ

5. W = ∫ Fdx = ∫ (7 − 2x + 3x 2 ) dx

constant) work done by all the forces is not zero. 5. Slope of S -t graph is increasing.

1 = m (v 2f − vi2 ) 2 v f = vt2 = Slope of S -t graph at t2 vi = vt1 = Slope of S -t graph at t1 These two slopes are not necessarily equal.

s f

If there is no slip between A and B then f is static and total work done by static friction on system is zero.. 4. W = F ⋅ S = F ⋅ (rf − ri ) 5

mg

A B

A

11. W = Mgh + mg

=

h  m =  M +  gh 2  2 h 2

m

M

h

Ans.

Chapter 9

50 100 = N 3 3 100 W = FS = ×3 3 = 100 J

12. (a) Velocity is decreasing. Therefore, acceleration (or net force) is opposite to the direction of motion. (b) and (c): some other forces (other than friction) may also act which retard the motion. 13. Let retarding force is F 1 Then, ...(i) Fx = mv 2 2 1 ...(ii) and F (x′ ) = m (2v )2 2 Solving these two equations, we get Ans. x′ = 4 x

14. K i + U i = K f + U f 1 1 mv02 + 0 = mv 2 + mgh 2 2

∴ ∴

15. a =

v = v02 − 2gh

Ans.

F m

P = F ⋅v =

16. a =

2F 2 m

= 50 −

19. S = 0

Ans.

F m

W = FS = 0 W Pav = =0 t

∴

1 l l of shorter part is less, therefore value of k is more 1 W = kx 2 2 ∴ W shorter part will be more. 21. Let θ 1 is the angle of F with positive x-axis.

20. k ∝

tan θ 1 =

F t m  F 2 P = F ⋅v =   t  m or P∝t i.e. P -t graph is a straight line passing through origin. 1 1 17. K = mv 2 = m (gt 2 ) 2 2 i.e. K ∝ t 2 i.e. during downward motion. K -t graph is a parabola passing through origin with K increasing with time. Then in upward journey K will decrease with time. 18. Upthrust = (Volume immersed) (density of liquid) g  5  =  (1000) (10)  3000 50 N 3 weight = 50 N ∴ Applied force (upwards) = weight − upthrust =

Fy Fx

=

15 3 = = m1 20 4

α 3 W = 0 if m1 m2 = − 1  α −  = −1  3

(say)

Slope of given line, m2 = −

or v = at =

Ans.

(in vertical direction)

⇒

∴

 F v = at =   (2)  m

Work, Energy and Power — 577

∴

 3    4

∴

F ⊥S

α=4

Ans.

22. Decrease in gravitational potential energy of block = increase in spring potential energy 1 mg (xm sin θ ) = K xm2 ∴ 2 2 mg sin θ Ans. ∴ xm = K dS 23. v = = (4 t ) i$ dt P = F ⋅ v = 12 t 2 ∴

2

2

0

0

W = ∫ Pdt = ∫ (12) t 2dt = 24 J

Ans.

24. Decrease in potential energy = Work done against friction ∴ mg (h + d ) = F ⋅ d Here F = average resistance h  F = mg 1 +  ⇒  d

Ans.

578 — Mechanics - I 25. F = kx ⇒ Now,

F k 1 2 U = kx 2

30. Speed (and hence the kinetic energy) will increase

x=

as long as mg sinθ > kx.

31. (a) W N = NS cos 90° = 0  h  (b) WT = TS cos 0° = (mg sin θ )    sin θ 

2

1  F = k  2 k 1 or U ∝ k k B is double. Therefore, U B will be half. 1 1 26. mv 2 = kX m2 2 2 m ∴ Xm = ⋅v k 0.5 = × 1.5 50 = 0.15 m

= mgh (c) W mg = (mg ) (h) cos 180° = − mgh (d) WTotal = ∆K = K f − K i = 0 As block is moved slowly or K f = K i 32. Displacement of floor = 0

33. W = area under F -x graph

Ans.

27. F = resistance is same 1 mv 2 = n (F ⋅ d ) 2 n ∝ v2

⇒

If v is doubled, n will become four times. 28. Ei = E f 0=

1 × 10 × (0.15)2 − 0.1 × 10 × 0.15 2 +

∴

v = 0.866 m/s

1 × 0.1 × v 2 2 Ans.

29. From F = k x d

From X = 1 to X = 3, force and displacement both are positive. Therefore work done is positive. W 1 = + Area = + 20 J From X = 3 to X = 4, force is negative but displacement is positive. Therefore work done is again positive. ∴ W 2 = − Area = − 5 J From X = 4 to X = 5, force and displacement both are positive. Therefore work done is again positive. ∴ W 3 = + Area = + 5J WTotal = W 1 + W 2 + W 3 = 20 J dx 34. v = = t2 dt 1 1 W = mv 2 = mt 4 2 2 1 Ans. = × 2 × (2)4 = 16 J 2 35. At highest point, 1 mux2 = K 2 ux =

∴ 30°

F 100 = x 1.0 = 100 N/m Decrease in gravitational potential energy = increase in spring potential energy 1 ⇒ 10 × 10 × (d + 2) sin 30° = × 100 × (2)2 2 Solving, we get d = 2 m ∴ Total distance covered before coming momentarily to rest Ans. = d + 2= 4 m

2K m

R = 4H

k=

2uxuy g ∴ Now,

=

4 u2y 2g

uy = ux =

2K m

1 1 mu2 = m (ux2 + u2y ) 2 2 1  2K 2K  = m +  2  m m

Ki =

= 2K

Ans.

Chapter 9 4

Work, Energy and Power — 579

4. For increase in gravitational potential energy of a

36. ∆K = W = ∫ Pdt = ∫ (3t 2 − 2t + 1) dt

rod we see the centre of the rod.

2

= 46 J

Ans.

37. K f − K i = W = ∫ Fdx

60°

⇒

C

C

30

∴

K f = Ki +

∫ ( − 0.1x ) dx

20

30

 1 x2  = × 10 × (10)2 − 0.1  2 2  20  = 475 J

Ans.

38. KE = decrease in potential energy = mgh KE ∝ m (KE)1 12 2 or = = (KE)2 6 1 1 39. W = U f − U i = K ( X f2 − X i2 ) 2 1 = × 5 × 103 [(0.1)2 − (0.05)2 ] 2 = 18.75 J

W = change in potential energy l = mg (1 − cos θ ) 2 Substituting the values, we have  1.0 W = (0.5) (9.8)   (1 − cos 60° )  2 = 1225 . J

or

5. WT = (T ) (l ) cos β T

2. p = 2 Km or P ∝ K

∴ Work done on one mass =

−2K x02 = − K x02 2

90° + α W

W N = (N ) (l ) cos 90° = 0 WW = (W ) (l ) cos (90 + α ) = − W l sin α W f = (F ) (l ) cos 180° = − F l 6. Decrease in potential energy of chain = increase in kinetic energy h= l 2

Ans.

1 2

For small % changes, 1 % change in p = (% change in K ) 2 1 = (1 %) = 0.5 % 2 1 3. Total work done = − K (2 x0 )2 = − 2 K x02 2

l

F

Ans.

K′ =

∴

S=

α

p2 2m p′ 2 ( 1.5 p )2 = 2m 2m p2 = (2.25) = 2.25 K 2m K′ − K % increase = × 100 K = 125 %

β

N

Subjective Questions 1. K =

Ans.

∴

mg

or

l 1 = mv 2 2 2 v = gl

7. T − mg = ma ∴ Ans.

(a)

C

T = m (g + a) = 72 (9.8 + 0.98) = 776.16 N WT = TS cos 0° = (776.16) (15) = 11642 J

a

T

mg

Ans.

(b) W mg = (mg ) (S ) cos 180° = (72 × 9.8 × 15) (−1) = − 10584 J

Ans.

580 — Mechanics - I (c) K = WTotal = 11642 − 10584

11. FBD of particle w.r.t. sphere

= 1058 J (d) K = ∴

Ans.

N

F = ma

1 mv 2 2

mg

h

Pseudo force = ma

x

2K 2 × 1058 v= = m 72 = 5.42 m/s

Ans.

vr

R

θ

a

2

8. (a) K = W = ∫ Fdx = ∫ (2.5 − x 2 ) dx 0

= 2.33 J

Ans.

(b) Maximum kinetic energy of the block is at a point where force changes its direction. or F =0 at X = 2.5 = 1.58 m K max =

∫ (2.5 − X

2

) dx Ans.

9. W f = f S cos 45° Here, f = mg sin θ, because uniform velocity means net acceleration = 0 or net force = 0 ∴ W f = (mg sin θ ) (S ) (cos 45° ) = (1) (10) (sin 45° ) (S ) (cos 45° ) If S = vt = 2 × 1 = 2 m S f 45° 45°

 1  1 W f = (1) (10)   (2)    2  2 = 10 J

10. vm1 = vm2 = v dm1 = hm2 = 4 m Using the equation, Ei − E f =Work done against friction 1  0 −  × (10 + 5) (v 2 ) − 5 × 10 × 4  2  = 0.2 × 10 × 10 × 4 Solving we get, v = 4 m/s

1 mvr2 = W All 2

1 mvr2 = W N + W F + W mg 2 = 0 + Fx + mgh = (ma) (R sin θ ) + mg [ R (1 − cos θ )] vr = 2gR (1 + sin θ − cos θ )

Ans.

12. From constraint relations, we can see that

0

= 2.635 J

or

∴

1.58

∴

Kf =

vA = 2 v B Therefore, vA = 2(0.3) = 0.6 m /s as vB = 0.3 m / s (given) ApplyingW nc = ∆U + ∆K we get 1 1 − µ mAgS A = − mB gS B + mAvA2 + mB vB2 2 2 Here, S A = 2S B = 2 m as S B = 1m (given) 1 ∴ − µ(4.0) (10) (2) = − (1) (10) (1) + (4 ) (0.6)2 2 1 + (1) (0.3)2 2 or −80µ = − 10 + 0. 72 + 0. 045 or

80µ = 9. 235 or µ = 0.115

Ans.

13. Let xmax = maximum extension of spring Ans. (say)

Decrease in potential energy of A = Increase in elastic potential energy of spring 1 2 mA g x max = kx max ∴ 2 2mA g ∴ xmax = k To just lift the block B, k x max = mB g ∴ 2mA g = mB g = mg m mA = ∴ 2

kxmax

mBg

Ans.

Work, Energy and Power — 581

Chapter 9

(b) W f = f S cos 180°

14. (a) K A + U A = K B + U B 1 ∴ 0 + × 500 × (0.5 − 0.1)2 2 1 1 = × 10 × v 2 + × 500 × (0.3 − 0.1)2 2 2 On solving, we get Ans. v = 2.45 m/s (b) CO = ( BO )2 + ( BC )2 ≈ 36 cm = 0.36 m Again applying the equation, K A + U A = KC + UC 1 0 + × 500 × (0.5 − 0.1)2 2 1 1 = × 10 × v 2 + × 500 (0.36 − 0.1)2 2 2 On solving, we get v = 2.15 m/s

∴

g

µm

or Now, or ∴

16. (a)

Ans.

E =U + K = − 4 J

∴

K max = E − U min = 12 J

Ans.

° 37

Ans.

(c) K = 0 ∴ U =E or (x − 4 )2 − 16 = − 4 x = (4 ± 2 3 ) m

or

Ans.

dU = (8 − 2x ) dx (e) Fx = 0 at x = 4 m

(d) Fx = −

Ans. Ans.

19. Decrease in potential energy of 1 kg = increase in kinetic energy of both

1 kX m2 = MgX m 2 2 Mg Xm = k kX m = mg sin 37° + µmg cos 37°  3  3   4 2 Mg = (mg )   +   mg    5  4   5 3 M = m 5 W mg = (mg ) (S ) cos 30°

1 × (4 + 1) × v 2 2 v = 2 m/s

1 × 10 × 1 =

∴

= 34.6 J

Ans.

speed of A at this instant is 2.5 m/s, then speed of B at this instant will be 5 m/s. Now, Decrease in potential energy of A = increase in potential energy of B + increase in kinetic energy of both 1  300 2 ∴ (300) x = (50) (2x ) +   (2.5) 2  9.8  +

Ans.

1 2

 50  2   (5.0)  9.8

Solving we get, x = 0.796 m

30°

Ans.

20. If A descends X then B will ascend 2x. Further if

= (20) (2) ( 3 / 2)

S

Ans.

= − 4 + 16

⇒ ∴

Ans.

(b) U min = − 16 J at x = 4 m

kxm

in 3

= − 10 J dU A 17. F = − = 2 dr r

K =8J

decrease in potential energy of M = increase in elastic potential energy of spring

s mg

S

U = (6 − 4 )2 − 16 = − 12 J

15. Let X m is maximum extension of spring. Then

cos

180°

 1 = −   (20) (cos 60° ) (2)  2

18. (a) At x = 6 m,

= (30)2 + (20)2

7°

f

= (µmg cos θ ) (S ) (−1)

Ans.

21. If speed of sphere is v downwards then speed of mg

wedge at this instant will be v cot α in horizontal direction.

582 — Mechanics - I Now, Decrease in potential energy of sphere = Increase in kinetic energy of both 1 1 ∴ mgR = mv 2 + m (v cot α )2 2 2 1 2 = mv cosec2 α 2 ∴ v = 2gR sin α = speed of sphere and speed of wedge = v cot α = 2gR cos α

v = 2a2 S 2 = 15 . ∴

S2 =

1.5 1.5 = = 0.2 m 2a2 2 × 3.7

h = S 2 sin 30° = 0.1 m Now work done by friction = −[ initial mechanical energy] = − mgh = − (0.05) (10) (0.1) = − 0.05 J

Ans.

Ans.

25. If A moves 1 m down the plane and its speed is v,

22. If block drops 12 mm, then spring will further

then B will move 2m upwards and its speed will be 2v.

stretch by 24 mm. Now, Ei = E f 1 × 1050 × (0.075)2 = − 45 × 10 × 0.012 ∴ 2 1 1 + × 45 × v 2 + × 1050 × (0.099)2 2 2 Solving we get, v = 0.37 m/s Ans.

23. If speed of block of 1.0 kg is 0.3 m/s then speed of 4.0 kg block at this instant would be 0.6 m/s. Applying, Ei − E f = work done against friction 1 1 ∴ 0 −  × 1.0 × (0.3)2 + × 4.0 × (0.6)2 2 2  − 1 × 10 × 1] = µ k × 4 × 10 × (2) Solving this equation we get, Ans. µ k = 0.12

d θ

Using the equation, Ei − E f =Work done against friction 1 1 ∴ 0 −  × 30 × v 2 + × 5 × (2v )2 2 2 3 + 5 × 10 × 2 − 30 × 10 ×  5 4 = 0.2 × 30 × 10 × × 1 5 Solving this equation we get, Ans. v = 1.12 m/s

24. Retardation on horizontal surface, a1 = µg = 0.15 × 10 = 1.5 m/s2 Velocity just entering before horizontal surface, v = 2a1S 1 = 2 × 1.5 × 0.5 a2 h

30°

Acceleration on inclined plane, a2 = g sin θ − µg cos θ 1 3 − 0.15 × 10 × 2 2 = 3.7 m/s = 10 ×

Note

3  3 h = d sin θ = ( 1 )   m = m  5 5 mA =

wA w = 30 kg and mB = B = 5 kg g g

26. (a) Thermal energy = Work done against friction

= 1.5 m/s

S2

h

= µ K mgd = (0.25) (3.5) (9.8) (7.8) J = 66.88 J (b) Maximum kinetic energy = work done against friction = 66.88 J 1 (c) k xm2 = maximum kinetic energy 2 1 × 640 × xm2 = 66.88 2 xm = 0.457 m = 45.7 cm

Ans.

Ans.

Ans.

Chapter 9

Work, Energy and Power — 583 2  = mgR 1 −   π

LEVEL 2 Single Correct Option 1. W All = ∴ ∴

Now,

1 mv 2 2

1 mv 2 2 1 1 1  (5 × 5) +  × 10 × 5 + 0 = × × v 2 2  2 2 v = 14.14 m/s

2 1  mgR 1 −  = 0 + mv 2  2 π

or

2  v = 2gR 1 −   π

P v P = v

F=

∴

 dv  mv    ds 

∴

s

∫ ds =

∴

0

m P

2v

∫ v dv 2

v

7mv 3 Solving we get, s = 3P 3. F = kx

Ans.

F 100 = x 1 = 100 N/m Ei = E f 1 1 2 ∴ × 10 × v = × 100 × (2)2 − (10) (10) (2 sin 30° ) 2 2 Solving we get, k=

∴

v = 20 m/s

Ans.

 m   2m  d θ =   dθ  π   π /2

4. dm = 

h = R (1 − cos θ ) θ

dθ

dm h

dU i = (dm) gh = π/2

Ui =

∫ dU i = 0

2mgR (1 − cos θ ) dθ π

2mgR  π   −1   π 2

Ans.

5. T = 2mg Ans.

2. P = F v = constant

∴

∴

W F + W mg + W N =

∴

U i + Ki = U f + K f

As soon as string is cut T (on A ) suddenly becomes zero. Therefore a force of 2mg acting on upward direction on A suddenly becomes zero. So net force on it will become 2mg downwards. 2mg (downwards) a1 = = 2g ∴ m Spring force does not become instantly zero. So acceleration of B will not change abruptly. or a2 = 0 6. Let X m is maximum elongation of spring. Then, increase in potential energy of spring = decrease in potential energy of C. 1 KX m2 = M 1 gX m ∴ 2 or KX m = maximum spring force = 2M 1g = µ min Mg 2M 1 Ans. µ min = ∴ M 7. Ti = mg ...(i) 2kx = 2mg ∴ kx = mg One kx force (acting in upward direction) is suddenly removed. So, net downward force on system will be kx or mg. Therefore, net downward acceleration of system, mg g a= = 2m 2 Free body diagram of lower block gives the equation, mg mg − T f = ma = 2 mg ..(ii) ∴ Tf = 2 From these two equations, we get mg Ans. ∆T = 2 8. Fnet = mg sin θ − µmg cos θ = mg sin θ − 0.3 xmg cos θ

...(i)

584 — Mechanics - I At maximum speed Fnet = 0. Because after this net force will become negative and speed will decrease. From Eq. (i), Fnet = 0 at tan θ 3/ 4 Ans. x= = = 2.5 m 0.3 0.3 9. At C, potential energy is minimum. So, it is stable equilibrium position. Further, dU F =− = − (Slope of U -r graph) dr Negative force means attraction and positive force means repulsion. 10. P = F ⋅ v = (− mg $j) ⋅ [ u $i + (u − gt ) $j ] x

We can see that initial velocity is in the direction of PO. So the particle will cross the X -axis at origin. Ki + U i = K f + U f ∴ 0 + (3 × 6 + 4 × 8) = K f + (3 × 0 + 4 × 0)

direction, force is positive i.e. in the opposite direction of displacement. Similarly, when displaced in positive direction, force is negative.

15. Fnet = mg sin θ − µmg cos θ

y

2

1 2 1  x x  kx − K   = µmg  x +   2 2  2 2 4µmg x= ∴ k

∴ or

∴

12. P = F ⋅ v = Fv cos θ = TV cos θ

Ans. Ans.

 ∂φ ∂φ $  13. F = −  i$ + j ∂y   ∂x = (−3i$ − 4 $j)

0

xm

0

0

∫ vdv =

∫ (g sin θ − µ 0xg cos θ ) dx

Solving this equation we get, 2 xm = tan θ µ0 (k 1x ) AC = (k 2x ) BC AC k 2 = BC k 1

...(i)

AC + BC = l Solving these two equation we get,  k2  AC =   l  k1 + k2

P

...(ii) Ans.

17. Work done by friction = E f − Ei

53° 6m

X

^ -3i 53°

1 × 1 × (2)2 − 1 × 10 × 1 2 =−8J dU 12a 6b 18. F = − = 13 − 7 dx x x =

At equilibrium, F = 0 O

Ans.

16. ∑ (Moments about C ) = 0 ∴

8m

O

= mg sin θ − µ 0 xg cos θ F a = net = g sin θ − µ 0 xg cos θ m dv v⋅ = g sin θ − µ 0 xg cos θ dx

∴

Y 53°

Ans.

14. F = 0 at x = x2. When displaced from x2 in negative

= (− mguy ) + mg 2t i.e. P versus t graph is a straight line with negative intercept and positive slope. 11. Ei − E f = Work done against friction

K f = 50 J

or

^ -4j

Since, particle was initially at rest. So, it will move in the direction of force.

or

 2a x=   b

Ans.

1/ 6

d 2U is positive. dx 2 So, potential energy is minimum or equilibrium is stable.

At this value of x, we can see that

Chapter 9

Work, Energy and Power — 585

19. ∑ (Moment about O ) = 0

kx = mg

mg  l (kx ) l = mg   or x =  2 2k

∴

U =

1 2 (mg ) kx = 2 8k

2

Ans.

or

dθ

1 (m1 + m2 ) v 2 2

 m − m2  v = 2gh  1   m1 + m2 

h

Ans.

∴

dU dx dU = − Fdx = (ax − bx 2 ) dx

θ

h=0

m  dU = (dm) gh =  dθ gr sin θ π  π

U i = ∫ dU = 0

AssumingU = 0 at x = 0, and integrating the above equation we get, ax 2 bx 3 U = − 2 3 3a U = 0 at x = 0 and x = 2b 3a x 3 x2 and U will become For x > ,b >a 2b 3 2 negative. So, option (c) is the most appropriate answer. 22. W = FS

Now, ∴ ∴

2mgr π

Ki + U i = K f + U f 2mgr 1  πr  0+ = mv 2 − mg    2 2 π  2 π v = 2gr  +  π 2

Ans.

26. Maximum speed is at equilibrium where F = kx ⇒ x = Now, or

F and S are same. Therefore, WA 1 = WB 1

F ⋅x =

F k

1 1 mv 2 + kx 2 2 2

1  F  F 1 F   = mv 2 + k   k 2 2 k

2

Solving we get, v=

From work energy theorem, K A WA 1 = = K B WB 1 1 1 or mAvA2 = mB vB2 2 2 vA mB 2 ∴ = = vB mA 1

F = vmax mk

Ans.

27. Increase in potential energy per unit time = decrease in kinetic energy of both d 1 1  = −  m1v12 + m2v22  dt  2 2 dv  dv    = v1  − m1 1  + v2  − m2 2    dt  dt 

23. Work done by conservative force = − ∆U =Ui −U f = [ k (1 + 1)] − [ k (2 + 3)] = − 3k

mg k

h = R sin θ

dm

0 = m2gh − m1gh +

21. F = −

x=

and this does not depend on h. m 25. dm =   d θ  π

20. Ei = E f

∴

or,

Ans.

24. After falling on plank downward force on block is mg and upward force is kx. Kinetic energy will increase when mg > kx and it will decrease when kx > mg. Therefore it is maximum when,

= v1 (− m1 a1 ) + v2 (− m2 a2 ) dU or ...(i) = v1 (− F1 ) + v2 (− F2 ) dt Here, − F1 = − F2 = kx = 200 × 0.1 = 20 N Substituting in Eq. (i) we have, dU = (4 ) (20) + (6) (20) dt Ans. = 200 J/s

586 — Mechanics - I = (3$i + 4 $j) m/s

28. There is slip, so maximum friction will act on A in the direction of motion (or towards right)

| v| = (3)2 + (4 )2 = 5 m/s

S

2. F = −

A f

f = µ mg = 0.2 × 45 × 10 = 90 N S = 40 − 10 − 10 = 20 cm = 0.2 m W = f S = 18J

∴

Ans.

29. Constant velocity means net force = 0 Using Lami's theorem in the figure, N 37°

f

dU = 5 − 200x dx At origin, x = 0 ∴ F = 5N F 5 a= = = 50 m/s2 m 01 . Mean position is at F = 0 5 or at, x = = 0.025 m 200 F 5 − 200 x a= = = (50 − 2000 x ) m 0.1 At 0.05 m from the origin,

Ans.

53° 53° S = vt = 20 m

...(i)

x = + 0.05 m or x = − 0.05 m Substituting in Eq. (i), we have,

37° mg = 100 N

| a | = 150 m/s2

We have, N 100 = sin (180° − 53° ) sin 90° ∴ Now,

At 0.05 m from the mean position means,

N = 100 sin 53° = 80 N W N = NS cos 53° = (80) (20) (0.6) = 960 J

x = 0.075 or x = − 0.025 m Substituting in Eq. (i) we have,

spring would be kx.

∆F kx (in both case) ∴ a= = m m In one case it is downwards and in other case it is upwards.

More than One Correct Options  ∂U $ ∂U i+ ∂y  ∂X a=

| a | = 100 m/s2

Ans.

30. In both cases sudden changes in force by cutting the

1. F = − 

= 50 m/s2

or

$j = (−7$i − 24 $j) N  

3. Spring force is always towards mean position. If displacement is also towards mean position, F and S will be of same sign and work done will be positive. 4. Work done by conservative force = − ∆U Work done by all the forces = ∆K Work done by forces other than conservative forces = ∆E 5. At equilibrium m

Equilibrium position

F  7 $ 24 $ = − i − j m/s m  5 5  2

2

 7  24  | a | =   +   = 5 m/s2  5  5 Since, a = constant, we can apply, v = u + at 24 $  7 = (8.6i$ + 23.2$j) +  − $i − j (4 )  5 5 

Ans.

Ans.

3m

k δ 0 = mg mg or δ0 = k where, δ 0 = compression

Work, Energy and Power — 587

Chapter 9 3mg k  3mg  =k   = 3mg  k 

Solving these two equations, we get a = 0 and T = 2 mg

(b) δTotal = δ + δ 0 = Fmax = k δ max

(downward) kx = 2 mg

Nmax

T

a m

2m

T

a

2 mg

3m 3 mg

∴

8. Work done by conservative forces Fmax

N max = 3mg + Fmax = 6 mg

Ans.

4mg , then upper block will move a k 4 mg 3mg from natural distance x > − δ 0 or x > k k length. Hence in this case, extension 3mg x> k

(d) If δ >

or F = kx > 3mg (upwards on lower block) So lower block will bounce up. 6. (a) Decrease in potential energy of B = increase in spring potential energy 1 2mg xm = k xm2 ∴ 2 4 mg Ans. xm = ∴ k (b) Ei = E f 1 1  2mg  0 = (m + 2m) v 2 + × k ×    k  2 2

2

v = 2g

m 3k

kx − 2mg (c) a = m 2m  4 mg  k  − 2mg  k  = =g 2m (d) T − 2mg = ma 2mg − T = 2ma

Comprehension Based Questions 1. U = E − K = 25 − K K ≥0 U ≤ 25 J

Since, ∴

2. U = E − K = − 40 − K K ≥0 U ≤ − 40 J

Since ∴

Match the Columns 1. S = rf − ri = (+ 2 $i ) − (+ 4 $i ) = − 2$i

 2mg  − (2mg )    k  ∴

= U i − U f = − 20 + 10 = − 10 J Work done by all the forces, = K f − K i = 20 − 10 = 10 J 9. (a) Work done by gravity in motion 1 is zero (θ = 90° ) and in motion 2 is negative (θ = 180° ). (b) In both cases angle between N and S is acute. (c) and (d) : Depending on the value of acceleration in motion 1, friction may act up the plane or down the plane. Therefore angle between friction and displacement may be obtuse or acute. So, work done by friction may be negative or positive.

Now apply W = F ⋅ S

2. Friction force = 0, as tension will serve that purpose N

S T

(upwards) mg

...(i) ...(ii)

Angle between N and S or between T and S is acute ∴ WT and W N are positive. Angle between S and mg is 180°. Therefore, W mg is negative.

588 — Mechanics - I 3. (a) Net force is towards the mean position x = 0

1 2 1 10 at = × × (0.3)2 = 0.15 m 2 2 3

S =

where, F = 0when displaced from this position. Therefore, equilibrium is stable. Net force +q

+q

x=0

(a) W mg on 2 kg block = (20) (S ) cos 0° = (20) (0.15) = 3 J

 40 =   (0.15) (−1)  3 = −2J

Ans.

(d) WT on 1 kg block = (T ) (S ) cos 0°  40 =   (0.15) (1)  3

Q q

x=0

=+2J

6. (a) increasing. Therefore net potential energy should decrease. (b) From A to B, a part of decrease in gravitational potential energy goes in increasing the kinetic energy and rest goes in increasing the potential energy of spring. (c) and (d) : From B to C kinetic and gravitational potential energy are decreasing and spring potential energy is increasing. ∴ (decrease in kinetic energy) + (decrease in gravitational potential energy) = increase in spring potential energy. Net pulling force Total mass

T a

∴

T − 10 = ma = 1 × T =

40 N 3

10 3

W f = f S cos 180° = negative S=0

(b) f

wf = 0

(c) and (d) : No solution is required.

Subjective Questions

1. m1

F

µ

1 2 kxm 2 or kxm = 2(F − µ m1g ) Second block will shift if kxm ≥ µm2g ∴ 2(F − µm1g ) > µm2g m  or F >  m1 + 2  µg  2 (F − µm1g )xm =

20 − 10 10 = = m/s2 1+ 2 3 1 kg

S

m2

5. Common acceleration of both blocks

Ans.

f

4. (a) From A to B speed (or kinetic energy) will be

a=

Ans.

(c) WT on 2 kg block = (T ) (S ) cos 180°

F1

q

Ans.

= (10) (0.15) (−1) = −1.5J

+q

(c) Same logic can be applied as was applied in part (b). (d) Net force is neither towards x = 0 nor away from x = 0. Therefore, equilibrium is none of the three. F2

20 N

(b) W mg on 1 kg block = (10) (S ) cos 180°

x=0

y

2 kg

10 N

S

=

1 kg

S

Net force

x

T

+Q S

(b) Net force is away from the mean position. Therefore, equilibrium is unstable.

+q

T S

10 N

Ans.

Work, Power and Energy — 589

Chapter 9 2.

x A dθ θ

θ

h=0

h Initial

Final

B

m (x + h) l Net pulling force ∴ Acceleration a = Total mass being pulled Total mass being pulled =

Initial PE, Ui = ∫

θ = π/2 θ = 0°

(r dθ ) (ρ ) (g ) (r cos θ ) π/2

= (ρg r2 ) [sin θ ] 0

= ρgr2

=

Final PE, π 2 r2ρg  πr   πr / 2  Uf =  × ρ (g )  − =−  2   8 2 

∴

 π 2  ∆U = r2ρ g 1 + 8 

or

⇒

 1 π v = 4 rg  +  π 8

or

 π 4 v = rg  +   2 π

∴

and

 20  v=  t  0.3

P = Fv = (53.3 t ) kW

Ans.

0.2 0

(53.3 t ) dt +

= 1.69 kJ

v2 l−h = gh [ln(x + h)] 0 2

or

v2  l = gh ln    h 2

0.3

∫0.2

v = 2gh ln(l / h)

5. (a) From energy conservation principle, Work done against friction = decrease in elastic PE 1 or f (x0 + a1 ) = k (x02 − a12 ) 2 2f or …(i) x0 − a1 = k

(160 t − 533 t 2 ) dt Ans.

4. At the instant shown in figure, net pulling force =

m gh l

Ans.

From Eq. (i), we see that decrease of amplitude 2f , which is constant and same for each (x0 − a1 ) is k cycle of oscillation.

P = Fv = (160 t − 533 t 2 ) kW W =∫

dx

∴

∴ Ans.

For t > 0.2 s 20  800 F = 800 −  t  (t − 0.2) and v =  0.1  0.3 ∴

0

∫ v ⋅ dv = − gh ∫(l − h) x + h 0

3. For t ≤ 0.2 s F = 800 N

gh  dv  v −  =  dx  x + h v

∆U = KE  π 2  1  πr   =   (ρ ) v 2 r2ρg 1 + 8 2  2 

or

gh x+h

x0

a1

590 — Mechanics - I (b) The block will come to rest when ka = f a=

or

k f

…(A)

In the similar manner, we can write 2f a1 − a2 = k 2f a2 − a3 = k …… 2f an − 1 − an = k Adding Eqs. (i), (ii),… etc., we get 2f x0 − an = n    k 

n=

or

2f k Number of cycles, m=

k f

=

 m g  m g 2m1 gH = k  2  + 2m1g  2   k   k 

…(ii) or …(n)

Thus,

H =

m2g  m2 + 2m1    k  2m1 

H min =

m2g  m2 + 2m1    k  2m1  2

…(B)

∴ or

1 1  v 1 mv 2 = m  + kx 2 2 2  2 2 k=

kx0 1 − 2f 2

Ans.

kxm

EA = EB 1 1 m1v 2 = kx 2 + m1 gx 2 2

mc g

…(i)

v=0 m1 B

x

m1

m2

ac

C

2m1 gH = kx 2 + 2m1 gx

A

Ans.

(mB + mC )g = (1 + 2) × 10 = 30 N Since (mB + mC )g > µmAg , aA = aB = 0. From conservation of energy principle we can prove that maximum distance moved by C or maximum extension in the spring would be 2m g 2 × 1 × 10 xm = C = k 1000 = 0.02 m

6. Conservation of mechanical energy gives,

or

3v 2m 4 x2

8. µmAg = 0.8 × 6 × 10 = 48 N

n kx0 1 1  kx0  = − =  − 1 2 4f 4 4  f 

or

Ans.

7. Ei = E f

Equating Eq. (A) and (B), we get k 2f = x0 − n    k  f x0 −

2

…(i)

2f an = x0 − n    k 

or

The lower block will rebounce when mg (kx = m2g) x> 2 k mg Substituting, x = 2 in Eq. (i), we get k

At maximum extension kx − mC g aC = m mC Substituting the values we have, aC = 10 m/s2.

Ans.

9. Rate at which kinetic plus gravitational potential

v = Ö2gH

m2

energy is dissipated at time t is actually the magnitude of power of frictional force at time t. | Pf | = f . v = (µmg cos α )(at ) = (µmg cos α )[(g sin α − µg cos α )t ] = µmg 2 cos α (sin α − µ cos α )t

Ans.

Work, Power and Energy — 591

Chapter 9 10. From work-energy principle, W = ∆KE 1 m (v 2 − u2 ) 2 m 2 t= (v − u2 ) 2P

∴

Pt =

or

(c)

(P = power)

1 mgR 2   l  l  mv 2 = sin  R  + sin θ − sin θ + R   2 l  

…(i) v2 =

Further F ⋅v = P dv 2 m⋅ ⋅ v = P ds v 2 P x ∫u v dv = m ∫0 ds 3P (v 3 − u3 ) = ⋅x m m 3x = P v 3 − u3

∴ or ∴ or

Substituting in Eq. (i) 3x (u + v ) t= 2 (u2 + v 2 + uv ) m (a) Mass per unit length = l

∴ 2v ⋅

Hence proved.

α

l   l  sin  R  + sin θ − sin θ + R    

2gR 2 l

dv 2gR 2 = dt l

m R dα l h = R cos α

l  R   Ans.

1 1 m1v12 + m2v22 2 2 Here, v1 = v2 cos θ 1 ∴ 2 × 10 × 1 = (0.5) (10) ( 5 − 1) + × 0.5 2 m2 gh2 = m1 gh1 +

2

1  2 × v22 ×   + × 2 × v22  5 2

mgR 2 cos α ⋅ dα l l/ R mgR 2  l sin   U = ∫ dU = 0  R l

dU = (dm) gh =

(b) KE = U i − U f Ui =

…(i)

12. From conservation of energy,

dm =

Here,

 dθ     dt 

t = 0, θ = 0° dv gR   l = 1 − cos     R  dt l 

Hence,

h= 0

∴

l   dθ   cos θ − cos θ + R   ⋅ dt  

2gR 2 l

At

αh

l   l  sin  R  + sin θ − sin θ + R    

l   cos θ − cos θ + R     2v  dθ     dt  ω 1 Here = = v v R Substituting in Eq. (i), we get dv gR   = cos θ − cos θ +  dt l  dv = dt

dm dα

11.

2gR 2 l

v=

or

∴

20 = 6.18 + 0.2 v22 + v22 2m θ

1m 2

mgR  l sin    R l

5m

v2 θ

m2 m1

v1

and Uf = ∫

l/ R + θ θ

∴ KE =

dU =

mgR 2 l

mgR 2   l   sin  R + θ − sin θ  l  

l   l  sin  R  + sin θ − sin θ + R     Ans.

v2 = 3.39 m/s and

2 v1 = v2 cos θ = × 3.39 5

or

v1 = 3.03 m/s

Ans.

Ans.

592 — Mechanics - I 13. Net retarding force = kx + bMgx

Substituting the values we have,

 k + bMg  ∴ Net retardation =   ⋅x   M

v f = 2 × 9.8 × 19 . + = 8.72 m/s

So, we can write v⋅

 k + bMg    M

∫ v ⋅ dv = − 

v0

x=

or

Work done by all forces = Change in kinetic energy 1 1 ∴ Fx − kx 2 = mv 2 2 2

x

∫ x dx 0

∴

M v0 k + bMg

v=

or

∆E =

 2 1 k  M Mv02 −   v0 2 2  k + bMg 

or

∆E =

v02 2

or ∴

Ei = E f 1 2 1 kxi + mghi = mv 2f 2 2 v f = 2ghi +

Ans.

(b) From conservation of mechanical energy, Ei = E f 1 2 1 2 1 2 or mvi + kxi = kx f 2 2 2 Ans.

14. From conservation of mechanical energy,

2 × 20 × 0.25 − 40 × 0.25 × 0.25 0.5

= 15 m/s = 3.87 m/s

   M  M − k   k + bMg   

v02bM 2g 2(k + bMg )

2Fx − kx 2 m

v=

Substituting the values we have,

Loss in mechanical energy 1 1 ∆E = Mv02 − kx 2 2 2

=

Ans.

15. (a) From work energy theorem,

dv  k + bMg  =−  ⋅x   dx M

0

or

2300 (0.045)2 012 .

or

xf = =

mvi2 + xi2 k 0.5 × 15 + (0.25)2 40

= 0.5 m (compression) k 2 xi m

∴ Distance of block from the wall = (0.6 − 0.5) m = 01 . m

Ans.

10. Circular Motion INTRODUCTORY EXERCISE

10.1

1. Direction of acceleration (acting towards centre) continuously keeps on changing. So, it is variable acceleration. 2. In uniform circular motion speed remains constant. In projectile motion (which is a curved path) acceleration remains constant. v 2 (2t )2 3. (a) ar = = = 4 t2 R 1.0 At t = 1 s, ar = 4 cm/s2 Ans. dv Ans. (b) at = = 2 cm/s2 dt

(b) ar =

v2 R

= 7.35 m/s

Ans.

s 2R = = t (T / 4 )

T/4

4 2R (2πR )/ v

R

INTRODUCTORY EXERCISE

1. v =

dv 6. at = = 8t dt v 2 (4 t 2 )2 8 4 ar = = = t R 54 27 At t = 3 s, at = 24 m/s

µ=

v2 (5)2 = gR 10 × 10 = 0.25

≈ 17 m/s

Ans.

at

a

Ans.

4. a ≠ 0 ⇒ Fnet = ma ≠ 0 Hence, particle is not in equilibrium.

5. If he applies the breaks, then to stop the car in a distance r Wall

Ans. v

θ

r

ar v

2

ar = 24 m/s tan θ =

Ans.

3. v = Rg tan θ = 50 × 10 × tan 30°

(αt )2 = α 1 1 t= = α 4 = 0.5 s

∴

v2 (5)2 1 = = Rg 10 × 10 4

 1 θ = tan −1    4

∴

O

ω2 = α

or

10.2

18 × 5 = 5 m/s 18

∴

S

Ans.

∴

Ans.

2. v = µRg

R

v 2 2 ∴ av = v π 5. (Rω 2 ) = R α

Ans.

 1 (c) at = a sin 30° = (25)   = 12.5 m/s2  2

tan θ =

= 2 5 cm/s2

=

v = (21.65) (2.5)

or

(c) a = ar2 + at2 = (4 )2 + (2)2

4. vav

⇒ v = ar R

0 = v − 2a1 r v2 ∴ a1 = = minimum retardation required. 2r (by friction) If he takes a turn of radius r, the centripetal acceleration required is v2 (provided again by friction) a2 = r Since a1 < a2, it is better to apply brakes. 2

2

at =1 ar

∴ θ = 45°

Ans. 3 2 = 21.65 m/s2

7. (a) ar = a cos 30° = (25)

Ans.

594 — Mechanics - I 6.

mv 2 R  v2 N 2 = m g +  R 

(c) N 2 − mg =

T1

∴

θ 5

θ

4

=

θ T2

3

...(i) INTRODUCTORY EXERCISE

ω = (2nπ ) Here, n = number of revolutions per second. Substituting the proper values in Eq. (i),  3  3 200 ×   + T2 ×   = (4 )(3)(2nπ )2  5  5 or

600 + 3T2 = 240 n2π 2

T1 sin θ = T2 sin θ + mg 4 4 or 200 × = T2 × + 4 × 10 5 5 or 800 = 4T2 + 200 Solving Eqs. (ii) and (iii) we get, T2 = 150 N and n = 0.66 rps = 39.6 rpm mv 2 7. (a) mg − N 1 = R

2500   10 + 250 

= 32 × 103 N = 32 kN

mg

T1 cos θ + T2 cos θ = mrω 2

16 × 103 10

Ans.

10.3

1. (a) String will slack at height h1, discussed in article 10.5 where, u2 + gR 5 h1 = = R 3g 3 ...(ii)

Further,

v = u2 − 2gh1 = 4 gR − 2g × =

...(iii)

N1 N2 mg mg

mg mv 2 or mg − = 2 R mv 2 mg 16 kN ∴ = = = 8 kN R 2 2 mv 2 mv 2 Now, N 2 − mg = or N 2 = mg + R R mg 3 = mg + = mg 2 2 3 Ans. = (16 kN) = 24 kN 2 mv 2 mv 2 (b) mg − N = or N = mg − R R 2 mvmax 0 = mg − R Ans. ∴ vmax = gR = 10 × 250 = 50 m /s

(as u2 = 4 gR ) 5 R 3

2 gR 3

2. (a) Velocity becomes zero at height h2 discussed in article 10.5 where, h2 =

u2 gR R = = 2g 2g 2

h2 = R (1 − cos θ ) R = R (1 − cos θ ) ⇒ θ = 60° ∴ 2 mg (b) T = mg cos θ = at θ = 60° 2 Now,

3. (a) v = u2 − 2gh = 7gR − 2g (2R ) = 3gR mv 2 m = (mgR ) R R ∴ T = 2 mg mu2 m (c) T − mg = = (7gR ) R R ∴ T = 8 mg l 4. h = l − l cos 60° = = 2.5 m 2 (b) T + mg =

60°

u=0

h v0

v0 = 2gh = 2 × 9.8 × 2.5 = 7 m/s

Ans.

Exercises LEVEL 1 Assertion and Reason 2. A to B

S = 2R

and

| ∆v| = v + v 2 − 2v ⋅ v cos 90° 2

= 2v | ∆v| 2v = t t S 2R and average velocity = = t t v The desired ratio is = ω R 3. Direction of acceleration keeps on changing. So, it is variable accelration. Further, it is accelerated so it is non-inertial. Average acceleration =

4.

v2 R is zero. But tangential acceleration g sin θ is non-zero. 8. For t < 3 sec, speed is negative which is not possible., 9. In circular motion, a ≠ constant, so we cannot apply v = u + at directly. In vertical circular motion, gravity plays an important role.

7. At A and C, v = 0. Therefore, radial acceleration

v

10.

N a

mg θ

v a

a

v = constant

⇒ v is increasing

v a

θ

N cos θ = mg N = mg sec θ

mv 2 R 11. Weight (plus normal reaction) provides the necessary centripetal force or weight is used in providing the centripetal force. and

N sin θ =

Single Correct Option 1. Magnitude of tangential acceleration is constant but

v is decreasing

In first figure, v⋅ a = 0 In second figure, v ⋅ a = positive and In third figure, v ⋅ a = negative Further, ω and v are always perpendicular, so ω⋅ v=0 5. Component of acceleration perpendicular to velocity is centripetal acceleration. v2 v 2 (2)2 or R = ∴ ac = = =2m R ac 2

6. At A

At θ = 360° , | ∆P | = 0 = minimum

3. h = l − l cos θ = l (1 − cos θ ) l

θ

h v

Tangential acceleration is g. Radial acceleration is v2 . R ∴

its direction keeps on changing.

2. At θ = 180° , | ∆P | = 2 mv = maximum

 v2 anet = g 2 +    R

2

2 v 2 = 2gh = 2gl (1 − cos θ ) = vmax 1 Ans. ∴ K max = mv 2 = mgl (1 − cos θ ) 2 4. Rod does not slack (like string). So, minimum velocity at topmost point may be zero also.

596 — Mechanics - I 5.

T

l

θ

T

u mg

r

2 mu2 m ( 5gR ) = R R T = 6 mg

mg

T − mg = ∴ 2

6.

v = at = a R (at )2 or =a R t=

∴

Ans.

(Here, at = a say)

R = a

20 =2s 5

Ans.

7. cos (dθ ) components of T are cancelled and sin (dθ ) components towards centre provide the necessary centripetal force to small portion PQ.

T

dθ P

dθ Q

Solving these two equations we get, g g cos θ = 2 = lω l (2π n)2 10 = 2 2  π 2 ×  π  θ = cos−1 (5/ 8) mg 0.1 × 10 8 10. T = = = N cos θ 5/ 8 5 ∴

T

= m (R α )2 + [ R (α t )2 ] 2 = 0.36 × 10

2 2  1  1    0.25 ×  + 0.25  × 2   3   3 

2

= 50 × 10−6 N

2T sin (dθ ) = (mPQ ) (R ) ω 2

= 50 µ N

For small angle, sin d θ ≈ d θ  m 2T dθ =   (2θ ) (R ) (2 n π )2 ∴  2π 

Ans.

12. v = 2gh 2

T

T = 2π mn2R

Substituting the values we get,

v

T = (2π ) (2π ) (300/ 60)2 (0.25) ≈ 250 N

Ans. mg

2

8.

Ans.

= m (Rα )2 + (Rω 2 )2

−3

∴

Ans.

11. f = ma = m at2 + ar2

dθ dθ

∴

(l = 1 m )

mv = µ mg R ∴ v = µ Rg = 0.3 × 300 × 10 = 30 m/s = 108 km/h

T − mg = Ans.

9. T cos θ = mg

...(i)

T sin θ = m rω = m (l sin θ ) ω 2

2

...(ii)

or

mv 2 l

m (2gh) l 2h  = mg 1 +   l

T = mg +

Ans.

Circular Motion — 597

Chapter 10 13. (ω 1 − ω 2 ) t = 2π

Taking moment about C

2π 2π t= = ω 1 − ω 2 (2π /T1 ) − (2π /T2 ) T1 T2 3×1 = T2 − T1 3 − 1 = 1.5 min

14.

N o (x ) = f (h) or Ans.

dB vB t 2.5 5 = = = dC vC t 2 4

Ans.

Subjective Questions 1. at = 8 m/s2 ar =

 mv 2  (mg ) x =   h  R 

∴

gRx h

v=

9.8 × 250 × 0.75 15 . = 35 m/s 5. Let ω be the angular speed of rotation of the bowl.Two forces are acting on the ball. =

ω

v 2 (16)2 = = 5.12 m/s2 R 50 R

a = at2 + ar2 = 9.5 m/s2

2.

Ans.

α

N

µN

A

r mg

N

mg

N = mRω

2

...(i)

µN = mg From these two equations we get, g ω= µR = v2 3. a = R

1. normal reaction N 2. weight mg The ball is rotating in a circle of radius r (= R sin α ) with centre at A at an angular speed ω. Thus, N sin α = mrω 2 = mRω 2 sin α

...(ii)

9.8 = 4.7 rad /s 0.15 × 3

Ans.

and N cos α = mg Dividing Eq. (i) by Eq. (ii), we get 1 ω 2R = cos α g ∴

v 2 (u cos θ )2 ⇒ R= = a g

Ans.

6. R = L sin θ

f = mv 2/R

θ

No

Ni = 0

In critical case, normal reaction on inner wheel N i will become zero. Normal reaction on outer wheel N o = mg. Friction will provide the necessary centripetal force. mv 2 R

T

mg

x

f =

L

h

R

mg

∴

…(ii)

g R cos α

ω=

4. C

…(i)

T sin θ = mRω 2 = m (L sin θ ) ω 2 ∴ Now, ∴

T = mLω 2 T cos θ = mg mg mg g cos θ = = = T mLω 2 Lω 2 Hence proved.

598 — Mechanics - I 7. t =

2h = g

2 × 2.9 = 0.77 s 9.8

Horizontal distance x = vt x 10 v= = ∴ t 0.77 ≈ 13 m/s v 2 (13)2 a= = R 1.5 ≈ 113 m/s2

At lowermost point, θ = 0° g ω= ∴ R Substituting ω = 2g / R in Eqs. (i), we have,

Ans.

centripetal force. ∴ mLω 2 = µmg

∴

10. (a)

Fnet = m at2 + an2 µmg = m (Lα )2 + (Lω 2 )2

…(ii)

at = Lα and an = Lω 2

Substituting ω = αt in Eq. (ii) we have,

 µ 2g 2 − L2α 2  t=  L2α 4  

1 kg T = 64 N f = 48 N

Fc = m1 r1 ω = (1) (1) (4 )2 = 16 N 2

But available tension is 64 N. So, the extra force of 48 N is balanced by friction acting radially outward from the centre. (b) fmax = µm1g = 0.8 × 1 × 10 = 8 N 2 kg

1 4

T (m2 r2 ω2)

Substituting the value of t in Eq. (i) we have, 1

 µg  2 4 ω =   − α 2   L  

T

1 kg

T = m2 r2 ω 2 = (2) (2) ω 2 = 4 ω 2

...(i)

T − 8 = m1 r1 ω 2 = (1) (1) (ω 2 ) = ω 2

...(ii)

(c) T = m2 r2 ω

ω

8N

(m1 r1 ω2)

Solving Eqs. (i) and (ii) we get, ω = 1.63 rad /s

9. N cos θ = mg

θ

T

Centripetal force required to 1 kg block is

µg = L2α 2 + L2α 4t 4 ∴

2 kg

Ans.

T = m2 r2ω 2 = (2) (2) (4 )2 = 64 N

µg or ω= L (b) Net force of circular motion will be provided by the friction …(i) ω = αt

Here,

1 2 θ = 60°

cos θ =

8. (a) The frictional force provides the necessary

∴

g R cos θ

ω=

∴

Ans.

2

2 kg

T m 2 r 2 ω2

R N

∴ 100 = (2) (2) ω 2 or ω = 5 rad /s 11. v = v02 + 2gh = (0.5 gr )2 + 2gr (1 − cos θ ) 2

mg r

N sin θ = mrω 2 = m (R sin θ ) ω 2 Dividing these two equations, g we get cos θ = R ω2

...(i)

= (2.25 gr − 2gr cos θ ) At the time of leaving contact, N = 0 mv 2 = 2.25 mg − 2 mg cos θ ∴ mg cos θ = r 2.25 3 ∴ cos θ = = 3 4 ∴ θ = cos−1 (3/ 4 )

Ans.

Chapter 10

Circular Motion — 599

LEVEL 2

h

Single Correct Option 1. EA = EB

v

θ mg

θ

∴

1 1 mvA2 = × 200 × (13 − 7)2 2 2

∴

mv 2 12. 2.5 mg − mg cos 30° = r

vA = 60 m/s

(2) (60)2 Ans. = 1440 N R S 2. Let us see the FBD with respect to rotating cone (non-inertial) At A,

θ

N =

mvA2

=

N T = 2.5 mg

mg

150° 30°

mg

∴

F = mRω2 = centrifugal force

R

v

∴

(m = 2 kg)

v2 v2 = r 2 v = 5.66 m/s

1.63 g =

Fnet = (2.5 mg )2 + (mg )2 + (2) (2.5 mg ) (mg ) cos 150° = 1.7 mg F Ans. anet = net = 1.7 g ≈ 16.75 m/s2 m

13. v 2 = v02 + 2 gh = v02 + 2 gR sin θ θ h

2 mg

v 2 2gh 2gR (1 − cos θ ) = = R R R Given, aA = aB ∴ sin θ = 2 − 2 cos θ Squaring these two equations we have, sin 2 θ = 4 + 4 cos2 θ − 8 cos θ aB =

1 − cos2 θ = 4 + 4 cos2 θ − 8 cos θ

On solving this equation, we get 3 or θ = cos−1 (3/ 5) cos θ = 5 5. At the time of leaving contact

mg

= (5) + 2 × 10 × 2 sin θ 2

= (25 + 40 sin θ ) mv 2 Now, 2 mg − mg sin θ = R v2 or 2g − g sin θ = R 25 + 40 sin θ or 2 × 10 − 10 sin θ = 2 Solving this equation we get,  1 θ = sin −1    4

4. aA = g sin θ

or

v

90° – θ

N , F and mg are balanced in the shown diagram. If displaced upwards, F will increase as R is increased. This will have a component up the plane. So, it will move upwards. Hence, the equilibrium is unstable. 3. µmg = mRω 2 µg (1/ 3) g 4 g Ans. ∴ ω2 = = = R 5a/ 4 15 a

Ans.

v θ

θ

mg

N =0 Ans.

∴

mg cos θ =

mv 2 m (2gh) = R R

600 — Mechanics - I  R 2  + R (1 − cos θ ) 2h 4   ∴ cos θ = = R R On solving this equation, we get  5 Ans. cos θ = 5/ 6 or θ = cos−1    6

9. v =

2πr (2π ) (0.5) = T 1.58 F1

F2

mg

6. hAB = R cos 37° − R cos 53° vB 37°

B

≈ 2 m/s F1 = mg = 100 N mv 2 10 × 4 F2 = = = 80 N r 0.5 ∴ Net force by rod on ball

37° g

a^ = radial acceleration

= 0.8 R − 0.6 R = 0.2 R vB = 2ghAB = 0.4 gR

∴

= F12 + F22 = 128 N

10. t = T

v 2 0.4 gR a⊥ = g cos 37° = (0.8g ) = B = r r ∴ r = radius of curvature at B R = 2 7. At the time of leaving contact at P,

3a/4 θ

P θ

v

mg

∴ ∴ ∴

∴

2H 2π = g ω

∴

ω = 2π

N =0 mv 2 m (u2 + 2gh) mg cos θ = = a a u2 + 2g (a/ 4 ) g (3a/ 4 ) = a ag u= 2

g =π 2H

2g H

Ans.

11. ω 1t − ω 2t = 2π ∴

u a/4

Ans.

2π ω1 − ω2 2π = (2π /T1 ) − (2π /T2 ) TT = 1 2 T2 − T1 (3600) (60) = (3600) − (60) 3600 = s 59

t=

12. Particle breaks off the sphere at cos θ = θ

Ans.

Ans. 2 3

θ g

2

8.

v 4 = 2 r r ∴

The tangential acceleration at this instant is

2 v= r P = mv =

g sin θ = g 1 − cos2 θ 2m r

Ans.

= g 1−

4 5 = g 9 3

Ans.

Circular Motion — 601

Chapter 10

∴ [cos θ 2gR cos θ ]

13. u = 5 gR

+ v ar at

h=R

∴

4 gR cos2 θ − 2gR sin 2 θ = 0

or

2 cos2 θ − (1 − cos2 θ ) = 0 1 cos θ = 3

or u

= 3gR v2 ar = = 3g R at = g

14. N = mRω 2

dv v 2 = dt R

∴ v

−2 ∫ v dv =

∴

v0

Ans.

or N

or mg

∴ or

h

°– θ 90

v0

Vertical component of velocity is, v = v0 sin θ = sin θ 2gR cos θ For v to be maximum dv =0 dθ

0

x

t

0

0

Rv0

v t  = − R ln 1 − 0   R ∴

θ

∫ dt

 1 t x = Rv0  −  [ ln (R − v0t )] 0  v0 

Ans.

v0 = 2gh = 2gR cos θ

R

t

∫ dx = ∫ R − v0t dt

15. h = R cos θ

θ

1 R

(v = speed )

 1 1 t v − v = R  0  1 1 t R − v0t = − = v v0 R Rv0 Rv0 v= R − v0t dx Rv0 (x = distance travelled) = dt R − v0t

µN

µN = mg From these two equations. we get g µ= Rω 2 10 = = 0.2 2 × (5)2

Ans.

16. at = ar

a = ar2 + at2 = g 10

 1 θ = cos−1    3

∴

v 2 = u2 − 2gh = (5gR ) − 2gR

∴

sin θ (−2gR sin θ ) = 0 2 2gR cos θ

or

1−

v0t = e− x/ R R R t=t= (1 − e−x/ R ) v0

Putting x = 2πR, we get R t= (1 − e−2π ) v0

Ans.

17. If u > 5gl , then T and a are in same direction. Hence, T ⋅ a is positive. If u = 5gl , then T = 0 ∴ T⋅a =0 18. At lowest point T and a are always in same direction (towards centre). ∴ T ⋅ a is always positive.

602 — Mechanics - I = (h cot α ) (g tan α )

More than One Correct Options

= gh

1. Radial acceleration is given by, C

N

α

R B

D

α mg

h

θ

α

P A

v2 R At A, speed is maximum. Therefore, ar is maximum. At C, speed is minimum. Therefore, ar is minimum. Tangential acceleration is g sin θ. At point B, θ = 90°. Therefore, tangential acceleration is maximum (= g ). mu2 2. T + mg = l ar =

u mg

T h = 2l

Now,

2πR 2π h cot α = v gh

T =

h cot α g

= 2π ∴

T ∝ h T ∝ cot α

and 5. N cos θ = mg N sin θ = mRω

...(i) 2

...(ii)

From Eq. (i), mg = constant cosθ as θ = constant Net force is the resultant of N and mg and both forces are constant. Hence, net force is constant. N =

ω

L

N

v

∴ or

2 mg + mg =

mu l

θ R

2

u = 3gl

θ

= 7 gl

N cos α = mg mv 2 R Solving these two equations, we get N sin α =

v = Rg tan α

g tan θ R h R= tan θ

ω=

v = 7gl

3. R = h cot α

mg

h

v 2 = u2 + 2 gh = 3 gl + 2g (2l ) ∴

θ

But ∴ or

g tan θ h 1 ω∝ h

ω=

(θ = constant)

Chapter 10 Comprehension Based Questions

(b) ar =

1. ∆U = ∆U rod + ∆U ball  l = Mg   + mgl  2 M  = + m gl  2 

Ans.

v l Now, decrease in rotational kinetic energy = increase in potential energy 1 M  ∴ I ω2 =  + m gl  2  2

∴

v2 v2 or ar R l (10)2 = = 100 m/s2 1 at = g = 10 m/s2 = 100.49 m/s2

mv 2 (1) (10)2 = = 100 N l 1 (d) at = g = 10 m/s2

(c) T =

2. ar =

  v 2  M 1  Ml 2  + m gl + ml 2    =       2 3 2 l 

∴

v=

at

a = ar2 + at2

2. ω =

or

Circular Motion — 603

v 2 (2t )2 = = 2t 2 R 2 at 45°

M  + m gl   2  M m   +    6 2

ar

v = 2 m/s a = Öar2 +at2 = 2Ö2 m/s2

3. Maximum velocity is at bottommost point and minimum velocity is at topmost point.

dv = 2 m/s2 dt v = 2t v 2t ω= = =t R 2

at =

2 umin + 2g (2L) 2 = umin 1

On solving, we get umin = 2

4. umax = 2 umin ∴

K max

gL 3

Ans.

gL =4 3 1 8 mgL 2 = m umax = 2 3

At 1 s, ar = 2 m/s2 , at = 2 m/s2 ,

Ans.

2 5. v = u max − 2g (L)

v = 2m/s and ω = 1 rad /s (a) a ⋅ v = av cos 45° = (2 2 ) (2) (1 2 ) = 4 m 2 /s3 (b) | a × ω | = aω sin 90° = (2 2 ) (1) (1) = 2 2 m/s3 (c) v ⋅ ω = 0 as θ = 90°

h=L umax

(d) | v × a | = va sin 45° = 4 m 2 /s3

v

16 gL 10 gL = − 2gL = 3 3

Ans.

mv 2 =F R

∴

Match the Columns 1. (a) v = u2 − 2gh = 12gl − 2gl = 10gl = 10 × 10 × 1 = 10 m/s

3. f =

F =1 f

With increase in speed f will increase but Ans.

remain same.

F will f

604 — Mechanics - I (c) | a av | = N1

C

N2

=

f = mv 2/R mg

Further, with increase in the value of v, friction f will increase. Therefore, anticlockwise moment of f about C will also increase. Hence, clockwise moment of N 2 should also increase. Thus, N 2 will increase. But N 1 + N 2 = mg. So, N 1 will decrease.

4. r = | r | = (3)2 + (−4 )2 = 5 m v = | v | = 16 + a2

4 2v 2π 2 2 = m/s π =

(T / 4 )

=

4 2v T

=

(4 2 ) v 2 2v 2 = (2πR / v ) πR

(as vi = v f = v )

= 2 m/s2 (d)

 2 S = 2R = 2    π =

2 2 m π

1. (a) Applying conservation of energy

In uniform circular motion, v is always perpendicular to a. ∴ v⋅ a = 0 or −24 − ab = 0 ...(i) ∴ ab = 24 2 v ar = r 16 + a2 2 ...(ii) 36 + b = ∴ 5 Solving these two equations, we can find the values of a and b. (d) r , v and a lie in same plane. But v × a is perpendicular to this plane. ∴ r⊥ v×a or r ⋅ (v × a) = 0 5. (a) Since, speed = constant ∴ Average speed = this constant value = 1 m/s. S (b) Average velocity = T/4 t 2R = R S T /4 4 2R (2πR / v )

v 2f + vi2 − 2v f vi cos 90°

Subjective Questions

ar = | a | = 36 + b2

=

| ∆v | t

R

1 m ( 3Lg )2 2 3L Ans. ∴ h= 2 (b) Since, 3Lg lies between 2Lg and 5Lg, the string will slack in upper half of the circle. Assuming that string slacks when it makes an angle θ with horizontal. We have mv 2 …(i) mg sin θ = L mgh =

v 2 = ( 3gL)2 − 2gL (1 + sin θ )

…(ii)

Solving Eq. (i) and Eq. (ii), we get 1 sin θ = 3 gL 2 and v = 3 Maximum height of the bob from starting point, ar

v

h = L (1 + sin θ)

O

u

H = L (1 + sin θ ) +

v 2 sin 2 (90° − θ ) 2g

Circular Motion — 605

Chapter 10 4 L  gL 4L 4L + +   cos2 θ = 3 3 27  6g  40 = L 27

Minimum Speed

=

N

Maximum height in part (b) is less than that in part (a), think why? 2. h = 0.8 sin 30° = 0.4 m

Note

∴ (a)

v 2 = 2gh Just before,

mv 2 R1 mg m (2g ) (0.4) T1 = + 2 0.8 3mg = 2 (b) Just after, mv 2 T2 − mg sin 30° = R2 mg m (2g ) (0.4) T2 = + 2 0.4 5 mg or T2 = 2 3. h = l (1 − cos θ ) T1 − mg sin 30° =

(R1 = 0.8 m)

Ans.

(R2 = 0.4 m )

θ

θ mg

Equation of motion are, N cos θ + µN sin θ = mg m 2 N sin θ − µN cos θ = vmin R Solving these two equations, we get vmin = 4.2 m/s

…(ii)

Maximum Speed Equations of motion are, N cos θ − µN sin θ = mg m 2 N sin θ + µN cos θ = vmax R

…(iii)

…(i)

Ans.

…(iv)

N θ

Ans.

v 2 = v02 − 2gh = 3gl − 2gl (1 − cos θ ) = gl (1 + 2 cos θ ) At 45° means radial and tangential components of acceleration are equal. v2 ∴ = g sin θ l or 1 + 2 cos θ = sin θ π Solving the equation we get, θ = 90° or Ans. 2  v2  4. Banking angle, θ = tan −1    Rg  ∴

µN

θ

Ans.

36 km/h = 10 m/s  100  θ = tan −1   = 27°  20 × 9.8

Angle of repose, θ r = tan −1 (µ ) = tan −1 (0.4 ) = 21.8° Since θ > θ r , vehicle cannot remain in the given position with v = 0. At rest it will slide down. To find minimum speed, so that vehicle does not slip down, maximum friction will act up the plane. To find maximum speed, so that the vehicle does not skid up, maximum friction will act down the plane.

θ

θ

µN mg

Solving these two equations, we have vmax = 15 m/s

Ans.

5. Let v be the velocity at that instant. Then, horizontal component of velocity remains unchanged. v θ/2

g

θ u cos θ = u cos θ or v = θ 2 cos 2 Tangential component of acceleration of this instant will be, at = g cos (π / 2 + θ/2) = − g sin θ/2 θ an = a2 − at2 = g 2 − g 2 sin 2 2 θ = g cos 2

∴

v cos

606 — Mechanics - I an =

Since,

v2 R 2

T′ T sin 30°

Ans. 30 °

   u cos θ     cos θ  2 2 2  v 2  = u cos θ or R = = θ an  θ g cos g cos3    2 2 6. After 1 s v = u + at = 20 $i + 10 $j,

T

v = 500 m/s = 10 5 m/s a = − 10 $j at = a cos θ a ⋅ v − 100 = = v 10 5 = − 2 5 m/s2

Now

Ans.

an = a2 − at2 = (10)2 − (2 5 )3 = 80 m/s2 = 4 5 m/s2 2

T − µm1g = m2Rω

Ans.

where, ∴

…(ii)

Substituting,

or Ans.

(b) T = m2Rω + µm1 g = (1) (0.5) (6.32)2 + (0.5) (2) (10) ≈ 30 N

8. Speed of bob in the given position, v = 2gh Here, ∴

h = (400 + 400 cos 30° ) mm = 746 mm = 0.746 m v = 2 × 9.8 × 0.746 = 3.82 m/s

mv 2 R or N + mg cos θ = 2mg (1 − cos θ ) or …(i) N = 2mg − 3mg cos θ The tube breaks its contact with ground when 2N cos θ > Mg Substituting, 2N cos θ = Mg N + mg cos θ =

or

2

Ans.

(from energy conservation) h = R (1 − cos θ ) v = 2gR (1 − cos θ )

…(i)

Solving Eqs. (i) and (ii), we have 2m1 µg ω= (m1 – m2 )R Substituting the values, we have ω min = 6.32 rad /s

Ans.

T ′ = T cos 30° v = 2gh

below

2

T = 90 N R = T sin 30° = 45 N

Ans.

7. (a) Force diagrams of m1 and m2 are as shown

(Only horizontal forces have been shown) Equations of motion are T + µm1g = m1Rω 2

mv 2 r 2 × (3.82)2 T = 2 × 9.8 × cos 30° + (0.4 ) T − mg cos θ =

9. Speed of each particle at angle θ is,

2

v (10 5) R= = = 25 5 m. an 4 5

or ∴

T cos 30°

4 mg cos θ − 6mg cos2 θ = Mg θ = 60° 3mg 2mg − = Mg 2 M 1 = m 2

Ans.

Initially normal reaction on each ball will be radially outward and later it will be radially inward, so that normal reactions on tube is radially outward to break it off from the ground. 10. At distance x from centre, Centrifugal force = mx ω 2

Note

∴ Acceleration a = x ω 2 dv or v. = x ω 2 dx

Circular Motion — 607

Chapter 10 v

L

∫0v dv = ω ∫ax dx

or

2

or

v2 ω2 2 = (L − a2 ) 2 2

or

v = ω L2 − a2

Substituting in Eq. (i) 1 1 2  Fmax = 2mg  2 × − 3 ×  = mg  3 9 3 Ans.

mv 2 11. N = R

Fmax > Mg 2 mg > Mg 3 3 m> M 2

or or

fmax = µN =

µmv R

2

f ∴ Retardation a = max m µv 2 = R 2  dv  µv ∴ −  =  dt  R v µ t dv or ∫v0 v 2 = − R ∫0 dt v0 or v= µv t 1+ 0 R 12. Let R be the radius of the ring h = R (1 − cos θ ) v 2 = 2gh = 2gR (1 − cos θ )

Hence proved.

13. Minimum velocity of particle at the lowest position to complete the circle should be 4gR inside a tube. u = 4 gR

So, ∴

h = R (1 − cos θ ) v 2 = u2 − 2gh

or

v 2 = 4 gR − 2gR (1 − cos θ ) = 2gR(1 + cos θ ) θ  v 2 = 2gR  2 cos2   2

or Ans.

R

θ

v

2

or

mv = N + mg cos θ R N = 2mg (1 − cos θ ) − mg cos θ N = 2mg − 3mg cos θ

h u

v = 2 gR cos

or

ds = v ⋅ dt

From θ θ

θ

v

In the critical condition, tension in the string is zero and net upward force on the ring F = 2N cos θ = 2mg (2 cos θ − 3 cos2 θ ) dF F is maximum when =0 dθ or −2 sin θ + 6 sin θ cos θ = 0 1 or cos θ = 3

R dθ = 2 gR cos

We get

h

…(i)

θ 2

R π/ 2  θ sec   dθ  2 g ∫0

1

t

θ ⋅ dt 2

or

∫0 dt = 2

or

t=

R g

or

t=

R ln (1 + g

π/2

θ θ    ln  sec 2 + tan 2   0 2)

Hence proved.

14. At position θ, v 2 = v02 + 2gh where, ∴

h = a(1 − cos θ ) v 2 = ( 2ag )2 + 2ag (1 − cos θ )

or

v 2 = 2ag (2 − cos θ )

…(i)

608 — Mechanics - I N + mg cos θ =

Particle leaves contact with the track where N = 0

mv 2 a

sin θ

or N + mg cos θ = 2mg (2 − cos θ ) or N = mg (4 − 3 cos θ ) Net vertical force, F = N cos θ + mg = mg(4 cos θ − 3 cos2 θ + 1)

vr 1m θ

x

a=

This force (or acceleration) will be maximum when dF =0 dθ or −4 sin θ + 6 sin θ cos θ = 0 So, either sin θ = 0, θ = 0° 2 or cos θ = , 3  2 θ = cos−1    3 θ = 0° is unacceptable Therefore, the desired position is at  2 θ = cos−1    3

(1 – cos θ)

2g 9

y

 2g  mg cos θ − m   sin θ = mvr2  9 2g 4g or g cos θ − sin θ = 2g (1 − cos θ ) + sin θ 9 9 6 or 3 cos θ − sin θ = 2 9 Solving this, we get Ans. θ ≈ 37° or

(b) From Eq. (i), we have 4g sin θ 9 or vr = 2.58 m/s at θ = 37° Vertical component of its velocity is v y = vr sin θ 3 = 2.58 × 5 = 1.55 m/s 1   Now, 1.3 = 1.55t + 5t 2 Q s = ut + gt 2  2  vr = 2g (1 − cos θ ) +

Ans.

15. (a) Let vr be the velocity of mass relative to track at angular position θ. From work energy theorem, KE of particle relative to track = Work done by force of gravity + work done by pseudo force 1  2g  mvr2 = mg (1 − cos θ ) + m   sin θ ∴  9 2 4g or …(i) vr2 = 2g (1 − cos θ ) + sin θ 9

or or

5t 2 + 1.55t − 1.3 = 0 t = 0.38 s

Ans.

JEE Main and Advanced Previous Years’ Questions (2018-13) JEE Main 1. The density of a material in the shape of a cube is determined by measuring three sides of the cube and its mass. If the relative errors in measuring the mass and length are respectively 1.5% and 1% , the maximum error in determining the density is (2018) (a) 2.5%

(b) 3.5%

(c) 4.5%

Distance Position

(c)

Time

(b)

Position

Time

(d)

3. Two masses m1 = 5 kg and m2 = 10 kg connected by an inextensible string over a frictionless pulley, are moving as shown in the figure. The coefficient of friction of horizontal surface is 0.15. The minimum weight m that should be put on top of m2 to stop the motion is (2018) m m2

k 2a 2 3 k (d) − 2 a2

k 4a 2

(b)

(c) zero

T

initial speed v0 strikes a stationary particle of the same mass. If the final total kinetic energy is 50% greater than the original kinetic energy, the magnitude of the relative velocity between the two particles after collision, is (2018) (a)

Velocity Time

(a) −

5. In a collinear collision, a particle with an

represent the same motion. One of them does it incorrectly. Pick it up. (2018)

(a)

radius a under the action of an attractive k potential U = − 2 . Its total energy is (2018) 2r

(d) 6%

2. All the graphs below are intended to

Velocity

4. A particle is moving in a circular path of

v0 4

(b) 2 v 0

(c)

v0 2

v0 2

(d)

6. A particle is moving with a uniform speed in a circular orbit of radius R in a central force inversely proportional to the nth power of R. If the period of rotation of the particle is T , then : (2018) n

(a)T ∝ R 3 / 2 for any n (c)T ∝ R ( n + 1) / 2

(b)T ∝ R 2

+1

(d)T ∝ R n / 2

7. A body is thrown vertically upwards. Which one of the following graphs correctly represent the velocity vs time? (2017) v

T

(a)

v t

(b)

t

m1 m 1g

(a) 18.3 kg (c) 43.3 kg

(b) 27.3 kg (d) 10.3 kg

(d) v

(c) v t

t

2

Mechanics Vol. 1 8. A body of mass m = 10−2kg is moving in a medium and experiences a frictional force F = − kv 2. Its initial speed is v0 = 10 ms −1 . If, after 10 s, its energy is 1/ 8 mv02, the value of k will be (2017) (a) 10−3 kgs−1 (c) 10−1 kgm−1 s−1

(b) 10−4 kgm−1 (d) 10−3 kgm−1

9. A time dependent force F = 6t acts on a particle of mass 1 kg. If the particle starts from rest, the work done by the force during the first 1 s will be (2017) (a) 22 J

(b) 9 J

(c) 18 J

(d) 4.5 J

10. A point particle of mass m, moves along the uniformly rough track PQR as shown in the figure. The coefficient of friction, between the particle and the rough track equals µ . The particle is released from rest, from the point P and it comes to rest at a point R. The energies lost by the ball, over the parts PQ and QR of the track, are equal to each other, and no energy is lost when particle changes direction from PQ to QR. The values of the coefficient of friction µ and the distance x (= QR), are respectively close to (2016)

12. The period of oscillation of a simple pendulum is T = 2π L/ g. Measured value of L is 20.0 cm known to 1 mm accuracy and time for 100 oscillations of the pendulum is found to be 90 s using a wrist watch of 1s resolution. The accuracy in the determination of g is (2015) (a) 3%

(b) 2%

(c) 1%

(d) 5%

13. Two stones are thrown up simultaneously from the edge of a cliff 240 m high with initial speed of 10 m/s and 40 m/s, respectively. Which of the following graph best represents the time variation of relative position of the second stone with respect to the first? (Assume stones do not rebound after hitting the ground and neglect air resistance, take g = 10m /s2 ) (2015) 240

(y2 – y1)m

(y2 – y1)m

240

(a)

(b) t

240

8 12 (y2 – y1)m

t(s) 240

8 12 (y2 – y1)m

t(s)

(d)

(c)

t(s)

12

P

8

12

t(s)

14. Given in the figure are two blocks A and h = 2m R

30º Horizontal surface

(a) 0.2 and 6.5 m (c) 0.29 and 3.5 m

Q

(b) 0.2 and 3.5 m (d) 0.29 and 6.5 m

B of weight 20 N and 100 N, respectively. These are being pressed against a wall by a force F as shown in figure. If the coefficient of friction between the blocks is 0.1 and between block B and the wall is 0.15, the frictional force applied by the wall in block B is (2015)

fat lifts a mass of 10 kg upto a height of 1 m 1000 times. Assume that the potential energy lost each time he lowers the mass is dissipated. How much fat will he use up considering the work done only when the weight is lifted up? Fat supplies 3.8 × 107 J of energy per kg which is converted to mechanical energy with a 20% efficiency (2016) rate. (Take, g = 9.8 ms−2) (a) 2.45 × 10−3 kg (c) 9.89 × 10−3 kg

(b) 6.45 × 10−3 kg (d) 12.89 × 10−3 kg

f2

f1

11. A person trying to loose weight by burning F

R

A

WA = 20 N

(a) 100 N

(b) 120 N

R

B

f1

R

WB = 200 N

(c) 80 N

(d) 150 N

15. The current voltage relation of diode is given by I = ( e1000V/T − 1) mA, where the applied voltage V is in volt and the temperature T is in kelvin. If a student makes an error measuring ± 0.01 V while

3

Previous Years’ Questions (2018-13) measuring the current of 5 mA at 300 K, what will be the error in the value of current in mA? (2014)

the maximum height above the ground at which the block can be placed without slipping is (2014)

(a) 0.2 mA (c) 0.5 mA

(a) 1/6 m (c) 1/3 m

(b) 0.02 mA (d) 0.05 mA

16. A student measured the length of a rod

and wrote it as 3.50 cm. Which instrument did he use to measure it? (a) A meter scale

(2014)

(b) A vernier caliper where the 10 divisions in vernier scale matches with 9 divisions in main scale and main scale has 10 divisions in 1 cm (c) A screw gauge having 100 divisions in the circular scale and pitch as 1 mm and pitch as 1 mm (d) A screw gauge having 50 divisions in the circular scale

17. From a tower of height H, a particle is thrown vertically upwards with a speed u. The time taken by the particle to hit the ground, is n times that taken by it to reach the highest point of its path. The relation between H, u and n is (2014) (a) 2gH = n 2 u 2 (c) 2gH = nu 2 (n − 2)

(b) gH = (n − 2)2 u 2 (d) gH = (n − 2)2 u 2

18. A block of mass m is placed on a surface with a vertical cross-section given by y = x 3 / 6. If the coefficient of friction is 0.5,

(b) 2/3 m (d) 1/2 m

19. When a rubber band is stretched by a distance x, it exerts a restoring force of magnitude F = a x + bx 2, where a and b are constants. The work done in stretching the unstretched rubber band by L is (2014) (a) aL2 + bL3 aL2 bL3 (c) + 2 3

(b) 1 / 2 (aL2 + bL3 ) 1  aL2 bL3  (d)  +  2 2 3 

20. Let [ε 0 ] denote the dimensional formula of the permittivity of vacuum. If M = mass, L = length, T = Time and (2013) A = electric current, then (a) (b) (c) (d)

[ε0 ] = [M−1 L−3 T 2 A] [ε0 ] = [M−1 L−3 T 4 A 2 ] [ε0 ] = [M−2 L2 T −1 A −2 ] [ε0 ] = [M−1 L2 T −1 A 2 ]

21. A projectile is given an initial velocity of ( i + 2 j) m/s, where, i is along the ground and j is along the vertical. If g = 10 m/s 2, then the equation of its trajectory is (2013) (a) y = x − 5 x 2 (c) 4 y = 2 x − 5 x 2

(b) y = 2 x − 5 x 2 (d) 4 y = 2 x − 25 x 2

Answer with Explanations Mass M or = 3 Volume L ∆ρ ∆M 3∆L ⇒ Error in density = + ρ M L

1. (c) ∴ Density, ρ =

ρ=

M L3

So, maximum % error in measurement of ρ is 3∆ L ∆ρ ∆M × 100 = × 100 + × 100 ρ M L or % error in density = 15 . + 3×1 % error = 4.5%

2. (b) If velocity versus time graph is a straight line with negative slope, then acceleration is constant and negative. With a negative slope distance-time graph will be 1 parabolic  s = ut − at 2  .   2 So, option (b) will be incorrect.

3. (b) Motion stops when pull due to m1 ≤ force of friction between m and m 2 and surface. ⇒ m1 g ≤ µ( m2 + m)g ⇒ 5 × 10 ≤ 015 . (10 + m) × 10 ⇒ m ≥ 23.33 kg Here, nearest value is 27.3 kg So, mmin = 27.3 kg dU 4. (c) ∴Force = − dr d  −k  k ⇒ F = −  2 = − 3 dr  2 r  r As particle is on circular path, this force must be centripetal force. mv 2 ⇒ |F |= r

4

Mechanics Vol. 1 k mv 2 = 3 r r 1 k 2 mv = 2 ⇒ 2 2r ∴ Total energy of particle = KE + PE k k = 2 − 2 =0 2r 2r Total energy = 0 So,

5. (b) Momentum is conserved in all type of collisions. Final kinetic energy is 50% more than initial kinetic energy 1 1 150 1 …(i) mv 22 + mv12 = × mv 02 ⇒ 2 2 100 2 m

m v0

⇒

1 = 0.5 + 5000 k 0.5 k= 5000 ⇒ k = 10 −4 kg/m ∆p 9. (d) From Newton’s second law, =F ∆t

Before collision m

⇒

m v2

v1 After collision

Conservation of momentum gives, mv 0 = mv1 + mv 2 v 0 = v 2 + v1 From Eqs. (i) and (ii), we have v12 + v 22 + 2 v1 v 2 = v 02 − v 02 ⇒ 2 v1 v 2 = 2 ∴ ( v1 − v 2 )2 = ( v1 + v 2 )2 − 4 v1 v 2 = 2 v 02 or vrel = 2 v 0

⇒ …(ii)

6. (c) ∴Force = Mass × Acceleration = mω 2 R 1 and given, F ∝ n R So, we have

k ⇒ F = n R 2

k 2π = m  R  T  Rn ⇒

T2 =

4 π 2m n +1 ⋅R k n +1

⇒

T ∝R

2

7. (b) Initially velocity keeps on decreasing at a constant rate, then it increases in negative direction with same rate.

8. (b) Given, force, F = − kv ∴ Acceleration, a = or

−k 2 dv = v dt m

2

−k 2 v m ⇒

Now, with limits, we have v dv k t ∫10 v 2 = − m ∫0 dt v  − 1 = − k t ⇒    v  10 m 1 kt = 01 . + ⇒ v m 1 1 ⇒ v = = kt 01 . + 1000k . + 01 m 1 1 × m × v 2 = × v 02 ⇒ 2 8 v ⇒ v = 0 =5 2 1 ⇒ =5 01 . + 1000 k

dv k =− . dt v2 m

⇒

∆p = F∆t ⇒ p = ∫ dp = p=

Also, ∆k =

∫

1 0

∫

1 0

F dt

m 6 t dt = 3 kg   s

∆p2 32 = = 4.5 2m 2 × 1

So, work done = ∆k = 4.5 J

10. (c) As energy loss is same, thus µ mg cos θ ⋅ ( PQ ) = µmg ⋅ (QR ) ∴

QR = ( PQ ) cosθ

3 = 2 3 ≈ 3.5 m 2 Further,decrease in potential energy = loss due to friction ∴ mgh = (µmg cos θ)d 1 + (µmg )d 2 3 m × 10 × 2 = µ × m × 10 × ×4 2 + µ × m × 10 × 2 3 ⇒ 4 3µ =2 1 ⇒ µ = = 0.288 = 0.29 2 3 ⇒

QR = 4 ×

11. (d) Work done in lifting mass = (10 × 9.8 × 1) × 1000 If m is mass of fat burnt, then energy 20 = m × 3.8 × 10 7 × 100 Equating the two, we get 49 ∴ m= ≈ 12.89 × 10 −3 kg 3.8

5

Previous Years’ Questions (2018-13) 12. (a) Time period is given by, T = T = 2π

Further, ⇒

g =

t n

L g

( 4 π 2 )L ( 4 π 2 )( L ) L = = ( 4 π 2 n2 ) 2 2 T2 t t     n

Percentage error in the value of ‘g’ will be ∆g ∆L ∆t × 100 =   × 100 + 2   × 100  L   t  g =

01 . 1 × 100 + 2 ×   × 100 = 2.72%  90  20

∴The nearest answer is 3%.

13. (b)

∴(y2 − y1 ) versus t graph is a straight line passing through origin At t = 8 s, y2 − y1 = 240 m From 8 s to 12 s y1 = 0 1 y2 = 240 + 40 t − × 10 × t 2 ⇒ 2 = 240 + 40 t − 5 t ∴ ( y2 − y1 ) = 240 + 40 t − 5 t 2 Therefore, ( y2 − y1 ) versus t graph is parabolic, substituting the values we can check that at t = 8 sec, y2 − y1 is 240 m and at t = 12 sec, y2 − y1 is zero.

14. (b) Note It is not given in the question, best assuming that both blocks are in equilibrium. The free body diagram of two blocks is as shown below, Reaction force, R = applied force F

10 m/s 40 m/s

For vertical equilibrium of A; 1

2

+ve

240 m

–ve

f1 = friction between two blocks = W A = 20 N For vertical equilibrium of B; f2 = friction between block B and wall = W B + f1 = 100 + 20 = 120 N

15. (a) Given, Let us first find, time of collision of two particles with ground in putting proper values in the equation 1 s = ut + at 2 2 1 − 240 = 10 t 1 − × 10 × t 12 ⇒ 2 Solving, we get the position value of t 1 = 8 sec Therefore, the first particle will strike the ground at 8 sec. 1 Similarly, − 240 = 40 t 2 − × 10 × t 22 2 Solving this equation, we get positive value of t 2 =2s Therefore, second particle strikes the ground at 12 sec. If y is measured from ground. Then, from 0 to 8 s y1 = 240 + s1 = 240 + u1t +

1 2 a1t 2

1 × 10 × t 2 2 1 Similarly, y2 = 240 + 40t − × 10 × t 2 2 ⇒ t 2 − y1 = 30 t or

y1 = 240 + 10 t −

I = (e1000 V /T − 1) mA, dV = ± 0.01 V

T = 300 K So, I = e1000 V /T − 1 ⇒ I + 1 = e1000 V /T Taking log on both sides, we get 1000 V log( I + 1) = T dI 1000 On differentiating, dV = I+1 T 1000 dI = × ( I + 1) dV T 1000 dI = × ( 5 + 1) × 0.01 ⇒ 300 = 0.2 mA So, error in the value of current is 0.2 mA.

16. (c) If student measures 3.50 cm, it means that there is an uncertainly of order 0.01 cm. 1 cm 10 and 9 MSD = 10 VSD LC of vernier caliper = 1MSD − 1VSD 1  9 = 1 −  10  10  1 = cm = 0.01 cm 100 For vernier scale with 1 MSD =

6

Mechanics Vol. 1

17. (c) Time taken to reach the maximum height, t 1 =

u g

So, the maximum height above the ground at which the block can be placed without slipping is 1/6 m.

19. (c) Thinking Process We know that change in

t1 u

potential energy of a system corresponding to a conservative internal force as

H t2

f

Uf − Ui = − W = − ∫ F ⋅d r i

i.e.

− H = ut 2 −

But

t 2 = nt 1 −H = u

So,

We know that work done in stretching the rubber band

1 2 gt 2 2

by L is|dW | = | Fdx | [Given]

nu 1 n2u 2 − g 2 g 2 g

nu 2 1 n2u 2 − g 2 g 2 gH = nu 2 ( n − 2 ) −H =

⇒

18. (a) A block of mass m is placed on a surface with a vertical cross-section, then y m y θ

x

 x3  d   dy  6  x2 tanθ = = 2 dx dx At limiting equilibrium, we get x2 µ = tanθ ⇒ 0 . 5 = 2 ⇒

x2 = 1 ⇒ x = ± 1 x3 Now, putting the value of x in y = , we get 6

When x = 1 (1)3 1 y= = 6 6

When x = − 1 y=

F = ax + bx 2

Given,

If t 2 is the time taken to hit the ground, then

(−1)3 −1 = 6 6

L

L

 ax 2   bx 3  + ∫0  2     0  3 0 2 3 2  aL a × ( 0)   b × L b × ( 0)3  = − +  −   2 3  2   3  2 3 aL bL = |W | = + 2 3 1 q 1q 2 20. (b) From Coulomb’s law, F = 4 πε0 R 2 qq ε0 = 1 2 2 4 πFR Substituting the units, we have C2 [AT]2 ε0 = = 2 [MLT −2 ] [L2 ] N-m |W | =

L

( ax + bx 2 ) dx =

= [M−1L−3 T 4 A 2 ] 21. (b) Initial velocity = ( $i + 2 $j ) m/s Magnitude of initial velocity, u = (1)2 + (2 )2 =

5 m/s

Equation of trajectory of projectile is gx 2 y 2 y = x tanθ − (1 + tan2 θ)  tanθ = = = 2    2u2 x 1 2 10 ( x ) [1 + (2 )2 ] ∴ y= x×2 − 2( 5 )2 10 ( x 2 ) = 2x − (1 + 4) 2×5 = 2 x − 5 x2

JEE Advanced 1. Two vectors A and B are defined as A = a$i and B = a (cos ωt$i + sin ωt$j), where a is a constant and ω = π /6 rad s −1 . If |A + B|= 3|A − B| at time t = τ for the

first time, the value of τ, in seconds, is ............. . [Numerical Value, 2018]

2. A spring block system is resting on a frictionless floor as shown in the figure. The spring constant is 2.0 Nm −1 and the mass of the block is 2.0 kg . Ignore the mass of the spring. Initially, the spring is in an unstretched condition. Another block of mass 1.0 kg moving with a speed of 2.0 ms−1 collides elastically with the first block. The collision is such that the 2.0 kg block does not hit the wall. The distance, in metres, between the two blocks when the spring returns to its unstretched position for the first time after the collision is ............ . [Numerical Value, 2018]

1kg

2 ms–1

2 kg

Paragraph X (Q. Nos. 3-4) In electromagnetic theory, the electric and magnetic phenomena are related to each other. Therefore, the dimensions of electric and magnetic quantities must also be related to each other. In the questions below, [E] and [ B ] stand for dimensions of electric and magnetic fields respectively, while [ ε 0 ] and [µ 0 ] stand for dimensions of the permittivity and permeability of free space, respectively. [ L ] and [ T ] are dimensions of length and time, respectively. All the quantities are given in SI units. (2018)

3. The relation between [E] and [B] is (a) [E ] = [B ] [L] [T] (c) [E ] = [B][L][T]−1

−1

(b) [E ] = [B] [L] [T] (d) [E ] = [B][L]−1[T]−1

4. The relation between [ε 0 ] and [µ 0 ] is (a) [µ 0 ] = [ε0 ][L]2 [T]−2 (c) [µ 0 ] = [ε0 ]−1 [L]2 [T]−2

(b) [µ 0 ] = [ε0 ][L]−2 [T]2 (d) [µ 0 ] = [ε0 ]−1[L]−2 [T]2

Paragraph A (Q. Nos. 5-6) If the measurement errors in all the independent quantities are known, then it is possible to determine the error in any dependent quantity. This is done by the use of series expansion and truncating the expansion at the first power of the error. For example, consider the relation z = x / y. If the errors in x, y and z are ∆x, ∆y and ∆z respectively, then z ± ∆z =

x ± ∆x x = y ± ∆y y

 ∆x   ∆y  1 ±  1 ±   x y

−1

−1

 ∆y The series expansion for 1 ±  , to first power in  y ∆y / y, is 1m ( ∆y / y ). The relative errors in independent variables are always added. So, the error in z will be  ∆x ∆y  ∆z = z  +   x y The above derivation makes the assumption that ∆x / x v2 (a)

F

F

m1

m2

(b) F

F

v1'

m1

m2

m1

v

v

m1

m2

Maximum compression =xm (c) v2' m2 v1' < v2' (e)

(d)

Fig. 11.32

Two blocks of masses m1 and m2 are moving with velocities v1 and v 2 ( < v1 ) along the same straight line in a smooth horizontal surface. A spring is attached to the block of mass m2 . Now, let us see what happens during the collision between two particles. Figure (a) Block of mass m1 is behind m2 . Since, v1 > v 2 , the blocks will collide after some time. Figure (b) The spring is compressed. The spring force F ( = kx ) acts on the two blocks in the directions shown in figure. This force decreases the velocity of m1 and increases the velocity of m2 . Figure (c) The spring will compress till velocity of both the blocks become equal. So, at maximum compression (say x m ) velocities of both the blocks are equal (say v). Figure (d) Spring force is still in the directions shown in figure, i.e. velocity of block m1 is further decreased and that of m2 is increased. The spring now starts relaxing. Figure (e) The two blocks are separated from one another. Velocity of block m2 becomes more than the velocity of block m1 , i.e. v 2′ > v1′ .

Equations Which can be Used in the Above Situation Assuming spring to be perfectly elastic following two equations can be applied in the above situation. (i) In the absence of any external force on the system the linear momentum of the system will remain conserved before, during and after collision, i.e. m1 v1 + m2 v 2 = ( m1 + m2 ) v = m1 v1′ + m2 v 2′

…(i)

28 — Mechanics - II (ii) In the absence of any dissipative forces, the mechanical energy of the system will also remain conserved, i.e. 1 1 1 1 m1 v12 + m2 v 22 = ( m1 + m2 ) v 2 + kx m2 2 2 2 2 1 1 …(ii) = m1 v1′ 2 + m2 v 2′ 2 2 2 Note In the above situation we have assumed, spring to be perfectly elastic, i.e. it regains its original shape and size after the two blocks are separated. In actual practice there is no such spring between the two blocks. During collision, both the blocks (or bodies) are slightly deformed. This situation is similar to the compression of the spring. Due to deformation, two equal and opposite forces act on both the blocks. These two forces redistribute their linear momentum in such a manner that both the blocks are separated from one another. The collision is said to be elastic if both the blocks regain their original shape and size completely after they are separated. On the other hand if the blocks do not return to their original form the collision is said to be inelastic. If the deformation is permanent and the blocks move together with same velocity after the collision, the collision is said to be perfectly inelastic.

Types of Collision Collision between two bodies may be classified in two ways: 1. Elastic collision and inelastic collision. 2. Head on collision or oblique collision. As discussed earlier also collision between two bodies is said to be elastic if both the bodies come to their original shape and size after the collision, i.e. no fraction of mechanical energy remains stored as deformation potential energy in the bodies. Thus in addition to the linear momentum, kinetic energy also remains conserved before and after collision. On the other hand, in an inelastic collision, the colliding bodies do not return to their original shape and size completely after collision and some part of the mechanical energy of the system goes to the deformation potential energy. Thus, only linear momentum remains conserved in case of an inelastic collision. Further, a collision is said to be head on (or direct) if the directions of the velocity of colliding objects are along the line of action of the impulses, acting at the instant of collision. If just before collision, at least one of the colliding objects was moving in a direction different from the line of action of the impulses, the collision is called oblique or indirect.

Head on Elastic Collision Let the two balls of masses m1 and m2 collide each other elastically with velocities v1 and v 2 in the directions shown in Fig. 11.33(a). Their velocities become v1 ′ and v 2 ′ after the collision along the same line. Applying conservation of linear momentum, we get m2

v2

m1

m2

v1

v2'

m1

v1'

(b) After collision

(a) Before collision

Fig. 11.33

m1 v1 + m2 v 2 = m1 v1 ′ + m2 v 2 ′

…(iii)

Centre of Mass, Linear Momentum and Collision — 29

Chapter 11

In an elastic collision kinetic energy before and after collision is also conserved. Hence, 1 1 1 1 m1 v12 + m2 v 22 = m1 v1 ′ 2 + m2 v 2 ′ 2 2 2 2 2 Solving Eqs. (iii) and (iv) for v1 ′ and v 2 ′, we get  m − m2   2m2  v1 ′ =  1  v1 +   v2  m1 + m2   m1 + m2   m − m1   2m1  v2 ′ =  2  v2 +   v1  m1 + m2   m1 + m2 

and

…(iv)

…(v) …(vi)

Special Cases 1. If m1 = m2 , then from Eqs. (v) and (vi), we can see that v1 ′ = v 2 and v 2 ′ = v1 i.e. when two particles of equal mass collide elastically and the collision is head on, they exchange their velocities., e.g. 4 m/s

3 m/s

3 m/s

m

4 m/s m

m

After collision

Before collision 2 m/s

m

v=0

v=0

m

m

m

2 m/s m

After collision

Before collision

Fig. 11.34

  m2 m ≈ 0 with these two substitutions  v1 = 0 and 2 = 0 m1 m1  

2. If m1 > > m2 and v1 = 0. Then

v1 = 0 m2

v2

v2

m1

v1′ ≈ 0

v2′ ≈ – v2

m1

m2

Before collision

After collision

Fig. 11.35

we get the following two results v1 ′ ≈ 0 and v 2 ′ ≈ − v 2 i.e. the particle of mass m1 remains at rest while the particle of mass m2 bounces back with same speed v 2 . 3. If m2 > > m1 and v1 = 0, m2

v2

v2′ ≈ v2

v1= 0 m1

m2

v1′ ≈ 2v2 m1

After collision

Before collision

Fig. 11.36

30 — Mechanics - II with the substitution

m1 ≈ 0 and v1 = 0, we get the results m2

v1 ′ ≈ 2v 2 and v 2 ′ ≈ v 2 i.e. the mass m1 moves with velocity 2v 2 while the velocity of mass m2 remains unchanged. Note It is important to note that Eqs. (v) and (vi) and their three special cases can be used only in case of a head on elastic collision between two particles. I have found that many students apply these two equations even if the collision is inelastic and do not apply these relations where clearly a head on elastic collision is given in the problem.

Head on Inelastic Collision As we have discussed earlier also, in an inelastic collision, the particles do not regain their shape and size completely after collision. Some fraction of mechanical energy is retained by the colliding particles in the form of deformation potential energy. Thus, the kinetic energy of the particles no longer remains conserved. However, in the absence of external forces, law of conservation of linear momentum still holds good. m2

v2

m1

v1

v2'

m2

m1

v1' +ve

After collision

Before collision

Fig. 11.37

Suppose the velocities of two particles of mass m1 and m2 before collision be v1 and v 2 in the directions shown in figure. Let v1 ′ and v 2 ′ be their velocities after collision. The law of conservation of linear momentum gives m1 v1 + m2 v 2 = m1 v1′ + m2 v 2′

v'

Fig. 11.38

…(vii)

Collision is said to be perfectly inelastic if both the particles stick together after collision and move with same velocity, say v′ as shown in Fig. 11.38. In this case, Eq. (vii) can be written as m1 v1 + m2 v 2 = ( m1 + m2 ) v ′ or

v′ =

m1 v1 + m2 v 2 m1 + m2

…(viii)

Newton's Law of Restitution When two objects are in direct (head on) impact, the speed with which they separate after impact is usually less than or equal to their speed of approach before impact. Experimental evidence suggests that the ratio of these relative speeds is constant for two given set of objects. This property formulated by Newton, is known as the law of restitution and can be written in the form separation speed …(ix) =e approach speed The ratio e is called the coefficient of restitution and is constant for two particular objects. In general

0 ≤ e ≤1

Centre of Mass, Linear Momentum and Collision — 31

Chapter 11

e = 0, for completely inelastic collision, as both the objects stick together. So, their separation speed is zero or e = 0 from Eq. (ix). e =1, for an elastic collision, as we can show from Eq. (v) and (vi), that v1 ′ − v 2 ′ = v 2 − v1 or separation speed = approach speed or e =1 m2

m1

v2

v2'

v1

v1'

After collision

Before collision

Fig. 11.39

Let us now find the velocities of two particles after collision if they collide directly and the coefficient of restitution between them is given as e. Applying conservation of linear momentum …(x) m1 v1 + m2 v 2 = m1 v1 ′ + m2 v 2 ′ Further, separation speed = e (approach speed) or …(xi) v1 ′ − v 2 ′ = e ( v 2 − v1 ) Solving Eqs. (x) and (xi), we get

and

 m − em2   m + em2  v1 ′ =  1  v1 +  2  v2  m1 + m2   m1 + m2 

…(xii)

 m − em1   m1 + em1  v2 ′ =  2  v2 +   v1  m1 + m2   m1 + m2 

…(xiii)

Special Cases 1. If collision is elastic, i.e. e =1, then  m − m2   2m2  v1 ′ =  1  v1 +   v2  m1 + m2   m1 + m2  and

 m − m1   2m1  v2 ′ =  2  v2 +   v1  m1 + m2   m1 + m2 

which are same as Eqs. (v) and (vi). 2. If collision is perfectly inelastic, i.e. e = 0, then

v1 ′ = v 2 ′ = which is same as Eq. (viii).

m1 v1 + m2 v 2 = v ′ (say ) m1 + m2

32 — Mechanics - II 3. If m1 = m2

m2

and

v2

v1 = 0, then

1 + e v1 ′ =  v  2  2

and

m1

1 − e v2 ′ =  v  2  2 v2'

v1 = 0

…(xiv) v1'

After collision

Before collision

Fig. 11.40

Note

(i) If mass of one body is very-very greater than that of the other, then after collision velocity of heavy body does not change appreciably. (Whether the collision is elastic or inelastic). (ii) In the situation shown in figure if e is the coefficient of restitution between the ball and the ground, then after nth collision with the floor the speed of ball will remain env 0 and it will go upto a height e2 n h or, v n = env 0 = en 2 gh and

hn = e2 n h

u=0 h

v0 = √2gh

Fig. 11.41

EXERCISE Derive the above two relations.

Oblique Collision x During collision between two objects a pair of equal and opposite impulses act at the moment of impact. If just before impact at least one of the objects was moving y in a direction different from the line of action of these impulses the collision is said v to be oblique. y In the figure, two balls collide obliquely. During collision impulses act in the direction xx. Henceforth, we will call this direction as common normal direction x and a direction perpendicular to it (i.e. yy) as common tangent. Fig. 11.42 Following four points are important regarding an oblique collision : 1. A pair of equal and opposite impulses act along common normal direction. Hence, linear momentum of individual bodies do change along common normal direction. If mass of the colliding bodies remain constant during collision, then we can say that linear velocity of the individual bodies change during collision in this direction. 2. No component of impulse act along common tangent direction. Hence, linear momentum or linear velocity of individual bodies (if mass is constant) remain unchanged along this direction. 3. Net impulse on both the bodies is zero during collision. Hence, net momentum of both the bodies remain conserved before and after collision in any direction. 4. Definition of coefficient of restitution can be applied along common normal direction, i.e. along common normal direction we can apply Relative speed of separation = e (relative speed of approach) Here, e is the coefficient of restitution between the particles.

Chapter 11 V

Centre of Mass, Linear Momentum and Collision — 33

Example 11.19 Two blocks A and B of equal mass m = 1.0 kg are lying on a smooth horizontal surface as shown in figure. A spring of force constant k = 200 N / m is fixed at one end of block A. Block B collides with block A with velocity v0 = 2.0 m/ s . Find the maximum compression of the spring.

2.0 m/s B

A

Fig. 11.43

At maximum compression ( x m ) velocity of both the blocks is same, say it is v. Applying conservation of linear momentum, we have

Solution

( m A + mB )v = mB v 0 or (1.0 + 1.0)v = (1.0) v 0 B v 0 2.0 or v= = = 1.0 m/s 2 2 Using conservation of mechanical energy, we have 1 1 1 mB v 02 = ( m A + mB )v 2 + kx m2 2 2 2 Substituting the values, we get 1 1 1 × (1) × ( 2.0) 2 = × (1.0 + 1.0) × (1.0) 2 + × (200) × x m2 2 2 2

xm

Fig. 11.44

2 = 1.0 + 100 x m2

or

x m = 0.1 m = 10.0 cm

or V

v A

Ans.

Example 11.20 Two balls of masses m and 2 m moving in opposite directions collide head on elastically with velocities v and 2v. Find their velocities after collision. Solution Here, v1 = − v, v 2 = 2v, m1 = m and m2 = 2m. 2v 2m

v

m

+ve

Fig. 11.45

Substituting these values in Eqs. (v) and (vi), we get

and

 m − 2m  4m  v1 ′ =   (− v ) +   ( 2v ) or  m + 2m  m + 2m

v1 ′ =

v 8v + = 3v 3 3

 2m − m  2m  v2 ′ =   ( 2v ) +   ( − v ) or  m + 2m  m + 2m

v2 ′ =

2 2 v− v=0 3 3

2m

m

3v

Fig. 11.46

i.e. the second ball (of mass 2m) comes to a rest while the first (of mass m) moves with velocity 3v in the direction shown in Fig. 11.46.

34 — Mechanics - II V

Example 11.21 Two pendulum bobs of masses m and 2m collide head on elastically at the lowest point in their motion. If both the balls are released from a height H above the lowest point, to what heights do they rise for the first time after collision? Solution Given, m1 = m, m2 = 2m, v1 = − 2gH and v 2 = 2gH

2

+ve

1 v2

v1

Fig. 11.47

Since, the collision is elastic. Using Eqs. (v) and (vi) discussed in the theory the velocities after collision are  m − 2m  4m  v1 ′ =   ( − 2gH ) +    m + 2m  m + 2m 2gH

= and

3

+

4 2gH 3

=

2gH

5 2gH 3

 2m − m  2m  v2 ′ =   ( 2gH ) +   ( − 2gH )  m + 2m  m + 2m 2gH

=

v2'

3

=

−

2 2gH

√2gH 3

3

v1'

=

=−

2gH 3

5 √2gH 3

Fig. 11.48

i.e. the velocities of the balls after the collision are as shown in Fig. 11.48. Therefore, the heights to which the balls rise after the collision are: h1 =

or

h1 =

( v1 ′ ) 2 2g

5  2gH   3  2g

(using v 2 = u 2 − 2gh)

2

or

h1 =

25 H 9

Centre of Mass, Linear Momentum and Collision — 35

Chapter 11

h2 =

and

or

h2 =

or

h2 =

( v2′ ) 2 2g  2gH     3   

2

2g H 9

Note Since, the collision is elastic, mechanical energy of both the balls will remain conserved, or Ei = Ef ⇒

( m + 2 m) gH = mgh1 + 2 mgh2  25   H 3 mgH = ( mg )  H + (2 mg )   9   9

⇒ ⇒ V

3 mgH = 3 mgH

Example 11.22 A ball of mass m moving at a speed v makes a head on inelastic collision with an identical ball at rest. The kinetic energy of the balls 3 after the collision is th of the original. Find the coefficient of restitution. 4 m

v

m

v2'

v1'

After collision

Before collision

Fig. 11.49

For the given conditions, we can use Eq. (xiv) or  1 + e  1 − e v1 ′ =   v and v 2 ′ =  v  2   2  3 Given that K f = Ki 4 1 1 3  1 2 2 2 or mv1 ′ + mv 2 ′ =  mv   2 2 4 2 Substituting the value, we get 2 2 3  1 + e  1 − e   +  =  2   2  4 Solution

or or or or

(1 + e ) 2 + (1 − e ) 2 = 3 2 + 2e 2 = 3 1 e2 = 2 1 e= 2

Ans.

36 — Mechanics - II V

Example 11.23 A ball is moving with velocity 2 m/s towards a heavy wall moving towards the ball with speed 1 m/s as shown in figure. Assuming collision to be elastic, find the velocity of ball immediately after the collision. 2 m/s

1 m/s

Fig. 11.50

The speed of wall will not change after the collision. So, let v be the velocity of the ball after collision in the direction shown in figure. Since, collision is elastic ( e = 1).

Solution

or

separation speed = approach speed v − 1= 2 + 1 v = 4 m/s

or V

Ans.

1 m/s

v

After collision Fig. 11.51

Example 11.24 A ball of mass m hits a floor with a speed v0 making an angle of incidence α with the normal. The coefficient of restitution is e. Find the speed of the reflected ball and the angle of reflection of the ball. The component of velocity v 0 along common tangent direction v 0 sin α will remain unchanged. Let v be the component along common normal direction after collision. Applying

Solution

v α v0 sin α

v0 sin α

α v0 v0 cos α Fig. 11.52

Relative speed of separation = e (relative speed of approach) along common normal direction, we get v = ev 0 cos α Thus, after collision components of velocity v′ are v 0 sin α and ev 0 cos α v′ = ( v 0 sin α ) 2 + ( ev 0 cos α ) 2

∴

tan β =

or

tan α tan β = e

Note For elastic collision, e = 1 ∴

v 0 sin α ev 0 cos α

and

v′ = v 0

and β = α .

ev0 cos α

v' β v0 sin α

Fig. 11.53

Chapter 11 V

Centre of Mass, Linear Momentum and Collision — 37

Example 11.25 After perfectly inelastic collision between two identical balls moving with same speed in different directions, the speed of the combined mass becomes half the initial speed. Find the angle between the two before collision. Solution

Let θ be the desired angle. Linear momentum of the system will remain conserved. v m

2m v 2

θ

m

v Fig. 11.54

p 2 = p12 + p 22 + 2 p 2 p 2 cos θ

Hence,

2

or or ∴ V

  v  2 2  2m    = ( mv ) + ( mv ) + 2( mv )( mv ) cos θ   2   1 1 = 1 + 1 + 2 cos θ or cos θ = − 2 θ = 120°

Ans.

Example 11.26 The coefficient of restitution between a snooker ball and the 1 side cushion is 3. If the ball hits the cushion and then rebounds at right angles to its original direction, show that the angles made with the side cushion by the direction of motion before and after impact are 60° and 30° respectively. Solution Let the original speed be u, in a direction making an angle θ with the side cushion. Using the law of restitution Relative speed of separation along common normal direction = e (relative speed of approach) u 3 1 or v = ( u sin θ ) ⇒ = 3 v sin θ θ

θ

u sin θ θ

u cos θ θ

u cos θ

v

Fig. 11.55

After impact, ⇒ ⇒

tan θ =

u cos θ 3 cos θ = v sin θ

tan 2 θ = 3 tan θ = 3

⇒ θ = 60° Therefore, the directions of motion before and after impact are at 60° and 30° to the cushion.

38 — Mechanics - II INTRODUCTORY EXERCISE

11.6

1. Two blocks of masses 3 kg and 6 kg respectively are placed on a smooth horizontal surface.

They are connected by a light spring of force constant k = 200 N/m. Initially the spring is unstretched. The indicated velocities are imparted to the blocks. Find the maximum extension of the spring. 1.0 m/s

2.0 m/s

3kg

6kg

Fig. 11.56

2. A moving body of mass m makes a head on elastic collision with another body of mass 2 m which is initially at rest. Find the fraction of kinetic energy lost by the colliding particle after collision.

3. What is the fractional decrease in kinetic energy of a body of mass m1 when it makes a head on elastic collision with another body of mass m 2 kept at rest?

4. In one dimensional elastic collision of equal masses, the velocities are interchanged. Can velocities in a one dimensional collision be interchanged if the masses are not equal.

5. After an head on elastic collision between two balls of equal masses, one is observed to have a speed of 3 m/s along the positive x-axis and the other has a speed of 2 m/s along the negative x-axis. What were the original velocities of the balls ?

6. A ball of mass 1 kg moving with 4 ms −1 along + x-axis collides elastically with an another ball of mass 2 kg moving with 6 m/s is opposite direction. Find their velocities after collision.

7. Three balls A, B and C are placed on a smooth horizontal surface. Given that m A = mC = 4m B . Ball B collides with ball C with an initial velocity v as shown in figure. Find the total number of collisions between the balls. All collisions are elastic. A

B

v

C

Fig. 11.57

8. Ball 1 collides directly with another identical ball 2 at rest. Velocity of second ball becomes two times that of 1 after collision. Find the coefficient of restitution between the two balls?

9. A sphere A of mass m, travelling with speed v, collides directly with a stationary sphere B. If A is brought to rest and B is given a speed V, find (a) the mass of B (b) the coefficient of restitution between A and B? 10. A smooth sphere is moving on a horizontal surface with velocity vector 2 $i + 2$j immediately before it hits a vertical wall. The wall is parallel to $j and the coefficient of restitution of the sphere and the wall is e =

1 . Find the velocity of the sphere after it hits the wall? 2

11. A ball falls vertically on an inclined plane of inclination α with speed v 0 and makes a perfectly elastic collision. What is angle of velocity vector with horizontal after collision.

Centre of Mass, Linear Momentum and Collision — 39

Chapter 11

Final Touch Points 1. Centre of mass frame of reference or C-frame of reference or zero momentum frame A frame of reference carried by the centre of mass of an isolated system of particles (i.e. a system not subjected to any external forces) is called the centre of mass or C-frame of reference. In this frame of reference. (i) Position vector of centre of mass is zero. (ii) Velocity and hence momentum of centre of mass is also zero.

2. A liquid of density ρ is filled in a container as shown in figure. The liquid comes out from the container through a orifice of area ‘a’ at a depth ‘h’ below the free surface of the liquid with a velocity v. This exerts a thrust force in the container in the backward direction. This thrust force is given by Ft h v

 dm  Ft = vr  −   dt  vr = v dm   −  = ρav  dt 

Here, and

(in forward direction)

 dV    = Volume of liquid flowing per second  dt 

as

= av  dm   dV  −  =ρ   = ρav  dt   dt 

∴ ∴

Ft = v (ρav )

or

Ft = ρav 2

(in backward direction)

Further, we will see in the chapter of fluid mechanics that v = 2gh .

3. Suppose, a chain of mass per unit length λ begins to fall through a hole in the ceiling as shown in Fig. (a) or the end of the chain piled on the platform is lifted vertically as in Fig. (b). In both the cases, due to increase of mass in the portion of the chain which is moving with a velocity v at certain moment of time a thrust force acts on this part of the chain which is given by

(a)

(b)

40 — Mechanics - II  dm  Ft = vr  ..  dt 

Here, vr = v

and

dm = λv dt

Here, vr is upwards in case (a) and downwards in case (b). Thus, Ft = λv 2 The direction of Ft is upwards in case (a) and downwards in case (b).

4. Suppose a ball is a projected with speed u at an angle θ with horizontal. It collides at some distance with a wall parallel to y-axis as shown in figure. Let v x and v y be the components of its velocity along x and y-directions at the time of impact with wall. Coefficient of restitution between the ball and the wall is e. Component of its velocity along y-direction (common tangent) v y will remain unchanged while component of its velocity along x-direction (common normal) v x will become ev x is opposite direction. v

vy vx u

vy

evx

y

θ

x

Further, since v y does not change due to collision, the time of flight (time taken by the ball to return to the same level) and maximum height attained by the ball will remain same as it would had been in the absence of collision with the wall. Thus, A E

u

D u

O

θ

hA = hE =

θ O

tOAB = tCD + tDEF = T = and

C

F

B

2 u sin θ g

u 2 sin2 θ 2g

Further,

CO + OF ≤ Range or OB

It collision is elastic, then

CO + OF = Range =

and if it is inelastic,

CO + OF < Range

u 2 sin 2 θ g

5. In the projectile motion as shown in figure, y

uy To :Ho : Ro O

ux

1 P

2 Q

x

Centre of Mass, Linear Momentum and Collision — 41

Chapter 11

T= H=

⇒

2u y g

⇒T ∝ u y

u y2

⇒

2g

H ∝ u y2

 2 uy  R = u xT = u x   ⇒ R ∝ uxuy  g  euy

P

ux

=

uy Just before collision

ux P

Just after collision

As shown in above figure, vertical component of velocity just after collision becomes eu y or e times, while horizontal component remains unchanged. Hence, the next time, T will become e times (as T ∝ u y ), H will become e 2 times (as H ∝ u y2 ) and R will also becomes e times (as R ∝ u x u y ). Thus, if T0, H0 and R 0 are the initial values then after first collision, T1 = eT0 , H1 = e 2H0 and R1 = eR 0 Similarly after n-collisions, Tn = enT0 , Hn = e 2nH0 and Rn = enR 0

6. Thrust force in variable mass system is nothing but a result of law of conservation of linear momentum. Let us take an example.

smooth

A boy is standing over a trolley kept over a smooth surface. If the boy throws a stone towards right, then mass of (trolley + boy) system is decreasing. Relative velocity of stone is towards right. So, a thrust force will act on (trolley + boy) system towards left (in a direction opposite to relative velocity, as mass is decreasing). Due to this thrust force (trolley + boy) system will move towards left. And this is nothing but law of conservation of linear momentum. Initial momentum of system was zero. Therefore final momentum should also be zero. If momentum of stone is towards right, then momentum of (trolley + boy) system should be towards left (of same magnitude), to make total momentum equal to zero.

7. In perfectly inelastic collision, all bodies stick together and they have a common velocity given by: v common =

Total momentum p Total = Total mass M Total

and this common velocity is also equal to the velocity of centre of mass of the system.

Solved Examples TYPED PROBLEMS Type 1. Based on law of conversation of linear momentum.

Concept (i) If net force on a system is zero, then linear momentum of the system (or centre of mass remains constant). (ii) Normally the ground is given smooth. So, the net force on a system in horizontal direction will be zero and momentum of the system in horizontal direction will remain constant. ∴Initial momentum = Final momentum or (in horizontal direction) pi = p f (iii) If all individual bodies/blocks of the system are initially at rest then, pi will be zero. Therefore, total momentum of the system at any instant or p f is also zero. (iv) Since, we are using only one conservation law i.e. law of conservation of linear momentum, so we have only one equation and only one unknown. V

Example 1 A trolley of mass M is at rest over a smooth horizontal surface as shown in figure. Two boys each of mass ‘m’ are standing over the trolley. They jump from the trolley (towards right) with relative velocity vr [relative to velocity of trolley just after jumping] (a) together (b) one after the other. Find velocity of trolley in both cases. Solution (a) Let velocity of trolley just after jumping is v1 (towards left). Relative velocity of boys towards right is vr. Therefore, there absolute velocity is vr − v1, towards right. M + 2m At rest Before jumping

+ve v1

M

2m

After jumping

(vr – v1)

Net force on the system in horizontal direction is zero. Hence, linear momentum of the system in horizontal direction is zero. or pi = pf ⇒ ⇒

0 = 2m (vr − v1 ) − Mv1 2m vr v1 = 2m + M

Ans.

Centre of Mass, Linear Momentum and Collision — 43

Chapter 11

(b) Let v1 be the velocity of trolley after the first boy jumps. Then, +ve

(M + 2m)

M+m

At rest

m

vr – v 1

v1

pi = pf 0 = m (vr − v1 ) − (M + m) v1 mvr v1 = M + 2m

⇒ ⇒

…(i)

Now, the second boy jumps from the moving trolley and let v2 be the velocity of trolley after the second boy also jumps. Then, +ve M+m

m

M

v1

v2

vr – v2

pi = pf − (M + m)v1 = m (vr − v2) − Mv2 mvr + (M + m) v1 v2 = (M + m)

⇒ ∴

 m  =  vr + v1  M + m Substituting the value of v1 from Eq. (i), we have   1 1 v2 = mvr  +  M + m M + 2 m   V

Ans.

Example 2 Two toy trains each of mass ‘M’ are moving in opposite directions with velocities v1 and v2 over two smooth rails. Two stuntmen of mass ‘m’ each are also moving with the trains (at rest w.r.t. trains). When trains are opposite to each other the stuntmen interchange their positions, then find the final velocities of the trains. Solution When the stuntmen are in air (after jumping) they also have horizontal velocities v1 and v2 in opposite directions. Initially (A) and (B) were together. Similarly (C) and (D) were together. m

v1 A

v2

v1

M B

M

+ve

v2 m

D

C V1 System 1

V2 System 2

Now, (A) will fall over (D). So this is one system and suppose velocity of this system is V1 (towards right or in positive direction) after jumping. Similarly, (C) will fall over (B) after jumping.

44 — Mechanics - II Let V 2 is the velocity of this system towards right after jumping. Applying conservation of linear momentum. System 1 pi = pf ⇒

m v1 − M v2 = (M + m) V1 m v1 − M v2 V1 = M +m

⇒

Ans.

pi = pf

System 2 ⇒

M v1 − m v2 = (M + m) V 2 Mv1 − mv2 V2 = M +m

⇒

Ans.

Type 2. Based on conservation on linear momentum and mechanical energy.

Concept In these type of problems, all surfaces are given smooth. So, linear momentum in horizontal direction and mechanical energy of the system remains conserved. Since, we are applying two conservation laws. Therefore, number of equations are two. Hence, number of unknowns are also two. V

Example 3 All surfaces shown in figure are smooth. Find velocity of wedge (of mass M) when the block (of mass m) reaches the bottom of the wedge. m

h

M θ

Solution When the block reaches to the bottom of the wedge, their velocities are as shown in figure. Here v is the absolute velocity (with respect to ground), but vr is the relative velocity (relative to wedge).

vr

v θ

Their absolute velocity components are as shown below: m v

M vr sin θ

+ve (vr cos θ – v)

v

v = (vr sin θ )2 + (vr cos θ − v)2

Chapter 11

Centre of Mass, Linear Momentum and Collision — 45

Linear momentum in horizontal direction is conserved. ∴ or Mechanical energy is also conserved.

pi = pf 0 = m (vr cos θ − v) − Mv

…(i)

∴

Ei = E f 1 1 or mgh = Mv2 + mv2 2 2 1 2 1 or …(ii) mgh = Mv + m [(vr sin θ )2 + (vr cos θ − v)2] 2 2 We have two equations and two unknowns v and vr. Solving these equations, we can find the value of v.

Exercise: Solve these two equations to find the value of v. V

Example 4 All surfaces shown in figure are smooth. Wedge of mass ‘M’ is free to move. Block of mass 'm’ is given a horizontal velocity v0 as shown. Find the maximum height 'h’ attained by ‘m’ (over the wedge or outside it).

v0

M

m

Solution At maximum height vertical component of velocity of ‘m’ will be zero. It will have only horizontal component of velocity and this is equal to the horizontal component of ‘M’ also (think why ?). So at the highest point figure is like this: v v

v h

h

or

Applying law of conservation of linear momentum in horizontal direction, pi = pf ⇒ mv0 = (M + m) v Now, applying law of conservation of mechanical energy, ∴

Ei = E f 1 1 mv02 = (M + m) v2 + mgh 2 2

We have two unknowns v and h. So, we can find the value of ‘h’ by solving these two equations. Exercise : Solve these two equations to find the value of ‘h’.

…(i)

…(ii)

46 — Mechanics - II Type 3. Based on following four equations. Σ FR = Σ FL , Σ m R a R = Σ m L a L , Σ m R v R = Σ m L v L , Σ m R x R = Σ m L x L In these equations, R stands for right hand side and L stands for left hand side. x is the absolute displacement with respect to ground.

Concept If net force on a system in a particular direction is zero (normally in horizontal direction). And this can done by giving the ground smooth. Initially, the system is at rest. So, in this case individual bodies can move towards right or towards left, but centre of mass will remain stationary. Further, net force in horizontal direction is zero. Hence, total force towards right is equal to the total force towards left or, …(i) Σ FR = Σ FL or …(ii) Σ mR a R = Σ mL a L Now, integrating ‘a’ we will get ‘v’ and by further integrating ‘v’, we will get ‘x’. …(iii) ∴ Σ mR vR = Σ mL vL and …(iv) Σ mR x R = Σ mL x L Note

V

If the system is initially not at rest but net force is zero, then Eqs. (i) and (ii) are still applicable but not the Eqs. (iii) and (iv). Think why ?

Example 5 A wooden plank of mass 20 kg is resting on a smooth horizontal floor. A man of mass 60 kg starts moving from one end of the plank to the other end. The length of the plank is 10 m. Find the displacement of the plank over the floor when the man reaches the other end of the plank.

10 m

Solution Here, the system is man + plank. Net force on this system in horizontal direction is zero and initially the centre of mass of the system is at rest. Therefore, the centre of mass does not move in horizontal direction. Let x be the displacement of the Plank. Assuming the origin, i.e. x = 0 at the position shown in figure.

x=0

x

10 m

10 – x

x Initial position

Final position

As we said earlier also, the centre of mass will not move in horizontal direction (x-axis). Therefore, for centre of mass to remain stationary, xi = xf

Chapter 11

Centre of Mass, Linear Momentum and Collision — 47

 10  10  (60)(0) + 20   (60)(10 − x) + 20  − x 2 2  = 60 + 20 60 + 20 5 = 4

or

  10 − x 6 (10 − x) + 2   2 8

=

60 − 6x + 10 − 2x 8

5 = 30 − 3x + 5 − x 4x = 30 30 x= m or x = 7.5 m 4

or or or

Note The centre of mass of the plank lies at its centre. Alternate Method

and Applying or or ∴ V

xL = displacement of plank towards left = x mL = mass of plank displaced towards left = 20 kg xR = displacement of man relative to ground towards right = 10 − x mR = mass of man displaced towards right = 60 kg xRmR = xLmL , we get (10 − x)(60) = 20x x = 30 − 3x 4x = 30 30 x= = 7.5 m 4

Example 6 A man of mass m1 is standing on a platform of mass m2 kept on a smooth horizontal surface. The man starts moving on the platform with a velocity vr relative to the platform. Find the recoil velocity of platform. Solution Absolute velocity of man = vr − v where v = recoil velocity of platform.

vr – v v

Taking the platform and the man as a system, net external force on the system in horizontal direction is zero. The linear momentum of the system remains constant. Initially both the man and the platform were at rest. Hence, 0 = m1 (vr − v) − m2v m1vr v= ∴ m1 + m2 Alternate Method we have, ∴

Using the equation, ΣmR vR = ΣmL vL m1 (vr − v) = m2v m1vr v= m1 + m2

48 — Mechanics - II V

Example 7 A block of mass m is released from the top of a wedge of mass M as shown in figure. Find the displacement of wedge on the horizontal ground when the block reaches the bottom of the wedge. Neglect friction everywhere. m

y

h M 30°θ

x

Solution Here, the system is wedge + block. Net force on the system in horizontal direction (x-direction) is zero, therefore, the centre of mass of the system will not move in x-direction so we can apply, xRmR = xLmL

…(i)

Let x be the displacement of wedge. Then, xL = displacement of wedge towards left = x mL = mass of wedge moving towards left = M xR = displacement of block with respect to ground towards right = h cot θ − x and

mR = mass of block moving towards right = m

Substituting in Eq. (i), we get ∴

m (h cot θ − x) = xM mh cot θ x= M +m

Ans.

Type 4. Explosion of a bomb or a projectile.

Concept If a bomb or a projectile explodes in two or more than two parts, then it explodes due to internal forces. Therefore, net force or net external force is zero. Hence, linear momentum of the system can be conserved just before and just after explosion. By this momentum conservation equation, we can find the velocity of some unknown part. If explosion takes place in air, then during the explosion, the external force due to gravity (= weight) can be neglected, as the time of explosion is very short. So, impulse of this force is negligible and impulse is change in linear momentum.Hence, change in linear momentum is also negligible. V

Example 8 A bomb of mass ‘5m’ at rest explodes into three parts of masses 2m, 2m and m. After explosion, the equal parts move at right angles with speed v each. Find speed of the third part and total energy released during explosion. Solution Let the two equal parts move along positive x and positive y directions and suppose the velocity of third part is V. From law of conservation of linear momentum, we have, pi = pf ⇒ 0 = 2m (v$i ) + 2 m (v $j) + mV

Chapter 11

Centre of Mass, Linear Momentum and Collision — 49 V = − 2 v$i − 2 v$j

Solving this equation we have, ∴Speed of this particle

= V or V = (− 2 v)2 + (− 2 v)2 = 2 2v

Ans.

Energy released during explosion, = kinetic energy of all three parts 1 1 1 = (2m)v2 + (2m)v2 + (m) (2 2v)2 2 2 2 = 6 mv2 V

Ans.

Example 9 A projectile of mass 3 kg is projected with velocity 50 m/s at 37° from horizontal. After 2 s, explosion takes place and the projectile breaks into two parts of masses 1 kg and 2 kg. The first part comes to rest just after explosion. Find, (a) the velocity of second part just after explosion. (b) maximum height attained by this part. Take g = 10 m/ s2

Solution

y 50 m/s

a=g

37°

x

O

u = (50 cos 37° ) $i + (50 sin 37° ) $j = (40 $i + 30 $j) m/s

(a)

a = (−10$j) m/s 2 = constant After t = 2 s

v = u + at = (40$i + 30$j) + (− 10$j) (2) = (40$i + 10$j) m/s S = ut +

1 2 at 2

1 = (40i$ + 30$j)(2) + (− 10$j)(2)2 = (80$i + 40$j)m 2 sy = 40 m Hence, the explosion takes place at a height of 40 m and the velocity just before explosion is (40$i + 10$j) m/s. Velocity of first part just after explosion is zero and let velocity of second part just after explosion is V , then from conservation of linear momentum, ⇒ ⇒

pi = pf $ 3 (40i + 10$j) = (1) (0) + 2 V V = (60 $i + 15$j) m/s

Ans.

50 — Mechanics - II (b) Vertical component of velocity of second part just after collision is 15 m/s and explosion has taken place at a height of 40 m. Therefore total height of this part from ground. vy2 (15)2 Ans. = 40 + = 40 + = 51.25 m 2g 2 × 10

Type 5. Based on variable mass system.

Concept dm has to be applied on dt the system, whose mass is changing. Direction of this force is in the direction of v r if mass is increasing and in the opposite direction of v r , if mass is decreasing, Further, v r is zero if a mass is just dropped from a moving body. Because the dropped body has the same velocity as of the moving body at the time of dropping. So, no thrust force will act in this case. In case of a variable mass system, a thrust force of magnitude v r ×

V

Example 10 A constant force F is applied on a trolley of initial mass m0 kept over a smooth surface. Sand is poured gently over the trolley at a constant rate of ( µ ) kg / s. After time t, find

F

Smooth

(a) mass of the trolley (with sand) (b) net force on the trolley (c) velocity of trolley Solution (a) Mass of the trolley after time t is m = mass of trolley + (mass of sand poured per second) (time) = m0 + µt (b) Let v is the velocity of trolley at time t. Sand is poured gently. So, velocity of sand is zero or, relative velocity of sand is v in the opposite direction of velocity of trolley. Mass of trolley is increasing. So, thrust force on the trolley is in the direction of relative velocity or in the opposite direction of motion of trolley. vr = v F v Thrust force = Ft

Chapter 11

Centre of Mass, Linear Momentum and Collision — 51

Net force on trolley = F − Ft = F − vr

dm = F − (v) (µ ) dt

Fnet = F − µv

or

Ans.

Note In the above expression of Fnet , velocity v is a function of time which has been asked in the next part. (c) Fnet = F − v µ ∴ ⇒ ⇒

ma = F − µv  dv (m0 + µt )   = F − µv  dt  dv

v

dt

t

∫0 F − µv = ∫0 m0 + µt

Solving this equation we get v= V

Ft m0 + µt

Ans.

Example 11 A trolley of initial mass m0 is kept over a smooth surface as shown in figure. A constant force F is applied on it. Sand kept inside the trolley drains out from its floor at a constant rate of ( µ ) kg / s. After time t find: F

Smooth

(a) total mass of trolley and sand. (b) net force on the trolley. (c) velocity of trolley. Solution (a) Total mass of trolley and sand m = initial mass − mass of sand drained out or m = m0 − µt (b) Let v is the velocity of trolley at time t. Then, velocity of sand drained out is also v or relative velocity is zero. Hence, no thrust force will act in this case. Therefore, Fnet = F (c) Fnet = F or

ma = F

or

(m0 − µt )

dv =F dt

v

t

or m

dv =F dt

Fdt

∫0 dv = ∫0 m0 − µt

∴ Solving this equation we get, v=

 m0  1 log e    m0 − µt  µ

Ans.

52 — Mechanics - II Type 6. Problems based on linear impulse (J) and coefficient of restitution (e).

Concept (i) During collision, equal and opposite impulses act on the two colliding bodies along common normal directions. This impulse changes the linear momentum of an individual body, but total momentum of the system remains constant, as the total linear impulse is zero. Further, this linear impulse is equal to change in linear momentum. (ii) Coefficient of restitution between two colliding bodies is defined along common normal direction and this is given by: relative speed of separation e= relative speed of approach V

Example 12 Two balls of masses m and 2m and 2p 4p momenta 4p and 2p (in the directions shown) collide as m 2m shown in figure. During collision, the value of linear impulse between them is J. In terms of J and p find coefficient of restitution ‘e’. Under what condition collision is elastic. Also find the condition of perfectly inelastic collision. Solution Directions of linear impulses on the two colliding bodies at the time of collision are shown in Fig. (b). 4p m

v1

v2

2p 2m

J

J

m

2m

J – 4p v 1¢

At the time to collision (b)

Before collision (a)

m

J – 2p 2m

v 2¢

After collision (c)

Linear impulse is equal to the change in linear momentum. Hence, momenta of the balls after collision are shown in Fig. (c). Let v1 and v2 are their velocities before collision [in the directions shown in figure (a)] and v1′ and v2′ are the velocities after collision as shown in figure (c). linear momentum v= mass relative speed of separation e= relative speed of approach J −4p J −2p + v1′ + v2′ 2m or e= = m 4p 2p v1 + v2 + m 2m 3J − 10 p 3J or Ans. e= = −1 10 p 10 p For elastic collision, e=1 ⇒

3J = 2 or 10 p

J=

3J = 1 or 10 p

J=

20 p 3

Ans.

10 p 3

Ans.

For perfectly inelastic collision, e=0

⇒

Chapter 11

Centre of Mass, Linear Momentum and Collision — 53

Type 7. Based on conservation of linear momentum and vertical circular motion.

Concept A ball of mass M is suspended from a massless string of length l as shown in figure. A bullet of mass m moving with velocity v 0 collides with the ball and sticks with the ball. l Now, velocity of the combined mass ( M + m) just after collision at the bottommost point (say it's u) can by obtained from law of conservation of v0 linear momentum. After finding the value of u we can use the theory of m M vertical circular motion. For example, if u ≥ 5 gl , then the combined mass will complete the vertical circular motion. If 2 gl < u < 5 gl , string will slack in upper half of the circle and if 0 < u ≤ 2 gl, the combined mass will oscillate in lower half of the circle. V

Example 13 In the situation discussed above, find (a) velocity of combined mass just after collision at the bottommost point (or u). (b) loss of mechanical energy during collision. (c) minimum value of v0 so that the combined mass completes the vertical circular motion. Solution (a) Applying conservation of linear momentum, just before and just after collision. pi = pf

⇒

mv0 = (M + m) u ⇒ u =

mv0 M +m

Ans.

l v0 m

u M

(M+m)

Just before collision

Just after collision

(b) Loss of mechanical energy during collision, = Ei − E f 1 1 = mv0 − (M + m) u 2 2 2 =

 mv0  1 1 mv02 − (M + m)   2 2 M + m

=

1 Mm v02 2 (M + m)

(E = mechanical energy)

2

Ans.

(c) For completing the vertical circular motion, velocity at bottommost point, u ≥ 5 gl mv0 M +m ∴ ≥ 5 gl ⇒ ∴ v0 ≥ 5 gl M +m m or

(v0 )min =

M +m 5 gl m

Ans.

54 — Mechanics - II Example 14 A pendulum bob of mass 10−2 kg is raised to a height 5 × 10−2 m and then released. At the bottom of its swing, it picks up a mass 10−3 kg. To what height will the combined mass rise ? Solution Velocity of pendulum bob at bottom most point v1 = 2 gh = 2 × 10 × 5 × 10−2 = 1 m/s When the bob picks up a mass 10−3 kg at the bottom, then by conservation of linear momentum the velocity of combined mass is given by 10 or

−2

m1v1 + m2v2 = (m1 + m2) v × 1 + 10−3 × 0 = (10−2 + 10−3 ) v 10−2 10 m/s v= = 1.1 × 10−2 11 h=

Now,

v2 (10 /11)2 = = 4.1 × 10−2 m 2g 2 × 10

Ans.

Miscellaneous Examples V

Example 15 The friction coefficient between the horizontal surface and each of the block shown in the figure is 0.2. The collision between the blocks is perfectly elastic. Find the separation between them when they come to rest. (Take g = 10 m/s 2 ). 1.0 m/s 2kg

4kg

16 cm

Solution Retardation, a =

µmg = µg = 0.2 × 10 = 2 m/s 2 m

Velocity of first block before collision, v12 = 12 − 2 (2) × 0.16 = 1 − 0.64 v1 = 0.6 m/s By conservation of momentum, 2 × 0.6 = 2v′1 + 4v′2 also v′2 − v1′ = v1 for elastic collision It gives v′2 = 0.4 m/s, v1′ = − 0.2 m/s Now distance moved after collision (0.4) 2 s1 = 2 ×2 and ∴

s2 =

(0.2) 2 2 ×2

s = s1 + s2 = 0.05m = 5 cm

Ans.

Chapter 11 V

Centre of Mass, Linear Momentum and Collision — 55

Example 16 Three identical balls, ball I, ball II and ball III are placed on a smooth floor on a straight line at the separation of 10 m between balls as shown in figure. Initially balls are stationary. Ball I is given velocity of 10 m/s towards ball II, collision between balls I and II is inelastic with coefficient of restitution 0.5 but collision between balls II and III is perfectly elastic. What is the time interval between two consecutive collisions between ball I and II ? I

II

10 m

III

10 m

Solution Let velocity of I ball and II ball after collision be v1 and v2 v2 − v1 = 0.5 × 10 K (i) mv2 + mv1 = m × 10 K (ii) ⇒ v2 + v1 = 10 Solving Eqs. (i) and (ii), we get v1 = 2.5 m/s, v2 = 7.5 m/s Ball II after moving 10 m collides with ball III elastically and stops. But ball I moves towards ball II. Time taken between two consecutive collisions 2.5 10 − 10 × 10 7.5 = 4 s Ans. − 7.5 2.5 V

Example 17 A plank of mass 5 kg is placed on a frictionless horizontal plane. Further a block of mass 1 kg is placed over the plank. A massless spring of natural length 2 m is fixed to the plank by its one end. The other end of spring is compressed by the block by half of spring’s natural length. They system is now released from the rest. What is the velocity of the plank when block leaves the plank ? (The stiffness constant of spring is 100 N/m) 1 kg 5 kg 4m

Solution Let the velocity of the block and the plank, when the block leaves the spring be u and v respectively. By conservation of energy

1 2 1 1 kx = mu 2 + Mv2 [M = mass of the plank, m = mass of the block] 2 2 2

⇒ By conservation of momentum

Solving Eqs. (i) and (ii)

100 = u 2 + 5v2 mu + Mv = 0 ⇒ u = − 5v 10 m/s 30v2 = 100 ⇒ v = 3

From this moment until block falls, both plank and block keep their velocity constant. 10 m/s. Thus, when block falls, velocity of plank = 3

K (i) K (ii)

Ans.

56 — Mechanics - II V

Example 18 A ball is projected from the ground with speed u at an angle α with horizontal. It collides with a wall at a distance a from the point of projection and returns to its original position. Find the coefficient of restitution between the ball and the wall. Solution As we have discussed in the theory, the horizontal component of the velocity of ball during the path OAB is u cos α while in its return journey BCO it is eu cos α . The time of flight T also remains unchanged. Hence, T = tOAB + tBCO 2u sin α a a = + g u cos α eu cos α

or or

a 2u sin α a = − eu cos α g u cos α

or

a 2u 2 sin α cos α − ag = eu cos α gu cos α

or

V

e=

ag 2u 2 sin α cos α − ag

e=

1  u 2 sin 2α  − 1  ag  

C

A O a

Example 19 A ball of mass m = 1 kg falling vertically with a velocity v0 = 2 m/s strikes a wedge of mass M = 2 kg kept on a smooth, horizontal surface as shown in figure. If impulse between ball and wedge during collision is J . Then make two equations which relate J with velocity components of wedge and ball. Also find impulse on wedge from ground during impact.

m v0

M 30°

Solution Given M = 2 kg and m = 1 kg Let, J be the impulse between ball and wedge during collision and v1 , v2 and v3 be the components of velocity of the wedge and the ball in horizontal and vertical directions respectively. v3 v1

J

J cos 30°

m

v2

M

m

J sin 30° 30°

J sin 30°

30° J

30°

J cos 30°

Applying we get or

impulse = change in momentum J sin 30° = Mv1 = mv2 J = 2v1 = v2 2

…(i)

Chapter 11

Centre of Mass, Linear Momentum and Collision — 57

J cos 30° = m (v3 + v0 ) 3 or J = (v3 + 2) 2 So these are two equations relating J and velocity components of wedge and ball.

…(ii)

Further, net vertical impulse on wedge should be zero. Therefore, impulse on wedge from J ground is J sin 30° or in upward direction. 2 V

Example 20 Two blocks of equal mass m are connected by an unstretched spring and the system is kept at rest on a frictionless horizontal surface. A constant force F is applied on one of the blocks pulling it away from the other as shown in figure. (a) Find the displacement of the centre of mass at time t. (b) If the extension of the spring is x 0 at time t, find the displacement of the two blocks at this instant. k m

m

F

Solution (a) The acceleration of the centre of mass is F 2m The displacement of the centre of mass at time t will be 1 Ft 2 x = a COM t 2 = 2 4m a COM =

Ans.

(b) Suppose the displacement of the first block is x1 and that of the second is x2. Then, mx1 + mx2 Ft 2 x1 + x2 or x= = 2m 4m 2 Ft 2 or x1 + x2 = 2m

…(i)

Further, the extension of the spring is x1 − x2. Therefore, x1 − x2 = x0 From Eqs. (i) and (ii),   1  Ft 2 1  Ft 2 x1 =  + x0 and x2 =  − x0 . 2  2m 2  2m   V

Example 21 A block of mass m is connected to another block of mass M by a massless spring of spring constant k. The blocks are kept on a smooth horizontal plane. Initially, the blocks are at rest and the spring is unstretched when a constant force F starts acting on the block of mass M to pull it. Find the maximum extension of the spring. k m

M

F

F1

k m

M

F2

58 — Mechanics - II Solution The centre of mass of the system (two blocks + spring) moves with an acceleration F . Let us solve the problem in a frame of reference fixed to the centre of mass of the m+ M system. As this frame is accelerated with respect to the ground, we have to apply a pseudo force ma towards left on the block of mass m and Ma towards left on the block of mass M. The net external force on m is mF (towards left) F1 = ma = m+ M a=

and the net external force on M is F2 = F − Ma = F −

MF mF = m+ M m+ M

(towards right)

As the centre of mass is at rest in this frame, the blocks move in opposite directions and come to instantaneous rest at some instant. The extension of the spring will be maximum at this instant. Suppose, the left block is displaced through a distance x1 and the right block through a distance x2 from the initial positions. The total work done by the external forces F1 and F2 in this period are mF W = F1x1 + F2x2 = (x1 + x2) m+ M This should be equal to the increase in the potential energy of the spring, as there is no change in the kinetic energy. Thus, mF 1 2mF Ans. (x1 + x2) = k (x1 + x2)2 or x1 + x2 = m+ M 2 k(m + M ) This is the maximum extension of the spring. V

Example 22 Two blocks A and B of masses m and 2m respectively are placed on a smooth floor. They are connected by a spring. A third block C of mass m moves with a velocity v0 along the line joining A and B and collides elastically with A, as shown in figure. At a certain instant of time t0 after collision, it is found that the instantaneous velocities of A and B are the same. Further, at this instant the compression of the spring is found to be x 0 . Determine (i) the common velocity of A and B at time t0 , and (ii) the spring constant. B C m

v0

A m

2m

Solution Initially, the blocks A and B are at rest and C is moving with velocity v0 to the right. As masses of C and A are same and the collision is elastic the body C transfers its whole momentum mv0 to body A and as a result the body C stops and A starts moving with velocity v0 to the right. At this instant the spring is uncompressed and the body B is still at rest. The momentum of the system at this instant = mv0 Now, the spring is compressed and the body B comes in motion. After time t0 , the compression of the spring is x0 and common velocity of A and B is v (say). As external force on the system is zero, the law of conservation of linear momentum gives v Ans. mv0 = mv + (2m)v or v = 0 3

Chapter 11

Centre of Mass, Linear Momentum and Collision — 59

The law of conservation of energy gives 1 1 mv02 = mv2 + 2 2 1 3 or mv02 = mv2 + 2 2

1 1 (2m)v2 + kx02 2 2 1 2 kx0 2 2

1 3 v  1 mv02 = m  0  + kx02 2 2  3 2 1 2 1 1 kx0 = mv02 − mv02 2 2 6 1 2 1 2 kx0 = mv0 2 3 2 mv02 k= 3 x02

∴ or

V

Ans.

Example 23 A uniform chain of mass m and length l hangs on a thread and touches the surface of a table by its lower end. Find the force exerted by the table on the chain when half of its length has fallen on the table. The fallen part does not form heap.

Solution Force exerted by the chain on the table consists of two parts: A

v=

√2g 2l

B

= √gl

C

1. Weight of the portion BC of the chain lying on the table, mg (downwards) W = 2 2. Thrust force Ft = λv2 Here,

∴ ∴

λ = mass per unit length of chain = v2 = ( gl )2 = gl  m Ft =   ( gl) = mg  l

Net force exerted by the chain on the table is mg 3 F = W + Ft = + mg = mg 2 2

m l

(downwards)

(downwards)

60 — Mechanics - II So, from Newton’s third law the force exerted by the table on the chain will be

3 mg 2

(vertically upwards).  dm   dt 

Note Here, the thrust force ( Ft ) applied by the chain on the table will be vertically downwards, as Ft = v r  and in this expression v r is downwards plus V

dm is positive. So, Ft will be downwards. dt

Example 24 A ball of mass m makes an elastic collision with another identical ball at rest. Show that if the collision is oblique, the bodies go at right angles to each other after collision. Solution In head on elastic collision between two particles, they exchange their velocities. In

this case, the component of ball 1 along common normal direction, v cos θ becomes zero after collision, while that of 2 becomes v cos θ. While the components along common tangent direction of both the balls remain unchanged. Thus, the components along common tangent and common normal direction of both the balls in tabular form are given below. v sin θ

v sin θ θ

v

1

1 v cos θ

2 2 Before collision

Ball

Component along common tangent direction

v cos θ After collision

Component along common normal direction

Before collision

After collision

Before collision

After collision

1

v sin θ

v sin θ

v cos θ

0

2

0

0

0

v cos θ

From the above table and figure, we see that both the balls move at right angles after collision with velocities v sin θ and v cos θ.

Exercises LEVEL 1 Assertion and Reason Directions : Choose the correct option. (a) If both Assertion and Reason are true and the Reason is correct explanation of the Assertion. (b) If both Assertion and Reason are true but Reason is not the correct explanation of Assertion. (c) If Assertion is true, but the Reason is false. (d) If Assertion is false but the Reason is true. 1. Assertion : Centre of mass of a rigid body always lies inside the body. Reason :

Centre of mass and centre of gravity coincide if gravity is uniform.

2. Assertion : A constant force F is applied on two blocks and one spring system as shown in figure. Velocity of centre of mass increases linearly with time. m

2m

F

Smooth

Reason :

Acceleration of centre of mass is constant.

3. Assertion : To conserve linear momentum of a system, no force should act on the system. Reason :

If net force on a system is zero, its linear momentum should remain constant.

4. Assertion : A rocket moves forward by pushing the surrounding air backwards. Reason : motion.

It derives the necessary thrust to move forward according to Newton’s third law of

5. Assertion : Internal forces cannot change linear momentum. Reason :

Internal forces can change the kinetic energy of a system.

6. Assertion : In case of bullet fired from gun, the ratio of kinetic energy of gun and bullet is equal to ratio of mass of bullet and gun. 1 Reason : Kinetic energy ∝ ; if momentum is constant. mass

7. Assertion : All surfaces shown in figure are smooth. System is released from rest. Momentum of system in horizontal direction is constant but overall momentum is not constant. A B

Reason :

A net vertically upward force is acting on the system.

8. Assertion : During head on collision between two bodies let ∆p1 is change in momentum of first body and ∆p2 the change in momentum of the other body, then ∆p1 = ∆p2. Reason : Total momentum of the system should remain constant.

62 — Mechanics - II 9. Assertion : In the system shown in figure spring is first stretched then left to oscillate. At some instant kinetic energy of mass m is K . At the same instant kinetic energy of mass 2m K should be . 2 Reason :

Their linear momenta are equal and opposite and K =

m

2m

F

Smooth

p2 1 or K ∝ . 2m m

10. Assertion : Energy can not be given to a system without giving it momentum. Reason :

If kinetic energy is given to a body it means it has acquired momentum.

11. Assertion : The centre mass of an electron and proton, when released moves faster towards proton. Reason :

Proton is heavier than electron.

12. Assertion : The relative velocity of the two particles in head-on elastic collision is unchanged both in magnitude and direction. Reason :

The relative velocity is unchanged in magnitude but gets reversed in direction.

13. Assertion : An object of mass m1 and another of mass m2( m2 > m1 ) are released from certain distance. The objects move towards each other under the gravitational force between them. In this motion, centre of mass of their system will continuously move towards the heavier mass m2. Reason : In a system of a heavier and a lighter mass, centre of mass lies closer to the heavier mass.

14. Assertion : A given force applied in turn to a number of different masses may cause the same rate of change in momentum in each but not the same acceleration to all. Reason :

F=

dp F and a = dt m

15. Assertion : In an elastic collision between two bodies, the relative speed of the bodies after collision is equal to the relative speed before the collision. Reason :

In an elastic collision, the linear momentum of the system is conserved.

Objective Questions Single Correct Option 1. A ball is dropped from a height of 10 m. Ball is embedded in sand through 1 m and stops. (a) only momentum remains conserved (b) only kinetic energy remains conserved (c) both momentum and kinetic energy are conserved (d) neither kinetic energy nor momentum is conserved

2. If no external force acts on a system (a) (b) (c) (d)

velocity of centre of mass remains constant position of centre of mass remains constant acceleration of centre of mass remains non-zero and constant All of the above

Chapter 11

Centre of Mass, Linear Momentum and Collision — 63

3. When two blocks connected by a spring move towards each other under mutual interaction (a) (b) (c) (d)

their velocities are equal their accelerations are equal the force acting on them are equal and opposite All of the above

4. If two balls collide in air while moving vertically, then momentum of the system is conserved because (a) (b) (c) (d)

gravity does not affect the momentum of the system force of gravity is very less compared to the impulsive force impulsive force is very less than the gravity gravity is not acting during collision

5. When a cannon shell explodes in mid air, then identify the incorrect statement (a) (b) (c) (d)

the momentum of the system is conserved at the time of explosion the kinetic energy of the system always increases the trajectory of centre of mass remains unchanged None of the above

6. In an inelastic collision (a) (b) (c) (d)

momentum of the system is always conserved velocity of separation is less than the velocity of approach the coefficient of restitution can be zero All of the above

7. The momentum of a system is defined (a) as the product of mass of the system and the velocity of centre of mass (b) as the vector sum of the momentum of individual particles (c) for bodies undergoing translational, rotational and oscillatory motion (d) All of the above

8. The momentum of a system with respect to centre of mass (a) is zero only if the system is moving uniformly (b) is zero only if no external force acts on the system (c) is always zero (d) can be zero in certain conditions

9. Three identical particles are located at the vertices of an equilateral triangle. Each particle moves along a meridian with equal speed towards the centroid and collides inelastically. (a) (b) (c) (d)

all the three particles will bounce back along the meridians with lesser speed. all the three particles will become stationary all the particles will continue to move in their original directions but with lesser speed nothing can be said

10. The average resisting force that must act on a 5 kg mass to reduce its speed from 65 to 15 ms−1 in 2s is (a) 12.5 N

(b) 125 N

(c) 1250 N

(d) None of these

11. In a carbon monoxide molecule, the carbon and the oxygen atoms are separated by a distance 1.2 × 10−10 m. The distance of the centre of mass from the carbon atom is (a) 0.48 × 10−10 m (c) 0.74 × 10−10 m

(b) 0.51 × 10−10 m (d) 0.68 × 10−10 m

64 — Mechanics - II 12. A bomb of mass 9 kg explodes into two pieces of masses 3 kg and 6 kg. The velocity of mass 3 kg is 16 ms−1. The kinetic energy of mass 6 kg is

(a) 96 J

(b) 384 J

(c) 192 J

(d) 768 J

13. A heavy ball moving with speed v collides with a tiny ball. The collision is elastic, then immediately after the impact, the second ball will move with a speed approximately equal to (a) v

(b) 2v

(c)

v 2

(d)

v 3

14. A loaded 20, 000 kg coal wagon is moving on a level track at 6 ms−1. Suddenly 5000 kg of coal is dropped out of the wagon. The final speed of the wagon is

(a) 6 ms−1

(b) 8 ms−1

(c) 4.8 ms−1

(d) 4.5 ms−1

15. A machine gun shoots a 40 g bullet at a speed of 1200 ms −1. The man operating the gun can bear a maximum force of 144 N. The maximum number of bullets shot per second is (a) 3

(b) 5

(c) 6

(d) 9

16. A projectile of mass m is fired with a velocity v from point P at an angle 45°. Neglecting air resistance, the magnitude of the change in momentum leaving the point P and arriving at Q is v 45° Q

P

(a) 2 mv

(b) 2mv

mv (c) 2

(d)

mv 2

17. A ball after freely falling from a height of 4.9 m strikes a horizontal plane. If the coefficient of restitution is 3/4, the ball will strike second time with the plane after (a)

1 s 2

(b) 1 s

(c)

3 s 2

(d)

3 s 4

18. The centre of mass of a non uniform rod of length L, whose mass per unit length varies as

k ⋅ x2 where k is a constant and x is the distance of any point from one end is (from the same L end)

ρ=

 3 (a)   L  4

 1 (b)   L  4

 1 (c)   L  6

 2 (d)   L  3

19. A boat of length 10 m and mass 450 kg is floating without motion in still water. A man of mass 50 kg standing at one end of it walks to the other end of it and stops. The magnitude of the displacement of the boat in metres relative to ground is (a) zero

(b) 1 m

(c) 2 m

(d) 5 m

20. A man of mass M stands at one end of a stationary plank of length L, lying on a smooth surface. The man walks to the other end of the plank. If the mass of the plank is M / 3, the distance that the man moves relative to the ground is (a)

3L 4

(b)

L 4

(c)

4L 5

(d)

L 3

21. A ball of mass m moving at a speed v collides with another ball of mass 3m at rest. The lighter block comes to rest after collision. The coefficient of restitution is (a)

1 2

(b)

2 3

(c)

1 4

(d) None of these

Chapter 11

Centre of Mass, Linear Momentum and Collision — 65

22. A particle of mass m moving with velocity u makes an elastic

F

one-dimensional collision with a stationary particle of mass m. They F come in contact for a very small time t0. Their force of interaction 0 increases from zero to F0 linearly in time 0.5t0, and decreases linearly to zero in further time 0.5 t0 as shown in figure. The magnitude of F0 is mu t0 mu (c) 2t0

(a)

(b)

2mu t0

t0

0.5t0

t

(d) None of these

23. Two identical blocks A and B of mass m joined together with a massless spring as shown in figure are placed on a smooth surface. If the block A moves with an acceleration a0, then the acceleration of the block B is m

m

A

B

(b) −a 0

(a) a 0

(c)

F

F − a0 m

(d)

F m

24. A ball of mass m moving with velocity v0 collides a wall as shown

3 v0. The m 4

in figure. After impact it rebounds with a velocity impulse acting on ball during impact is

v0 X

37°

m ^ (a) − v0 j 2 3 ^ (b) − mv0 i 4 −5 ^ (c) mv0 i 4 (d) None of the above

53° 3v 0 4

Y

25. A steel ball is dropped on a hard surface from a height of 1 m and rebounds to a height of 64 cm. The maximum height attained by the ball after n th bounce is (in m) (a) (0.64)2n (c) (0.5)2n

(b) (0.8)2n (d) (0.8)n

26. A car of mass 500 kg (including the mass of a block) is moving on a smooth road with velocity 1.0 ms−1 along positive x-axis. Now a block of mass 25 kg is thrown outside with absolute velocity of 20 ms−1 along positive z-axis. The new velocity of the car is (ms−1) v

Y

X Z

^

(a) 10 i + 20 k 20 ^ 20 ^ (c) i− k 19 19 ^

^

^

(b) 10 i − 20 k 20 ^ ^ (d) 10 i − k 19

66 — Mechanics - II 27. The net force acting on a particle moving along a straight line varies with time as shown in the diagram. Force is parallel to velocity. Which of the following graph is best representative of its speed with time ? (Initial velocity of the particle is zero) F

t v

v

(a)

v

(b) t

v

(c)

(d)

t

t

t

28. In the figure shown, find out centre of mass of a system of a uniform circular plate of radius 3R from O in which a hole of radius R is cut whose centre is at 2R distance from the centre of large circular plate 2R O R 3R

R 2 R (c) 4

(a)

(b)

R 5

(d) None of these

29. From the circular disc of radius 4R two small discs of radius R are cut

y

off. The centre of mass of the new structure will be at R ^R + j 5 5 ^ R ^ R (b) − i + j 5 5 ^ R ^ R (c) − i − j 5 5 (d) None of the above ^

(a) i

4R

30. A block of mass m rests on a stationary wedge of mass M. The wedge can slide freely on a smooth horizontal surface as shown in figure. If the block starts from rest (a) the position of the centre of mass of the system will change (b) the position of the centre of mass of the system will change along the vertical but not along the horizontal (c) the total energy of the system will remain constant (d) All of the above

m M

O

R

x

Chapter 11

Centre of Mass, Linear Momentum and Collision — 67

31. A bullet of mass m hits a target of mass M hanging by a string and gets embedded in it. If the block rises to a height h as a result of this collision, the velocity of the bullet before collision is (a) v = 2 gh

m  (b) v = 2 gh 1 +  M 

M  (c) v = 2 gh 1 +  m 

m  (d) v = 2 gh 1 −  M 

32. A loaded spring gun of mass M fires a bullet of mass m with a velocity v at an angle of elevation θ. The gun is initially at rest on a horizontal smooth surface. After firing, the centre of mass of the gun and bullet system v m M vm in the horizontal direction (b) moves with velocity M cos θ (c) does not move in horizontal direction v (M − m) (d) moves with velocity in the horizontal direction M +m

(a) moves with velocity

33. Two bodies with masses m1and m2 ( m1 > m2 ) are joined by a string passing over fixed pulley. Assume masses of the pulley and thread negligible. Then the acceleration of the centre of mass of the system ( m1 + m2 ) is 2

 m − m2  (a)  1 g  m1 + m2 m1 g (c) m1 + m2

 m − m2  (b)  1  g  m1 + m2 m2g (d) m1 + m2

34. A rocket of mass m0 has attained a speed equal to its exhaust speed and at that time the mass of the rocket is m. Then the ratio (a) 2.718

m0 is ( neglect gravity ) m

(b) 7.8

(c) 3.14

(d) 4

35. A jet of water hits a flat stationary plate perpendicular to its motion. The jet ejects 500g of water per second with a speed of 1 m/ s. Assuming that after striking, the water flows parallel to the plate, then the force exerted on the plate is (a) 5 N (c) 0.5 N

(b) 1.0 N (d) 10 N

36. Two identical vehicles are moving with same velocity v towards an intersection as shown in figure. If the collision is completely inelastic, then (a) the velocity of separation is zero

θ

θ (b) the velocity of approach is 2vsin 2 (c) the common velocity after collision is v cos

v v

θ 2

(d) All of the above

37. A ball of mass m = 1 kg strikes a smooth horizontal floor as shown in figure. The impulse exerted on the floor is m 5 ms–1 53°

(a) 6.25 Ns

(b) 1.76 Ns

m 37°

(c) 7.8 Ns

(d) 2.2 Ns

68 — Mechanics - II 38. A small block of mass m is placed at rest on the top of a smooth wedge of mass M, which in turn is placed at rest on a smooth horizontal surface as shown in figure. It h be the height of wedge and θ is the inclination, then the distance moved by the wedge as the block reaches the foot of the wedge is m

h

M θ

(a)

Mh cot θ M +m

(b)

mh cot θ M +m

(c)

Mh cosecθ M +m

(d)

mh cosecθ M +m

39. A square of side 4 cm and uniform thickness is divided into four squares. The square portion A′ AB′ D is removed and the removed portion is placed over the portion DB′ BC ′. The new position of centre of mass is Y C

D'

O

(a) (2 cm, 2 cm)

C'

B

D

B'

A

A'

(b) ( 2 cm, 3 cm)

X

(c) (2 cm, 2.5 cm)

(d) (3 cm, 3 cm)

40. A boy having a mass of 40 kg stands at one end A of a boat of length 2 m at rest. The boy walks to the other end B of the boat and stops. What is the distance moved by the boat? Friction exists between the feet of the boy and the surface of the boat. But the friction between the boat and the water surface may be neglected. Mass of the boat is 15 kg.

B

A 2m

(a) 0.49 m

(b) 2.46 m

(c) 1.46 m ^

^

(d) 3.2 m ^

41. Three identical particles with velocities v0 i , − 3v0 j and 5v0 k collide successively with each other in such a way that they form a single particle. The velocity vector of resultant particle is (a)

v0 ^ ^ ^ (i + j + k ) 3

(b)

v0 ^ ^ ^ (i − j + k ) 3

(c)

v0 ^ ^ ^ (i − 3 j + k ) 3

(d)

v0 ^ ^ ^ (i − 3 j + 5 k ) 3

M 4M and at the top of 5 5 the trajectory. The smaller mass falls very close to the mortar. In the same time the bigger piece lands a distance D from the mortar. The shell would have fallen at a distance R from the mortar if there was no explosion. The value of D is (neglect air resistance)

42. A mortar fires a shell of mass M which explodes into two pieces of mass

(a)

3R 2

(b)

4R 3

(c)

5R 4

(d) None of these

Chapter 11

Centre of Mass, Linear Momentum and Collision — 69

Subjective Questions 1. Consider a rectangular plate of dimensions a × b. If this plate is considered to be made up of

a b × and we now remove one out of four rectangles. Find the 2 2 position where the centre of mass of the remaining system will be zero. four rectangles of dimensions

y

b

x

O

a

2. The uniform solid sphere shown in the figure has a spherical hole in it. Find the position of its centre of mass. y b

O

x

a

R

3. A gun fires a bullet. The barrel of the gun is inclined at an angle of 45° with horizontal. When the bullet leaves the barrel it will be travelling at an angle greater than 45° with the horizontal. Is this statement true or false?

4. Two blocks A and B of masses mA and mB are connected together by means of a spring and are resting on a horizontal frictionless table. The blocks are then pulled apart so as to stretch the spring and then released. Show that the kinetic energies of the blocks are, at any instant inversely proportional to their masses.

5. Show that in a head on elastic collision between two particles, the transference of energy is maximum when their mass ratio is unity.

6. A particle moving with kinetic energy K makes a head on elastic collision with an identical particle at rest. Find the maximum elastic potential energy of the system during collision.

7. A ball is projected from the ground at some angle with horizontal. Coefficient of restitution between the ball and the ground is e. Let a, b and c be the ratio of times of flight, horizontal range and maximum height in two successive paths. Find a, b and c in terms of e. 1 2

70 — Mechanics - II 8. x-y is the vertical plane as shown in figure. A particle of mass 1 kg is at (10 m, 20 m) at time t = 0. It is released from rest. Another particle of mass 2 kg is at (20 m, 40 m) at the same instant. It is projected with velocity (10 i$ + 10 $j) m/s. After 1 s. Find (a) acceleration, (b) velocity and (c) position of their centre of mass.

y

g = 10 m/s2

x

O

9. A system consists of two particles. At t = 0, one particle is at the origin; the other, which has a

mass of 0.60 kg, is on the y-axis at y = 80 m. At t = 0, the centre of mass of the system is on the y-axis at y = 24 m and has a velocity given by ( 6.0 m/ s3 ) t 2$j . (a) Find the total mass of the system. (b) Find the acceleration of the centre of mass at any time t. (c) Find the net external force acting on the system at t = 3.0 s.

10. A particle of mass 2 kg moving with a velocity 5$i m/s collides head-on with another particle of

mass 3 kg moving with a velocity − 2$i m/s. After the collision the first particle has speed of 1.6 m/ s in negative x direction. Find (a) velocity of the centre of mass after the collision, (b) velocity of the second particle after the collision, (c) coefficient of restitution.

11. A rocket of mass 40 kg has 160 kg fuel. The exhaust velocity of the fuel is 2.0 km/s. The rate of consumption of fuel is 4 kg/s. Calculate the ultimate vertical speed gained by the rocket. ( g = 10 m/ s2 )

12. A boy of mass 60 kg is standing over a platform of mass 40 kg placed over a smooth horizontal

surface. He throws a stone of mass 1 kg with velocity v = 10 m/s at an angle of 45° with respect to the ground. Find the displacement of the platform (with boy) on the horizontal surface when the stone lands on the ground. Take g = 10 m/ s2.

13. A man of mass m climbs to a rope ladder suspended below a balloon of mass M. The balloon is stationary with respect to the ground. (a) If the man begins to climb the ladder at speed v (with respect to the ladder), in what direction and with what speed (with respect to the ground) will the balloon move ? (b) What is the state of the motion after the man stops climbing ?

14. Find the mass of the rocket as a function of time, if it moves with a constant acceleration a, in absence of external forces. The gas escapes with a constant velocity u relative to the rocket and its mass initially was m0.

15. A particle of mass 2 m is projected at an angle of 45° with horizontal with a velocity of 20 2 m/ s. After 1 s explosion takes place and the particle is broken into two equal pieces. As a result of explosion one part comes to rest. Find the maximum height attained by the other part. Take g = 10 m/ s2.

16. A ball of mass 1 kg is attached to an inextensible string. The ball is released from the position shown in figure. Find the impulse imparted by the string to the ball immediately after the string becomes taut. (Take g = 10 m/ s2)

1m

Chapter 11

Centre of Mass, Linear Momentum and Collision — 71

17. Two balls shown in figure are identical. Ball A is moving towards right with a speed v and the second ball is at rest. Assume all collisions to be elastic. Show that the speeds of the balls remain unchanged after all the collisions have taken place.

v B

A

18. A particle of mass 0.1 kg moving at an initial speed v collides with another particle of same mass kept initially at rest. If the total energy becomes 0.2 J after the collision, what would be the minimum and maximum values of v ?

19. A particle of mass m moving with a speed v hits elastically another stationary particle of mass 2 m on a smooth horizontal circular tube of radius r. Find the time when the next collision will take place?

20. In a one-dimensional collision between two identical particles A and B, B is stationary and A has momentum p before impact. During impact, B gives an impulse J to A. Find the coefficient of restitution between A and B ?

21. Two billiard balls of same size and mass are in contact on a billiard table. A third ball of same mass and size strikes them symmetrically and remains at rest after the impact. Find the coefficient of restitution between the balls?

22. Two identical blocks each of mass M = 9 kg are placed on a rough

horizontal surface of frictional coefficient µ = 0.1. The two blocks are joined by a light spring and block B is in contact with a vertical fixed wall as shown in figure. A bullet of mass m = 1 kg and v0 = 10 m/s hits block A and gets embedded in it. Find the maximum compression of spring. (Spring constant = 240 N/m, g = 10 m/s 2)

v0

A

B

m

M

M

23. Block A has a mass of 5 kg and is placed on top of a smooth triangular block, B having a mass of 30 kg. If the system is released from rest, determine the distance, B moves when A reaches the bottom. Neglect the size of block A. A

B 30° 0.5 m

24. A trolley was moving horizontally on a smooth ground with velocity v with respect to the earth. Suddenly a man starts running from rear end of the trolley with a velocity (3/2) v with respect to the trolley. After reaching the other end, the man turns back and continues running with a velocity (3/2) v with respect to trolley in opposite direction. If the length of the trolley is L, find the displacement of the man with respect to earth when he reaches the starting point on the trolley. Mass of the trolley is equal to the mass of the man.

25. A 4.00 g bullet travelling horizontally with a velocity of magnitude 500 m/s is fired into a wooden block with a mass of 1.00 kg, initially at rest on a level surface. The bullet passes through the block and emerges with speed 100 m/s. The block slides a distance of 0.30 m along the surface from its initial position. (a) What is the coefficient of kinetic friction between block and surface ? (b) What is the decrease in kinetic energy of the bullet ? (c) What is the kinetic energy of the block at the instant after the bullet has passed through it ? Neglect friction during collision of bullet with the block.

72 — Mechanics - II 26. A bullet of mass 0.25 kg is fired with velocity 302 m/s into a block of wood of mass m1 = 37.5 kg. It gets embedded into it. The block m1 is resting on a long block m2 and the horizontal surface on which it is placed is smooth. The coefficient of friction between m1 and m2 is 0.5. Find the displacement of m1 on m2 and the common velocity of m1 and m2. Mass m2 = 1.25 kg.

27. A wagon of mass M can move without friction along horizontal rails. A simple pendulum consisting of a sphere of mass m is suspended from the ceiling of the wagon by a string of length l. At the initial moment the wagon and the pendulum are at rest and the string is deflected through an angle α from the vertical. Find the velocity of the wagon when the pendulum passes through its mean position.

28. A block of mass M with a semicircular track of radius R rests on a horizontal frictionless surface shown in figure. A uniform cylinder of radius r and mass m is released from rest at the point A. The cylinder slips on the semicircular frictionless track. How far has the block moved when the cylinder reaches the bottom of the track ? How fast is the block moving when the cylinder reaches the bottom of the track ?

A R

M

29. A ball of mass 50 g moving with a speed 2 m/s strikes a plane surface at an angle of incidence 45°. The ball is reflected by the plane at equal angle of reflection with the same speed. Calculate (a) The magnitude of the change in momentum of the ball. (b) The change in the magnitude of the momentum of the wall.

30. A uniform rope of mass m per unit length, hangs vertically from a support so that the lower end just touches the table top shown in figure. If it is released, show that at the time a length y of the rope has fallen, the force on the table is equivalent to the weight of a length 3y of the rope.

31. Sand drops from a stationary hopper at the rate of 5 kg/s on to a conveyor belt moving with a constant speed of 2 m/s. What is the force required to keep the belt moving and what is the power delivered by the motor, moving the belt ?

32. A 3.0 kg block slides on a frictionless horizontal surface, first moving to the left at 50 m/s. It collides with a spring as it moves left, compresses the spring and is brought to rest momentarily. The body continues to be accelerated to the right by the force of the compressed spring. Finally, the body moves to the right at 40 m/s. The block remains in contact with the spring for 0.020 s. What were the magnitude and direction of the impulse of the spring on the block? What was the spring’s average force on the block?

33. Block A has a mass 3 kg and is sliding on a rough horizontal surface with a velocity vA = 2 m/s when it makes a direct collision with block B, which has a mass of 2 kg and is originally at rest. The collision is perfectly elastic. Determine the velocity of each block just after collision and the distance between the blocks when they stop sliding. The coefficient of kinetic friction between the blocks and the plane is µ k = 0.3. (Take g = 10 m/ s2)

LEVEL 2 Objective Questions Single Correct Option 1. A pendulum consists of a wooden bob of mass m and length l. A bullet of mass m1 is fired towards the pendulum with a speed v1 and it emerges from the bob v with speed 1 . The bob just completes motion along a vertical circle. Then v1 3 is (a)

m m1

(c)

2  m   3  m1 

(b)

5 gl

3m 2m1

m  (d)  1   m

5 gl

O

5 gl

v1 3

m m1 v1 m1

gl

2. A bob of mass m attached with a string of length l tied to a point on ceiling is released from a position when its string is horizontal. At the bottom most point of its motion, an identical mass m gently stuck to it. Find the maximum angle from the vertical to which it rotates.  2 (a) cos −1    3  1 (c) cos −1    4

 3 (b) cos −1    4 (d) 60°

3. A train of mass M is moving on a circular track of radius R with constant speed v. The length of the train is half of the perimeter of the track. The linear momentum of the train will be (a) zero

(b)

2Mv π

(c) MvR

(d) Mv

4. Two blocks A and B of mass m and 2 m are connected together by a light spring of stiffness k. The system is lying on a smooth horizontal surface with the block A in contact with a fixed vertical wall as shown in the figure. The block B is pressed towards the wall by a distance x0 and then released. There is no friction anywhere. If spring takes time ∆t to acquire its natural length then average force on the block A by the wall is (a) zero

(b)

2mk x0 ∆t

(c)

mk x0 ∆t

m

2m

A

B

(d)

5. A striker is shot from a square carom board from a point A exactly at

3mk x0 ∆t

y

−1

midpoint of one of the walls with a speed of 2 ms at an angle of 45° with the x-axis as shown in the figure. The collisions of the striker with the walls of the fixed carom are perfectly elastic. The coefficient of kinetic friction between the striker and board is 0.2. The coordinate of the striker when it stops (taking point O to be the origin) is (in SI units) O 1 1 (a) , 2 2 2 1 (c) ,0 2 2

1 (b) 0, 2 2 1 1 (d) , 2 2 2

45° A 1 L= m √2

x

74 — Mechanics - II m = 4 kg

6. A ball of mass 1 kg is suspended by an inextensible string 1 m long attached to a point O of a smooth horizontal bar resting on fixed smooth supports A and B. The ball is released from A rest from the position when the string makes an angle 30° with the vertical. The mass of the bar is 4 kg. The displacement of bar when ball reaches the other extreme position (in m ) is (a) 0.4 (c) 0.25

(b) 0.2 (d) 0.5

O B 30°

M = 1kg

7. A ball falls vertically onto a floor with momentum p and then bounces repeatedly. If coefficient of restitution is e, then the total momentum imparted by the ball to the floor is p 1−e  1 + e (d) p    1 − e

(a) p(1 + e)

(b)

 1 − e (c) p    1 + e

8. A bullet of mass m penetrates a thickness h of a fixed plate of mass M. If the plate was free to move, then the thickness penetrated will be Mh M +m mh (c) 2(M + m)

2Mh M +m Mh (d) 2(M + m)

(a)

(b)

9. Two identical balls of equal masses A and B, are lying on a smooth surface as shown in the figure. Ball A hits the ball B (which is at rest) with a velocity v = 16 ms−1. What should be the minimum value of coefficient of restitution e between A and B so that B just reaches the highest point of inclined plane? ( g = 10 ms−2 )

A

(a)

2 3

(b)

1 4

v

5m

B

(c)

1 2

(d)

1 3

10. The figure shows a metallic plate of uniform thickness and density. The value L

of l in terms of L so that the centre of mass of the system lies at the interface of the triangular and rectangular portion is L 3 L (c) l = 3

(a) l =

L 2 2 (d) l = L 3

(b) l =

l b

11. Particle A makes a head on elastic collision with another stationary particle B. They fly apart in opposite directions with equal speeds. The mass ratio will be 1 3 1 (c) 4

(a)

1 2 2 (d) 3

(b)

Chapter 11

Centre of Mass, Linear Momentum and Collision — 75 v

12. A particle of mass 4m which is at rest explodes into four equal fragments. All four fragments scattered in the same horizontal plane. Three fragments are found to move with velocity v as shown in the figure. The total energy released in the process is 1 mv2(3 − 2 ) 2 1 (d) mv2 (1 + 2 ) 2

(a) mv2(3 − 2 )

90° v 135°

(b)

(c) 2mv2

v

13. In figures (a), (b) and (c) shown, the objects A, B and C are of same mass. String, spring and pulley are massless. C strikes B with velocity u in each case and sticks it. The ratio of velocity of B in case (a) to (b) to (c) is

A C (a)

(a) 1 : 1 : 1 (c) 3 : 2 : 2

C

B A

B

C

A

(b)

B (c)

(b) 3 : 3 : 2 (d) 1 : 2 : 3

14. A ladder of length L is slipping with its ends against a vertical wall and a horizontal floor. At a certain moment, the speed of the end in contact with the horizontal floor is v and the ladder makes an angle θ = 30° with horizontal. Then, the speed of the ladder’s centre of mass must be v 2 (d) 2v

3 v 2 (c) v

(b)

(a)

15. A body of mass 2 g, moving along the positive x-axis in gravity free space with velocity 20 cms −1

explodes at x = 1 m, t = 0 into two pieces of masses 2/3 g and 4/3 g. After 5s, the lighter piece is at the point (3m, 2m, − 4 m). Then the position of the heavier piece at this moment, in metres is

(a) (1.5, − 1, − 2) (c) (1.5, − 1, − 1)

(b) (1.5, − 2, − 2) (d) None of these

16. A body of mass m is dropped from a height of h. Simultaneously another body of mass 2m is h . If the collision is 2 perfectly inelastic, the velocity of combined mass at the time of collision with the ground will be

thrown up vertically with such a velocity v that they collide at height

(a)

5 gh 4

(b)

gh

(c)

gh 4

(d) None of these

17. A man is standing on a cart of mass double the mass of man. Initially cart is at rest. Now, man jumps horizontally with velocity u relative to cart. Then work done by man during the process of jumping will be mu 2 2 (c) mu 2

(a)

3mu 2 4 (d) None of these

m

(b)

2m

76 — Mechanics - II 18. Two balls of equal mass are projected upwards simultaneously, one from the ground with initial

velocity 50 ms−1 and the other from a 40m tower with initial velocity of 30 ms−1. The maximum height attained by their COM will be

(a) 80 m (c) 100 m

(b) 60 m (d) 120 m

19. A particle of mass m and momentum p moves on a smooth horizontal table and collides directly and elastically with a similar particle (of mass m) having momentum − 2 p. The loss ( − ) or gain ( + ) in the kinetic energy of the first particle in the collision is (a) +

p2 2m

(b) −

p2 4m

(c) +

3 p2 2m

(d) zero

20. An equilateral triangular plate of mass 4m of side a is kept as shown.

y P

Consider two cases : (i) a point mass 4m is placed at the vertex P of the plate (ii) a point mass m is placed at the vertex R of the plate. In both cases the x-coordinate of centre of mass remains the same. Then x coordinate of centre of mass of the plate is a 3 6a (c) 7

a 6 2a (d) 3

(b)

(a)

O (0, 0)

R

x

21. Four cubes of side a each of mass 40 g, 20 g, 10 g and 20 g are arranged in XY plane as shown in the figure. The coordinates of COM of the combination with respect to point O is y

40 20 10 20

x

O

(a)

19a 17a , 18 18

(b)

17a 11a , 18 18

(c)

17a 13a , 18 18

(d)

13a 17a 18, 18

22. A particle of mass m0, travelling at speed v0, strikes a stationary particle of mass 2m0. As a result the particle of mass m0 is deflected through 45° and has a final speed of speed of the particle of mass 2m0 after this collision is (a)

v0 2

(b)

v0 2 2

(c) 2v0

(d)

v0

2

. Then the

v0 2

23. Two bars of masses m1 and m2, connected by a weightless spring of stiffness k, rest on a smooth horizontal plane. Bar 2 is shifted by a small distance x0 to the left and released. The velocity of the centre of mass of the system when bar 1 breaks off the wall is

1 m1

(a) x0

km2 m1 + m2

(b)

x0 m1 + m2

km2

2 m2

(c) x0k

m1 + m2 m2

(d) x0

km1 (m1 + m2)

Chapter 11

Centre of Mass, Linear Momentum and Collision — 77

24. n elastic balls are placed at rest on a smooth horizontal plane which is circular at the ends with m m m , 2 .............. n − 1 2 2 2 respectively. What is the minimum velocity which should be imparted to the first ball of mass m such that n th ball completes the vertical circle

radius r as shown in the figure. The masses of the balls are m,

m

 3 (a)    4

n −1

 3 (c)    2

n −1

 4 (b)    3

n −1

5 gr

 2 (d)    3

n −1

5 gr

5 gr 5 gr

More than One Correct Options 1. A particle of mass m, moving with velocity v collides a stationary particle of mass 2m. As a result of collision, the particle of mass m deviates by 45° and has final speed of situation mark out the correct statement (s). (a) The angle of divergence between particles after collision is

π 2

(b) The angle of divergence between particles after collision is less than

v . For this 2

π 2

(c) Collision is elastic (d) Collision is inelastic

2. A pendulum bob of mass m connected to the end of an ideal string of length l

m

l

is released from rest from horizontal position as shown in the figure. At the lowest point the bob makes an elastic collision with a stationary block of mass 5m, which is kept on a frictionless surface. Mark out the correct statement(s) for the instant just after the impact. 17 mg 9 (b) Tension in the string is 3 mg. 2 gl (c) The velocity of the block is 3 (a) Tension in the string is

5m

(d) The maximum height attained by the pendulum bob after impact is (measured from the lowest 4l position) 9

3. A particle of mass m strikes a horizontal smooth floor with a velocity u making an angle θ with the floor and rebound with velocity v making an angle φ with the floor. The coefficient of restitution between the particle and the floor is e. Then (a) the impulse delivered by the floor to the body is mu (1 + e) sin θ (b) tan φ = e tan θ (c) v = u 1 − (1 − e2) sin 2 θ (d) the ratio of the final kinetic energy to the initial kinetic energy is cos 2 θ + e2 sin 2 θ

78 — Mechanics - II 4. A particle of mass m moving with a velocity ( 3 i + 2 j) ms−1 collides with another body of mass M ^

^

and finally moves with velocity ( −2 i + j) ms−1. Then during the collision ^

^

^

^

(a) impulse received by m is m (5 i + j) ^

^

(b) impulse received by m is m (−5 i − j) ^

^

(c) impulse received by M is m (−5 i − j) ^

^

(d) impulse received by M is m (5 i + j)

5. All surfaces shown in figure are smooth. System is released from rest. x and y components of acceleration of COM are m1

x y m2

(a) (a cm )x =

m1m2g m1 + m2

(b) (a cm )x = 2

 m2  (c) (a cm )y =   g  m1 + m2

m1m2g (m1 + m2)2

 m2  (d) (a cm )y =   g  m1 + m2

6. A block of mass m is placed at rest on a smooth wedge of mass M placed

m

at rest on a smooth horizontal surface. As the system is released (a) (b) (c) (d)

the COM of the system remains stationary the COM of the system has an acceleration g vertically downward momentum of the system is conserved along the horizontal direction acceleration of COM is vertically downward (a < g )

M

1 2

7. In the figure shown, coefficient of restitution between A and B is e = , then A m

v

B m

Smooth

v (a) velocity of B after collision is 2 3 mv 4 3 (c) loss of kinetic energy during the collision is mv2 16 1 (d) loss of kinetic energy during the collision is mv2 4 (b) impulse between two during collision is

8. In case of rocket propulsion, choose the correct options. (a) Momentum of system always remains constant (b) Newton’s third law is applied (c) If exhaust velocity and rate of burning of mass is kept constant, then acceleration of rocket will go on increasing (d) Newton’s second law can be applied

Chapter 11

Centre of Mass, Linear Momentum and Collision — 79

Comprehension Based Questions Passage 1 (Q. Nos. 1 to 2) A block of mass 2 kg is attached with a spring of spring constant 4000 Nm −1 and the system is kept on smooth horizontal table. The other end of the spring is attached with a wall. Initially spring is stretched by 5 cm from its natural position and the block is at rest. Now suddenly an impulse of 4 kg-ms−1 is given to the block towards the wall.

1. Find the velocity of the block when spring acquires its natural length (a) 5 ms −1 (c) 6 ms −1

(b) 3 ms −1 (d) None of these

2. Approximate distance travelled by the block when it comes to rest for a second time (not including the initial one) will be (Take 45 = 6.70)

(a) 30 cm (c) 40 cm

(b) 25 cm (d) 20 cm

Passage 2 (Q. Nos. 3 to 7) A uniform bar of length 12L and mass 48 m is supported horizontally on two fixed smooth tables as shown in figure. A small moth (an insect) of mass 8m is sitting on end A of the rod and a spider (an insect) of mass 16m is sitting on the other end B. Both the insects moving towards each other along the rod with moth moving at speed 2v and the spider at half this speed (absolute). They meet at a point P on the rod and the spider eats the moth. After this the spider v moves with a velocity relative to the rod towards the end A. The spider takes negligible time 2 L in eating on the other insect. Also, let v = where T is a constant having value 4 s. T B

A

3L

4L

3. Displacement of the rod by the time the insect meet the moth is (a)

L 2

(b) L

(c)

3L 4

(d) zero

4. The point P is at (a) the centre of the rod (b) the edge of the table supporting the end B (c) the edge of the table supporting end A (d) None of the above

5. The speed of the rod after the spider eats up the moth and moves towards A is (a)

v 2

(b) v

(c)

v 6

(d) 2v

6. After starting from end B of the rod the spider reaches the end A at a time (a) 40 s

(b) 30 s

(c) 80 s

(d) 10 s

7. By what distance the centre of mass of the rod shifts during this time? (a)

8L 3

(b)

4L 3

(c) L

(d)

L 3

80 — Mechanics - II Match the Columns 1. Two identical blocks A and B are connected by a spring as shown in

y

figure. Block A is not connected to the wall parallel to y-axis. B is compressed from the natural length of spring and then left. Neglect friction. Match the following two columns. Column I

A

B

x

Column II

(a)

Acceleration of centre of mass of two blocks

(p) remains constant

(b)

Velocity of centre of mass (q) first increases then of two blocks becomes constant

(c)

x-coordinate of centre of mass of two blocks

(r) first decreases then becomes zero

(d)

y-coordinate of centre of mass of two blocks

(s) continuously increases

2. One particle is projected from ground upwards with velocity 20 ms−1. At the same time another identical particle is dropped from a height of 180 m but not along the same vertical line. Assume that collision of first particle with ground is perfectly inelastic. Match the following two columns for centre of mass of the two particles (g = 10 ms−2) Column I

Column II

(a) Initial acceleration

(p) 5 SI units

(b) Initial velocity

(q) 10 SI units

(c) Acceleration at t = 5 s (r) 20 SI units (d) Velocity at t = 5 s

Note

(s) 25 SI units

Only magnitudes are given in column-II.

3. Two identical blocks of mass 0.5 kg each are shown in figure. A massless elastic spring is connected with A. B is moving towards A with kinetic energy of 4 J. Match the following two columns. Neglect friction. A

Column I

B

Column II

(a) Initial momentum of B (p) zero (b) Momentum of centre of mass of (q) 1 kg-ms −1 two blocks (c) Momentum of A at maximum (r) 2 kg-ms −1 compression (d) Momentum of B when spring is (s) 4 kg-ms −1 relaxed after compression

Chapter 11

Centre of Mass, Linear Momentum and Collision — 81

4. Two identical balls A and B are kept on a smooth table as shown. B collides with A with speed v. For different conditions mentioned in Column I, match with speed of A after collision given in Column II. Column I

B m

A m

v

Column II

(a) Elastic collision

3 v 4 5 v (q) 8

(p)

(b) Perfectly inelastic collision 1 2 1 (d) Inelastic collision with e = 4

(c) Inelastic collision with e =

(r)

v

(s)

v 2

5. Two boys A and B of masses 30 kg and 60 kg are standing over a plank C of mass 30 kg as shown. Ground is smooth. Match the displacement of plank of Column II with the conditions given in Column I. A

B

C Smooth

Note

Column I

Column II

(a) A moves x towards right (b) B moves x towards left (c) A moves x towards right and B moves x towards left (d) A and B both move x towards right

(p) x towards right (q) 2x towards left x towards left (r) 3 (s) None

All displacements mentioned in two columns are with respect to ground.

6. A man of mass M is standing on a platform of mass m1 and holding a string passing over a system of ideal pulleys. Another mass m2 is hanging as shown. (m2 = 20kg, m1 = 10kg, g = 10 ms−2) Column I (a) Weight of man for equilibrium (b) Force exerted by man on string to accelerate the COM of system upwards (c) Force exerted by man on string to accelerate the COM of system downwards (d) Normal reaction of platform on man in equilibrium

Column II (p) 100 N (q) 150 N

M m2

(r) 500 N m1

(s) 600 N

82 — Mechanics - II 7. Two blocks of masses 3 kg and 6 kg are connected by an ideal spring and are placed on a frictionless horizontal surface. The 3 kg block is imparted a speed of 2 ms−1 towards left. (consider left as positive direction) 2m/s

3kg

6kg

Column I

Column II

(a) When the velocity of 3 kg 2 block is ms −1 3

(p) velocity of centre of mass is 2 ms −1 3

(b) When the velocity of 6 kg 2 block is ms −1 3

(q) deformation of the spring is zero

(c) When the speed of 3 kg block is minimum

(r) deformation of the spring is maximum

(d) When the speed of 6 kg block is maximum

(s) both the blocks are at rest with respect to each other

8. In a two block system shown in figure match the following Rough 1kg 5 m/s

10 m/s

2kg Smooth

Column I

Column II

(a) Velocity of centre of mass

(p) Keep on changing all the time

(b) Momentum of centre of mass

(q) First decreases then become zero

(c) Momentum of 1 kg block

(r) Zero

(d) Kinetic energy of 2 kg block

(s) Constant

Subjective Questions 1. A ladder AP of length 5 m inclined to a vertical wall is slipping over a horizontal surface with velocity of 2 m/s, when A is at distance 3 m from ground. What is the velocity of COM at this moment? P

2 m/s A

O 3m

Chapter 11

Centre of Mass, Linear Momentum and Collision — 83

2. A ball of negligible size and mass m is given a velocity v0 on the centre of the cart which has a mass M and is originally at rest. If the coefficient of restitution between the ball and walls A and B is e. Determine the velocity of the ball and the cart just after the ball strikes A. Also, determine the total time needed for the ball to strike A, rebound, then strike B, and rebound and then return to the centre of the cart. Neglect friction.

v0

B

d

A

d

3. Two point masses m1 and m2 are connected by a spring of natural length l0. The spring is compressed such that the two point masses touch each other and then they are fastened by a string. Then the system is moved with a velocity v0 along positive x-axis. When the system reached the origin, the string breaks ( t = 0). The position of the point mass m1 is given by x1 = v0t − A (1 − cos ωt ) where A and ω are constants. Find the position of the second block as a function of time. Also, find the relation between A and l0.

4. A small sphere of radius R is held against the inner surface of larger sphere of radius 6R (as shown in figure). The masses of large and small spheres are 4M and M respectively. This arrangement is placed on a horizontal table. There is no friction between any surfaces of contact. The small sphere is now released. Find the coordinates of the centre of the large sphere, when the smaller sphere reaches the other extreme position. y

6R M (L,O) R

x

4M

5. A chain of length l and mass m lies in a pile on the floor. If its end A is raised vertically at a constant speed v0, express in terms of the length y of chain which is off the floor at any given instant.

P A

(a) The magnitude of the force P applied to end A. (b) Energy lost during the lifting of the chain.

6. A is a fixed point at a height H above a perfectly inelastic smooth horizontal plane. A light inextensible string of length L(> H ) has one end attached to A and other to a heavy particle. The particle is held at the level of A with string just taut and released from rest. Find the height of the particle above the plane when it is next instantaneously at rest. A

L

H

7. A particle of mass 2 m is projected at an angle of 45° with horizontal with a velocity of 20 2 m/ s. After 1 s, explosion takes place and the particle is broken into two equal pieces. As a result of explosion one part comes to rest. Find the maximum height attained by the other part. (Take g = 10 m/ s2)

84 — Mechanics - II 8. A sphere of mass m, impinges obliquely on a sphere, of mass M, which is at rest. Show that, if m = eM , the directions of motion of the spheres after impact are at right angles.

9. A gun of mass M (including the carriage) fires a shot of mass m. The gun along with the carriage is kept on a smooth horizontal surface. The muzzle speed of the bullet vr is constant. Find (a) The elevation of the gun with horizontal at which maximum range of bullet with respect to the ground is obtained. (b) The maximum range of the bullet.

10. A ball is released from rest relative to the elevator at a distance h1 above the floor. The speed of the elevator at the time of ball release is v0. Determine the bounce height h2 relative to elevator g of the ball (a) if v0 is constant and (b) if an upward elevator acceleration a = begins at the 4 instant the ball is released. The coefficient of restitution for the impact is e.

a= h1

g 4

h2 v0

11. A plank of mass 5 kg is placed on a frictionless horizontal plane. Further a block of mass 1 kg is placed over the plank. A massless spring of natural length 2 m is fixed to the plank by its one end. The other end of spring is compressed by the block by half of spring’s natural length. The system is now released from the rest. What is the velocity of the plank when block leaves the plank? (The stiffness constant of spring is 100 N/m).

1 kg 5 kg 4m

12. To test the manufactured properties of 10N steel balls, each ball is released from rest as shown

and strikes a 45° inclined surface. If the coefficient of restitution is to be e = 0.8, determine the distance s, where the ball must strike the horizontal plane at A. At what speed does the ball strike at A?(g = 9.8 m/s 2 ) B 1.5 m

1.0 m

45° A s

Chapter 11

Centre of Mass, Linear Momentum and Collision — 85

13. Two particles A and B of equal masses lie close together on a horizontal table and are connected by a light inextensible string of length l. A is projected vertically upwards with a velocity 10gl. Find the velocity with which it reaches the table again.

14. A small cube of mass m slides down a circular path of radius R cut into a large block of mass M, as shown in figure. M rests on a table, and both blocks move without friction. The blocks are initially at rest, and m starts from the top of the path. Find the horizontal distance from the bottom of block where cube hits the table. m

R

M

R/2

15. A thin hoop of mass M and radius r is placed on a horizontal plane. At the initial instant, the hoop is at rest. A small washer of mass m with zero initial velocity slides from the upper point of the hoop along a smooth groove in the inner surface of the hoop. Determine the velocity u of the centre of the hoop at the moment when the washer is at a certain point A of the hoop, whose radius vector forms an angle φ with the vertical (figure). The friction between the hoop and the plane should be neglected.

φ A

16. A shell of mass 1 kg is projected with velocity 20 m/s at an angle 60° with horizontal. It collides inelastically with a ball of mass 1 kg which is suspended through a thread of length 1 m. The 4 other end of the thread is attached to the ceiling of a trolley of mass kg as shown in figure. 3 Initially the trolley is stationary and it is free to move along horizontal rails without any friction. What is the maximum deflection of the thread with vertical? String does not slack. Take g = 10 m/ s2. 1m

20 m/s

16 m

60°

17. A small ball is projected at an angle α between two vertical walls such that in the absence of the wall its range would have been 5d. Given that all the collisions are perfectly elastic, find (a) maximum height attained by the ball. (b) total number of collisions with the walls before the ball comes back to the ground, and (c) point at which the ball finally falls. The walls are supposed to be very tall.

u α d/2

86 — Mechanics - II 18. Two large rigid vertical walls A and B are parallel to each other and separated by 10 m. A particle of mass 10 g is projected with an initial velocity of 20 m/s at 45° to the horizontal from point P on the ground, such that AP = 5 m. The plane of motion of the particle is vertical and perpendicular to the walls. Assuming that all the collisions are perfectly elastic , find the maximum height attained by the particle and the total number of collisions suffered by the particle with the walls before it hits ground. Take g = 10 m/ s2.

45° P

A

B

19. Two blocks of masses 2 kg and M are at rest on an inclined plane and are separated by a distance of 6.0 m as shown. The coefficient of friction between each block and the inclined plane is 0.25. The 2 kg block is given a velocity of 10.0 m/s up the inclined plane. It collides with M , comes back and has a velocity of 1.0 m/s when it reaches its initial position. The other block M after the collision moves 0.5 m up and comes to rest. Calculate the coefficient of restitution between the blocks and the mass of the block M. [Take sin θ ≈ tan θ = 0.05 and g = 10 m/ s2] M 2kg

m 6.0

θ

20. A small block of mass m is placed on top of a smooth hemisphere also of mass m which is placed on a smooth horizontal surface. If the block begins to slide down due to a negligible small impulse, show that it will loose contact with the hemisphere when the radial line through vertical makes an angle θ given by the equation cos3 θ − 6 cos θ + 4 = 0.

21. A ball is projected from a given point with velocity u at some angle with the horizontal and after hitting a vertical wall returns to the same point. Show that the distance of the point from the eu 2 wall must be less than , where e is the coefficient of restitution. (1 + e)g

Answers Introductory Exercise 11.1 n

1. rCOM =

Σ mi ri

i =1 n

Σ mi

n

Σ w i ri

i =1 n

while rCG =

w = weight (mg ),

Here,

Σ wi

i =1

rCOM = rCG

when

g = constant

i =1

2. False

3. True

4. True 2 2 4  a + ab + b    3π  a+ b  3l 12. xCOM = 4

19 m 6  π  11.   a  π + 4

6. False

r 2

5a 5a  9.  ,   6 6

8.

7.

10. 0.74 m2

5. less than

13.

2 L 3

Introductory Exercise 11.2 1. xCM = 12.67 m

2. zero

3.

ml 11 + m2 l2 m1 + m2

4. (a) 0.30 kg (b) (2.4 kg-ms −1)$j

(c) (8.0 ms −1) $j

 3 − 1  g 6.    4 2 

5. (a) 28 cm (b) 2.3 ms −1

Introductory Exercise 11.3 $ ) cm/s 1. (2.5 $i + 15$j + 5k 4. 2.4 × 10 5 m/s

2. 12.5 m/s in opposite direction, 17.5 m/s

5. 1.5 × 10 −23 m/s

3. 1 : 2

6. 9 ms −1, 1.08 kJ

7. 60 m

Introductory Exercise 11.4 1. 1.225 kgs −1, (i) 2.8 kms −1, (ii) 3.6 kms −1 d 2x 3. (m0 − µt ) 2 = µu − (m0 − µt ) g dt

2. 1232.6 ms −1 3 4. u ln   − g  2

Introductory Exercise 11.5 1. 4 × 103 N 2. − $i + 3$j

4. (a) 1.2 × 10 −3 s

3. 200 m/s

(b) 0.5 N-s

(c) 417 N

Introductory Exercise 11.6 1. 30 cm

2.

8 9

3.

4mm 1 2 (m1 + m2 )2

4. No

5. 2 ms −1 in negative x-axis, 3 m/s in positive x-axis. 28 2 6. v1 = ms −1 (in negative x-direction) and v2 = ms −1 (in positive x-direction) 3 3 1 mv V 7. Two 8. 9. (a) (b) 10. − $i + 2$j 3 V v

11. 90° − 2α

Exercises LEVEL 1 Assertion and Reason 1. (d)

2. (a)

3. (d)

4. (d)

5. (b)

11. (d)

12. (d)

13. (d)

14. (a)

15. (d)

6. (a)

7. (c)

8. (d)

9. (a)

10. (d)

88 — Mechanics - II Single Correct Option 1. 11. 21. 31. 41.

2. 12. 22. 32. 42.

(a) (d) (d) (c) (d)

3. 13. 23. 33.

(a) (c) (b) (c) (c)

(c) (b) (c) (b)

4. 14. 24. 34.

(b) (a) (c) (a)

5. 15. 25. 35.

6. 16. 26. 36.

(b) (a) (b) (c)

(d) (a) (c) (d)

7. 17. 27. 37.

(d) (c) (a) (a)

8. 18. 28. 38.

9. 19. 29. 39.

(c) (a) (c) (b)

(d) (b) (d) (c)

10. 20. 30. 40.

(b) (b) (d) (c)

Subjective Questions a b 1.  − ,−   12 12  1 1 1 7. , , e e e2 8. (a) (−10$j ) ms −2

2. xCOM = −

a3 b R − a3

3. True

3

6.

K 2

10 $ $ 70 $ (2 i − j ) ms −1 (c)  i + 35$j  m  3  3 4 9. (a) 2.0 kg (b) (12.0 ms −2 )t $j (c) (72.0 N)$j 10. (a) 0.8 $i m/s (b) 2.4 $i m/s (c) 7 mv 11. 2.82 kms −1 12. 10 cm 13. (a) (b) balloon will also stop moving M + m (b)

14. m = m0e − (a/u )t

15. 35 m

−1

18. 2 ms , 2 2 ms

16. 2 10 N-s 2 πr 19. v 2 21. 3

−1

2J −1 P 1 22. m 6 20.

23. 71.4 mm

24.

25. (a) 0.43 (b) 480 J (c) 1.28 J

α 27. 2m sin    2

26. 0.011mm, 1.94 m/s 28. 29. 31. 32. 33.

m (R − r ) M + m

,m

4L 3 gl M (M + m)

2 g (R − r ) M (M + m)

(a) 0.14 kg-ms −1 (b) zero 10 N, 20 W 270 N-s (to the right), 13.5 kN (to the right) 0.4 ms −1, 2.4 ms −1, 0.93 m

LEVEL 2 Single Correct Option 1. (b) 11. (a) 21. (a)

2. (b) 12. (a) 22. (b)

3. (b) 13. (b) 23. (b)

4. (b) 14. (c) 24. (a)

5. (a) 15. (d)

6. (b) 16. (d)

7. (d) 17. (d)

8. (a) 18. (c)

9. (b) 19. (c)

10. (c) 20. (b)

More than One Correct Options 1. (b,d)

2. (a,c,d)

3. (all)

4. (b,d)

5. (b,c)

6. (c,d)

7. (b,c)

5. (c)

6. (c)

7. (a)

Comprehension Based Questions 1. (b)

2. (b)

3. (d)

4. (b)

8. (b,c,d)

Chapter 11

Centre of Mass, Linear Momentum and Collision — 89

Match the Columns 1. (a) → (r) (b) → (q) (c) → (s) (d) → (p) 2. (a) → (q) (b) → (q)

(c) → (p) (d) → (s)

3. (a) → (r) (b) → (r) (c) → (q) (d) → (p) 4. (a) → (r) (b) → (s) (c) → (p) (d) → (q) 5. (a) → (r) (b) → (p) (c) → (p) (d) → (s) 6. (a) → (r) (b) → (r,s) (c) → (p,q) (d) → (p) 7. (a) → (p,r,s)

(b) → (p,r,s)

(c) → (p) (d) → (p,q)

8. (a) → (r,s) (b) → (r,s) (c) → (q) (d) → (q)

Subjective Questions  eM − m   e+1  d  1 2 2. vball =  + 2   v0 (leftwards), vcart =   mv0 (rightwards); t = 1 + v0  e e   M + m  M + m

1. 1.25 ms −1 3. x2 = v0 t + 5. (a)

m  m1 A (1– cos ω t), l0 =  1 + 1 A m2  m2 

myv02 m ( gy + v02 ) (b) 2l l

4. (L + 2R , 0) 6.

H5 L4

 M  vr2 9. (a) 45° (b)    M + m g

7. 35 m 10. (a) e 2 h1 (b) e 2 h1 11.

10 ms −1 3

12. 0.93 m, 6.6 ms −1 14. R

13. 2 gl 15. v = m cos φ 17. (a)

u2 sin2 α 2g

18. 10 m, Four

2 gr (1 + cos φ) (M + m)(M + m sin2 φ)

2(M + m) M

16. 60°

(b) nine (c) point O 19. 0.84, 15.12 kg

12

Rotational Mechanics Chapter Contents 12.1

Introduction

12.2

Moment of Inertia

12.3

Angular Velocity

12.4

Torque

12.5

Rotation of a Rigid Body about a Fixed Axis

12.6

Angular Momentum

12.7

Conservation of Angular Momentum

12.8

Combined Translational and Rotational Motion of a Riggid Body

12.9

Uniform Pure Rolling

12.10 Instantaneous Axis of Rotation 11.11 Accelerated Pure Rolling 12.12 Angular Impulse 12.13 Toppling

92 — Mechanics - II

12.1 Introduction A particle means mass with negligible volume. A rigid body is made up of too many particles but distance between any two particles is always constant. In any type of motion of a rigid body this distance always remains constant. A particle has only translational motion. Even if a particle is 1 rotating in a circle it has only translational motion and it has only translational kinetic energy mv 2 . 2 A rigid body may have either of the following three types of motions : (i) Translational motion (ii) Rotational motion (iii) Translational plus rotational motion In translational motion of the rigid body all particles of the rigid body have same linear displacement, same linear velocity and same linear acceleration. In rest two motions, different particles have different linear displacement, different linear velocity and different linear acceleration. As far as translational motion is concerned we do not differentiate between a particle and a rigid body. This motion is already discussed in the chapter of kinematics. This is the reason, in the chapter of kinematics, sometimes we write: a particle is moving and sometimes we write: a block (or a body) is moving. Rest two motions are only defined for a rigid body. In the present chapter, we shall discuss these two motions of a rigid body.

12.2 Moment of Inertia Like the centre of mass, the moment of inertia is a property of an object that is related to its mass distribution. The moment of inertia (denoted by I ) is an important quantity in the study of system of particles that are rotating . The role of the moment of inertia in the study of rotational motion is analogous to that of mass in the study of linear motion. Moment of inertia gives a measurement of the resistance of a body to a change in its rotational motion. If a body is at rest, the larger the moment of inertia of a body, the more difficult it is to put that body into rotational motion. Similarly, the larger the moment of inertia of a body, the more difficult it is to stop its rotational motion. The moment of inertia is calculated about some axis (usually the rotational axis) and it depends on the mass as well as its distribution about that axis.

Moment of Inertia of a Single Particle For a very simple case the moment of inertia of a single particle about an axis is given by,

r

m

Fig. 12.1

I = mr 2 Here, m is the mass of the particle and r its distance from the axis under consideration.

…(i)

Chapter 12

Rotational Mechanics — 93

Moment of Inertia of a System of Particles The moment of inertia of a system of particles about an axis is given by, r1

m1 r2

m3

m2

r3

Fig. 12.2

I = Σ mi ri2

…(ii)

i

where, ri is the perpendicular distance from the axis to the ith particle, which has a mass mi . For example, in Fig. 12.2: I = m1 r12 + m2 r22 + m3 r32

Moment of Inertia of Rigid Bodies For a continuous mass distribution such as found in a rigid body, we replace the summation of Eq. (ii) by an integral. If the system is divided into infinitesimal elements of mass dm and if r is the distance from a mass element to the axis of rotation, the moment of inertia is, I = ∫ r 2 dm

r dm

Fig. 12.3

where the integral is taken over the system.

Moment of Inertia of a Uniform Cylinder Let us find the moment of inertia of a uniform cylinder about an axis through its centre of mass and perpendicular to its base. Mass of the cylinder is M and radius is R. We first divide the cylinder into annular shells of width dr and length l as shown in figure. The moment of inertia of one of these shells is

R l

(a)

r

dr

(b)

Fig. 12.4

dI = r 2 dm = r 2 (ρ ⋅ dV ) Here, and ∴

ρ = density of cylinder dV = volume of shell = 2πrl dr dI = 2πρl r 3 dr

94 — Mechanics - II The cylinder’s moment of inertia is found by integrating this expression between 0 and R, R πρl 4 So, I = 2πρl ∫ r 3 dr = R 0 2 The density ρ of the cylinder is the mass divided by the volume. M ρ= ∴ πR 2 l

…(iii)

…(iv)

From Eqs. (iii) and (iv), we have I=

1 MR 2 2

Proceeding in the similar manner we can find the moment of inertia of certain rigid bodies about some given axis. Moments of inertia of several rigid bodies with symmetry are listed in Table. 12.1. Table 12.1 Thin rod

3

2 θ

5

I1 = 0, I2 =

4

θ

ml 2 ml 2 2 , I4 = sin θ 3 12 ml 2 2 I5 = sin θ, I6 = mx2 3

1

I3 =

x 6

Circular disc

⊗

I1 = I2 =

1

⊗

5

Circular ring

mR 2 4 mR 2 I3 = I1 + I2 = 2 5 2 I4 = I2 + mR = mR 2 4 3 I5 = I3 + mR 2 = mR 2 2

2

4

3

4

mR 2 2 I3 = I1 + I2 = mR 2 3 I4 = I2 + mR 2 = mR 2 2 I5 = I3 + mR 2 = 2 mR 2 I1 = I2 =

2

⊗

1

⊗

3

5

Rectangular slab

b a

1 3

mb 2 12 ma2 I2 = 12 I3 = I1 + I2 m 2 (a + b 2 ) = 12 I1 =

2

⊗

ml 2 12

Chapter 12 Square slab

3

a

ma2 12 ma2 I4 = I1 + I3 = 6

I1 = I2 = I3 =

2 θ

⊗

Rotational Mechanics — 95

1

4 a

Solid sphere

2

I1 =

1

2 mR 2 5

I2 = I1 + mR 2 7 = mR 2 5 m = mass of sphere

Hollow sphere

2

2 mR 2 3 I2 = I1 + mR 2 5 = mR 2 3 I1 =

1

Theorems on Moment of Inertia There are two important theorems on moment of inertia, which, in some cases, enable the moment of inertia of a body to be determined about any general axis, if its moment of inertia about some other axis is known. Let us now discuss both of them.

Theorem of Parallel Axes A very useful theorem, called the parallel axes theorem relates the moment of inertia of a rigid body about two parallel axes, one of which passes through the centre of mass.

COM

r

Fig. 12.5

Two such axes are shown in figure for a body of mass M. If r is the distance between the axes and I COM and I are the respective moments of inertia about them then, these two are related by, I = I COM + Mr 2

96 — Mechanics - II We now present a proof of the above theorem. Proof To prove this theorem, we consider two axes, both parallel to the z-axis, one through the center of mass and the other through a point P (Fig 12.6). First we take a very thin slice of the body, parallel to the xy-plane and perpendicular to the z-axis. We take the origin of our coordinate system to be at the centre of mass of the body; the coordinates of the centre of mass are then x cm = ycm = z cm = 0. The axis through the centre of mass passes through this thin slice at point O and the parallel axis passes through point P, whose x and y coordinates are ( a, b). The distance of this axis from the axis through the centre of mass is r, where r 2 = a 2 + b 2 . Axis of rotation passing through COM and perpendicular to the plane of the figure y

Mass element mi

yi

mi yi - b x - a P i

a r

b

xi COM Second axis of rotation parallel to the one through the COM

O

x

Slice of a body of mass M

Fig. 12.6

We can write an expression for the moment of inertia I P about the axis through point P. Let mi be a mass element in our slice, with coordinates ( x i , yi , z i ). Then, the moment of inertia I COM of the slice about the axis through the centre of mass (at O) is I COM = Σ mi ( x i2 + yi2 ) i

The moment of inertia of the slice about the axis through P is I P = Σ mi [( x i − a ) 2 + ( yi − b) 2 ] i

These expressions don’t involve the coordinates z i measured perpendicular to the slice, so we can extend the sums to include all particles in all slices. Then, I P becomes the moment of inertia of the entire body for an axis through P. We then expand the squared terms, regroup and obtain I P = Σ mi (x i2 + yi2 ) − 2a Σ mi x i − 2b Σ mi yi + ( a 2 + y 2 ) Σ mi i

i

i

i

The first sum is I COM . From the, definition of the centre of mass the second and third sums are proportional to x cm and ycm . These are zero because we have taken our origin to be the centre of mass. The final term is r 2 multiplied by the total mass or r 2 . This completes our proof that I P = I COM + Mr 2

Chapter 12

Rotational Mechanics — 97

Note From the above theorem we can see that among several parallel axes, moment of inertia is least about an axis which passes through centre of mass. e. g. I2 is least among I1 , I2 and I3 . Similarly, I5 is least among I4 , I5 and I6 . 1

2 3

4 5 6

COM

COM

Fig. 12.7

Theorem of Perpendicular Axes This theorem is applicable only to the plane bodies (two dimensional). The theorem states that the moment of inertia of a plane lamina about an axis perpendicular to the plane of the lamina is equal to the sum of the moments of inertia of the lamina about two axes perpendicular to each other, in its own plane and intersecting each other, at the point where the perpendicular axis passes through it. Let x and y axes be chosen in the plane of the body and z-axis perpendicular, to this plane, three axes being mutually perpendicular, then the theorem states that

y

ri

z

P yi x

Ä

xi

I z = Ix + I y

Fig. 12.8

Consider an arbitrary particle P of mass mi , distant ri fromO and x i and yi are the perpendicular distances of point P from the x and y-axes respectively, we have

Proof

I z = Σ mi ri2 , I x = Σ mi yi2 and I y = Σ mi x i2 i

So that,

Ix + I y = Σ i

i

mi yi2

+Σ i

i

mi x i2

= Σ mi ( yi2 + x i2 ) = Σ mi ri2 = I z i

i.e.

i

I z = Ix + I y

Hence Proved.

Radius of Gyration Radius of gyration ( K ) of a body about an axis is the effective distance from this axis where the whole mass can be assumed to be concentrated so that the moment of inertia remains the same. Thus, I = MK

2

or

K=

I M

e.g. radius of gyration of a disc about an axis perpendicular to its plane and passing through its centre of mass is 1 MR 2 R 2 K= = M 2

M

⇒

Fig. 12.9

K

M

98 — Mechanics - II Extra Points to Remember ˜

˜

˜

Theorem of parallel axes is applicable for any type of rigid body whether it is a two dimensional or three dimensional, while the theorem of perpendicular axes is applicable for laminar type or two dimensional bodies only. In theorem of perpendicular axes, the point of intersection of the three axes (x, y and z) may be any point on the plane of body (it may even lie outside the body also). This point may or may not be the centre of mass of the body. If whole mass of the rigid body is kept at same distance x or R from the axis, then moment of intertia is mx2 or mR 2 , where m is the mass of whole body. Rod x Axis x = mx 2

R Axis

Ring I = mR 2

Fig. 12.10 ˜

If a portion is symmetrically cut about an axis and mass of remaining portion is M. Then, moment of inertia of the remaining portion is same as the moment of inertia of the whole body of O same mass M. e.g. in figure 12.11(a) moment of inertia of the section shown (a part of circular disc) about an axis 1 perpendicular to its plane and passing through point O is MR 2 2 1 as the moment of inertia of the complete disc is also MR 2 . 2

R M

M

(a)

(b)

Fig. 12.11

1 th part of the disc, then mass of the disc will be nM. n 1 Idisc = (nM )R 2 2 1 1 ∴ Isection = Idisc = MR 2 n 2 Rod If whole mass of the rigid body is kept over the axis then, moment of inertia I=0 is zero. For example, moment of inertia of a thin rod about an axis passing Fig. 12.12 through the rod is zero.

Proof : Suppose the given section is

˜

V

Axis

Example 12.1 Three particles of masses 1g, 2g and 3g are kept at points (2cm, 0), (0,6 cm), (4cm, 3cm). Find moment of inertia of all three particles (in gm - cm2 ) about, (a) x-axis (b) y-axis (c) z-axis y

Solution (a) About x-axis

2

I x = I1 + I 2 + I 3 = m1 r12 + m2 r22 + m3 r32 Here r = perpendicular distance of the particle from x-axis ∴

I 1 = (1) ( 0) 2 + ( 2)( 6) 2 + ( 3)( 3) 2 = 99g -cm 2

3

6 cm

Ans.

(b) About y-axis I y = m1 r12 + m2 r22 + m3 r32

3 cm O

r0 1 2 cm 4 cm Fig. 12.13

x

Chapter 12

Rotational Mechanics — 99

Here, r = perpendicular distance of the particle from y-axis ∴ (c) About z-axis

I y = (1)( 2) 2 + ( 2)( 0) 2 + ( 3)( 4 ) 2 = 52 g -cm 2

Ans.

I z = m1 r12 + m2 r22 + m3 r32

Here, r = perpendicular distance of the particle from z-axis. r0 = ( 3) 2 + ( 4 ) 2 = 5 cm ∴

I z = (1)( 2) 2 + ( 2)( 6) 2 + 3 ( 5) 2 = 151g -cm 2

Ans.

Note In the above example, from theorem of perpendicular axes, we can see that Iz = Ix + I y V

Example 12.2 Three rods each of mass m and length l are joined together to form an equilateral triangle as shown in figure. Find the moment of inertia of the system about an axis passing through its centre of mass and perpendicular to the plane of the triangle.

A

COM

Moment of inertia of rod BC about an axis perpendicular to plane of triangle ABC and passing through the mid-point of rod BC ( i.e. D ) is ml 2 I1 = 12 Solution

B

C

Fig. 12.14 A

r = BD tan 30° l  l  1  r =    =  2  3  2 3

or

From theorem of parallel axes, moment of inertia of this rod about the asked axis is

COM r B

30°

C

D Fig. 12.15

2

 l  ml 2 ml 2 I 2 = I 1 + mr = + m  =  2 3 6 12 2

∴ Moment of inertia of all the three rods is  ml 2  ml 2 I = 3I 2 = 3  = 2  6  V

Ans.

Ans.

Example 12.3 Find the moment of inertia of a solid sphere of mass M and radius R about an axis XX shown in figure. Also find radius of gyration about the given axis. x

x Fig. 12.16

100 — Mechanics - II Solution

x

From theorem of parallel axis, IXX = I COM + Mr 2 = =

K=

Radius of gyration,



Note If whole mass M is kept at a distance K  = 

V

2 MR 2 + MR 2 5

I = M

Ans. 7 MR 2 5 = M

x r=R

7 R 5

Fig. 12.17

7  7 R as a particle, then moment of inertia is again MR 2 . 5  5

Example 12.4 Consider a uniform rod of mass m and length 2l with two particles of mass m each at its ends. Let AB be a line perpendicular to the length of the rod and passing through its centre. Find the moment of inertia of the system about AB. Solution

A

I AB = I rod + I both particles m( 2l ) 2 + 2 ( ml 2 ) 12 7 = ml 2 3

=

V

COM

7 MR 2 5

l

l

m

m

Ans.

B Fig. 12.18

Example 12.5 Find the moment of inertia of the rod AB about an axis yy as shown in figure. Mass of the rod is m and length is l. B

y

A

α

y Fig. 12.19

m Mass per unit length of the rod = l  m Mass of an element PQ of the rod is, dm =   dx  l Perpendicular distance of this elemental mass about yy is, r = x sin α ∴ Moment of inertia of this small element of the rod (can be assumed as a point mass) about yy is, m  m  dI = ( dm )r 2 =  dx ( x sin α ) 2 =  sin 2 α x 2 dx l  l  Solution

y

B r a

A

Q P dx

x

y

Fig. 12.20

Chapter 12

Rotational Mechanics — 101

∴ Moment of inertia of the complete rod, l x=l m ml 2 I=∫ dI = sin 2 α ∫ x 2 dx = sin 2 α 0 x=0 l 3 Note (i) I = 0 if α = 0

(ii) I =

Ans.

ml 2 π if α = or 90° 3 2

INTRODUCTORY EXERCISE

12.1

1. Find the radius of gyration of a rod of mass m and length 2l about an axis passing through one of its ends and perpendicular to its length.

2. A mass of 1 kg is placed at (1 m, 2 m, 0). Another mass of 2 kg is placed at (3 m, 4 m, 0). Find moment of inertia of both the masses about z-axis.

3. Four thin rods each of mass m and length l are joined to make a square. Find moment of inertia of all the four rods about any side of the square.

4. About what axis would a uniform cube have its minimum moment of inertia? 5. There are four solid balls with their centres at the four corners of a square of side a. The mass of each sphere is m and radius is r. Find the moment of inertia of the system about (i) one of the sides of the square (ii) one of the diagonals of the square.

6. A non-uniform rod AB has a mass M and length 2l . The mass per unit length of the rod is mx at a point of the rod distant x from A. Find the moment of inertia of this rod about an axis perpendicular to the rod (a) through A (b) through the mid-point of AB.

7. The uniform disc shown in the figure has a moment of inertia of 0.6 kg -m 2 around the axis that passes through O and is perpendicular to the plane of the page. If a segment is cut out from the disc as shown, what is the moment of inertia of the remaining disc?

O 60°

Fig. 12.21

8. If two circular disks of the same weight and thickness are made from metals having different densities. Which disk, if either will have the larger moment of inertia about its central axis.

9. Particles of masses 1g, 2 g, 3 g,K 100 g are kept at the marks 1cm, 2cm, 3cm,K , 100 cm respectively on a metre scale. Find the moment of inertia of the system of particles about a perpendicular bisector of the metre scale.

10. If I1 is the moment of inertia of a thin rod about an axis perpendicular to its length and passing through its centre of mass and I 2 the moment of inertia of the ring formed by the same rod about an axis passing through the centre of mass of the ring and perpendicular to the plane of the ring. I Then find the ratio 1 . I2

102 — Mechanics - II

12.3 Angular Velocity Angular velocity is a vector quantity. It is represented by ω. Its SI unit is radian per second. It can be defined for following three situations : (i) Angular velocity of a particle (in motion) about a fixed point. (ii) Angular velocity of a rigid body in pure rotational motion. (iii) Angular velocity of a rigid body in rotational and translational motion.

Angular Velocity of a Particle (in motion) about a Fixed Point At a given instant a particle P has velocity v. It has position vector r with respect to a fixed point O as shown in figure. After some time position vector has become r′. We can see two changes in its position vector. First, its magnitude | r | has changed, second its direction has changed or we can say, its position vector has been rotated. If we O resolve v along r and perpendicular to r then its two components v|| and v⊥ have the following meanings. v⋅r v|| = v cos θ = r d |r | = component of v along r = dt = rate by which magnitude of r changes = rate by which distance of P from O changes. θ = angle between r and v v⊥ v v = ⊥ = ⊥ =ω |r| r OP

θ

P

v r dθ

r'

Fig. 12.22

v

P

v

r v^

O

Fig. 12.23

= angular velocity of particle P about point O at this instant = rate by which r rotates If θ is acute, cos θ or v|| is positive i.e. distance of P from O is increasing. If θ is obtuse, cos θ or v|| is negative i.e. distance of P from O is decreasing. If θ is 90°, r P then cos θ or v|| is zero or distance of P from O is constant. For example, when O particle rotates in a circle then with respect to centre, θ is always 90°. This is the 90° reason, why distance of particle from centre always remains constant.With v respect to any other point θ is sometimes acute and sometimes Fig. 12.24 obtuse.Therefore, distance sometimes increases and sometimes decreases. v ⊥ = ω × r i.e. perpendicular component of velocity in vector form is the cross product of ω and r. Direction of ω is given by right hand screw law. For example: P v r in Fig. 12.24 rotation is clockwise. So, ω is perpendicular to O Fig. 12.25 paper inwards. If a particle moves in a straight line, then about any point lying on this line angle between r and v is 0° or180°. Hence, v ⊥ = 0 or ω = 0.

Chapter 12

Rotational Mechanics — 103

Angular Velocity of a Rigid Body in Pure Rotational Motion Consider a rigid body rotating about a fixed line AB. Consider a particle P. Draw a perpendicular PO to the axis of rotation. In time ∆t, this particle moves to point Q.

A ω

∠QOP = ∆θ

Let

Then, we say that the particle has rotated through an angle ∆θ. In fact, all the particles of the rigid body have rotated the same angle ∆θ or we can say that the whole body has rotated through an angle ∆θ. The average angular velocity of the rigid body during the time interval ∆t is ∆θ ω av = ∆t

O

r

Q ∆θ

P

B

Fig. 12.26

The instantaneous angular velocity of the rigid body is ∆θ dθ ω = lim = ∆t → 0 ∆ t dt Direction of angular velocity is given by right hand rule. The direction of angular velocity is defined to be the direction in which the thumb of your right hand points when you curl your fingers in the direction of rotation. For example, direction of ω in the given figure is along the axis of rotation from B to A. The magnitude of angular velocity is called angular speed. However, we shall continue to use the word angular velocity.

Angular Velocity of a Rigid Body in Rotational and Translational Motion Consider two particles A and B on a rigid body (in translational and rotational motion). In general, velocity of A is not equal to the velocity of B.

A

β

α

vA

vB

B

Fig. 12.27

or v A ≠ vB Find their components along AB and perpendicular to AB. Now, (a) Along AB their components are always equal or, v A cos α = v B cos β This is because, in a rigid body distance between two particles (here A and B) is always constant.

104 — Mechanics - II If the components are not equal then the distance AB will either increase (when v A cos α > v B cos β) or decrease (when v B cos β > v A cos α). A vA sin α B

vB

sin

β

Fig. 12.28

(b) Find relative component perpendicular to AB and divide it by the distance AB to find angular velocity of the rigid body. In the given figure, the perpendicular components are v A sin α and v B sin β (in the same direction). Suppose v A sin α > v B sin β. Then, the relative component perpendicular to AB is v r = v A sin α − v B sin β v v sin α − v B sin β ∴ ω= r = A AB AB As, v A sin α > v B sin β, so rotation of the body is clockwise and according to right hand rule, angular velocity vector is perpendicular to paper inwards. This direction is also shown like ⊗.

Extra Points to Remember ˜

If a particle P is moving in a circle, its angular velocity about centre of the circle (ωC ) is two times the angular velocity about any point on the circumference of the O circle (ωO ) or ωC = 2 ωO C

This is because ∠P′CP = 2 ∠P′OP (by property of a circle) ∠P ′CP ∠P ′OP ωC = , ωO = t pp ′ t pp ′ ˜

˜

P′

From these relations we can see that ωC = 2 ωO . In pure translational motion of a rigid body its angular velocity will be zero.

P

Fig. 12.29

According to right hand rule direction of angular velocity in some cases (in pure rotational motion) have been shown below :

ω

ω is perpendicular to paper inwards or in direction

Fig. 12.30

ω is perpendicular to paper outwards or in direction

Chapter 12 V

Rotational Mechanics — 105

Example 12.6 Rod AB has a length L. Velocity of end A of the rod has velocity v0 at the given instant. A

v0 60° B

Fig. 12.31

(a) Which type of motion the rod has? (b) Find velocity of end B at the given instant. (c) Find the angular velocity of the rod. Solution (a) The rod has rotational plus translational motion. (b) Let velocity of end B is v B in the direction shown in figure. A 30°

v0 2

√ 3v0 2 √ 3vB 2

v0 B

60°

60°

vB vB 2

Fig. 12.32

Components of v A and vB along AB and perpendicular to AB are also shown in the same figure. Rod is a rigid body. So, distance AB should remain constant or the components along AB should be same. 3v 0 v B = 2 2

∴

or

v B = 3v 0

Ans.

(c) Perpendicular to AB, components are in opposite directions. So, the relative component will be A v0 2 √ 3 vB 2 B

Fig. 12.33

vr = Substituting v B = 3v 0 we get,

v0 3v B + 2 2

v r = 2v 0

106 — Mechanics - II vr AB 2v o Ans. or ω= L Rotation of the rod is anticlockwise. Therefore, from right hand rule ω is perpendicular to paper outwards or in O. direction. Now angular velocity of the rod is

ω=

INTRODUCTORY EXERCISE

12.2

1. Find angular speed of second’s clock. 2. Two points P and Q, diametrically opposite on a disc of radius R have linear velocities v and 2v as shown in figure. Find the angular speed of the disc. v

P

2v Q Fig. 12.34

3. A particle is located at (3 m, 4 m) and moving with v = (4$i − 3$j) m /s. Find its angular velocity about origin at this instant. 4. In the figure shown, the instantaneous speed of end A of the rod is v to the left. Find angular velocity of the rod at given instant. B

A

30°

60°

Fig. 12.35

12.4 Torque Suppose a force F is acting on a particle P and let r be the position vector of this particle about some reference point O. The torque of this force F, about O is defined as, τ =r × F

r

P

O

This is a vector quantity having its direction perpendicular to both r and F, according to the rule of cross product. Note Here, r = rP − rO rP = position vector of point, where force is acting and rO = position vector of point about which torque is required.

Fig. 12.36

F

Chapter 12

Rotational Mechanics — 107

Torque of a Force about a Line Consider a rigid body rotating about a fixed axis AB. Let F be a force acting on the body at point P. Take the origin O somewhere on the axis of rotation. The torque of F about O is

A

τ=r×F Its component along AB is called the torque of F about AB. This is also equal to

O

τ = F × r⊥ Here,

P r F

B

Fig. 12.37

r⊥ = perpendicular distance of point of application of force from the line AB

Extra Points to Remember When a rigid body is rotating about a fixed axis and a force is applied on it at some point then we are concerned with the component of torque of this force about the axis of rotation not with the net torque.

˜

The component of torque about axis of rotation is independent of the choice of the origin O, so long as it is chosen on the axis of rotation, i.e. we may choose point O anywhere on the line AB.

˜

˜

Component of torque along axis of rotation AB is zero if (a) F is parallel to AB (b) F intersects AB at some point

˜

˜

If F is perpendicular to AB, but does not intersect it, then component of torque about line AB = magnitude of force F × perpendicular distance of F from the line AB (called the lever arm or moment arm) of this torque. If there are more than one force F1, F2 ,K acting on a body, the total torque will be τ = r1 × F1 + r2 × F2 + K

But if the forces act on the same particle, one can add the forces and then take the torque of the resultant force, or τ = r × (F1 + F2 + . . . ) V

Example 12.7 Find the torque of a force F = ( $i + 2$j − 3k$ ) N about a point O. The position vector of point of application of force about O is r = ( 2$i + 3$j − k$ ) m . Solution

$i Torque τ = r × F = 2 1

$j k$ 3 −1 2 −3

= $i ( −9 + 2) + $j( −1 + 6) + k$ ( 4 − 3) or V

τ = ( −7$i + 5$j + k$ ) N-m

Ans.

Example 12.8 A small ball of mass 1.0 kg is attached to one end of a 1.0 m long massless string and the other end of the string is hung from a point O. When the resulting pendulum is making 30° from the vertical, what is the magnitude of net torque about the point of suspension? [Take g = 10 m/ s2 ]

108 — Mechanics - II Two forces are acting on the ball (i) tension (T ) (ii) weight ( mg ) Torque of tension about point O is zero, as it passes through O. Solution

Here,

= (1)(10)(0.5) = 5 N-m V

30°

τ mg = F × r⊥ r⊥ = OP = 1.0 sin 30° = 0.5 m τ mg = ( mg )(0.5)

∴

r⊥

O

1.0 m T

Ans.

P

30°

mg

Fig. 12.38

Example 12.9 A force F = (2$i + 3$j − 4k$ ) N is acting at point P (2 m, −3 m, 6 m). Find torque of this force about a point O whose position vector is ( 2$i − 5$j + 3k$ ) m. Solution τ = r × F Here, r = r P − rO = ( 2$i − 3$j + 6k$ ) − ( 2$i − 5$j + 3k$ ) = ( 2$j + 3k$ ) m Now,

 $i τ = r × F = 0  2

$j 2 3

k$  3= ( −17i$ + 6$j − 4k$ ) N-m  − 4

Ans.

12.3

INTRODUCTORY EXERCISE

1. A force F = (2$i + 3$j − 2k$ ) N is acting on a body at point (2 m, 4 m, –2 m). Find torque of this force about origin.

2. A particle of mass m = 1kg is projected with speed u = 20 2 m/s at angle θ = 45° with horizontal. Find the torque of the weight of the particle about the point of projection when the particle is at the highest point.

3. Point C is the centre of mass of the rigid body shown in figure. Find the total torque acting on the

20

N

body about point C.

C 45° 10 cm

60° 30 N

5 cm 15 cm 10 N

Fig. 12.39

4. Find the net torque on the wheel in figure about the point O if a = 10 cm and b = 25 cm. 10.0 N 60°

a O

12 N b 9.0 N

Fig. 12.40

Chapter 12

Rotational Mechanics — 109

12.5 Rotation of a Rigid Body about a Fixed Axis When a body is rotating about a fixed axis, any point P located in the body travels along a circular path. Before, analyzing the circular motion of point P, we will first study the angular motion properties of a rigid body.

dθ α

Angular Motion

ω

Since, a point is without dimension, it has no angular motion. Only lines or bodies undergo angular motion. Let us consider the angular motion of a radial line r located with the shaded plane.

Angular Position The angular position of r is defined by the angle θ, measured between a fixed reference line OA and r. O r

Angular Displacement The change in the angular position, often measured as a differential dθ is called the angular displacement. (Finite angular displacements are not vector quantities, although differential rotations dθ are vectors). This vector has a magnitude dθ and the direction of dθ is along the axis.

P

A

θ

dθ

Fig. 12.41

Specifically, the direction of dθ is determined by right hand rule; that is, the fingers of the right hand are curled with the sense of rotation, so that in this case the thumb or dθ points upward.

Angular Velocity The time rate of change in the angular position is called the angular velocity ω. Thus, α

ω O r A P

dθ

θ

Fig. 12.42

ω=

dθ dt

…(i)

It is expressed here in scalar form, since its direction is always along the axis of rotation, i.e. in the same direction as dθ.

Angular Acceleration The angular acceleration α measures the time rate of change of the angular velocity. Hence, the magnitude of this vector may be written as,

110 — Mechanics - II α=

dω dt

…(ii)

It is also possible to express α as, α=

d 2θ

dt 2 The line of action of α is the same as that for ω, however its sense of direction depends on whether ω is increasing or decreasing with time. In particular, if ω is decreasing, α is called an angular deceleration and therefore, has a sense of direction which is opposite to ω.

Torque and Angular Acceleration for a Rigid Body The angular acceleration of a rigid body is directly proportional to the sum of the torque components along the axis of rotation. The proportionality constant is the inverse of the moment of inertia about that axis, or Στ α= I Thus, for a rigid body we have the rotational analog of Newton’s second law : Στ = Iα

…(iii)

Following two points are important regarding the above equation : (i) The above equation is valid only for rigid bodies. If the body is not rigid like a rotating tank of water, the angular acceleration α is different for different particles. (ii) The sum Στ in the above equation includes only the torques of the external forces, because all the internal torques add to zero.

Rotation with Constant Angular Acceleration If the angular acceleration of the body is constant then Eqs. (i) and (ii) when integrated yield a set of formulae which relate the body’s angular velocity, angular position and time. These equations are similar to equations used for rectilinear motion. Table given ahead compares the linear and angular motion with constant acceleration. Table 12.2 Straight line motion with constant linear acceleration a = constant v = u + at 1 2 at 2 v 2 = u 2 + 2 a ( s − s0 ) s = s0 + ut +

Fixed axis rotation with constant angular acceleration α = constant ω = ω0 + αt

1 2 αt 2 ω2 = ω20 + 2α (θ − θ0 ) θ = θ0 + ω0t +

Here,θ 0 and ω 0 are the initial values of the body’s angular position and angular velocity respectively. Note If α is not constant then we will have to take help of either differentiation or integration (with limits). The equations involved are :

Chapter 12

Equations of Differentiation ω=

Rotational Mechanics — 111

dθ dω dω and α = =ω dt dt dθ

Equations of Integration

∫ dθ = ∫ ω dt, ∫ dω = ∫ α dt

and

∫ ω dω = ∫ α dθ

Kinetic Energy of a Rigid Body Rotating about a Fixed Axis ω

Suppose a rigid body is rotating about a fixed axis with angular speed ω. Then, kinetic energy of the rigid body will be

ri

1 1 mi v i2 = Σ mi (ωri ) 2 i 2 i 2 1 2 1 = ω Σ mi ri2 = Iω 2 (as Σ mi ri2 = I ) i i 2 2 1 2 Thus, KE = Iω 2 Sometimes it is called the rotational kinetic energy. K =Σ

mi

Fig. 12.43

Extra Points to Remember ˜

˜

˜

˜

ω

In pure rotational motion of a rigid body all particles rotate in circular paths except the particles lying on the axis.The particles lying on the axis are at rest. Planes of all circular paths are mutually parallel with their centres lying on the axis.

r1 C1

P1

Linear velocity of any particle is tangential to its own circle and its linear r2 P2 C2 speed is, v = rω In this equation, ω is same for all particles (and that is also called ω of the rigid body). ∴ v ∝r Fig. 12.44 In the figure shown, r2 > r1 ⇒ ∴ v 2 > v1 Further, r for the particles lying on the axis is zero. Therefore, their linear velocity is zero or they are at rest. Since every particle is rotating in a circle (except the particles lying on the axis). Hence acceleration of the particle has two components : (i) Radial component or centripetal acceleration given by v2 ar = r ω2 = r This component is always towards the centre and this can't be zero. dv dω (ii) Tangential component given by at = = r α where, α = dt dt This component is tangential and it may be zero, positive or negative. If v or ω is constant then at is zero. If v or ω is increasing then this component is positive and in the direction of linear velocity. If v or ω is decreasing then this component is negative and in the opposite direction of linear velocity. (iii) Net acceleration of the particle is the resultant of these two perpendicular components.

112 — Mechanics - II Note From the above discussions, we have seen that pure rotational motion of a rigid body is nothing but circular motion of its different particles. So, before solving the problems of this topic make yourself expert in circular motion. V

Example 12.10 A solid sphere of mass 2 kg and radius 1 m is free to rotate about an axis passing through its centre. Find a constant tangential force F required to rotate the sphere with 10 rad/s in 2 s. Also find the number of rotations made by the sphere in that time interval. Solution Since, the force is constant, the torque produced by it and the angular acceleration α will be constant. Hence, we can apply ω = ω 0 + αt ⇒ 10 = 0 + (α )( 2) ⇒ α = 5 rad /s 2

Further, the force is tangential. Therefore, the perpendicular distance from the axis of rotation will be equal to the radius of the sphere. τ F ⋅R 5F 2mRα or F = α= = = ∴ 2 I 2mR 5 mR 2 5 ( 2)( 2)(1)( 5) Substituting the value, we have Ans. F= =4N ( 5) 1 2 1 Further, angle rotated θ = αt = ( 5)( 2) 2 = 10 rad 2 2 θ 10 5 Ans. ∴ Number of rotations n = = = 2π 2π π V

Example 12.11 The angular position of a point on the rim of a rotating wheel is given by θ = 4t − 3t2 + t3 , where θ is in radians and t is in seconds. What are the angular velocities at (b) t = 4.0s ? (a) t = 2.0 s and (c) What is the average angular acceleration for the time interval that begins at t = 2.0 s and ends at t = 4.0 s? (d) What are the instantaneous angular acceleration at the beginning and the end of this time interval? dθ d Solution Angular velocity ω = = ( 4t − 3t 2 + t 3 ) or ω = 4 − 6t + 3t 2 dt dt (a) At t = 2.0 s, ω = 4 − 6 × 2 + 3( 2) 2 or ω = 4 rad /s Ans. (b) At t = 4.0 s, ω = 4 − 6 × 4 + 3( 4 ) 2 or ω = 28 rad /s ω f − ω i 28 − 4 or α av = 12 rad /s 2 (c) Average angular acceleration α av = = 4−2 t f − ti

Ans. Ans.

(d) Instantaneous angular acceleration is, dω d α= = ( 4 − 6t + 3t 2 ) or α = − 6 + 6t dt dt At t = 2.0 s,

α = − 6 + 6 × 2 = 6 rad/s 2

Ans.

At t = 4.0 s,

α = − 6 + 6 × 4 = 18 rad/s

Ans.

2

Chapter 12

y

Example 12.12 A circular disc is rotating with an angular speed (in radian per sec) ω = 2t

y

2

P

Given, CP = 2 m $ $ $ at t = 1 s In terms of i, j and k, find, (a) ω (b) α (c) linear velocity of the particle lying at P (d) linear acceleration of the particle lying at P dω Solution ω = 2t 2 ⇒ α = = 4t dt At t = 1s,

37° C

x

x

ω

Fig. 12.45

ω = 2 rad /s and α = 4 rad /s 2

For the particle at P, r = CP = 2m (a) Rotation is clockwise. So, according to right hand rule, ω is perpendicular to paper inwards along negative z-direction. ∴ ω = ( − 2 k$ ) rad /s (b) ω is increasing. So, α is also in the direction of ω. ∴ α = ( − 4 k$ ) rad /s 2 (c) v = r ω = ( 2) ( 2) = 4 m/s This velocity is tangential to the doted circle of P as shown in figure. ∴ v = ( 4 cos 53° ) i$ − ( 4 sin 53° ) $j

y

ω

v = ( 2.4 i$ − 3.2 $j ) m/s

or

C

(d) Acceleration of the particle has two components (i) a r = r ω 2 (radial component) This components is towards centre C. (ii) a t = rα (tangential component) 2 = ( 2) ( 4 ) = 8 m/s

or

a = ( − 1.6 i$ − 11.2 $j ) m/s 2

53° x v = 4 m/s

37°

y

/s 2

This component is in the direction of linear velocity, as ω or v is increasing. ∴ a = ( 8cos 53° − 8cos 37° )( $i ) + ( 8sin 53° + 8sin 37° ) ( − $j )

P

Fig. 12.46

= ( 2) ( 2) 2 = 8 m/s 2

ar = 8m

V

Rotational Mechanics — 113

P 37°

53°

C

x a t = 8 m/s2

Ans. Fig. 12.47

114 — Mechanics - II

INTRODUCTORY EXERCISE

12.4

1. A wheel rotating with uniform angular acceleration covers 50 rev in the first five seconds after the start. Find the angular acceleration and the angular velocity at the end of five seconds.

2. A body rotates about a fixed axis with an angular acceleration 1 rad/s 2. Through what angle does it rotates during the time in which its angular velocity increases from 5 rad/s to 15 rad/s?

3. A flywheel of moment of inertia 5.0 kg -m 2 is rotated at a speed of 10 rad/s. Because of the friction at the axis it comes to rest in 10 s. Find the average torque of the friction.

4. A wheel starting from rest is uniformly accelerated at 4 rad/s 2 for 10 s. It is allowed to rotate uniformly for the next 10 s and is finally brought to rest in the next 10 s. Find the total angle rotated by the wheel.

5. A wheel of mass 10 kg and radius 0.2 m is rotating at an angular speed of 100 rpm, when the motion is turned off. Neglecting the friction at the axis, calculate the force that must be applied tangentially to the wheel to bring it to rest in 10 rev. Assume wheel to be a disc.

6. A solid body rotates about a stationary axis according to the law θ = 6t − 2 t 3. Here, θ is in radian and t in seconds. Find (a) the mean values of the angular velocity and angular acceleration averaged over the time interval between t = 0 and the complete stop, (b) the angular acceleration at the moment when the body stops. t2

Hint

If y = y (t ), then mean/average value of y between t1 and t 2 is < y > =

∫t

y (t ) dt

1

t 2 − t1

.

7. A body rotating at 20 rad/s is acted upon by a constant torque providing it a deceleration of 2 rad/s 2. At what time will the body have kinetic energy same as the initial value if the torque continues to act ?

8. A wheel whose moment of inertia is 0.03 kg m 2, is accelerated from rest to 20 rad/s in 5 s. When the external torque is removed, the wheel stops in 1 min. Find (a) the frictional torque, (b) the external torque.

9. A flywheel whose moment of inertia about its axis of rotation is 16 kg -m 2 is rotating freely in its own plane about a smooth axis through its centre. Its angular velocity is 9 rad s −1 when a torque is applied to bring it to rest in t 0 seconds. Find t 0 if (a) the torque is constant and of magnitude 4 N-m, (b) the magnitude of the torque after t seconds is given by kt.

10. A shaft is turning at 65 rad/s at time zero. Thereafter, angular acceleration is given by α = − 10 rad/s 2 − 5 t rad/s 2 where t is the elapsed time. (a) Find its angular speed at t = 3.0 s. (b) How much angle does it turn in these 3s ?

11. The angular velocity of a gear is controlled according to ω = 12 − 3t 2 where ω, in radian per second, is positive in the clockwise sense and t is the time in seconds. Find the net angular displacement ∆θ from the time t = 0 to t = 3 s. Also, find the number of revolutions N through which the gear turns during the 3 s.

Rotational Mechanics — 115

Chapter 12

12.6 Angular Momentum A mass moving in a straight line has linear momentum ( P ) . When a mass rotates about some point/axis, there is momentum associated with rotational motion called the angular momentum ( L) . Just as net external force is required to change the linear momentum of an object a net external torque is required to change the angular momentum of an object. Keeping in view the problems asked in JEE, the angular momentum is classified in following three types.

Angular Momentum of a Particle about a Fixed Point

Suppose a particle A of mass m is moving with linear momentum P = mv. Its angular momentum L about point O is defined as L = r × P = r × ( mv ) = m( r × v ) Here, r is the radius vector of particle A about O at that instant of time. The magnitude of L is L = mvr sin θ = mvr⊥ A

θ P=mv r⊥ = r sin θ

r θ

O

Fig. 12.48

Here, r⊥ = r sin θ is the perpendicular distance of line of action of velocity v from point O. The direction of L is same as that of r × v. Direction of angular momentum can also be by right hand rule as shown below: P

v

v

r⊥

P

r⊥

O (a)

O (b) v

P r⊥

P

v

O

O (c)

(d)

Fig. 12.49

In figures (a) and (c), rotation is clockwise. Hence, according to right hand rule angular momentum is perpendicular to paper inwards or in ⊗ direction. In figures (b) and (d), rotation is anticlockwise. Hence, direction of angular momentum is . direction. perpendicular to paper outwards or in O

116 — Mechanics - II Extra Points to Remember ˜

If line of action of velocity passes through the point O, then r⊥ = 0. Therefore, angular momentum is zero. v

O

P

Fig. 12.50 ˜

If a particle rotates in a circle then r⊥ from centre is always equal to R (= radius of circle). Or, θ between r and v (or P) is always 90°. Therefore, angular momentum about centre is L = mv R v

v R

R

C

C

(a)

(b) Fig. 12.51

˜

In figure (a), direction of angular momentum is perpendicular to paper inwards ⊗ and in figure (b) outwards . . or O O3 If a particle is moving with a constant velocity (speed and direction of velocity both are constant) then angular c momentum about any point always remains constant. But this constant value will be different about different points. v O1 v v In the figure shown, b v = constant LO1 = 0 as r⊥ = 0 LO 2 = mvb (= constant), perpendicular to paper inwards O2 LO 3 = mvc (= constant), perpendicular to paper outwards. Fig. 12.52

Angular Momentum of a Rigid Body Rotating about a Fixed Axis Suppose a particle P of mass m is going in a circle of radius r and at some instant the speed of the particle is v. For finding the angular momentum of the particle about the axis of rotation, the origin may be chosen anywhere on the axis. We choose it at the centre of the circle. In this case r and P are perpendicular to each other and r × P is along the axis. Thus, component of r × P along the axis is mvr itself. The angular momentum of the whole rigid body about AB is the sum of components of all particles, i.e.

A

v i = ri ω L = Σ mi ri2ω

or

L = ω Σ mi ri2

P

B

i

Here, ∴

r

O

L = Σ mi ri v i

ω

Fig. 12.53

i

i

or

L = Iω

(as Σ mi ri2 = I ) i

Here, I is the moment of inertia of the rigid body about AB. Thus, L = Iω is the component of angular momentum of the whole rigid body about axis of rotation AB. Direction of this component is again given by right hand rule. For example, in the given figure L = Iω is upwards or along the axis of rotation from B to A.

Chapter 12

Rotational Mechanics — 117

Extra Points to Remember The vector relation L = Iω is not correct in the above case because L and ω do not point in the same direction, but we could write LAB = Iω. If however the body is symmetric about the axis of rotation L and ω are parallel and we can write (L = Iω) in vector form as L = Iω..

˜

By symmetric we mean that for every mass element in the body there must be an identical mass element diametrically opposite the first element and at the same distance from the axis of rotation.

˜

Thus, remember that L = Iω applies only to bodies that have symmetry about the (fixed) rotational axis. Here, L stands for total angular momentum. However the relation LAB = Iω holds for any rigid body symmetrical or not that is rotating about a fixed axis.

˜

Angular Momentum of a Rigid Body in Combined Rotation and Translation Let O be the fixed point in an inertial frame of reference.

θ

r⊥ r

θ

v

ω

r sin θ = r⊥

O

Fig. 12.54

Angular momentum of the rigid body about O is the vector sum of two terms (as discussed above). (i) mvr sin θ or mvr⊥ (ii) Iω Here, v is the velocity of centre of mass. Moment of inertia I is about an axis passing through centre of mass. Directions of above two terms can be determined by right hand rule. If both the terms are in the same direction then these two terms are additive and if they are in opposite directions, then they are subtractive. V

Example 12.13 A particle of mass m is moving along the line y = b, z = 0 with constant speed v. State whether the angular momentum of particle about origin is increasing, decreasing or constant. Solution

| L | = mvr sin θ = mvr⊥ = mvb

∴

| L | = constant as m, v and b all are constants. y P

r

v

r⊥ = b

θ

x

O

Fig. 12.55

Direction of r × v also remains the same. Therefore, angular momentum of particle about origin remains constant with due course of time. Note In this problem | r | is increasing, θ is decreasing but r sin θ , i.e. b remains constant. Hence, the angular momentum remains constant.

118 — Mechanics - II V

Example 12.14 A particle of mass m is projected from origin O with speed u at an angle θ with positive x-axis. Positive y-axis is in vertically upward direction. Find the angular momentum of particle at any time t about O before the particle strikes the ground again. Solution

Here, and

L = m( r × v)

y

1 r ( t ) = x$i + y$j = ( u cos θ )t$i + ( ut sin θ − gt 2 ) $j 2 $ $ $ v( t ) = v i + v j = ( u cos θ )i + ( u sin θ − gt )$j x

θ

y

$i ∴

u

L = m ( r × v) = m ( u cos θ )t

$j 1 2 gt 2 u sin θ − gt

( u sin θ )t −

u cos θ

k$

x

O

Fig. 12.56

0 0

1   = m ( u 2 sin θ cos θ )t − ( u cos θ )gt 2 − ( u 2 sin θ cos θ )t + ( u cos θ )gt 2  k$ 2   1 Ans. = − m( u cos θ ) gt 2 k$ 2 V

Example 12.15 A rod of mass 2kg and length 2m is O rotating about its one end O with an angular velocity ω ω = 4 rad/s. Find angular momentum of the rod about the Fig. 12.57 axis rotation. Solution In pure rotational motion of a rigid body, component of total angular momentum about axis of rotation is given by  ml 2   ml 2  L= I ω =  ω I 0 =  3   3   Substituting the values we have, ( 2) ( 2) 2 (4 ) 3 32 = kg -m 2 /s 3

L=

Ans.

Direction of this component is perpendicular to paper inwards (from right hand rule), as the rotation is clockwise. V

Example 12.16 A circular disc of mass m and radius R is set into motion on a horizontal floor with a linear speed v in v the forward direction and an angular speed ω = in R clockwise direction as shown in figure. Find the magnitude of the total angular momentum of the disc about bottommost point O of the disc.

v ω O Fig. 12.58

Solution As we have discussed, angular momentum about O is the vector sum of two terms:

Chapter 12

Here,

Iω and mvr⊥ 1   v 1 Iω = I cmω =  mR 2    = mvR 2   R 2

Rotational Mechanics — 119

(perpendicular to paper inwards)

and (perpendicular to paper inwards) mvr⊥ = mv c r⊥ = mRv Since, both the terms are in the same direction. 1 LTotal = mvR + mvR v 2 3 or (perpendicular to paper inwards) LTotal = mvR 2 O Fig. 12.59

12.5

INTRODUCTORY EXERCISE

1. A uniform rod of mass m is rotated about an axis passing through point O as shown. Find angular momentum of the rod about rotational axis. ω

2l

l O

Fig. 12.60

2. A particle of mass 1 kg is moving along a straight line y = x + 4. Both x and y are in metres. Velocity of the particle is 2 m/s. Find magnitude of angular momentum of the particle about origin.

3. A particle of mass m is projected from the ground with an initial speed u at an angle α. Find the magnitude of its angular momentum at the highest point of its trajectory about the point of projection.

4. If the angular momentum of a body is zero about some point. Is it necessary that it will be zero about a different point?

5. A solid sphere of mass m and radius R is rolling without slipping as shown in figure. Find angular momentum of the sphere about z-axis. y

v R x

O

Fig. 12.61

6. In example number 12.16 suppose the disc starts rotating anticlockwise with the same angular v , then what will be the angular momentum of the disc about bottommost point in R this new situation?

velocity ω =

7. Two particles each of mass m and speed v, travel in opposite directions along parallel lines separated by a distance d. Show that the vector angular momentum of this system of particles is the same about any point taken as origin.

120 — Mechanics - II

12.7 Conservation of Angular Momentum As we have seen in Article 12.6, the angular momentum of a particle about some reference point O is defined as, …(i) L=r × p Here, P is the linear momentum of the particle and r its position vector with respect to the reference point O. Differentiating Eq. (i) with respect to time, we get dL d p dr …(ii) =r× + ×p dt dt dt dp dr (velocity of particle) Here, = F and =v dt dt Hence, Eq. (ii) can be rewritten as, dL =r×F+ v× p dt Now, v × p = a null vector, because v and p are parallel to each other and the cross product of two parallel vectors is a null vector. Thus, dL dL …(iii) = r × F = τ or τ = dt dt Which states that the time rate of change of angular momentum of a particle about some reference point in an inertial frame of reference is equal to the net torques acting on it. This result is rotational dp analog of the equation F = , which states that the time rate of change of the linear momentum of a dt particle is equal to the force acting on it. Eq. (iii), like all vector equations, is equivalent to three scalar equations, namely  dL   dL   dL  τ x =   , τ y =   and τ z =    dt  x  dt  y  dt  z dL . According to which the dt time rate of change of the total angular momentum of a system of particles about some reference point in an inertial frame of reference is equal to the sum of all external torques (of course the vector sum) acting on the system about the same reference point. dL Now, suppose that τ ext = 0, then = 0, so that L = constant. dt When the resultant external torque acting on a system is zero, the total vector angular momentum of the system remains constant. This is the principle of the conservation of angular momentum. For a rigid body rotating about an axis (the z-axis, say) that is fixed in an inertial reference frame, we have L z = Iω It is possible for the moment of inertia I of a rotating body to change by rearrangement of its parts. If no net external torque acts, then L z must remain constant and if I does change, there must be a The same equation can be generalised for a system of particles as, τ ext =

Chapter 12

Rotational Mechanics — 121

compensating change in ω. The principle of conservation of angular momentum in this case is expressed as Iω = constant

or

I 1ω 1 = I 2ω 2

…(iv)

Note According to above equation, if some mass moves away from the axis of rotation, its moment of inertia will increase. So, to conserve angular momentum or Iω its angular speed ω should decrease. Therefore, time 2π   period of rotation T =  should also increase.  ω V

Example 12.17 A wheel of moment of inertia I and radius R is rotating about its axis at an angular speed ω 0 . It picks up a stationary particle of mass m at its edge. Find the new angular speed of the wheel. Net external torque on the system is zero. Therefore, angular momentum will remain conserved. Thus, I ω I 1ω 1 = I 2ω 2 or ω 2 = 1 1 I2 Solution

Here,

I 1 = I , ω 1 = ω 0 , I 2 = I + mR 2 ω2 =

∴

Iω 0 I + mR 2

INTRODUCTORY EXERCISE

Ans.

12.6

1. A thin circular ring of mass M and radius R is rotating about its axis with an angular speed ω 0. Two particles each of mass m are now attached at diametrically opposite points. Find the new angular speed of the ring.

2. If the ice at the poles melts and flows towards the equator, how will it affect the duration of day-night?

3. When tall buildings are constructed on earth, the duration of day night slightly increases. Is this statement true or false?

4. If radius of earth is increased, without change in its mass, will the length of day increase, decrease or remain same?

12.8 Combined Translational and Rotational Motion of a Rigid Body This is the most complex motion of a rigid body. But this can be simplified by splitting this motion into following two parts: (i) pure translational motion with the velocity ( v ) and acceleration ( a ) of centre of mass. (ii) pure rotational motion about an axis passing through centre of mass with angular velocity ω and angular acceleration α. As we have discussed earlier also, different particles of the rigid body in this type of motion have different linear velocity and linear acceleration. So, now the question is how to find linear velocity and linear acceleration of a general particle P on the rigid body?

122 — Mechanics - II Method of Finding Linear Velocity of a General Particle P Suppose linear velocity v (of centre of mass) and angular velocity ω (of the rigid body) are known to us and we wish to find linear velocity of a general particle P at a distance r from C. v

P r rω C

v

ω

vP

Fig. 12.62

As we know that v PC = v P − v C ∴ v P = v C + v PC or, absolute velocity of P is the vector sum of v C and v PC . Here, v C = v and motion of P with respect to C is a circle (dotted circle shown in figure). In circular motion, velocity of a particle is rω, tangential to its circle (in the direction of rotation). So, v PC is rω in the tangential direction as shown in Fig. 12.62. Thus net velocity of P is the vector sum of following two terms: (i) v and (ii) rω For different particles, values of v and ω are same. But values of r and therefore rω and direction of rω are different . This is the reason why different particles have different linear velocities.

Method of Finding Acceleration of a General Particle P Suppose linear acceleration a (of centre of mass) angular velocity ω and angular acceleration α (of the rigid body) are known to us and we wish to find linear acceleration of a general particle P at a distance r from C. As we know that a PC = a P − a C ∴ a P = a C + a PC Thus, absolute acceleration of P is the vector sum of a C and a PC . Here, a C = a and motion of P with respect to C is a circle (dotted circle shown in figure). In circular motion, acceleration of a particle has two components: (i) tangential component, a t (ii) radial component, a r

P r ar

a at a

C ω, α

Fig. 12.63

dω ) dt This component is tangential in the direction of linear velocity if ω is increasing and in the opposite direction of linear velocity if ω is decreasing. Further, this component is zero if ω is constant. a r = rω 2 Here,

a t = rα

(where, α =

Chapter 12

Rotational Mechanics — 123

and this component is always towards centre C. Hence, net acceleration of the particle P is the vector sum of the following three terms: (i) a (ii) a t = rα (iii) a r = rω 2 For different particles, values of a, ω and α are same. But value of r is different. Therefore rα, rω 2 and their directions are different. This is the reason why different particles have different linear accelerations.

Kinetic Energy of Rigid Body in Combined Translational and Rotational Motion  1  2 Here, two energies are associated with the rigid body. One is translational  = m vCOM  and another  2    1 is rotational  = I COM ω 2  . Thus, total kinetic energy of the rigid body is   2 K= V

1 1 2 mvCOM + I COM ω 2 2 2

Example 12.18

y P C

37°

v

ω

x

Fig. 12.64

In the figure shown v = 2 m/ s , ω = 5rad/ s and CP = 1 m In terms of $i and $j find linear velocity of particle P. Solution For particle P, P 37° C

53° v = 2 m/s r ω = 5 m/s v

y

x

Fig. 12.65

r = CP = 1 m ⇒ rω = (1)( 5) = 5 m/s Net velocity of P is the vector sum of v and rω as shown in figure. ∴ v = 2 $i + ( 5 cos 53° $i − 5sin 53° $j ) P

= 2i$ + 3i$ − 4$j = ( 5i$ − 4$j ) m/s

Ans.

124 — Mechanics - II V

Example 12.19 A disc of radius R has linear velocity v and angular velocity ω as shown in the figure. Given v = rω. Find velocity of points A, B, C and D on the disc. C

B

D

v

R A

Fig. 12.66

As stated in above article, velocity of any point of the rigid body in rotation plus translation is the vector sum of v (the velocity of centre of mass) and rω. Here, r is the distance of the point under consideration from the centre of mass of the body. Direction of this rω is perpendicular to the line joining the point with centre of mass in the sense of rotation. Based on this, velocities of points A, B, C and D are as shown below Solution

R ω =v Rω

A vA = 0 as v = Rω

vB = √2v = √2 Rω v + Rω = 2v

v

C 45°

B

v v

D 45°

R ω =v

√2v = vD

Fig. 12.67

Thus, v A is zero, velocity of B and D is 2v or 2Rω and velocity of C is 2v or 2Rω in the directions shown in figure. V

Example 12.20 In the shown figure, a = 2 m/ s 2 , ω = ( 2t) rads −1 and CP = 1 m. y P C

ω

53° a x

Fig. 12.68

In terms of $i and $j , find linear acceleration of the particle at P at t = 1 s.

Chapter 12

Rotational Mechanics — 125

Solution For particle at P

r = CP = 1m dω d α= = ( 2t ) = 2 rad /s 2 dt dt

⇒ At t = 1s,

ω = 2rad /s α = 2 rad /s 2 a t = rα = 2 m/s 2 a r = rω 2 = 4 m/s 2 a = 2 m/s 2

and

Net acceleration of P is the vector sum of three terms a, a r and a t as shown in figure below. y P ar x

C

a = 2 m/s2 37°

2 53° at = 2 m/s a

37° ar = 4m/s2

Fig. 12.69

∴

a P = 2$i + ( 2cos 37° $i − 2sin 37° $j ) + ( − 4 sin 37° $i − 4 cos 37° $j ) = 2 i$ + 1.6 i$ − 1.2 $j − 2.4 $i − 3.2 $j = (1.2 $i − 4.4 $j ) m/s 2

INTRODUCTORY EXERCISE

Ans.

12.7

v 1. In the figure shown,ω = . In terms of $i and $j, find linear velocities of particles M, N, R and S. 2R y

R

N

S

C

v

x

ω

M

Fig. 12.70

2. In the same figure, if v and ω both are constant, then find linear accelerations of points M, N, R and S in terms of R, ω, $i and $j, where R is the radius of disc.

126 — Mechanics - II

12.9 Uniform Pure Rolling Pure rolling means no relative motion (or no slipping) at point of contact between two bodies. For example, consider a disc of radius R moving with linear velocity v and angular velocity ω on a horizontal ground. The disc is said to be moving without slipping if velocities of points P and Q (shown in figure 12.71 (b) are equal, i.e.

⇒

v

COM ω

Rω

P

v

Q (b)

(a)

Fig. 12.71

vP = vQ v − Rω = 0

or

v = Rω

or

If v P > v Q or v > Rω, the motion is said to be in forward slipping and if v P < v Q or v < Rω, the motion is said to be in backward slipping (or sometimes called forward English). v

⇒

ω v0

Rω

P v0

v

Q

Fig. 12.72

In pure rolling ( v = Rω ) over a stationary ground net velocity of bottommost point of the body is zero. In forward slip condition ( v > Rω ) net velocity is in the direction of motion. In backward slip condition ( v < Rω ) net velocity is in the opposite direction of motion. Thus, v = Rω is the condition of pure rolling on a stationary ground. Sometimes it is simply said rolling. Suppose the base over which the disc in rolling, is also moving with some velocity (say v 0 ) then in that case condition of pure rolling is different. For example, in the Fig. 12.72, vP = vQ or v − Rω = v 0 Thus, in this case v − Rω ≠ 0, but v − Rω = v 0 . By uniform pure rolling we mean that v and ω are constant. They are neither increasing nor decreasing.

Rotational Mechanics — 127

Chapter 12 Extra Points to Remember

˜

In case of pure rolling on a stationary horizontal ground (when v = Rω), following points are important to note: Distance moved by the centre of mass of the rigid body in one full rotation is 2 πR. v ω s = 2πR

Fig. 12.73

2π s = v ⋅ T = (ωR )   = 2 πR  ω

This is because

˜

In forward slipping s > 2 πR and in backward slipping s < 2 πR The speed of a point on the circumference of the body at the instant shown in θ θ figure is 2 v sin or 2 Rω sin . i.e. 2 2 θ θ |v P| = v p = 2 v sin = 2 Rω sin 2 2

(as v > ωR) (as v < ωR)

v P

This can be shown as : vPC = Rω = v

θ

ω

Fig. 12.74

vP

C P

θ

180°– θ vC = v

Fig. 12.75

Fig. 12.76

v P = vC + v PC |v P| =

∴ ˜

From the above expression we can see that : as vA = 0 and

˜

v 2 + v 2 + 2 v ⋅ v cos (180° − θ) = 2 v sin

θ 2

C

θ = 0°

vB = 2v

as

θ = 90°

vC = 2 v

as

θ = 180°

B

The path of a point on circumference is a cycloid and the distance moved by this point in one full rotation is 8R.

A Fig. 12.77

A3 A2

A4

A1

A5

Fig. 12.78

128 — Mechanics - II In the figure, the dotted line is a cycloid and the distance A1 A2 K A5 is 8R. This can be proved as under. In Fig. 12.79 c

c

v

ω

A

A t=0

θ

t=t

Fig. 12.79

θ = ωt Speed of point A at this moment is, Distance moved by it in time dt is,

θ ωt = 2 Rω sin   2  2 ωt ds = v A dt = 2 Rω sin   dt 2  v A = 2 Rω sin

Therefore, total distance moved in one full rotation is, s=

T = 2π / ω

∫0

ds or

s=

T = 2π / ω

∫0

ωt 2 Rω sin   dt 2 

s = 8R. KR = 1 for a ring KT 1 for a disc = 2 2 for a solid sphere = 5 2 for a hollow sphere etc. = 3 1 1 Here, K R stands for rotational kinetic energy  = Iω2  and KT for translational kinetic energy  = mv 2  .  2   2  For example, for a disc

On integration we get, ˜

∴

V

2

1 2 1 1 v Iω =  mR 2      R 2 2 2 1 1 = mv 2 and KT = mv 2 4 2 KR 1 = KT 2 KR =

Example 12.21 A solid disc is rolling without slipping on a horizontal ground as shown in figure. Its total kinetic energy is 100 J. What is its translational and rotational kinetic energy. Solution In case of pure rolling, ratio of rotational kinetic

1 energy and translational kinetic energy is . 2 or

KR 1 = KT 2

 1  ⇒ ∴ KR =  (Total kinetic energy)  1+ 2

Fig. 12.80

Chapter 12 =

1 100 (100 J ) = J 3 3

Ans.

 2  KT =  (Total kinetic energy)  1+ 2

Similarly,

= V

Rotational Mechanics — 129

2 200 (100 J ) = J 3 3

Ans.

Example 12.22 A disc of radius R starts at time t = 0 moving along the positive x-axis with linear speed v and angular speed ω. Find the x and y coordinates of the bottommost point at any time t. y

v

C ω

x

O

Fig. 12.81

At time t the bottommost point will rotate an angle θ = ωt with respect to the centre of the disc C. The centre C will travel a distance s = vt.

Solution

y

C

C

R P

O

θ

Q

M

s = vt

x

Fig. 12.82

In the figure, PQ = R sin θ = R sin ωt and CQ = R cos θ = R cos ωt Coordinates of point P at time t are, and ∴

x = OM − PQ = vt − R sin ωt y = CM − CQ = R − R cos ωt ( x, y ) ≡ ( vt − R sin ωt , R − R cos ωt )

INTRODUCTORY EXERCISE

12.8

1. A solid sphere is rolling without slipping on a horizontal ground. Its rotational kinetic energy is 10 J. Find its translational and total kinetic energy.

2. Under forward slip condition, translational kinetic energy of a ring is greater than its rotational kinetic energy. Is this statement true or false?

3. In backward slip condition, translational kinetic energy of a disc may be equal to its rotational kinetic energy. Is this statement true of false?

130 — Mechanics - II

12.10 Instantaneous Axis of Rotation As we have seen that combined rotational and translational motion of a rigid body is the most complex motion. But this motion can be simplified and may be assumed to be in pure rotational motion (with same ω) about an axis called instantaneous axis of rotation. Further as we know that, in pure rotational motion, points lying on the axis of rotation are at rest. Therefore, we can say that, instantaneous axis of rotation passes through those points which are at rest. For example, in pure rolling over ground instantaneous axis of rotation (IAOR) passes through the bottommost point, as it is a point of zero velocity. Thus, the combined motion of rotation and translation can be assumed to be pure rotational motion about bottommost point with same angular speed ω.

⇒

v ω

ω

Fig. 12.83

Now, there are two uses of the concept of instantaneous axis of rotation. (i) Velocity of any point P can obtained by a single term v = rω as in pure rotational motion this is the expression of velocity of any point P. Here r is the distance of P from instantaneous axis of rotation. With respect to O, point P is rotating in a circle with centre at O and radius as r. Velocity of P is tangential to this circle (or perpendicular to OP), in the direction of rotation. (ii) We can find total kinetic energy of the body by a single term. 1 K = Iω 2 2

v P

90° r

O ω Fig. 12.84

But here, I is the moment of inertia about instantaneous axis of rotation. V

Example 12.23 Using the concept of instantaneous axis of rotation. Find speed of particle P as shown in figure, under pure rolling condition. v P

90° r O

ω

Fig. 12.85

Chapter 12

Rotational Mechanics — 131

Solution In pure rolling, combined rotation and translation

motion may be assumed to be a pure rotational motion about an axis passing through bottommost point (with same ω) or instantaneous axis of rotation. θ | vP | = (OP ) ω Here, OP = 2R sin 2 θ θ   θ | vP | =  2R sin  ω = 2Rω sin   = 2v sin ∴   2 2 2 V

Example 12.24 A disc is rolling (without slipping) on a horizontal surface. C is its centre and Q and P are two points equidistant from C. Let vP , vQ and vC be the magnitude of velocities of points P ,Q and C respectively, then (JEE 2004) (a) vQ > vC > v P (b) vQ < vC < v P 1 (c) vQ = v P , vC = v P (d) vQ < vC > v P 2 Solution In case of pure rolling bottom most point is the instantaneous centre of zero velocity. Velocity of any point on the disc, v = r ω, where r is the distance of point from O. rQ > rC > rP ⇒ ∴ v Q > vC > v P Therefore, the correct option is (a).

INTRODUCTORY EXERCISE

P

v v = Rω

θ ω

Fig. 12.86

C

Q

P

Fig. 12.87

C

Q

P ω O Fig. 12.88

12.9

1. A disc is rolling without slipping with linear velocity v as shown in figure. With the concept of instantaneous axis of rotation, find velocities of points A, B, C and D. C

B

v D A Fig. 12.89

2. A solid sphere is rolling without slipping as shown in figure. Prove that

C

ω

v

O Fig. 12.90

1 1 1 mv 2 + IC ω 2 = I 0 ω 2 2 2 2

132 — Mechanics - II

12.11 Accelerated Pure Rolling Till now we were discussing the uniform pure rolling in which v and ω were constants. Now, suppose an external force is applied to the rigid body, the motion will no longer remain uniform. The condition of pure rolling on a stationary ground is, v = Rω Differentiating this equation with respect to time, we have dv dω =R. dt dt or a = Rα Thus, in addition to v = Rω at every instant of time there is one additional F condition, linear acceleration = R × angular acceleration or a = Rα for pure rolling to take place. Here, friction plays an important role in maintaining the pure rolling. The friction may sometimes act in forward direction, sometimes in backward direction or under certain conditions it may be zero. Here, we should not forget the basic nature of friction, which is a self adjusting force (upto a certain maximum limit) and which has a tendency to Fig. 12.91 stop the relative motion between two bodies in contact and here the relative motion stops when at every instant v = Rω. To satisfy this equation all the time, a = Rα equation should also be satisfied. Let us take an example illustrating the above theory. a Suppose a force F is applied at the topmost point of a rigid body of radius F R, mass M and moment of inertia I about an axis passing through the centre of mass. Now, the applied force F can produce by itself: a C (i) a linear acceleration a and (ii) an angular acceleration α. f If a = Rα, then there is no need of friction and force of friction f = 0. If a < Rα, then to support the linear motion the force of friction f will act in Fig. 12.92 forward direction. Similarly, if a > Rα, then to support the angular motion the force of friction will act in backward direction. So, in this case force of friction will be either backward, forward or even zero also. It all depends on M, I and R. For calculation purpose initially we can choose any direction of friction. Let we assume it in forward direction, Let, a = linear acceleration, α = angular acceleration F F+f …(i) then, a = net = M M τ (F − f )R …(ii) α= c = I I For pure rolling to take place, Solving Eqs. (i), (ii) and (iii), we get f =

a = Rα ( MR − I )

…(iii)

2

( MR 2 + I )

.F

…(iv)

Rotational Mechanics — 133

Chapter 12

From Eq. (iv) following conclusions can be drawn (i) If I = MR 2 (e.g. in case of a ring) f = 0 i.e. if a force F is applied on the top of a ring, the force of friction will be zero and the ring will roll without slipping. (ii) If I < MR 2 , (e.g. in case of a solid sphere or a hollow sphere), f is positive, i.e. force of friction will be forward. (iii) If I > MR 2 , f is negative, i.e. force of friction will be backwards. Although under no condition I > MR 2 . (Think why?). So, force of friction is either in forward direction or zero. Here, it should be noted that the force of friction f obtained in Eq. (iv) should be less than the limiting friction (µMg ) for pure rolling to take place. Further, we have seen that for I < MR 2 force of τ  friction acts in forward direction. This is because α is more if I is small  α =  i.e. to support  I the linear motion force of friction is in forward direction. Note It is often said that rolling friction is less than the sliding friction. This is because the force of friction calculated by equation number (iv) is normally less than the sliding friction (µ k N ) and sometimes it is in forward direction, i.e. it supports the motion.

Extra Points to Remember ˜

In accelerated pure rolling, v = Rω and a = Rα are not the only conditions to be satisfied, sense of rotation is also important as per the direction of linear acceleration. Sense of rotation (or direction of α) should be as shown below in following two figures: α

α a

a

Fig. 12.93 ˜

In following two figures accelerated pure rolling is not possible.

α

α

a

a

Fig. 12.94 ˜

There are certain situations in which the direction of friction is fixed. For example in the following situations the force of friction is backwards. This is because linear acceleration 'a' due to the applied force (= mg sinθ in first case) is in the direction shown in figure. So, direction of α should also be in the shown direction. We have only friction force which can provide 'α' in that direction. Hence, it should be in backward direction. a a

a

a Rough

F

a F

θ

Fig. 12.95

a

134 — Mechanics - II Rolling on Rough Inclined Plane A body of mass M, radius R and moment of inertia I is kept over a rough ground as shown in figure. In this case no external force is applied for accelerated pure rolling. But mg sin θ is already acting at centre. As we said earlier also, force of friction in this case will be backward. Equations of motion are Mg sin θ − f a= M fR α= I

M,R,I a

Mg

θ sin

α

f

θ

Fig. 12.96

…(i) …(ii)

For pure rolling to take place, a = Rα

…(iii)

Solving Eqs. (i), (ii) and (iii), we get f =

and

Mg sin θ

MR 2 1+ I g sin θ a= I 1+ MR 2

…(iv)

…(v)

From Eq. (v), we can see that if a solid sphere and a hollow sphere of same mass and radius are released from a rough inclined plane and pure rolling is taking place, then the solid sphere reaches the bottom first because I solid < I hollow or a solid > a hollow ⇒ ∴ t solid < t hollow Further, the force of friction calculated in Eq. (iv) for pure rolling to take place should be less than or equal to the maximum available friction µMg cos θ. tan θ Mg sin θ or ≤ µ Mg cos θ or µ ≥ 2 MR 2 MR 1+ 1+ I I Thus, minimum value of friction required for pure rolling is tan θ µ min = 1 + MR 2 / I If given value of µ > µ min , then friction acting on the body is Mg sin θ f = 1 + MR 2 / I and in this case linear acceleration of the body is g sin θ a= 1 + I / MR 2

Chapter 12 V

Rotational Mechanics — 135

Example 12.25 In the shown figure, accelerated pure rolling will takes place, if a = Rα. Find the case if (a) a > Rα

(b) a < Rα α a

Fig. 12.97

Solution (a) If a > Rα, then at any instant v > Rω. So, it is a case of forward slipping.

(b) If a < Rα, then at any instant v < Rω. So, it is a case of backward slipping. V

Example 12.26 If accelerated pure rolling is taking place on a stationary ground, then work done by friction is always zero. Comment on this. Solution In pure rolling on stationary ground the bottommost point of the rigid body (where

force of friction is acting) is at rest. Therefore, work done by friction is zero. V

Example 12.27 In the shown figure, M is mass of the body, R its radius and I the moment of inertia about an axis passing through centre. Find force of friction ‘f’ acting on the body (upwards), its linear acceleration ‘a’ (down the plane) and type of motion if:

M, R , I µ

θ (b) µ < µ min (c) µ > µ min (a) µ = 0 where, µ min is the minimum value of coefficient of friction Fig. 12.98 required for pure rolling. Solution (a) If µ = 0 then, f = 0 and a = g sin θ = a1 (say) and the motion is only translational. (b) If µ < µ min , then maximum value of friction will act, as friction is insufficient to provide accelerated pure rolling or to stop the relative motion.

∴ and

f = f max = µ mg cos θ Mg sin θ − µ Mg cos θ a= m

= g sin θ − µg cos θ = a 2 (say) In this case, motion is rotation + translation with forward slip (as a > Rα). (c) If µ > µ min , Then we have discussed in the above article that Mg sin θ g sin θ and a = f= = a 3 (say ) 2 I MR 1 + 1+ MR 2 I Motion in this case is rotation + translation with accelerated pure rolling. Note In the above example, we can see that a1 and a2 are independent of moment of inertia I, but a3 depends on it.

136 — Mechanics - II V

Example 12.28 A tangential force F acts at the top of a thin spherical shell of mass m and radius R. Find the acceleration of the shell if it rolls without slipping. F R

Fig. 12.99

Let f be the force of friction between the shell and the horizontal surface. Solution

F

For translational motion, F + f = ma

… (i) f

For rotational motion, FR − f R = Iα = I [Q a = Rα for pure rolling ] ⇒

F − f =I

Fig. 12.100

a R

a

K (ii)

R2

Adding Eqs. (i) and (ii), we get I   2F =  m + 2  a  R  2  5  =  m + m a = ma   3 3

V

or

F=

5 ma 6

⇒

a=

6F 5m

2  2 Q I shell = 3 mR  Ans.

Example 12.29 A horizontal force F acts on the sphere at its centre as shown. Coefficient of friction between ground and sphere is µ. What is maximum value of F, for which there is no slipping ? Solution

F − f = Ma

…(i)  a τ = Iα = I    R

⇒ ⇒

2 a MR 2 5 R 2 f = Ma 5

f ⋅R =

F

Fig. 12.101

…(ii)

Chapter 12

Rotational Mechanics — 137

Solving Eqs. (i) and (ii) we get, f=

⇒

2 F 7

F

2 F ≤ µmg 7 7 F ≤ µmg 2

INTRODUCTORY EXERCISE

f

Fig. 12.102

Ans.

12.10

1. Work done by friction in pure rolling is always zero. Is this statement true or false? 2. In the figure shown, a force F is applied at the top of a disc of mass 4 kg and radius 0.25 m. Find maximum value of F for no slipping. F

µ = 0.6

Fig. 12.103

3. In the figure shown a solid sphere of mass 4 kg and radius 0.25 m is placed on a rough surface. Find ( g = 10 ms 2 ) (a) minimum coefficient of friction for pure rolling to take place. (b) If µ > µmin, find linear acceleration of sphere. µ (c) If µ = min , find linear acceleration of cylinder. 2 Here, µmin is the value obtained in part (a).

30°

Fig. 12.104

4. A ball of mass M and radius R is released on a rough inclined plane of inclination θ. Friction is not sufficient to prevent slipping. The coefficient of friction between the ball and the plane is µ.Find: (a) the linear acceleration of the ball down the plane, (b) the angular acceleration of the ball about its centre of mass.

5. A spool is pulled by a force in vertical direction as shown in figure. What is the direction of friction in this case? The spool does not loose contact with the ground. F

Fig. 12.105

138 — Mechanics - II

12.12 Angular Impulse In the previous chapter, we have learnt that linear impulse = F ⋅ ∆t = J = change in linear momentum ∆P or, J = ∆P = P f − Pi = m ( v f − v i ) In one dimension, we can simply write as: J = ∆P = P f − Pi = m ( v f − v i ) If v i =0 and v f = v, then J J = mv or v = m In the similar manner, angular impulse = τ ⋅ ∆t = A ⋅ I = change in angular momentum ∆L or A ⋅ I = τ ∆t = ∆L = L f − Li But τ = F × r⊥ ∴ A ⋅ I = F × r⊥ × ∆t (as F × ∆t = J ) = J × r⊥ Thus, A ⋅ I = J × r⊥ = L f − Li If Li = 0, then L f = L = I ω J × r⊥ ∴ A ⋅ I = J × r⊥ = Iω or ω = I J

J

C

J v= m

P

C

(a)

CP = r⊥ ω

J v= m J × r⊥ ω= I

(b)

Fig. 12.106

In Fig. (a) A linear impulse J is applied at centre of mass C of the rigid body. Just after hitting, it will have only translational motion and its linear velocity will be given by J v= m In Fig. (b) A linear impulse J is applied at point P, at a perpendicular distance r⊥ = CP . Just after hitting it will have both translational and rotational motion. Its linear velocity v and angular velocity ω will be given by J × r⊥ J v = and ω = m I If r⊥ is increased (keeping J to be constant) then v will remain same but ω will increase. So, the translational kinetic energy will have the same value but rotational kinetic energy will be more.

Chapter 12

Rotational Mechanics — 139

Extra Points to Remember ˜

Angular impulse A ⋅ I = τ × ∆t = ∆L Now, there are following three cases: (i) If torque is constant, then angular impulse can be obtained by directly multiplying this constant torque with the given time interval. (ii) If torque is a function of time then angular impulse can be obtained by integration. tf

A ⋅ I = ∫ τ dt

∴

ti

(iii) If torque versus time graph is given then angular impulse can be obtained by the area under that graph. In all three cases, angular impulse is equal to the change in angular momentum. V

Example 12.30 A solid sphere of mass M and radius R is hit by a cue at a height h above the centre C. For what value of h the sphere will roll without slipping ? h C

Fig. 12.107

Solution For rolling without slipping, h C ω

v

Fig. 12.108

v = Rω Here, v and ω are the values obtained just after hitting. J  J × r⊥  =R ∴   I  M

…(i)

2 Here, r⊥ = h and I = I C = MR 2 5 Substituting these values in Eq. (i), we have 2 h= R 5

Ans.

140 — Mechanics - II V

Example 12.31 A uniform sphere of mass m and radius R starts rolling without slipping down an inclined plane. Find the time dependence of the angular momentum of the sphere relative to the point of contact at the initial moment. How will the result be affected in the case of a perfectly smooth inclined plane? The angle of inclination of the plane is θ. Solution

Applying the equation (about bottommost point)

m

f

in θ gs θ

Fig. 12.109

Angular impulse = change in angular momentum about point of contact we have, ∴

τ = ( mg sin θ ) R = constant Angular impulse = τ × t = ( mg sin θ ) Rt L = ( mg sin θ ) Rt

or

Ans.

Note There will be no change in the result even if body pure rolls or slides, as the torque of force of friction is zero about point of contact. So, it hardly matters whether the surface is rough or smooth.

INTRODUCTORY EXERCISE

12.11

1. A cylinder is rolling down a rough inclined plane. Its angular momentum about the point of contact remains constant. Is this statement true or false?

2. A solid sphere and a hollow sphere both of same mass and same radius are hit by a cue at a height h above the centre C. In which case,

h

h

C

C

Hollow sphere

Solid sphere

Fig. 12.110

(a) linear velocity will be more ? (b) angular velocity will be more ? (c) rotational kinetic energy will be more ?

Note Linear impulse in both cases is same.

Chapter 12

Rotational Mechanics — 141

12.13 Toppling You might have seen in your practical life that if a force F is applied to a block A of smaller width and greater height it is A more likely to topple down before sliding while if the same force F is applied to an another block B of broader base, chances of its sliding are more compared to its toppling. Have you ever thought why it happens so. To understand it in a better way let us take an example. Suppose a force F is applied at a height b above the base AE of the block. Further, suppose the friction f is sufficient to prevent sliding. In this case, if the normal reaction N also passes through C, then despite the fact that the block is in translational equilibrium (F = f and N = mg), an f unbalanced torque (due to the couple of forces F and f) is there. This torque has a tendency to topple the block about point E. To cancel the effect of this unbalanced torque the normal reaction N is shifted towards right a distance ‘a’ such that, net anticlockwise torque is equal to the net clockwise torque or Fb = ( mg ) a Fb or a= mg

F

F B

Fig. 12.111

N D

B

F b

C E W = mg Fig. 12.112

A

Now, if F or b (or both) is increased, distance a also increases. But it can not go beyond the right edge of the block. So, in extreme case (beyond which the block will topple down), the normal reaction passes through E as shown in Fig. 12.113 (b). N

N B

D C

f

A

mg (a)

B

F

a

D

F b

C

b f

E

A

mg (b)

E

Fig. 12.113

Now, if F or b is further increased, the block will topple down. This is why the block having the broader base has less chances of toppling in comparison to a block of smaller base. Because the block of larger base has more margin for the normal reaction to shift. On the similar ground we can see why the rolling is so easy. Because in this case the normal reaction has zero margin to shift. So even if the body is in translational equilibrium (F = f , N = mg) an unbalanced torque is left behind and the body starts toppling and here the toppling means motion. Under ideal conditions, the body will start moving by a very small force F tending to zero also.

N F

C f mg

Fig. 12.114

142 — Mechanics - II V

Example 12.32 A uniform cube of side a and mass m rests on a rough horizontal table. A horizontal force F is applied normal to one of the faces at a 3a point directly above the centre of the face, at a height above the base. What is 4 the minimum value of F for which the cube begins to tip about an edge? In the limiting case normal reaction will pass through O. The cube will tip about O if torque of F exceeds the torque of mg.  3a   a Hence, F   > mg    4  2 Solution

F>

or Therefore, minimum value of F is V

2 mg 3

N F C O f

3a 4

a mg 2

Fig. 12.115

2 mg. 3

Example 12.33 A uniform cylinder of height h and radius r is placed with its circular face on a rough inclined plane and the inclination of the plane to the horizontal is gradually increased. If µ is the coefficient of friction, then under what condition the cylinder will (a) slide before toppling (b) topple before sliding. Solution

(a) The cylinder will slide if N f

s mg

in θ

mg

cos

θ

θ

Fig. 12.116

mg sin θ > µmg cos θ tan θ > µ h The cylinder will topple if ( mg sin θ ) > ( mg cos θ )r 2 2r or tan θ > h or

Thus, the condition of sliding is tan θ > µ and condition of toppling is tan θ > cylinder will slide before toppling if µ


2r h

…(i)

…(ii) 2r . Hence, the h

Chapter 12

Rotational Mechanics — 143

Final Touch Points 1. Whether a particle is in translational motion, rotational motion or in both it merely depends on the reference point with respect to which the motion of the particle is described. v cos θ P

P

v

A

v

P

θ

v

90° r B (a)

v sin θ

r C

(b)

(c)

For example: a particle P of mass m is moving in a straight line as shown in figures (a), (b) and (c). Refer Fig. (a) With respect to point A, the particle is in pure translational motion. Hence, kinetic energy of the particle can be written as 1 KE = mv 2 2 Refer Fig. (b) With respect to point B, the particle is in pure rotational motion. Hence, the kinetic energy of the particle can be written as 2

KE =

1 2 1 1 v  Iω = (mr 2 )   = mv 2   2 2 2 r

Refer Fig. (c) With respect to point C, the particle can be assumed to be in rotational as well as translational motion. Hence, the kinetic energy of the particle can be written as 1 1 KE = m (v cos θ )2 + I ω 2 2 2 1 1  v sin θ  m (v cos θ )2 + (mr 2 )    r  2 2 1 = mv 2 2

2

=

Thus, in all the three cases, the kinetic energy of the particle comes out to be the same.

2. In cases where pulley is having some mass and friction is sufficient enough to prevent slipping, the tension on two sides of the pulley will be different and rotational motion of the pulley is also to be considered. 3. Finite angular displacements are not vector quantities, the reason being that they do not obey the law of vector addition. This law asserts that the order in which vectors are added does not affect their sum. A + B =B + A It can be seen applying two successive 90° rotations-one about the x-axis, and the other about the z-axis to a six-sided dice. In the first case, the z-rotation is applied before the x-rotation and vice versa in the second case. It can be seen that the dice ends up in two completely different states. Clearly, the z-rotation plus the x-rotation does not equal the x-rotation plus the z-rotation. This non-commutative algebra cannot be represented by vectors. We conclude that, rotations are not, in general, vector quantities. However infinitesimal angles do commute under addition, making it possible to treat them as vectors.

144 — Mechanics - II 4. Rotation plus translation motion of a rigid body is simplified by splitting this motion into two parts. (i) pure translation motion with the linear velocity and acceleration of the centre of mass. (ii) pure rotational motion about an axis passing through centre of mass and perpendicular to the plane of motion of the particles. But this motion may be considered as pure rotational motion about an axis called instantaneous axis of rotation (say IAOR). If this IAOR is non-inertial, then we cannot apply, τ ext = Iα about this axis. This is because in the derivation of this equation we use F = ma for each particle. If IAOR has an accelerationa, we have to apply a pseudo force − ma to each particle. These pseudo forces produce a pseudo torque about this axis. But this equation can be applied about an axis passing through centre of mass even if this is non-inertial. Let us prove this: Take the origin at the centre of mass. The total torque of the pseudo force is Σri × ( − mia i ) = − ( Σmi ri ) × a  Σmiri  = −M   ×a  M  Σmi ri is the position vector of the centre of mass and that is zero as the centre of mass is at the M origin. Hence, the torque of pseudo forces acting on all particles of the rigid body is zero.

But

Solved Examples TYPED PROBLEMS Type 1. Based on rotational equilibrium about a fixed axis.

Concept Given that a rigid body can only rotate about a fixed axis, (i.e. hinged at some point) still it is not rotating or it is at rest or it is in equilibrium.

How to Solve? l

l

Net torque about hinge point should be zero. But torque of hinge force about the same point is already zero. Net force on the rigid body is also zero.

V

Example 1 A uniform L shaped rod of mass 3m is hinged at point O. Length OB is two times the length OA. It is in equilibrium. Find A (a) relation between α and β (b) net hinge force. Solution (a) Length OB is two times the length OA. Therefore, mass of OB is 2m and that of OA is m and their weight will act at their centres (as the rod is uniform). If total length is 3l then, A OA l OA = l ⇒ OC1 = = 2 2 l and r1 = OC1 sin α = sin α 2 OB OB = 2l ⇒ OC 2 = =l 2

O α

α + β ≠ 90°

β

B

O C1

r1

α

β r2

C2 B

mg

2 mg

and r2 = OC 2 sin β = l sin β Net torque about O = 0 ⇒ anticlockwise torque of mg = clockwise torque of 2 mg ⇒ ⇒ or

(mg )r1 = (2mg )r2 l  mg  sin α  = (2mg ) l sin β 2  sin α = 4 is the required relation between α and β. sin β

(b) Net force on the rod is also zero. Therefore, hinge force is 3 mg (= mg + 2mg ) in upward direction.

146 — Mechanics - II Type 2. To find total kinetic energy of a system of rigid bodies in rotation plus translation.

Concept A particle has only translation kinetic energy

1 1 mv 2 . But a rigid body may have mv 2 and 2 2

1 2 Iω . Here, v is velocity of centre of mass, ω is angular speed of the rigid body and I is the 2 moment of inertia about an axis passing through centre of mass and perpendicular to the plane of motion of the body. V

Example 2 A ring of mass ‘m’ is rolling without slipping with linear speed v as shown in figure. Four particles each of mass ‘m’ are also attached at points A, B, C and D. Find total kinetic energy of the system.

Solution Earlier we have learned that in case of pure rolling,

C

B

v

D

A

v , vA = 0, vB = vD = 2v and vC = 2v R Now, total kinetic energy = [translational kinetic energy of four particles]+[translational kinetic energy of ring + rotational kinetic energy of ring] 1 1 1 1 1  1  (KE)total = m(0)2 + m ( 2 v)2 + m ( 2 v)2 + m (2v)2 + mv2 + I ω 2 ∴ 2 2 2 2 2  2  ω=

Substituting I = mR2

and

ω=

v R (KE)total = 5mv2

we get,

Ans.

Type 3. Energy conservation in pure rotational motion.

Concept A rigid body (suppose a rod) is hinged at O as shown in figure. It can have only rotational motion in a vertical plane about a smooth horizontal axis passing through O and perpendicular to plane of paper. O

O ⇒ ω

It is released from the horizontal position. As the rod rotates downwards its gravitational potential energy decreases and rotational kinetic energy increases.

Chapter 12

Rotational Mechanics — 147

How to Solve? l

l l l

1  Put decrease in gravitational potential energy (mgh) equal to increase in rotational kinetic energy  I ω 2  . 2  Here, h is the fall of height of centre of mass of the rigid body. In case of particle, ‘h’ is decrease in height of the particle. From this equation, we can find the value of ω. In pure rotational motion velocity of any point is v = rω Here, r is the distance of that point from the axis of rotation.

V

Example 3 A uniform circular disc has radius R and A mass m. A particle, also of mass m, is fixed at a point A on the edge of the disc as shown in the figure. The disc R can rotate freely about a horizontal chord PQ that is at C a distance R/4 from the centre C of the disc. The line AC is perpendicular to PQ. Initially the disc is held P vertical with the point A at its highest position. It is then allowed to fall, so that it starts rotation about PQ. Find the linear speed of the particle as it reaches its lowest position. Solution Initial and final positions are shown below. Decrease in potential energy of mass  5R  5 mgR = mg 2 × = 4  2  m

C

P

R 5R 4 R/4 Q 3R 4

P C

3R R/4 4 Q R/4 ω 5R R 4

m

Decrease in potential energy of disc  R  mgR = mg 2 ×  = 4 2  Therefore, total decrease in potential energy of system 5 mgR mgR = + = 3mgR 2 2 1 Gain in kinetic energy of system = I ω 2 2 where, I = moment of inertia of system (disc + mass) about axis PQ = moment of inertia of disc + moment of inertia of mass 2 2 mR2  5 R  R  = + m   + m    4  4   4 I=

15 mR2 8

R/4 Q

(JEE 1998)

148 — Mechanics - II From conservation of mechanical energy, Decrease in potential energy = Gain in kinetic energy 1  15 mR2 2 3 mgR =  ∴ ω 2 8  ω =

⇒

16 g 5R

Therefore, linear speed of particle at its lowest point  5 R v = rω =    4 ω= ⇒

5R 4

16 g 5R

v = 5 gR

Ans.

Type 4. Based on Angular Impulse.

Concept Angular impulse A.I = τ ∆t If torque is a function of time, then A.I = ∫ τ dt and this angular impulse is equal to the change in angular momentum. V

Example 4 A solid sphere of mass ‘m’ and radius ‘R’ is kept over a rough ground. A time varying force F =2t is acting at the topmost point as shown in figure. (a) Find angular momentum of the sphere about the bottommost point as a function of time ‘t’. (b) Does this result depend on the fact whether the ground is rough or smooth? Solution (a) Suppose ‘f ’ is the force of friction acting on the sphere in forward direction as shown in figure. Taking torque about bottommost point O. Torque of friction is already zero, as this force already passes through O. Torque of applied force is only there. ∴

τ = F × r⊥ = (2t ) (2R) = 4 RT

As, this toque is a function of time. t

t

0

0

F = 2t

Rough F = 2t

2R

f O

C

∴ Angular impulse = ∫ τ dt = ∫ (4Rt ) dt = 2 Rt 2 This angular impulse is equal to change in angular momentum. Hence, angular momentum at time ‘t’ is 2Rt 2. (b) This result is independent of the nature of surface (smooth or rough), as the torque of friction about bottommost point is already zero. It does not contribute in angular impulse and therefore in angular momentum.

Chapter 12

Rotational Mechanics — 149

Type 5. Role of friction in acceleration pure rolling.

Concept Friction has a tendency to stop the relative motion. In case of rolling (rotation + translation) over a stationary ground relative motion is stopped when v = Rω equation is satisfied at all instants. To satisfy this equation, a should be equal to ‘Rα’. Hence, friction has a tendency to satisfy the equation,a = Rα . Sometimes friction is sufficient and sometimes not. If friction is sufficient, then accelerated pure rolling will takes place. Otherwise forward or backward slipping occurs.

How to Solve? l l l l

l

l l

l l

l l l

l

Find N, µ sN and µ k N (or µN) Find requirement of friction to satisfy the equation a = Rα. For this, apply a general value of f in either forward or backward direction. Put a = Rα Fnet τ  or = R  net   I  m and find the required value of f for accelerated pure rolling. If f ≤ µ sN i.e. requirement ≤ availability then pure rolling will take place and a = Rα. In this case the obtained value of friction in step (iii) will act. If f > µ sN then either forward or backward slip will take place and a ≠ Rα. But, F τ a = net and α = net m I are still applicable. In this case, µ k N friction will act. If required value of friction in step (iii) comes out to be negative then just change the direction of friction which was initially assumed. Otherwise, magnitude wise all calculations are same. After calculation, if force of friction comes in forward direction, then if there is slip, it is backward slip and a < Rα.

V

Example 5 A solid sphere of mass 5 kg and radius 1 m is kept over a rough surface as shown in figure. A force F =30N is acting at the topmost point. (a) Check whether the pure rolling will take place or not. (b) Find direction and magnitude of friction actually acting on the sphere. (c) Find linear acceleration ‘a’ and angular acceleration ‘α’. Take g = 10 m/ s2 Solution (a) Availability of friction. N = mg = 5 × 10 = 50 N µ sN = 0.3 × 50 = 15 N µ kN = 0.2 × 50 = 10 N

F = 30 N

µs = 0.3 µk = 0.2

150 — Mechanics - II Requirement of friction for accelerated pure rolling (or to satisfy a = Rα). F a

C a

f

Let the friction f acts in forward direction. a = Rα Fnet τ  = R  net   I  m

⇒

  F+f  FR − fR  =R  2 m 2  mR   5 

or

Solving this equation, we get 3 F 7 3 90 = (30) = N 7 7 Since, this value of f is less than µ sN . Therefore, friction is sufficient for accelerated pure rolling to take place. Hence, pure rolling is taking place. 90 (b) The above value of f is positive, hence direction of friction is forward and N friction will 7 be acting. (c) The actual forces acting on the sphere are as under f=

F = 30 N a

α

f=

F a = net = m

30 + 5

90 F 7

90 7 = 8.57 m/s 2

Ans.

Since, a = Rα ∴ V

α=

a 8.57 = = 8.57 rad /s 2 R 1

Ans.

Example 6 Repeat all parts of above problem for F =40N . Solution (a) and (b): We have already calculated that, requirement of friction for pure rolling is f= For,

3 F 7

F = 40 N ⇒

f=

120 N 7

Chapter 12

Rotational Mechanics — 151

Now, this value of f is more than µ sN , so backward slip (a < Rα ) will take place because f is positive so kinetic friction or 10 N will act in forward direction. (c) The actual forces acting on the sphere are as under : F = 40 N

a a f = mK N = 10 N

Fnet 40 + 10 = = 10 m/s 2 m 5 a α≠ R τ FR − fR α = net = 2 I mR2 5 2.5 (F − f ) 2.5 (40 − 10) = = mR (5) (1) a=

But

= 15 rad /s 2.

Note

Ans.

Ans.

We can see that a < Rα.

Type 6. Based on conservation of mechanical energy and no change in rotational kinetic energy.

Concept If a body is kept in translational plus rotational motion condition over a smooth inclined plane then two forces acting smooth on the body (weight and normal reaction) pass through centre of mass. So, they cannot provide the torque and therefore angular acceleration. Therefore, its angular speed and θ rotational kinetic energy remains constant. Same is the case, when the body moves freely under gravity. Here, the only force acting on the body is ‘mg’ which also passes through centre of mass. When the body moves up, its translational kinetic energy decreases and gravitational potential energy increases but rotational kinetic energy remains constant. Opposite is the case when the body moves down. Further, in case of accelerated pure rolling over a stationary rough ground, work done by friction is zero. Hence, mechanical energy will remain constant. In this case, KR 1 = 1 for ring = for disc KT 2 2 = for solid sphere etc. 5 Thus, total mechanical energy ( K R + K T ± mgh ) remains constant in case of accelerated pure rolling, over a smooth ground or under gravity.

152 — Mechanics - II V

Example 7 A solid cylinder of mass m and radius r starts rolling down an inclined plane of inclination θ. Friction is enough to prevent slipping. Find the speed of its centre of mass when its centre of mass has fallen a height h. Solution

Considering the two shown positions of the cylinder. 1 As it does not slip hence total mechanical energy will be h conserved. 2 Energy at position 1 is E1 = mgh 1 1 2 Energy at position 2 is E 2 = mvCOM + I COM ω 2 2 2 vCOM mr 2 = ω and I COM = r 2 3 2 E 2 = mvCOM ⇒ 4 3 2 From conservation of energy, E1 = E 2 or mgh = mvCOM 4 4 ⇒ vCOM = gh 3 V

q

Example 8 A small solid cylinder of radius r is released coaxially from point A inside the fixed large cylindrical bowl of radius R as shown in figure. If the friction between the small and the large cylinder is sufficient enough to prevent any slipping, then find :

Ans. A R

(a) What fractions of the total energy are translational and rotational, when the small cylinder reaches the bottom of the larger one? (b) The normal force exerted by the small cylinder on the larger one when it is at the bottom. Solution

(a) K trans =

1 mv2 2

O 2

∴

1 1 1 1   v K rot = Iω 2 =  mr 2   = mv2   r 2 2 2 4 3 K = K trans + K rot = mv2 4 K trans 2 K rot 1 = ⇒ = K 3 K 3

R–r N v mg

(b) From conservation of energy, 3 mv2 4 mv2 4 = mg R−r 3

mg (R − r ) = ∴ Now, ∴

N − mg = N =

mv2 4 = centripetal force = mg R−r 3 7 mg 3

Ans.

Chapter 12 V

Rotational Mechanics — 153

Example 9 A small object of uniform density rolls up a curved surface with an 3v2 initial velocity v. It reaches up to a maximum height of with respect to the 4g initial position. The object is (JEE 2007)

v

(a) ring

(b) solid sphere

(c) hollow sphere

Solution

(d) disc

2

 3v 1 1  v mv2 + I   = mg   2 2  R  4g  2

I=

∴

1 mR2 2

∴ Body is disc. The correct option is (d). V

A

Example 10 A solid ball rolls down a parabolic path ABC from a height h as shown in figure. Portion AB of the path is rough while BC is smooth. How high will the ball climb in BC ? Solution

C

h

At B, total kinetic energy = mgh

B

Here, m = mass of ball The ratio of rotational to translational kinetic energy would be, KR 2 2 5 = ⇒ ∴ K R = mgh and K T = mgh KT 5 7 7 In portion BC, friction is absent. Therefore, rotational kinetic energy will remain constant and translational kinetic energy will convert into potential energy. Hence, if H be the height to which ball climbs in BC, then 5 5 mgH = K T or mgH = mgh or H = h 7 7 V

Example 11 A ball moves over a fixed track as shown in the figure. From A to B the ball rolls without slipping. If surface BC is frictionless and K A , K B and K C are kinetic energies of the ball at A, B and C respectively, then (JEE 2006)

A

C hC

hA B

(a) hA > hC ; K B > K C (b) hA > hC ; K C > K A (c) hA = hC ; K B = K C (d) hA < hC ; K B > K C Solution On smooth part BC, due to zero torque, angular velocity and hence the rotational kinetic energy remains constant. While moving from B to C translational kinetic energy converts into gravitational potential energy. ∴ The correct option is (a)

154 — Mechanics - II V

Example 12 A small solid sphere of mass ‘m’ is released from point A. Portion AB is sufficiently rough (to provide accelerated pure rolling), BC is smooth and after C, the ball moves freely under gravity. Find gravitational potential energy (U), rotational kinetic energy ( K R ) and translational kinetic energy ( K T ) at points A, B C, D and E. A

D C

60°

3h h B

E

Solution AB is sufficiently rough BC is smooth and after C motion is under gravity. So, total mechanical energy (U + K R + K T ) is always constant. On AB: KR 2 = (for solid sphere) KT 5 as accelerated pure rolling is taking place. After B, rotational kinetic energy will become constant. After C, centre of mass of the ball will follow a projectile motion. v vD = vC cos 60° = C = half of vC ∴ 2 1 (K T )D = (K T )C ∴ 4 At point A U = 3 mgh KR = 0 ⇒ KT = 0 ∴Total mechanical energy E = 3 mgh = constant. At point B ∴ But ∴

U = 0 ⇒ E = 3 mgh K = E = 3 mgh KR 2 = KT 5 2 6 K = mgh 7 7 5 15 KT = K = mgh 7 7

KR =

At point C U = mgh K R = (K R )B = ∴

6 mgh 7

K T = E −U − K R = 3 mgh − mgh −

6 8 mgh = mgh 7 7

(K = total K.E)

Chapter 12 At point D

Rotational Mechanics — 155

6 mgh 7 1 2 K T = (K T )C = mgh 4 7

K R = (K R )B =

∴

U = E − KR − KT 6 2 = 3mgh − mgh − mgh 7 7 13 = mgh 7

At point E

U =0 K R = (K R )B =

∴

6 mgh 7

K T = E − K R −U 6 15 = 3 mgh − mgh − 0 = mgh 7 7

Type 7. Based on instantaneous axis of rotation (IAOR).

Concept We have seen that in case of pure of rolling ( v = Rω ) over a stationary ground IAOR passes through the bottommost point of the rigid body. Following are three more cases where we can locate the position of IAOR. In the following cases IAOR is written as IC. (i) Given the velocity of a point (normally the centre of mass) on the body and the angular velocity of the body If v and ω are known, the IC is located along the line drawn perpendicular to v at P, v such that the distance from P to IC is, r = . Note that IC lie on that side of P which ω causes rotation about the IC, which is consistent with the direction of motion caused by ω and v. v

P r

ω IC

(ii) Given the lines of action of two non-parallel velocities Consider the body shown in figure where the line of action of the velocities v A and v B are known. Draw perpendiculars at A and B to these lines of action. The point of intersection of these perpendiculars as shown locates the IC at the instant considered.

vB vA ω

156 — Mechanics - II (iii) Given the magnitude and direction of two parallel velocities When the velocities of points A and B are parallel and have known magnitudes v A and v B then the location of the IC is determined by proportional triangles as shown in figure. A

d

IC

vA

IC

vA

A

d vB

B

B (a)

In both the cases,

vB

(b)

rA , IC =

vA ω

and rB, IC =

vB ω

In Fig. (a) rA , IC + rB, IC = d and in Fig. (b) rB, IC − rA , IC = d As a special case, if the body is translating, v A = v B and the IC would be located at infinity, in which case ω = 0. V

Example 13 A rotating disc moves in the positive direction of the x-axis. Find the equation y( x ) describing the position of the instantaneous axis of rotation if at the initial moment the centre c of the disc was located at the point O after which it moved with constant velocity v while the disc started rotating counterclockwise with a constant angular acceleration α. The initial angular velocity is equal to zero. y

O

Solution

t=

c v

x αx and ω = αt = v v

y IC

The position of IC will be at a distance v y= ω or

x

y=

or

v2 or αx

v αx v v2 xy = = constant α y=

y O

c v ω x

This is the desired x-y equation. This equation represents a rectangular hyperbola.

x

Chapter 12 V

Rotational Mechanics — 157

Example 14 A uniform thin rod of mass m and length l is standing on a smooth horizontal surface. A slight disturbance causes the lower end to slip on the smooth surface and the rod starts falling. Find the velocity of centre of mass of the rod at the instant when it makes an angle θ with horizontal. Solution

As the floor is smooth, mechanical energy of the rod will remain conserved. Further, no horizontal force acts on the rod, hence the centre of mass moves vertically downwards in a straight line. Thus velocities of COM and the lower end B are in the directions shown in figure. The location of IC at this instant can be found by drawing perpendiculars to vC and vB at respective points. Now, the rod may be assumed to be in pure rotational motion about IAOR passing through IC with angular speed ω. Applying conservation of mechanical energy. Decrease in gravitational potential energy of the rod = increase in rotational kinetic energy about IC 1 mgh = IIC ω 2 ∴ 2 l 1  ml2 or mg (1 − sin θ ) =  + 2 2  12

A

C l sin θ 2

C′ vC B

IC

h=

θ

l (1 – sin θ) 2 ω vB

l cos θ 2

 ml2 cos 2 θ ω 2 4 

Solving this equation, we get ω= Now,

12 g (1 − sin θ ) l(1 + 3 cos 2 θ )

l  |vC| =  cos θ ω 2  =

3 gl(1 − sin θ ) cos 2 θ (1 + 3 cos 2 θ )

Ans.

Type 8. Based on rotational pulleys.

Concept If the pulley is not massless and is sufficiently rough then tension on a string passing over it on its both sides will be different. In this case, the string does not slip over the pulley but the pulley also rotates.

a a

T

T

158 — Mechanics - II In the above figure, pulley is massless, smooth and stationary. String slips over the pulley. α a

R a (If T1 < T2 )

T1

T2

In this figure, pulley is neither massless nor smooth. It rotates with the string. If there is no slip then, a = Rα. V

Example 15 In the arrangement shown in figure the mass of the uniform solid cylindrical pulley of radius R is equal to m and the masses of two bodies are equal to m1 and m2 . The thread slipping and the friction in the axle of the pulley are supposed to be absent. Find the angular acceleration of the cylinder and the T ratio of tensions 1 of the vertical sections of the thread in the process of motion. T2

1 2

m1 m2

Solution

Let α = angular acceleration of the pulley and a = linear acceleration of two bodies α

T1 m1

T1

T2

Equations of motion are For mass m1 , For mass m2, For pulley,

For no slipping condition Solving these equations, we get and

T2 a

m 1g

m2

m 2g

T1 − m1 g = m1a m2g − T2 = m2a (T − T1 )R α= 2 1 mR2 2 a = Rα 2(m2 − m1 ) g α= (2m1 + 2m2 + m)R T1 m1 (m + 4m2) = T2 m2(m + 4m1 )

a

…(i) …(ii) …(iii)

…(iv) Ans. Ans.

Chapter 12

Rotational Mechanics — 159

Type 9. When friction converts forward or backward slip into pure rolling.

Concept If a body is kept in forward or backward slip condition ( v ≠ Rω ) over a rough horizontal ground, then kinetic friction will act in the opposite direction of slip. This friction provides both linear acceleration ‘a’ and angular acceleration ‘α’. They are in the directions so as the equation v = Rω (with proper sense of rotation) is satisfied so that pure rolling may start. For example, if initially v > rω, then ‘a’ is in the opposite direction of ‘v’ (to decrease it) and ‘α’ is in the direction of ‘ω’ (to increase it). Once pure rolling starts, relative motion (or slipping) is stopped and friction becomes zero. During slip, mechanical energy is not conserved. But, some part of mechanical energy is used up in doing work against friction. Once pure rolling starts, mechanical energy becomes constant. If only one coefficient of friction or µ is given in the question then, instead of kinetic friction, apply µN during the slip. V

Example 16 A solid sphere of radius r is gently placed on a rough horizontal ground with an initial angular speed ω 0 and no linear velocity. If the coefficient of friction is µ, find the time t when the slipping stops. In addition, state the linear velocity v and angular velocity ω at the end of slipping.

ω0

Solution

Let m be the mass of the sphere. Since, it is a case of backward slipping, force of friction is in forward direction. Limiting friction will act in this case. v ω0

ω

fmax

f µmg = = µg m m τ f ⋅r 5 µg Angular retardation α = = = 2 I mr 2 2 r 5

Linear acceleration a =

Slipping is ceased when v = rω or or ∴

(at ) = r (ω 0 − αt ) 5 µgt   µgt = r ω 0 −   2 r  t=

2 rω 0 7 µg

v = at = µgt = and

ω=

v 2 = ω0 r 7

or

7 µgt = rω 0 2 Ans.

2 rω 0 7

Ans. Ans.

160 — Mechanics - II Alternate Solution Net torque on the sphere about the bottommost point is zero as friction is passing through that point. Therefore, angular momentum of the sphere will remain conserved about the bottommost point. Li = L f Iω 0 = Iω + mrv 2 2 mr 2ω 0 = mr 2ω + mr (ωr ) 5 5 2 2 ω = ω 0 and v = rω = rω 0 7 7

∴ or ∴ V

Example 17 A billiard ball, initially at rest, is given a sharp impulse by a cue. The cue is held horizontally a distance h above the centre line as shown in figure. The ball leaves the cue with a speed v0 and because of its forward english (backward slipping) eventually acquires a final 4 9 speed v0 . Show that h = R 5 7

Ans. F h

where R is the radius of the ball. Let ω 0 be the angular speed of the ball just after it leaves the cue. The maximum friction acts in forward direction till the slipping continues. Let v be the linear speed and ω the angular speed when slipping is ceased.

Solution

F h R

w0

v0

w

v

v = 9 v0 7

fmax

∴ Given, ∴

v = Rω or ω =

v R

9 v0 7 9 v0 ω= 7 R v=

Applying, Linear impulse = change in linear momentum ∴ F dt = mv0 Angular impulse = change in angular momentum 2 ∴ τ dt = Iω 0 or (Fh )dt = mR2 ω 0 5 During the slip, angular momentum about bottommost point will remain conserved. i.e. Li = L f or Iω 0 + mRv0 = Iω + mRv 2 2  9 v0  9 ∴ mR2 ω 0 + mRv0 = mR2   + mRv0  7 R 7 5 5 4 Solving Eqs. (iii), (iv) and (v), we get h= R 5

…(i) …(ii)

…(iii) …(iv)

…(v) Proved.

Chapter 12

Rotational Mechanics — 161

Type 10. Based on critical value of µ.

Concept If a force is applied on a block as shown in figure and the force is increased then in some cases the block slides before toppling and in other cases it topples before sliding. It mainly depends on: F

(i) base length a (ii) height of point of application of force h h (iii) coefficient of friction µ. For example, if µ is small then chances of sliding are more. If a a is small then chances of toppling are more. Rough For given values of a and h it only depends on the value of µ. In such problems, there is a critical value µ cr . If given value of µ > µ cr , then the block topples before sliding and if µ < µ cr then the block slides before toppling.

How to Solve? l l l l

Make two conditions: Condition of sliding Condition of toppling, when the normal reaction (just before toppling) shifts to the right side edge). From these, two conditions we can find µ cr .

V

Example 18 For the given dimensions shown in figure, find critical value of coefficient of friction µ. F 2a f 2a

Solution Condition of sliding The block will slide if, F > µN N = mg F > µ mg

but ∴ Condition of toppling

(m = mass of the block) …(i) N

F

2a a f mg

O

162 — Mechanics - II Block will topple about an axis passing through O and perpendicular to plane of paper if: clockwise torque of F > anticlockwise torque of mg ∴

F (2a ) > (mg )(a ) 1 F > mg 2

or

…(ii)

From Eqs. (i) and (ii), we can see that, µ cr =

1 2

Ans.

1 If given value of µ is less than µ cr (say it is ) then Eq. (i) is, 4 1 F > mg 4 So, Eq. (i) is satisfied before Eq. (ii). Therefore, the block will slide before toppling. 3 If given value of µ is greater than µ cr(say it is ) then Eq. (i) is, 4 3 F > mg 4 So, Eq. (ii) is satisfied before Eq. (i). Hence, the block will topple before sliding.

Type 11. When pure translational motion of a rigid body converts into pure rotational motion by a jerk (or linear impulse).

Concept A block in pure translational motion (with linear velocity v) meets an obstacle at O. A linear impulse will act on the block at point O. Just after the impact, the block starts rotating about point O (with an angular speed say ω). The value of this ‘ω’ can be found by conservation of angular momentum about O because during the impact, the angular impulse of the linear impulse about point O will be zero (as r⊥ = 0). Just before impact motion is pure translational. So, angular momentum is m vr sinθ or mvr⊥ . Here, v is the velocity of centre of mass. Just after impact motion is pure rotational. So, angular momentum is Iω, where, I is the moment of inertia passing through O and perpendicular to plane of paper. ⇒

v O

ω O

Furthermore, there will be loss of mechanical energy during impact 1 1 2  2  = Ei − E f = mv − Iω  . But after impact, the mechanical energy remains constant. As   2 2 the block moves up, its potential energy increases and rotational kinetic energy decreases. V

Example 19 Find

In the figure shown in the text, if the block is a cube of side ‘a’.

(a) ω just after impact (b) loss of mechanical energy during impact (c) minimum value of v so as the block overcomes the obstacle and does not turn back.

Chapter 12 Solution

C

v

C r⊥ r⊥ =

Rotational Mechanics — 163

ω

⇒ a 2

O CO= r = a √2

O

(a) From conservation of angular momentum about O, Li = L f mvcr⊥ = I oω = (I c + mr 2) ω 2 2  a   a  ma mv   =  + m   ω  2  6  2    3  v ω=   4  a

⇒ ⇒

Ans.

(b) Loss of mechanical energy, = Ei − E f 1 1 = mv2 − I oω 2 2 2 2 2 1 1 ma 2  a    3 v = mv2 −  +m     2  4 a 2 2  6  5 = mv2 16

Ans.

(c) C

Ca √2

⇒

ω O

O

Block overcomes the obstacle at O if centre of mass rises upto a height figure (from the initial height

a as shown in 2

a ). 2

Because after that torque of ‘mg’ about O will itself rotate the block on other side as shown in figure. C

O

∴ ∴

mg

Decrease in rotational kinetic energy = increase in gravitational potential energy 1  a a I oω 2 = mg  −   2 2 2

or

2 2 1 ma 2  a   3 v  + m = mg        2   4 a  2  6 

∴

v = 1.1 g a

 a a −    2 2 Ans.

164 — Mechanics - II Type 12. To identify number of unknowns and then make equations corresponding to that.

Concept T c2 A disc and block system are released from rest as shown in figure. Ground is sufficiently rough, so that a2 there is no slip anywhere. We have to find accelerations of both, tension in the string and force α of friction. c1 f T In such problems, first of all find the number of unknowns: a1 (i) Block can have only translational motion. So, it has some linear acceleration say a1 (ii) Disc can have translational as well as rotational motion. So, it has linear acceleration a 2 and angular acceleration ‘α’. (iii) Ground is rough. So, there is one unknown friction ‘f ’ (iv) One more unknown is tension in the string T. Therefore, there are total five unknowns in this problem, a1 , a 2 , α, T and f .

How to Solve? l l

l l

V

We will make three acceleration equations for a1,a2 and α by using the equations F τ a = net and α = net m I There are two contact equations at c1 and c 2 . At c1, disc is in contact with ground and at c 2 with string. At the other end string is connected to the block. So, there should not be any slip at c1 between disc and ground and at c 2 between disc and string. Following two examples illustrate the method of making these equations.

Example 20 Consider the arrangement shown in figure. The string is wrapped around a uniform cylinder which rolls without slipping. The other end of the string is passed over a massless, frictionless pulley to a falling weight. Determine the acceleration of the falling mass m in terms of only the mass of the cylinder M, the mass m and g. Solution Let T be the tension in the string and f the force of (static) friction, between the cylinder and the surface. α a1 = acceleration of centre of mass of cylinder towards right a 2 = downward acceleration of block m α = angular acceleration of cylinder (clockwise) Acceleration equations For block,

mg − T = ma 2

M

m

T T

a2

a1 f mg

…(i)

Chapter 12

Rotational Mechanics — 165

T + f = Ma1 (T − f )R α= 1 MR2 2

For cylinder,

…(ii) …(iii)

Contact equations The string is attached to the mass m at the highest point of the cylinder, hence vm = vCOM + Rω Differentiating, we get We also have (for rolling without slipping) Solving these equations, we get

a 2 = a1 + Rα

…(iv)

a1 = Rα 8mg a2 = 3M + 8m

…(v) Ans.

Alternate Solution (Energy Method) Since, there is no slipping at all contacts mechanical energy of the system will remain conserved. ∴ Decrease in gravitational potential energy of block m in time t = increase in translational kinetic energy of block + increase in rotational as well as translational kinetic energy of cylinder. 1 1 1 mgh = mv22 + Iω 2 + Mv12 ∴ 2 2 2 1 1 1 1 1    or …(vi) mg  a 2t 2 = m (a 2t )2 +  MR2 (αt )2 + M (a1t)2 2  2  2 2 2 Solving Eqs. (iv), (v) and (vi), we get the same result. V

Example 21 A thin massless thread is wound on a reel of mass 3 kg and moment of inertia 0.6 kg-m2 . The hub radius is R = 10 cm and peripheral radius is 2R = 20 cm. The reel is placed on a rough table and the friction is enough to prevent slipping. Find the acceleration of the centre of reel and of hanging mass of 1 kg.

2R R

A

Solution

Here, number of unknowns are five: a1 = acceleration of centre of mass of reel a 2 = acceleration of 1 kg block α = angular acceleration of reel (clockwise) T = tension in the string f and f = force of friction Acceleration equations : Free body diagram of reel is as shown in figure: (only horizontal forces are shown). Equations of motion are T − f = 3a1 α=

τ f (2R) − T . R 0.2 f − 0.1T f T = = = − I I 0.6 3 6

α a1 T

…(i) …(ii)

166 — Mechanics - II Free body diagram of mass is, Equation of motion is,

T

10 − T = a 2

…(iii)

Contact equations : For no slipping condition,

a2

a1 = 2Rα

or

a1 = 0.2 α

…(iv)

and a 2 = a1 − Rα or a 2 = a1 − 0.1 α Solving the above five equations, we get

10 N

…(v)

a1 = 0.27 m/s 2 and a 2 = 0.135 m/s 2

Ans.

Type 13. Energy method of solving problems of accelerated pure rolling.

Concept In accelerated pure rolling over a stationary ground work done by friction is zero. So, mechanical energy remains constant. Therefore, some problems of accelerated pure rolling can also be solved by using energy conservation principle. V

Example 22 A body of mass m, radius R and moment of inertia I (about an axis passing through the centre of mass and perpendicular to plane of motion) is released from rest over a sufficiently rough ground (to provide accelerated pure rolling). Find linear acceleration of the body. Solution Let linear acceleration is ‘a’ and angular acceleration ‘α’.

a

c α

m, R , I

θ

For accelerated pure rolling, α=

a R

After time t, displacement of centre of mass along the plane, s = ∴

1 2 at 2

Height fallen by centre of mass h = (s)(sin θ ) =

1 2 at sin θ 2

linear velocity v = at at R From energy conservation principle, decrease in potential energy = increase in translational and rotational kinetic energy. 1 1 or mgh = mv2 + Iω 2 2 2 angular velocity ω = α t =

Substituting the value we have, 1  at  1  1 mg  at 2 sin θ = m (at )2 + I   2  2 2  R g sin θ Solving this equation we get, a= I 1+ mR2

2

Ans.

Chapter 12

Rotational Mechanics — 167

Type 14. Problems of accelerated pure rolling by finding angular acceleration ‘α’ about the bottommost axis.

Concept We have discussed in final touch points that : τ α = ext I can be applied from an inertial frame or about an axis passing through centre of mass (even if it is accelerated). Same is the case about an axis passing through bottommost axis if accelerated pure rolling is taking place on a stationary ground. Although the bottommost point is accelerated (= Rω 2 , towards centre), yet for symmetrical bodies net torque of pseudo forces on all particles of the rigid body about bottommost axis is zero. So, τ α = ext I can be applied about this axis also. V

Example 23 In example 22, find linear acceleration ‘a’ of the body by calculating α about bottommost axis. α=

Solution But I c = I

α=

∴ Now, a c or a = Rα =

τ 0 (mg sin θ )R = I 0 (I c + MR2)

mg R sin θ I + mR2

mg

sin

mg R2 sin θ I + mR2

or

a=

f

C R

θ

f = friction O α θ

g sin θ I 1+ mR2

Ans.

Type 15. Based on hinge force.

Concept A rod OA is hinged at O. It is released from the horizontal position as shown in figure. We have to find hinge force acting on the rod at a general angle θ. We can see that motion of the rod is pure rotational about an axis passing through O. As the rod rotates downwards, its gravitational potential energy decreases and rotational kinetic energy increases. If the hinge is smooth, then we can apply the equation.

O α

ω

C

A

θ C A

168 — Mechanics - II 1 …(i) I ω2 2 Here, h = height fallen by centre of mass C of the rod. and I = I0 2 From here we can find ω . At the same time, we can see that ω is increasing. So, there is an angular acceleration ‘α’ about O. Only two forces are acting on the rod, hinge force and weight. Torque of hinge force about O is zero. Therefore, torque of ‘mg’ will only provide ‘α’. Thus, τ mg …(ii) α= I Now, only two forces are acting on the rod, hinge force (say F) and weight ( mg). ∴ F + mg = ma COM or ma C or …(iii) F = ma c − mg By finding a c we can find the hinge force F from Eq. (iii). mgh =

How to Solve a c ? l

l l l

.

l In the figure, we can see that C is rotating in a circle with centre at O and radius r = OC = , where l is the 2 length of rod. In a circular motion, acceleration of a particle has two components. (i) radial ar = rω 2 (ii) tangential at = rα ω 2 can be obtained from Eq. (i), α can be obtained from Eq. (ii). Writing all vector quantities in proper vector notations and then substituting in Eq.(iii) we can find the hinge force F.

V

Example 24 In the figure given in the text if mass of the rod is ‘m’ then find hinge force. (a) Just after the rod is released from the horizontal position. (b) When the rod becomes vertical. Solution (a) r y

⊥

C

O α

A at

x

mg

Just after the release, ω = 0 ∴

a r = rω 2 = 0 τ mg α= I

(about O)

 l (mg )    2 3 g (mg ) (r⊥ ) = = = 2 I0 (ml / 3) 2 l ∴

 l a t = rα =    2

 3 g 3  = g 2 l  4

Rotational Mechanics — 169

Chapter 12 From Eq. (iii), the hinge force is

F = m (a c − g ) = m [a (− $j) − (− g$j)] t

 3  = m − g$j + g$j  4  mg $ = j 4 Therefore, hinge force is (b)

Ans.

mg , in vertically upward direction. 4 C

O ω

y h= l 2

ar

x

C mg

When the rod becomes vertical height fallen by centre of mass is h =

l 2

Therefore, from Eq. (i), ω2 =

= ∴

 ml2  I = I0 =  3  

2mgh I  l 2mg    2 (ml2 /3)

 3 g =   l 

 l   3 g 3 a r = rω 2 =     = g  2  l  2

(towards O)

At this moment 'mg' also passes through O. Therefore, its torque about O is also zero. So, from Eq. (ii), α = 0 ⇒ a t = rα = 0 Now, substituting proper values in Eq. (iii), the hinge force is, F = m (a c − g ) = m [a $j − (− g$j)] r

 3   = m  g $j + ( g )$j  2   5  =  mg $j 2  Therefore, hinge force is

5 mg in vertically upward direction. 2

Ans.

170 — Mechanics - II Type 16. Collision Problems.

Concept There are mainly three types of collisions.

Type 1.

O l

O ⇒

u m

v M+m

M

A ball of mass M is suspended by a light string of length ‘l’. A bullet of mass ‘m’ strikes the ball with velocity u and sticks. This type of collision we have already discussed in the previous chapter. The three important points in this collision are (i) Velocity of combined mass after collision (say v) can be obtained by conservation of linear momentum or p f = pi ⇒ ( M + m) v = mu mu or v= M+m (ii) Mechanical energy is lost only during collision. This loss is given by (E = mechanical energy) Ei − E f 1 1 2 2 = mu − ( M + m) v 2 2 (iii) After collision now the mechanical energy remains constant and the combined mass executes vertical circular motion. If v ≥ 5gl circle is completed. If 2 gl < v < 5 gl , string slacks in upper half of the circle and if 0 < v ≤ 2 gl, combined mass oscillates in lower half of the circle.

Type 2.

O

M L

m

u

O

⇒

ω

M

m

A rod of mass M and length L is hinged at point O. A bullet of mass m moving with velocity u strikes the rod at its bottommost point and sticks. Just after collision the combined system (rod + bullet) starts rotating about the hinge point O with an angular speed ω.

Rotational Mechanics — 171

Chapter 12

So, the translational motion converts into rotational motion. The three important points in this collision are (i) At the time of collision, a linear impulse acts on the system at point O from the hinge. Angular impulse of this linear impulse about O is zero (as r⊥ = 0). Therefore, angular momentum of the system about O remains constant or Li = L f ∴ mu r⊥ = I ω or mu L = ( I rod + I Bullet ) ω  ML2  = + mL2  ω  3  From this equation, we can find ‘ω’. (ii) Mechanical energy is lost only during collision (not after that) and this loss is 1 1 Ei − E f = mu 2 − Iω 2 2 2 (iii) After the collision, as the system moves upwards its gravitational potential energy increases and rotational kinetic energy decreases. So, we can write : decrease in rotational kinetic energy = increase in gravitational potential energy.

Type 3.

m

u

A

A ⇒ C1

B

M, L

C2

ω

v

C1

B

A rod of mass M and length L is lying on a smooth table. A bullet of mass ‘m’ and speed ‘u’ strikes the rod at A and sticks. C1 is the centre of rod and C2 is the centre of mass of system (rod + bullet). Just after collision motion of the combined system is rotation (with angular velocity ω) and translation (with linear velocity v). Thus, by the collision, translation motion converts into rotation plus translation motion. The three important points in this type of collision are: (i) System is kept over a smooth horizontal table. So, net linear impulse on the system is zero. Therefore, v can be obtained by conservation of linear momentum or, p f = pi ⇒ ( M + m) v = mu mu v= ∴ M+m Since, net linear impulse on the system is zero. Therefore, angular impulse about any point is also zero. Hence, angular momentum of the system can be conserved about any point and ω can be obtained. But normally, we conserve it about point of impact (or A).

172 — Mechanics - II ∴ or Here,

(about A) L f = Li ( m + M ) vr⊥ ± I ω = 0 ⇒ Li = mu r⊥ (as r⊥ = 0from A) =0 I = I rod + I Bullet To find r⊥ of v from A first we will have to find position of COM or C2 of the combined system. (ii) Mechanical energy is lost only during collision (not after that) and this loss is: 1  1  1 Ei − E f =  mu 2  −  ( M + m) v 2 + I ω 2  2  2 2  (iii) Since, the system is kept over a smooth table, v and ω remain constant after the collision. V

Example 25 Two uniform rods A and B of length 0.6 m each and of masses 0.01 kg and 0.02 kg respectively are rigidly joined end to end. The combination is pivoted at the lighter end, P as shown in figure. Such that it can freely rotate about point P in a vertical plane. A small object of mass 0.05 kg, moving horizontally, hits the lower end of the combination and sticks to it. What should be the velocity of the object, so that the system could just be raised to the horizontal position? (JEE 1994)

P

A

B

m v

Solution System is free to rotate but not free to translate. During collision, net torque P on the system (rod A + rod B + mass m) about point P is zero. Therefore, angular momentum of system before collision = angular momentum of system just after collision (about P). Let ω be the angular velocity of system just after collision, then …(i) Li = L f ⇒ mv(2l) = Iω Here, I = moment of inertia of system about P 2  l2  l   = m (2l)2 + mA (l2/ 3) + mB  +  + l  2  12  Given, l = 0.6 m, m = 0.05 kg, mA = 0.01 kg and mB = 0.02 kg. ω

ω=0

l

l v

ω

A

B

Chapter 12

Rotational Mechanics — 173

Substituting the values, we get I = 0.09 kg- m2 Therefore, from Eq. (i) ω=

2mvl (2) (0.05) (v)(0.6) = I 0.09

ω = 0.67v Now, after collision, mechanical energy will be conserved. Therefore, decrease in rotational KE = increase in gravitational PE 1 l  l or Iω 2 = mg (2l) + mA g   + mB g (l + )  2 2 2 gl (4m + mA + 3 mB ) or ω2 = I (9.8) (0.6) (4 × 0.05 + 0.01 + 3 × 0.02) = 0.09

…(ii)

= 17.64 (rad /s)2 ∴ ω = 4.2 rad /s Equating Eqs. (ii) and (iii), we get 4.2 v= m/s or v = 6.3 m/s 0.67 V

…(iii)

Ans.

Example 26 A rod AB of mass M and length L is lying on a horizontal frictionless surface. A particle of mass m travelling along the surface hits the end A of the rod with a velocity v0 in a direction perpendicular to AB. The collision is elastic. After the collision the particle comes to rest. (JEE 2000) (a) Find the ratio m/ M . (b) A point P on the rod is at rest immediately after collision. Find the distance AP. (c) Find the linear speed of the point P a time πL/ 3v0 after the collision. Solution (a) Suppose velocity of COM of the rod just after collision is v and angular velocity

about COM is ω . Applying following three laws: m

v0

m L 2 COM L 2

Before collision

COM

ω

v

x

After collision

(1) External force on the system (rod + mass) in horizontal plane along x-axis is zero . ∴ Applying conservation of linear momentum in x-direction. mv0 = Mv (2) Net torque on the system about COM of rod is zero.

…(i)

174 — Mechanics - II  L ∴ Applying conservation of angular momentum about COM of rod, we get mv0   = I ω  2 L ML2 = ω 2 12 MLω or mv0 = 6 (3) Since, the collision is elastic, kinetic energy is also conserved. 1 1 1 ∴ mv02 = Mv2 + Iω 2 2 2 2 2 ML or mv02 = Mv2 + ω2 12 From Eqs. (i), (ii) and (iii), we get the following results m 1 = M 4 mv0 6 mv0 and ω = v= M ML (b) Point P will be at rest if x ω = v v mv0/M or x = L /6 or x= = ω 6mv0/ML or

mv0

...(ii)

…(iii)

Ans.

A

v

⇒

ω

x

v ω

P

A

P

∴

AP =

L L + 2 6

t=

(c) After time angle rotated by rod, θ = ωt =

or

AP =

πL 3 v0

2 L 3

Ans.

6mv0 πL . ML 3v0  m  1 = 2π   = 2π    M  4

∴

θ=

π 2

Therefore, situation is as shown in figure. ∴ Resultant velocity of point P will be  m | vP | = 2v = 2   v0  M

or

2 v = v0 = 0 4 2 2 v0 | vP| = 2 2

xω = v

√2 v

P

v

Ans.

Miscellaneous Examples V

Example 27 A thread is wound around two discs on either sides. The pulley and the two discs have the same mass and radius. There is no slipping at the pulley and no friction at the hinge. Find out the accelerations of the two discs and the angular acceleration of the pulley. Solution Let R be the radius of the discs and T1 and T2 be the tensions in the left

1

2

and right segments of the rope. T1

T2 α2

α1

a1

mg

mg

a2

Acceleration of disc 1, a1 =

α

mg − T1 m

…(i)

Acceleration of disc 2, mg − T2 m τ TR 2T α1 = = 1 = 1 I 1 mR2 mR 2 2T2 α2 = mR a2 =

Angular acceleration of disc 1,

Similarly, angular acceleration of disc 2,

…(ii)

T1

T2

…(iii)

…(iv)

Both α 1 and α 2 are clockwise. Angular acceleration of pulley, α=

For no slipping, Solving these equations, we get α =0

(T2 − T1 )R 2(T2 − T1 ) = 1 mR mR2 2

Rα 1 – a1 = a 2 − Rα 2 = Rα and

a1 = a 2 =

…(v)

…(vi)

2g 3

Alternate Solution As both the discs are in identical situation, T1 = T2 and α = 0. i.e. each of the discs falls independently and identically. Therefore, this is exactly similar to the problem shown in figure.

Ans.

176 — Mechanics - II V

Example 28 Determine the maximum horizontal force F that may be applied to the plank of mass m for which the solid sphere does not slip as it begins to roll on the plank. The sphere has a mass M and radius R. The coefficient of static and kinetic friction between the sphere and the plank are µ s and µ k respectively. M R F

m Smooth

Solution

The free body diagrams of the sphere and the plank are as shown below: α

µs Mg a1

a2 F

µs Mg

Writing equations of motion For sphere Linear acceleration Angular acceleration

µ sMg = µsg M (µ Mg )R α= s 2 MR2 5 5 µsg = 2 R

a1 =

For plank Linear acceleration a2 =

F − µ sMg m

…(i)

…(ii)

…(iii)

For no slipping a 2 = a1 + Rα

…(iv)

Solving the above four equations, we get 7   F = µ s g  M + m  2  Thus, maximum value of F can be 7   µ s g  M + m  2  V

Ans.

Example 29 A uniform disc of radius r0 lies on a smooth horizontal plane. A similar disc spinning with the angular velocity ω 0 is carefully lowered onto the first disc. How soon do both discs spin with the same angular-velocity if the friction coefficient between them is equal to µ? Solution

From the law of conservation of angular momentum. Iω 0 = 2Iω

Chapter 12

Rotational Mechanics — 177

I = moment of inertia of each disc relative to common rotation axis ω ∴ ω = 0 = steady state angular velocity 2 The angular velocity of each disc varies due to the torque τ of the friction forces. To calculate τ, let us take an elementary ring with radii r and r + dr. The torque of the friction forces acting on the given ring is equal to Here,

dr r

Friction force

ω

mass = dm

dτ = (friction force) (r⊥ ) = [ µ (dm) g ](r )  mg  = µ  2 (2πrdr )r  πr0   2µmg  =  2  r 2 dr  r0  where, m is the mass of each disc. Integrating this with respect to r between 0 and r0, we get 2 τ = µmgr0 = constant 3 τ (2 µmgr0 / 3) ∴ α= = I (mr02 / 2) =

4 µg = constant 3r0

Now, angular speed of lower disc increases with this α from O to ∴ or

ω0 = αt 2 ω 3r ω t= 0 = 0 0 2α 8 µg

ω0 and α is constant. 2

Ans.

Exercises LEVEL 1 Assertion and Reason Directions : Choose the correct option. (a) If both Assertion and Reason are true and the Reason is correct explanation of the Assertion. (b) If both Assertion and Reason are true but Reason is not the correct explanation of Assertion.

(c) If Assertion is true, but the Reason is false. (d) If Assertion is false but the Reason is true.

1. Assertion : Moment of inertia of a rigid body about any axis passing through its centre of mass is minimum. Reason : From theorem of parallel axis, I = I cm + Mr 2

2. Assertion : A ball is released on a rough ground in the condition shown in figure. It will start pure rolling after some time towards left side. Reason : Friction will convert the pure rotational motion of the ball into pure rolling.

3. Assertion : A solid sphere and a hollow sphere are rolling on ground with same total kinetic energies. If translational kinetic energy of solid sphere is K , then translational kinetic energy of hollow sphere should be greater than K . Reason : In case of hollow sphere rotational kinetic energy is less than its translational kinetic energy.

4. Assertion : A small ball is released from rest from point A as shown. If bowl is smooth, than ball will exert more pressure at point B, compared to the situation if bowl is rough. Reason : Linear velocity and hence, centripetal force in smooth situation is more.

5. Assertion : A cubical block is moving on a rough ground with velocity v0.

During motion net normal reaction on the block from ground will not pass through centre of cube. It will shift towards right. Reason : It is to keep the block in rotational equilibrium.

6. Assertion : A ring is rolling without slipping on a rough ground. It strikes elastically with a smooth wall as shown in figure. Ring will stop after some time while travelling in opposite direction. Reason : After impact net angular momentum about an axis passing through bottommost point and perpendicular to plane of paper is zero.

A

B

v0

Rotational Mechanics — 179

Chapter 12

7. Assertion : There is a thin rod AB and a dotted line CD. All the axes we are talking about are perpendicular to plane of paper. As we take different axes moving from A to D, moment of inertia of the rod may first decrease then increase. D

A

B

C

Reason : Theorem of perpendicular axis cannot be applied here.

8. Assertion : If linear momentum of a particle is constant, then its angular momentum about any point will also remain constant. Reason : Linear momentum remains constant, if Fnet = 0 and angular momentum remains constant if τ net = 0.

9. Assertion : In the figure shown, A, B and C are three points on the circumference of a disc. Let vA , vB and vC are speeds of these three points, then

vC > v B > v A A

B

ω

v

C

Reason : In case of rotational plus translational motion of a rigid body, net speed of any point (other than centre of mass) is greater than, less than or equal to the speed of centre of mass.

10. Assertion : There is a triangular plate as shown. A dotted axis is lying in the plane of slab. As the axis is moved downwards, moment of inertia of slab will first decrease then increase.

Reason : Axis is first moving towards its centre of mass and then it is receding from it.

11. Assertion : A horizontal force F is applied at the centre of solid sphere placed over a plank. The minimum coefficient of friction between plank and sphere required for pure rolling is µ 1 when plank is kept at rest and µ 2 when plank can move, then µ 2 < µ 1. F

Reason : Work done by frictional force on the sphere in both cases is zero.

180 — Mechanics - II

Objective Questions Single Correct Option 1. The moment of inertia of a body does not depend on (a) (b) (c) (d)

mass of the body the distribution of the mass in the body the axis of rotation of the body None of the above

2. The radius of gyration of a disc of radius 25 cm about a centroidal axis perpendicular to disc is (a) 18 cm

(b) 12.5 cm

(c) 36 cm

(d) 50 cm

3. A shaft initially rotating at 1725 rpm is brought to rest uniformly in 20s. The number of revolutions that the shaft will make during this time is (a) 1680

(b) 575

(c) 287

(d) 627

4. A man standing on a platform holds weights in his outstretched arms. The system is rotated about a central vertical axis. If the man now pulls the weights inwards close to his body, then (a) (b) (c) (d)

the angular velocity of the system will increase the angular momentum of the system will remain constant the kinetic energy of the system will increase All of the above

5. The moment of inertia of a uniform semicircular disc of mass M and radius r about a line perpendicular to the plane of the disc through the centre is (a) Mr 2

(b)

1 Mr 2 2

(c)

1 Mr 2 4

(d)

2 Mr 2 5

6. Two bodies A and B made of same material have the moment of inertial in the ratio I A : I B = 16 : 18 . The ratio of the masses mA : mB is given by (a) cannot be obtained (c) 1 : 1

(b) 2 : 3 (d) 4 : 9

7. When a sphere rolls down an inclined plane, then identity the correct statement related to the work done by friction force (a) (b) (c) (d)

The friction force does positive translational work The friction force does negative rotational work The net work done by friction is zero All of the above

8. A circular table rotates about a vertical axis with a constant angular speed ω. A circular pan rests on the turn table (with the centre coinciding with centre of table) and rotates with the table. The bottom of the pan is covered with a uniform small thick layer of ice placed at centre of pan. The ice starts melting. The angular speed of the turn table (a) (b) (c) (d)

remains the same decreases increases may increase or decrease depending on the thickness of ice layer

9. If R is the radius of gyration of a body of mass M and radius r, then the ratio of its rotational to translational kinetic energy in the rolling condition is (a)

R2 R2 + r 2

(b)

R2 r2

(c)

r2 R2

(d) 1

Chapter 12

Rotational Mechanics — 181

10. A solid sphere rolls down two different inclined planes of the same height but of different inclinations (a) (b) (c) (d)

in both cases the speeds and time of descend will be same the speeds will be same but time of descend will be different the speeds will be different but time of descend will be same speeds and time of descend both will be different

11. For the same total mass, which of the following will have the largest moment of inertia about an axis passing through the centre of mass and perpendicular to the plane of the body (a) a disc of radius R (c) a square lamina of side 2R

(b) a ring of radius R (d) four rods forming a square of side 2R

12. A disc and a solid sphere of same mass and radius roll down an inclined plane. The ratio of the friction force acting on the disc and sphere is 7 6 3 (c) 2

(b)

(a)

5 4

(d) depends on angle of inclination

13. A horizontal disc rotates freely with angular velocity ω about a vertical axes through its centre. A ring, having the same mass and radius as the disc, is now gently placed coaxially on the disc. After some time, the two rotate with a common angular velocity. Then (a) no friction exists between the disc and the ring (b) the angular momentum of the system is conserved 1 (c) the final common angular velocity is ω 2 (d) All of the above

14. A solid homogeneous sphere is moving on a rough horizontal surface, partly rolling and partly sliding. During this kind of motion of the sphere (a) (b) (c) (d)

total kinetic energy of the sphere is conserved angular momentum of the sphere about any point on the horizontal surface is conserved only the rotational kinetic energy about the centre of mass is conserved None of the above

15. A particle of mass m = 3 kg moves along a straight line 4 y − 3x = 2 where x and y are in metre, with constant velocity v = 5 ms −1. The magnitude of angular momentum about the origin is (a) 12 kg m2s −1

(b) 6.0 kg m2s −1

(c) 4.5 kg m2s −1

(d) 8.0 kg m2s −1

16. A solid sphere rolls without slipping on a rough horizontal floor, moving with a speed v. It makes an elastic collision with a smooth vertical wall. After impact, (a) it will move with a speed v initially (b) its motion will be rolling with slipping initially and its rotational motion will stop momentarily at some instant (c) its motion will be rolling without slipping only after some time (d) All of the above B'

17. The figure shows a square plate of uniform mass distribution. AA′ and BB′ are the two axes lying in the plane of the plate and passing A through its centre of mass. If I 0 is the moment of inertia of the plate about AA′ then its moment of inertia about the axis BB′ is (a) I 0 (c) I 0 cos 2 θ

(b) I 0 cos θ (d) None of these

θ cm B

A'

182 — Mechanics - II 18. A spool is pulled horizontally on rough surface by two equal and opposite forces as shown in the figure. Which of the following statements are correct? (a) (b) (c) (d)

The centre of mass moves towards left The centre of mass moves towards right The centre of mass remains stationary The net torque about the centre of mass of the spool is zero

F F Rough

19. Two identical discs are positioned on a vertical axis as shown in the figure. The

bottom disc is rotating at angular velocity ω 0 and has rotational kinetic energy K 0. The top disc is initially at rest. It then falls and sticks to the bottom disc. The change in the rotational kinetic energy of the system is (b) − K 0 /2 (d) K 0 /4

(a) K 0 /2 (c) − K 0 /4

20. The moment of inertia of hollow sphere (mass M) of inner radius R and outer radius 2R, having material of uniform density, about a diametric axis is (a) 31 MR2 /70 (c) 19 MR2 /80

(b) 43 MR2 /90 (d) None of these

21. A rod of uniform cross-section of mass M and length L is hinged about an end to swing freely in a vertical plane. However, its density is non uniform and varies linearly from hinged end to the free end doubling its value. The moment of inertia of the rod, about the rotation axis passing through the hinge point is (a)

2ML2 9

(b)

3ML2 16

(c)

7ML2 18

(d) None of these 1

22. Let I1 and I 2 be the moment of inertia of a uniform square plate about axes shown in the figure. Then, the ratio I1: I 2 is 1 7 7 (c) 1 : 12

(a) 1 :

2

(b) 1 :

12 7

(d) 1 : 7

23. Moment of inertia of a uniform rod of length L and mass M, about an axis passing through L/ 4 from one end and perpendicular to its length is (a)

7 ML2 36

(b)

7 ML2 48

(c)

11 ML2 48

(d)

ML2 12

24. A uniform rod of length L is free to rotate in a vertical plane about a fixed

A

A'

horizontal axis through B. The rod begins rotating from rest. The angular velocity ω at angle θ is given as (a)

θ  6 g   sin  L 2

(b)

θ  6 g   cos  L 2

(c)

 6 g   sin θ  L

(d)

 6 g   cos θ  L

θ

L

B

25. Two particles of masses 1 kg and 2 kg are placed at a distance of 3m. Moment of inertia of the particles about an axis passing through their centre of mass and perpendicular to the line joining them is (in kg-m2). (a) 6 (c) 8

(b) 9 (d) 12

Chapter 12

Rotational Mechanics — 183 Axis

26. Find moment of inertia of a thin sheet of mass M in the shape of an equilateral triangle about an axis as shown in figure. The length of each side is L (a) ML2 /8 (c) 7ML2 /8

(b) 3 3 ML2 /8 (d) None of these

27. A square is made by joining four rods each of mass M and length L. Its moment of inertia about an axis PQ, in its plane and passing through one of its corner is 2

(a) 6 ML (c)

8 ML2 3

figure. What is the moment of inertia of the rod about the axis passing through O and perpendicular to the plane of the paper? ml2 3 ml2 (c) 12

(a)

45°

4 (b) ML2 3 10 (d) ML2 3

28. A thin rod of length 4l, mass 4 m is bent at the points as shown in the

P

L Q l

O l 90°

90°

l

l

10 ml2 3 ml2 (d) 24

(b)

29. The figure shows two cones A and B with the conditions : hA < hB ; ρA > ρB ; RA = RB mA = mB . Identify the correct statement about their axis of symmetry.

(a) (b) (c) (d)

Both have same moment of inertia A has greater moment of inertia B has greater moment of inertia Nothing can be said

(A) (B)

30. Linear mass density of the two rods system, AC and CB is x. Moment of

A

inertia of two rods about an axis passing through AB is (a)

xl3 4 3

(b)

xl3 2

(c)

xl3 4

(d)

xl3 6 2

C

45° l/2 90° 45° l/2 B

Subjective Questions 1. If radius of the earth contracts to half of its present value without change in its mass, what will be the new duration of the day?

2. The radius of gyration of a uniform disc about a line perpendicular to the disc equals its radius R. Find the distance of the line from the centre.

3. Find the moment of inertia of a uniform square plate of mass M and edge a about one of its diagonals. 7 ml 2 about a line perpendicular 12 to the rod. Find the distance of this line from the middle point of the rod.

4. Moment of inertia of a uniform rod of mass m and length l is

184 — Mechanics - II 5. Two point masses m1 and m2 are joined by a weightless rod of length r. Calculate the moment of inertia of the system about an axis passing through its centre of mass and perpendicular to the rod.

6. Radius of gyration of a body about an axis at a distance 6 cm from its centre of mass is 10 cm. Find its radius of gyration about a parallel axis through its centre of mass.

7. A wheel rotates around a stationary axis so that the rotation angle θ varies with time as θ = at 2 , where a = 0.2 rad/s2. Find the magnitude of net acceleration of the point A at the rim at the moment t = 2.5 s if the linear velocity of the point A at this moment is v = 0.65 m/s.

8. Particle P shown in figure is moving in a circle of radius R = 10 cm with linear speed v = 2 m/s. Find the angular speed of particle about point O. P O

9. A particle of mass m is projected with velocity v at an angle θ with the horizontal. Find its angular momentum about the point of projection when it is at the highest point of its trajectory.

10. Linear mass density (mass/length) of a rod depends on the distance from one end (say A) as λ x = (αx + β ). Here, α and β are constants. Find the moment of inertia of this rod about an axis passing through A and perpendicular to the rod. Length of the rod is l.

11. When a body rolls, on a stationary ground, the acceleration of the point of contact is always zero. Is this statement true or false?

12. A solid sphere of mass m rolls down an inclined plane a height h. Find rotational kinetic energy of the sphere.

13. The topmost and bottommost velocities of a disc are v1 and v2 (< v1 ) in the same direction. The radius is R. Find the value of angular velocity ω.

14. A circular lamina of radius a and centre O has a mass per unit area of kx 2, where x is the distance from O and k is a constant. If the mass of the lamina is M, find in terms of M and a, the moment of inertia of the lamina about an axis through O and perpendicular to the lamina.

15. A solid body starts rotating about a stationary axis with an angular acceleration

α = ( 2.0 × 10−2 ) t rad/ s2, here, t is in seconds. How soon after the beginning of rotation will the total acceleration vector of an arbitrary point of the body form an angle θ = 60° with its velocity vector?

16. A ring of radius R rolls on a horizontal ground with linear speed v and angular speed ω. For what value of θ the velocity of point P is in vertical direction. ( v < Rω )

v θ P

ω

Chapter 12

Rotational Mechanics — 185 F1

17. Two forces F1 and F2 are applied on a spool of mass M and moment of inertia I about an axis passing through its centre of mass. Find the ratio F1 , so that the force of friction is zero. Given that I < 2 Mr 2. F2

2r r

F2

18. A disc is placed on the ground. Friction coefficient is µ. What is the minimum force required to move the disc if it is applied at the topmost point?

19. A cube is resting on an inclined plane. If the angle of inclination is gradually increased, what must be the coefficient of friction between the cube and plane so that, (a) cube slides before toppling?

(b) cube topples before sliding?

20. A uniform disc of mass 20 kg and radius 0.5 m can turn about a smooth axis through its centre and perpendicular to the disc. A constant torque is applied to the disc for 3 s from rest and the 240 angular velocity at the end of that time is rev/min. Find the magnitude of the torque. If the π torque is then removed and the disc is brought to rest in t seconds by a constant force of 10 N applied tangentially at a point on the rim of the disc, find t.

21. A uniform disc of mass m and radius R is rotated about an axis passing through its centre and perpendicular to its plane with an angular velocity ω 0. It is placed on a rough horizontal plane with the axis of the disc keeping vertical. Coefficient of friction between the disc and the surface is µ. Find (a) the time when disc stops rotating, (b) the angle rotated by the disc before stopping.

22. A solid body rotates about a stationary axis according to the law θ = at − bt3 , where a = 6 rad/s and b = 2 rad /s 3 . Find the mean values of the angular velocity and acceleration over the time interval between t = 0 and the time, when the body comes to rest.

23. A rod of mass m and length 2R is fixed along the diameter of a ring of same mass m and radius R as shown in figure. The combined body is rolling without slipping along x-axis. Find the angular momentum about z-axis. y

v O

x

24. The figure shows a thin ring of mass M = 1 kg and radius R = 0.4 m spinning

1 MR 2 ). A small bead of mass m = 0.2 kg can 2 slide without friction along the ring. When the bead is at the top of the ring, the angular velocity is 5 rad/s. What is the angular velocity when the bead slips halfway to θ = 45°?

about a vertical diameter. (Take I =

m θ

R M

25. A horizontal disc rotating freely about a vertical axis makes 100 rpm. A small piece of wax of mass 10 g falls vertically on the disc and adheres to it at a distance of 9 cm from the axis. If the number of revolutions per minute is thereby reduced to 90. Calculate the moment of inertia of disc.

186 — Mechanics - II 26. A man stands at the centre of a circular platform holding his arms extended horizontally with 4 kg block in each hand. He is set rotating about a vertical axis at 0.5 rev/s. The moment of inertia of the man plus platform is 1.6 kg-m2, assumed constant. The blocks are 90 cm from the axis of rotation. He now pulls the blocks in toward his body until they are 15 cm from the axis of rotation. Find (a) his new angular velocity and (b) the initial and final kinetic energy of the man and platform. (c) how much work must the man do to pull in the blocks ?

27. A horizontally oriented uniform disc of mass M and radius R rotates freely about a stationary vertical axis passing through its centre. The disc has a radial guide along which can slide without friction a small body of mass m. A light thread running down through the hollow axle of the disc is tied to the body. Initially the body was located at the edge of the disc and the whole system rotated with an angular velocity ω 0. Then, by means of a force F applied to the lower end of the thread the body was slowly pulled to the rotation axis. Find : (a) the angular velocity of the system in its final state, (b) the work performed by the force F.

28. Consider a cylinder of mass M and radius R lying on a rough horizontal plane. It has a plank lying on its top as shown in figure. A force F is applied on the plank such that the plank moves and causes the cylinder to roll. The plank always remains horizontal. There is no slipping at any point of contact. Calculate the acceleration of the cylinder and the frictional forces at the two contacts.

F m

M

θ

R

29. Find the acceleration of the cylinder of mass m and radius R and that of plank of mass M placed on smooth surface if pulled with a force F as shown in figure. Given that sufficient friction is present between cylinder and the plank surface to prevent sliding of cylinder. m

M

F

30. A uniform rod AB of length 2l and mass m is rotating in a horizontal plane about a vertical axis

through A, with angular velocity ω, when the mid-point of the rod strikes a fixed nail and is brought immediately to rest. Find the impulse exerted by the nail.

31. A uniform rod of length L rests on a frictionless horizontal surface. The rod is pivoted about a fixed frictionless axis at one end. The rod is initially at rest. A bullet travelling parallel to the horizontal surface and perpendicular to the rod with speed v strikes the rod at its centre and becomes embedded in it. The mass of the bullet is one-sixth the mass of the rod. (a) What is the final angular velocity of the rod ? (b) What is the ratio of the kinetic energy of the system after the collision to the kinetic energy of the bullet before the collision ?

32. A uniform rod AB of mass 3m and length 2l is lying at rest on a smooth horizontal table with a smooth vertical axis through the end A. A particle of mass 2m moves with speed 2u across the table and strikes the rod at its mid-point C. If the impact is perfectly elastic. Find the speed of the particle after impact if (a) it strikes the rod normally,

(b) its path before impact was inclined at 60° to AC.

LEVEL 2 Objective Questions Single Correct Option 1. In the given figure a ring of mass m is kept on a horizontal surface while a body of equal mass m is attached through a string, which is wounded on the ring. When the system is released, the ring rolls without slipping. Consider the following statement and choose the correct option. (i)

acceleration of the centre of mass of ring is

2g 3

m

4g 3 (iii) frictional force (on the ring) acts in forward direction (iv) frictional force (on the ring) acts in backward direction (ii) acceleration of hanging particle is

(a) only statements (i) and (ii) are correct

(b) only statements (ii) and (iii) are correct

(c) only statements (iii) and (iv) are correct

(d) None of these

2. A solid sphere of mass 10 kg is placed on a rough surface having coefficient of friction µ = 0.1. A constant force F = 7 N is applied along a F = 7N line passing through the centre of the sphere as shown in the figure. The value of frictional force on the sphere is (a) 1 N

(b) 2 N

(c) 3 N

(d) 7 N

µ = 0.1

3. From a uniform square plate of side a and mass m, a square portion DEFG of

A

B

a is removed. Then, the moment of inertia of remaining portion about the 2 E axis AB is

side

7ma 2 16 3ma 2 (c) 4

3ma 2 16 9ma 2 (d) 16

(b)

(a)

F

G

D

4. A small solid sphere of mass m and radius r starting from rest from the rim of a fixed hemispherical bowl of radius R(> > r ) rolls inside it without sliding. The normal reaction exerted by the sphere on the hemisphere when it reaches the bottom of hemisphere is (a) (3 /7) mg

(b) (9 /7) mg

(c) (13 /7) mg

(d) (17 /7) mg

5. A uniform solid cylinder of mass m and radius R is placed on a rough horizontal surface. A horizontal constant force F is applied at the top point P of the cylinder so that it starts pure rolling. The acceleration of the cylinder is (a) F /3m (c) 4F /3m

(b) 2F /3m (d) 5F /3m

F

P

C

188 — Mechanics - II 6. In the above question, the frictional force on the cylinder is (a) F /3 towards right (c) 2F /3 towards right

(b) F /3 towards left (d) 2F /3 towards left

7. A small pulley of radius 20 cm and moment of inertia 0.32 kg-m 2 is used to hang a 2 kg mass with the help of massless string. If the block is released, for no slipping condition acceleration of the block will be (a) 2 m/s 2 (c) 1 m/s 2

(b) 4 m/s 2 (d) 3 m/s 2 2kg O

8. A uniform circular disc of radius R is placed on a smooth horizontal surface with its plane horizontal and hinged at circumference through point O as shown. An impulse P is applied at a perpendicular distance h from its centre C. The value of h so that the impulse due to hinge is zero, is (a) R

(b) R/2

(c) R/3

(d) R/4

C h

P

9. A rod is supported horizontally by means of two strings of equal length as shown in figure. If one of the string is cut. Then tension in other string at the same instant will (a) (b) (c) (d)

remain unaffected increase decrease become equal to weight of the rod

Mg

10. The figure represents two cases. In first case a block of mass M is attached to a string which is tightly wound on a disc of mass M and radius R. In second case F = Mg. Initially, the disc is stationary in each case. If the same length of string is unwound from the disc, then (a) (b) (c) (d)

same amount of work is done on both discs angular velocities of both the discs are equal both the discs have unequal angular accelerations All of the above

M

R

R

M

F = Mg

M

11. A uniform cylinder of mass M and radius R is released from rest on a rough inclined surface of inclination θ with the horizontal as shown in figure. As the cylinder rolls down the inclined surface, the maximum elongation in the spring of stiffness k is 3 Mg sin θ (a) 4 k Mg sin θ (c) k

k M,R

2 Mg sin θ (b) k

θ

(d) None of these

12. A uniform rod of mass m and length l rotates in a horizontal plane with an angular velocity ω about a vertical axis passing through one end. The tension in the rod at a distance x from the axis is (a)

1 mω 2x 2

(b)

 1 x2 mω 21 −  2 l 

(c)

 1 x2 mω 2l 1 − 2  2 l  

(d)

1 x  mω 2l 1 −  2 l 

Chapter 12

Rotational Mechanics — 189

13. A rod of length 1 m rotates in the xy plane about the fixed point O in

y

the anticlockwise sense, as shown in figure with velocity ω = a + bt where a = 10 rad s − 1 and b = 5 rad s − 2. The velocity and acceleration of O the point A at t = 0 is (a) + 10i$ ms−1 and + 5i$ ms− 2 (c) − 10$j ms− 1 and (100 $i + 5$j) ms− 2

A x

(b) + 10$j ms−1 and (− 100 i$ + 5$j) ms− 2 (d) − 10$j ms− 1 and − 5$j ms− 2

14. A ring of radius R rolls on a horizontal surface with constant acceleration a of the centre of mass as shown in figure. If ω is the instantaneous angular velocity of the ring, then the net acceleration of the point of contact of the ring with ground is

a

ω

(b) ω 2R (d) a 2 + (ω 2R)2

(a) zero (c) a

15. The density of a rod AB increases linearly from A to B. Its midpoint is O and its centre of mass is at C. Four axes pass through A, B, O and C, all perpendicular to the length of the rod. The moments of inertia of the rod about these axes are I A , I B , IO and IC respectively. Then (a) I A > IB

(b) IC < IB

(c) IO > IC

(d) All of these

16. The figure shows a spool placed at rest on a horizontal rough surface. A tightly wound string on the inner cylinder is pulled horizontally with a force F. Identify the correct alternative related to the friction force f acting on the spool (a) (b) (c) (d)

r F

f acts leftwards with f < F f acts leftwards but nothing can be said about its magnitude f < F but nothing can be said about its magnitude None of the above

17. A circular ring of mass m and radius R rests flat on a horizontal smooth

m

surface as shown in figure. A particle of mass m, and moving with a velocity v, collides inelastically ( e = 0) with the ring. The angular velocity with which the system rotates after the particle strikes the ring is m

(a)

R

v 2R

(b)

v 3R

(c)

2v 3R

(d)

3v 4R

R

v

18. A stationary uniform rod in the upright position is allowed to fall on a smooth horizontal surface. The figure shows the instantaneous position of the rod. Identify the correct statement. (a) (b) (c) (d)

normal reaction N is equal to Mg N does positive rotational work about the centre of mass a couple of equal and opposite forces acts on the rod All of the above

θ Mg Smooth N

19. A thin uniform rod of mass m and length l is free to rotate about its upper end. When it is at rest. It receives an impulse J at its lowest point, normal to its length. Immediately after impact (a) (b) (c) (d)

the angular momentum of the rod is Jl the angular velocity of the rod is 3J /ml the kinetic energy of the rod is 3J 2 /2m All of the above

190 — Mechanics - II 20. A rectangular block of size ( b × h ) moving with velocity vo enters on a rough

b

surface where the coefficient of friction is µ as shown in figure. Identify the correct statement. (a) The net torque acting on the block about its COM is µ mg

v0

h

Smooth Rough (µ)

h (clockwise) 2

(b) The net torque acting on the block about its COM is zero (c) The net torque acting on the block about its COM is in the anticlockwise sense (d) None of the above

21. A uniform rod of length L and mass m is free to rotate about a frictionless pivot at one end as shown in figure. The rod is held at rest in the horizontal position and a coin of mass m is placed at the free end. Now the rod is released. P The reaction on the coin immediately after the rod starts falling is (a)

3 mg 2

(c) zero

L 2 C

(b) 2 mg (d)

mg 2

22. A spool is pulled at an angle θ with the horizontal on a rough horizontal surface as shown in the figure. If the spool remains at rest, the angle θ is equal to  R (a) cos −1    r  r (c) π − cos −1    R

 r2  (b) sin −1  1 − 2   R    r (d) sin −1    R

θ

r R

23. Uniform rod AB is hinged at end A in horizontal position as shown in the figure. The other end is connected to a block through a massless string as shown. The pulley is smooth and massless. Mass of block and rod is same and is equal to m. Then acceleration of block just after release from this position is (a) 6 g/13 (c) 3 g/8

A

(b) g/4 (d) None of these

m,l B m

24. A cylinder having radius 0.4 m, initially rotating (at t = 0) with ω 0 = 54 rad /s is placed on a rough inclined plane with θ = 37° having friction coefficient µ = 0.5. The time taken by the cylinder to start pure rolling is ( g = 10 m/s 2 ) (a) 5.4 s (c) 1.4 s

(b) 2.4 s (d) None of these

37°

25. A disc of mass M and radius R is rolling purely with center’s velocity v0 on a flat horizontal floor when it hits a step in the floor of height R/ 4. The corner of the step is sufficiently rough to prevent any slipping of the disc against itself. What is the velocity of the centre of the disc just after impact?

(a) 4v0 /5 (c) 5v0 /6

(b) 4v0 /7 (d) None of these

Chapter 12

Rotational Mechanics — 191

26. A solid sphere is rolling purely on a rough horizontal surface (coefficient of kinetic friction = µ) with speed of centre = u. It collides inelastically with a smooth vertical wall at a certain moment, the coefficient of 1 restitution being . The sphere will begin pure rolling after a time 2 (a)

3u 7 µg

(b)

2u 7 µg

(c)

3u 5 µg

(d)

2u 5 µg

27. A thin hollow sphere of mass m is completely filled with non viscous liquid of mass m. When the sphere roll-on horizontal ground such that centre moves with velocity v, kinetic energy of the system is equal to (a) mv2

(b)

4 mv2 3

(c)

4 mv2 5

(d) None of these

28. A solid uniform disc of mass m rolls without slipping down a fixed inclined plank with an acceleration a. The frictional force on the disc due to surface of the plane is (a)

1 ma 4

(b)

3 ma 2

(c) ma

(d)

1 ma 2

29. A uniform slender rod of mass m and length L is released from rest, with its lower end touching

a frictionless horizontal floor. At the initial moment, the rod is inclined at an angle θ = 30° with the vertical. Then the value of normal reaction from the floor just after release will be (a) 4 mg/7 (c) 2 mg/5

(b) 5 mg/9 (d) None of these

30. In the above problem, the initial acceleration of the lower end of the rod will be (a) g 3 /4 (c) 3 g 3 /7

(b) g 3 /5 (d) None of these

31. A disc of radius R is rolling purely on a flat horizontal surface, with a constant angular velocity. The angle between the velocity and acceleration vectors of point P P is (a) zero (c) tan −1 (2)

(b) 45° (d) tan −1 (1 /2)

32. A straight rod AB of mass M and length L is placed on a frictionless horizontal surface. A force having constant magnitude F and a fixed direction starts acting at the end A. The rod is initially perpendicular to the force. The initial acceleration of end B is (a) zero (c) 4 F /M

(b) 2 F /M (d) None of these

33. A particle moves parallel to x-axis with constant velocity v as shown in

y

the figure. The angular velocity of the particle about the origin O (a) (b) (c) (d)

remains constant continuously increases continuously decreases oscillates

v θ O

x

34. A thin uniform rod of mass M and length L is hinged at its upper end, and released from rest from a horizontal position. The tension at a point located at a distance L/ 3 from the hinge point, when the rod becomes vertical, will be (a) 22 Mg/27 (c) 6 Mg/11

(b) 11 Mg/13 (d) 2 Mg

192 — Mechanics - II 35. A uniform rod AB of length L and mass m is suspended freely at A and hangs

A vertically at rest when a particle of same mass m is fired horizontally with speed v to strike the rod at its mid point. If the particle is brought to rest after the impact. Then m m the impulsive reaction at A in horizontal direction is v

(a) mv/4 (c) mv

(b) mv/2 (d) 2 mv

B v

36. A child with mass m is standing at the edge of a merry go round having moment of inertia I , radius R and initial angular velocity ω as shown in the figure. The child jumps off the edge of the merry go round with tangential velocity v with respect to the ground. The new angular velocity of the merry go round is Iω 2 − mv2 I Iω − mvR (c) I

(a)

(I + mR2) ω 2 − mv2 I (I + mR2) ω − mvR (d) I (b)

37. A disc of radius R is spun to an angular speed ω 0 about its axis and then

ω 0R . The coefficient of 4 friction is µ. The sense of rotation and direction of linear velocity are shown in the figure. The disc will return to its initial position imparted a horizontal velocity of magnitude

(a) (b) (c) (d)

v0 ω0

if the value of µ < 0.5 irrespective of the value of µ if the value of 0.5 < µ < 1 if µ > 1

38. A racing car is travelling along a straight track at a constant velocity of 40m/s

40 m/s. A fixed TV camera is recording the event as shown in figure. In order to keep the car in view, in the position shown, the angular velocity of 30m camera should be 30° (a) 3 rad/s (c) 4 rad/s

(b) 2 rad/s (d) 1 rad/s A

39. A uniform rod OA of length l, resting on smooth surface is slightly distributed from its vertical position. P is a point on the rod whose locus is a circle during the subsequent motion of the rod. Then the distance OP is equal to (a) (b) (c) (d)

l

l/2 l/3 l/4 there is no such point

P O

40. In the above question, the velocity of end O when end A hits the ground is (a) (b) (c) (d)

zero along the horizontal along the vertical at some inclination to the ground (≠ 90° )

41. In the above question, the velocity of end A at the instant it hits the ground is (a)

3 gl

(b)

12 gl

(c)

6 gl

(d) None of these

Chapter 12

Rotational Mechanics — 193

42. A solid sphere of mass m and radius R is gently placed on a conveyer belt moving with constant velocity v0. If coefficient of friction between belt and sphere is 2/7, the distance traveled by the centre of the sphere before it starts pure rolling is v0

v02 7g 2v02 (c) 5g

2v02 49 g 2v02 (d) 7g (b)

(a)

More than One Correct Options 1. A mass m of radius r is rolling horizontally without any slip with a linear speed v. It then rolls up to a height given by

3 v2 4 g

(a) the body is identified to be a disc or a solid cylinder (b) the body is a solid sphere 3 mr 2 2 7 (d) moment of inertia of the body about instantaneous axis of rotation is mr 2 5 (c) moment of inertia of the body about instantaneous axis of rotation is

2. Four identical rods each of mass m and length l are joined to form a rigid square frame. The frame lies in the xy plane, with its centre at the origin and the sides parallel to the x and y axes. Its moment of inertia about 2 2 ml 3 4 (b) the z-axis is ml2 3

(a) the x-axis is

(c) an axis parallel to the z-axis and passing through a corner is (d) one side is

5 2 ml 3

10 2 ml 3

B

3. A uniform circular ring rolls without slipping on a horizontal surface. At any instant, its position is as shown in the figure. Then (a) (b) (c) (d)

section ABC has greater kinetic energy than section ADC section BC has greater kinetic energy than section CD section BC has the same kinetic energy as section DA the sections CD and DA have the same kinetic energy

4. A cylinder of radius R is to roll without slipping between two planks as

A

v O

C

D

v

shown in the figure. Then v counter clockwise R 2v (b) angular velocity of the cylinder is clockwise R (c) velocity of centre of mass of the cylinder is v towards left (d) velocity of centre of mass of the cylinder is 2v towards right (a) angular velocity of the cylinder is

3v

194 — Mechanics - II 5. A uniform rod of mass m = 2 kg and length l = 0.5 m is sliding along two mutually perpendicular smooth walls with the two ends P and Q having velocities vP = 4 m/ s and vQ = 3 m/ s as shown. Then

(a) The angular velocity of rod, ω = 10 rad /s, counter clockwise (b) The angular velocity of rod, ω = 5.0 rad /s, counter clockwise (c) The velocity of centre of mass of rod, vcm = 2.5 m/s 25 joule (d) The total kinetic energy of rod, K = 3

Q vQ = 3 m/s

6. A wheel is rolling without slipping on a horizontal plane with velocity v and

P vP = 4 m/s C

acceleration a of centre of mass as shown in figure. Acceleration at (a) (b) (c) (d)

A is vertically upwards B may be vertically downwards C cannot be horizontal a point on the rim may be horizontal leftwards

B

A

7. A uniform rod of length l and mass 2 m rests on a smooth horizontal table. A point mass m moving horizontally at right angles to the rod with velocity v collides with one end of the rod and sticks it. Then 2v 5 l v (b) angular velocity of the system after collision is 2l

(a) angular velocity of the system after collision is

3 mv2 10 7mv2 (d) the loss in kinetic energy of the system as a whole as a result of the collision is 24 (c) the loss in kinetic energy of the system as a whole as a result of the collision is

8. A non-uniform ball of radius R and radius of gyration about geometric centre = R/ 2, is kept on a frictionless surface. The geometric centre coincides with the centre of mass. The ball is struck horizontally with a sharp impulse = J. The point of application of the impulse is at a height h above the surface. Then (a) (b) (c) (d)

the ball with slip on surface for all cases the ball will roll purely if h = 5R /4 the ball will roll purely if h = 3R /2 there will be no rotation if h = R

9. A hollow spherical ball is given an initial push, up an incline of inclination angle α. The ball rolls purely. Coefficient of static friction between ball and incline = µ. During its upwards journey (a) friction acts up along the incline (c) friction acts down along the incline

(b) µ min = (2 tan α)/5 (d) µ min = (2 tan α )/7

10. A uniform disc of mass m and radius R rotates about a fixed vertical axis passing through its centre with angular velocity ω. A particle of same mass m and having velocity of 2ωR towards centre of the disc collides with the disc moving horizontally and sticks to its rim. Then,

(a) the angular velocity of the disc will become ω/3 (b) the angular velocity of the disc will become 5ω/3 37 (c) the impulse on the particle due to disc is mωR 3 (d) the impulse on the particle due to disc is 2mωR

Chapter 12

Rotational Mechanics — 195

11. The end B of the rod AB which makes angle θ with the floor is being pulled

y

with a constant velocity v0 as shown. The length of the rod is l. (a) At θ = 37° velocity of end A is

4 v0 downwards 3 5v (b) At θ = 37° angular velocity of rod is 0 3l

A

(c) Angular velocity of rod is constant (d) Velocity of end A is constant

O

θ

v0

x

B

Comprehension Based Questions Passage 1 (Q. Nos. 1 to 4) y A uniform rod of mass m and length l is pivoted at point O. The rod is initially in vertical position and touching a block of mass M which is at rest on a horizontal surface. The rod is given a slight jerk and it starts rotating about point O. This causes the block to move forward as shown. The rod loses contact with the block at θ = 30°. All surfaces ω θ are smooth. Now answer the following questions. O 1. The value of ratio M/m is

(a) 2 : 3

(b) 3 : 2

(c) 4 : 3

M x

(d) 3 : 4

2. The velocity of block when the rod loses contact with the block is (a)

3 gl 4

(b)

5 gl 4

(c)

6 gl 4

7 gl 4

(d)

3. The acceleration of centre of mass of rod, when it loses contact with the block is (a) 5 g/4

(b) 5 g/2

(c) 3 g/2

(d) 3 g/4

4. The hinge reaction at O on the rod when it loses contact with the block is (a)

3 mg $ $ (i + j) 4

 mg  $ (b)   j  4 

 mg  $ (c)   i  4 

(d)

mg $ $ (i + j) 4

Passage 2 (Q. Nos. 5 to 7) Consider a uniform disc of mass m, radius r, rolling without slipping on a rough surface with linear acceleration ‘a’ and angular acceleration α due to an external force F as shown in the figure. Coefficient of friction is µ

F α

5. The work done by the frictional force at the instant of pure rolling is (a)

µmgat 2 2

(b) µmgat 2

(c) µmg

at 2 α

(d) zero

6. The magnitude of frictional force acting on the disc is (a) ma

(b) µmg

(c)

ma 2

7. Angular momentum of the disc will be conserved about (a) centre of mass (b) point of contact (c) a point at a distance 3R/2 vertically above the point of contact (d) a point at a distance 4R/3 vertically above the point of contact

(d) zero

196 — Mechanics - II Passage 3 (Q. No. 8 to 10) A tennis ball, starting from rest, rolls down the hill in the drawing. At the end of the hill the ball becomes airborne, leaving at an angle of 37° with respect to the ground. Treat the ball as a thin-walled spherical shell. v

h

H 37°

x

8. The velocity of projection v is (a)

(b)

2 gh

10 gh 7

(c)

5 gh 7

(d)

6 gh 5

9. Maximum height reached by ball H above ground is (a)

9h 35

(b)

18h 35

(c)

18h 25

(d)

27h 125

(b)

48 h 25

(c)

48 h 35

(d)

24 h 7

10. Range x of the ball is (a)

144 h 125

Match the Columns 1. A solid sphere, a hollow sphere and a disc of same mass and same radius are released from a rough inclined plane. All of them rolls down without slipping. On reaching the bottom of the plane, match the two columns. Column I (a) (b) (c) (d)

Column II

time taken to reach the bottom total kinetic energy rotational kinetic energy translational kinetic energy

(p) (q) (r) (s)

maximum for solid sphere maximum for hollow sphere maximum for disc same for all

2. A solid sphere is placed on a rough ground as shown. E is the centre of sphere and DE > EF . We have to apply a linear impulse either at point A, B or C. Match the following two columns. A

D

B

E F

C

Column I (a) Sphere will acquire maximum angular speed it impulse is applied at (b) Sphere will acquire maximum linear speed it impulse is applied at (c) Sphere can roll without slipping if impulse is applied at (d) Sphere can roll with forward slipping if impulse is applied at

Column II (p) A (q) B (r) C (s) at any point A, B or C

Rotational Mechanics — 197

Chapter 12

3. The inclined surfaces shown in Column I are sufficiently rough. In Column II direction and magnitudes of frictional forces are mentioned. Match the two columns. Column I

Column II

(a)

(p) upwards Rolling upwards

(b)

(q) downwards Kept in rotating position

(c)

(r) maximum friction will act Kept in translational position

(d)

(s) required value of friction will act

Kept in translational position

4. A rectangular slab ABCD have dimensions a × 2a as shown in figure. Match the following two columns.

1

2 2a

A 4 3

D

Column I

B a C

Column II a 12

(a) Radius of gyration about axis-1

(p)

(b) Radius of gyration about axis-2

(q)

(c) Radius of gyration about axis-3

a 3 (s) None

(d) Radius of gyration about axis-4

(r)

2a 3

198 — Mechanics - II 5. A small solid ball rolls down along sufficiently rough surface from 1 to 3 as shown in figure. From point-3 onwards it moves under gravity. Match the following two columns. 1 3

h 2

h/2 4

Column I

Column II

(a) Rotational kinetic energy of ball at point-2 (b) Translational kinetic energy of ball at point-3 (c) Rotational kinetic energy of ball at point-4 (d) Translational kinetic energy of ball at point-4

1 mgh 7 2 (q) mgh 7 5 (r) mgh 7 (s) None

(p)

6. A uniform disc of mass 10 kg, radius 1 m is placed on a rough horizontal surface. The coefficient of friction between the disc and the surface is 0.2. A horizontal time varying force is applied on the centre of the disc whose variation with time is shown in graph. F(N)

80N t (s) 8s

Column I (a) (b) (c) (d)

Disc rolls without slipping Disc rolls with slipping Disc starts slipping at Friction force is 10 N at

Column II (p) (q) (r) (s)

at t = 7 s at t = 3 s at t = 4 s None

7. Match the columns. Column I (a) Moment of inertia of a circular disc of mass M and radius R about a tangent parallel to plane of disc (b) Moment of inertia of a solid sphere of mass M and radius R about a tangent (c) Moment of inertia of a circular disc of mass M and radius R about a tangent perpendicular to plane of disc (d) Moment of inertia of a cylinder of mass M and radius R about its axis

Column II MR2 2 7 (q) MR2 5 5 (r) MR2 4 3 (s) MR2 2 (p)

Chapter 12

Rotational Mechanics — 199

Subjective Questions 1. Figure shows three identical yo-yos initially at rest on a horizontal surface. For each yo-yo the string is pulled in the direction shown. In each case there is sufficient friction for the yo-yo to roll without slipping. Draw the free-body diagram for each yo-yo. In what direction will each yo-yo rotate ? F F F

2. A uniform rod of mass m and length l is held horizontally by two vertical strings of negligible mass, as shown in the figure. (a) Immediately after the right string is cut, what is the linear acceleration of the free end of the rod ? (b) Of the middle of the rod ? (c) Determine the tension in the left string immediately after the right string is cut.

3. A solid disk is rolling without slipping on a level surface at a constant speed of 2.00 m/s. How far can it roll up a 30° ramp before it stops? (Take g = 9.8 m/ s2)

4. A lawn roller in the form of a thin-walled hollow cylinder of mass M is pulled horizontally with a constant horizontal force F applied by a handle attached to the axle. If it rolls without slipping, find the acceleration and the friction force.

5. Due to slipping, points A and B on the rim of the disk have the velocities shown. Determine the velocities of the centre point C and point F at this instant. vB = 3 m/s

B

C A

30° F

vA = 1.5 m/s

6. A uniform cylinder of mass M and radius R has a string wrapped around it. The string is held fixed and the cylinder falls vertically, as in figure. T

(a) Show that the acceleration of the cylinder is downward with magnitude a = (b) Find the tension in the string.

2g . 3

200 — Mechanics - II 7. A uniform disc of mass M and radius R is pivoted about the horizontal axis through its centre C. A point mass m is glued to the disc at its rim, as shown in figure. If the system is released from rest, find the angular velocity of the disc when m reaches the bottom point B. M m

C

B

8. A disc of radius R and mass m is projected on to a horizontal floor with a backward spin such that its centre of mass speed is v0 and angular velocity is ω 0. What must be the minimum value of ω 0 so that the disc eventually returns back ?

9. A ball of mass m and radius r rolls along a circular path of radius R. Its speed at the bottom (θ = 0° ) of the path is v0. Find the force of the path on the ball as a function of θ. θR

10. A heavy homogeneous cylinder has mass m and radius R. It is accelerated by a force F, which is applied through a rope wound around a light drum of radius r attached to the cylinder (figure). The coefficient of static friction is sufficient for the cylinder to roll without slipping.

F R

r

(a) Find the friction force. (b) Find the acceleration a of the centre of the cylinder. F (c) Is it possible to choose r, so that a is greater than ? How ? m (d) What is the direction of the friction force in the circumstances of part (c) ?

11. A man pushes a cylinder of mass m1 with the help of a plank of mass m2 as

F

m2

shown. There is no slipping at any contact. The horizontal component of the force applied by the man is F. Find :

m1

(a) the acceleration of the plank and the centre of mass of the cylinder and (b) the magnitudes and directions of frictional forces at contact points. 12. For the system shown in figure, M = 1 kg, m = 0.2 kg, r = 0.2 m.

M r

Calculate: (g = 10 m/s 2 )

(a) the linear acceleration of hoop, (b) the angular acceleration of the hoop of mass M and (c) the tension in the rope.

Hoop Smooth m

Note Treat hoop as the ring. Assume no slipping between string and hoop.

13. A cylinder of mass m is kept on the edge of a plank of mass 2m and length 12 m, which in turn is kept on smooth ground. Coefficient of friction between the plank and the cylinder is 0.1. The cylinder is given an impulse, which imparts it a velocity 7 m/s but no angular velocity. Find the time after which the cylinder falls off the plank. ( g = 10 m/ s2 )

m

7 m/s 2m 12 m

Chapter 12

Rotational Mechanics — 201

14. The 9 kg cradle is supported as shown by two uniform disks that roll

30 N

without sliding at all surfaces of contact. The mass of each disk is m = 6 kg and the radius of each disk is r = 80 mm. Knowing that the system is initially at rest, determine the velocity of the cradle after it has moved 250 mm.

A

B

15. The disc of the radius r is confined to roll without slipping at A and B. If the plates have the velocities shown, determine the angular velocity of the disc. v A r B

2v

16. A thin uniform rod AB of mass m = 1 kg moves transnationally with acceleration a = 2 m/ s2 due to two antiparallel forces F1 and F2. The distance between the points at which these forces are applied is equal to l = 20 cm. Besides, it is known that F2 = 5 N. Find the length of the rod. F1

A l F2

B

17. The assembly of two discs as shown in figure is placed on a rough horizontal surface and the

front disc is given an initial angular velocity ω 0. Determine the final linear and angular velocity when both the discs start rolling. It is given that friction is sufficient to sustain rolling in the rear wheel from the starting of motion.

R

ω0

18. A horizontal plank having mass m lies on a smooth horizontal surface.

A sphere of same mass and radius r is spined to an angular frequency ω 0 and gently placed on the plank as shown in the figure. If coefficient of friction between the plank and the sphere is µ. Find the distance moved by the plank till the sphere starts pure rolling on the plank. The plank is long enough.

19. A ball rolls without sliding over a rough horizontal floor with velocity v0 = 7 m/s towards a

smooth vertical wall. If coefficient of restitution between the wall and the ball is e = 0.7. Calculate velocity v of the ball long after the collision.

20. A uniform rod of mass m and length l rests on a smooth horizontal surface. One of the ends of the rod is struck in a horizontal direction at right angles to the rod. As a result the rod obtains velocity v0. Find the force with which one-half of the rod will act on the other in the process of motion.

202 — Mechanics - II 21. A sphere, a disk and a hoop made of homogeneous materials have the same radius (10 cm) and mass (3 kg). They are released from rest at the top of a 30° incline and roll down without slipping through a vertical distance of 2 m. (g = 9.8 m/s 2 ) (a) What are their speeds at the bottom ? (b) Find the frictional force f in each case (c) If they start together at t = 0, at what time does each reach the bottom ?

22. ABC is a triangular framework of three uniform rods each of mass m and length 2l. It is free to rotate in its own plane about a smooth horizontal axis through A which is perpendicular to ABC. If it is released from rest when AB is horizontal and C is above AB. Find the maximum velocity of C in the subsequent motion.

23. A uniform stick of length L and mass M hinged at one end is released from rest at an angle θ 0

with the vertical. Show that when the angle with the vertical is θ, the hinge exerts a force Fr 1 along the stick and Ft perpendicular to the stick given by Fr = Mg ( 5 cos θ − 3 cos θ 0 ) and 2 1 Ft = Mg sin θ 4

24. A uniform rod AB of mass 3m and length 4l, which is free to turn in a vertical plane about a smooth horizontal axis through A, is released from rest when horizontal. When the rod first becomes vertical, a point C of the rod, where AC = 3l, strikes a fixed peg. Find the linear impulse exerted by the peg on the rod if (a) the rod is brought to rest by the peg, (b) the rod rebounds and next comes to instantaneous rest inclined to the downward vertical at an π angle radian. 3

25. A uniform rod of length 4l and mass m is free to rotate about a horizontal axis passing through a point distant l from its one end. When the rod is horizontal, its angular velocity is ω as shown in figure. Calculate : l ω

(a) reaction of axis at this instant, (b) acceleration of centre of mass of the rod at this instant, (c) reaction of axis and acceleration of centre mass of the rod when rod becomes vertical for the first time, (d) minimum value of ω , so that centre of rod can complete circular motion.

26. A stick of length l lies on horizontal table. It has a mass M and is free to move in any way on the table. A ball of mass m, moving perpendicularly to the stick at a distance d from its centre with speed v collides elastically with it as shown in figure. What quantities are conserved in the collision ? What must be the mass of the ball, so that it remains at rest immediately after collision?

m

v

d

Chapter 12

Rotational Mechanics — 203

27. A rod of length l forming an angle θ with the horizontal strikes a frictionless floor at A with its centre of mass velocity v0 and no angular velocity. Assuming that the impact at A is perfectly elastic. Find the angular velocity of the rod immediately after the impact.

A

θ

v0

28. Three particles A, B and C, each of mass m, are connected to each other by three massless rigid rods to form a rigid, equilateral triangular body of side l. This body is placed on a horizontal frictionless table (x-y plane) and is hinged to it at the point A, so that it can move without friction about the vertical axis through A (see figure). The body is set into rotational motion on the table about A with a constant angular velocity ω. y x

A

ω

F l

B

C

(a) Find the magnitude of the horizontal force exerted by the hinge on the body. (b) At time T, when the side BC is parallel to the x-axis, a force F is applied on B along BC (as shown). Obtain the x-component and the y-component of the force exerted by the hinge on the body, immediately after time T.

29. A semicircular track of radius R = 62.5 cm is cut in a block. Mass of block, having track, is

M = 1 kg and rests over a smooth horizontal floor. A cylinder of radius r = 10 cm and mass m = 0.5 kg is hanging by thread such that axes of cylinder and track are in same level and surface of cylinder is in contact with the track as shown in figure. When the thread is burnt, cylinder starts to move down the track. Sufficient friction exists between surface of cylinder and track, so that cylinder does not slip. m R M

Calculate velocity of axis of cylinder and velocity of the block when it reaches bottom of the track. Also find force applied by block on the floor at that moment. ( g = 10 m/ s2 )

30. A uniform circular cylinder of mass m and radius r is given an initial angular velocity ω 0 and no

initial translational velocity. It is placed in contact with a plane inclined at an angle α to the horizontal. If there is a coefficient of friction µ for sliding between the cylinder and plane. Find the distance the cylinder moves up before sliding stops. Also, calculate the maximum distance it travels up the plane. Assume µ > tan α.

204 — Mechanics - II 31. Show that if a rod held at angle θ to the horizontal and released, its lower end will not slip if the friction coefficient between rod and ground is greater than

3 sin θ cos θ 1 + 3 sin2 θ

.

32. One-fourth length of a uniform rod of mass m and length l is placed on a rough horizontal surface and it is held stationary in horizontal position by means of a light thread as shown in the figure. The thread is then burnt and the rod starts rotating about the edge. Find the angle between the rod and the horizontal when it is about to slide on the edge. The coefficient of friction between the rod and the surface is µ.

l/4

33. In figure the cylinder of mass 10 kg and radius 10 cm has a tape wrapped round it. The pulley weighs 100 N and has a radius 5 cm. When the system is released, the 5 kg mass comes down and the cylinder rolls without slipping. Calculate the acceleration and velocity of the mass as a function of time. 10 cm

20 cm Rough

5 kg

34. A cylinder is sandwiched between two planks. Two constant horizontal forces F and 2F are applied on the planks as shown. Determine the acceleration of the centre of mass of cylinder and the top plank, if there is no slipping at the top and bottom of cylinder. M

F

R

M 2M

2F smooth

35. A ring of mass m and radius r has a particle of mass m attached to it at a point A. The ring can rotate about a smooth horizontal axis which is tangential to the ring at a point B diametrically opposite to A. The ring is released from rest when AB is horizontal. Find the angular velocity π and the angular acceleration of the body when AB has turned through an angle . 3

B

A

Chapter 12

Rotational Mechanics — 205

36. A hoop is placed on the rough surface such that it has an angular velocity ω = 4 rad/s and an

angular deceleration α = 5 rad/s 2. Also, its centre has a velocity of v0 = 5 m/s and a deceleration a0 = 2 m/ s2 . Determine the magnitude of acceleration of point B at this instant. ω = 4 rad/s A a0 = 2 m/s2

O 0.3m

α = 5 rad/s2 45°

v0 = 5 m/s

B

37. A boy of mass m runs on ice with velocity v0 and steps on the end of a plank of length l and mass M which is perpendicular to his path. M

l

m v0

(a) Describe quantitatively the motion of the system after the boy is on the plank. Neglect friction with the ice. (b) One point on the plank is at rest immediately after the collision. Where is it?

38. A thin plank of mass M and length l is pivoted at one end. The plank is released at 60° from the vertical. What is the magnitude and direction of the force on the pivot when the plank is horizontal?

l

Answers Introductory Exercise 12.1 1.

2 l 3

2. 55 kg-m2

3.

5 ml 2 3

4. About a diagonal, because the mass is more concentrated about a diagonal 8 8 1 5. (i) 6. (a) 2Ml 2 (b) mr 2 + 2ma2 (ii) mr 2 + ma2 Ml 2 5 5 3 7. 0.5 kg- m2

9. 0.43 kg- m2

8. The one having the smaller density

10.

π2 3

Introductory Exercise 12.2 1.

π rad s −1 30

2. ω =

v 2R

$ ) rad s −1 3. (−k

4.

v , perpendicular to paper inwards l

Introductory Exercise 12.3 $ ) N-m 1. (−2$i − 2k

2. 400 N-m (perpendicular to the plane of motion) 3. 2.71 N-m

4.

83 N-m 20

Introductory Exercise 12.4 1. 8 π rad- s −2 , (40 π ) rad- s −1

2. 100 rad

6. (a) 4 rad-s −1, −6 rad-s −2 (b) −12 rad-s −2 10. (a) 12.5 rad

3. 5 N-m 7. 20 s

4. 800 rad

5. 0.87 N

8. (a) 0.01 N-m (b) 10.13 N-m

9. (a) 36 s (b) 12

2 k

11. 9 rad, 1.43

(b) 127.5 rad

Introductory Exercise 12.5 1. ml 2ω

2. 4 2 kg- m2 s −1

3.

mu3 cos α sin2 α 2g

7 $ 6. 1 mRv (clockwise) 4. No 5. −  mRv  k 5  2

Introductory Exercise 12.6 1.

ω 0M M + 2m

2. Duration of day night will increase

3. True

4. Increase

Introductory Exercise 12.7 3 $ v v v v i, vN = v$i − $j, vR = $i, v S = v$i + $j 2 2 2 2 2. aM = (Rω 2 )$j, aN = (Rω 2 )$i, aR = − (Rω 2 )$j, aS = − (Rω 2 )$i

1. vM =

Introductory Exercise 12.8 1. 25 J, 35 J

2. True

3. True

Introductory Exercise 12.9 1. v A is zero. Rest three velocities are : | vC| = 2v, | vB| = | vD| =

2v

Introductory Exercise 12.10 1. False 2. 72 N

3. (a)

2 25 30 (b) ms −2 (c) ms −2 7 7 7 3

4. (a) g sin θ − µg cos θ (b)

Introductory Exercise 12.11 1. False 2. (a) Same in both cases (b) Solid sphere (c) Solid sphere

5 µg cos θ 2R

5. Leftwards

Exercises LEVEL 1 Assertion and Reason 1. (d)

2. (b)

3. (d)

4. (a)

5. (a)

6. (a)

7. (c)

8. (b)

9. (b)

10. (a)

11. (c)

Objective Questions 1. (d)

2. (a)

3. (c)

4. (d)

5. (b)

6. (a)

7. (c)

8. (b)

9. (b)

10. (b)

11. (d)

12. (a)

13. (b)

14. (b)

15. (b)

16. (d)

17. (a)

18. (b)

19. (b)

20. (d)

21. (c)

22. (d)

23. (b)

24. (a)

25. (a)

26. (d)

27. (c)

28. (b)

29. (a)

30. (d)

Subjective Questions Ma2 12

1. 6h 2.

R 2

6. 8 cm

7. 07 . ms −2

12. 18. 20.

3.

4.

l 2

5. I = µr 2 , where µ =

8. 10 rad- s −1

v − v2 2 mgh 13. 1 2R 7

14.

2 Ma2 3

9.

mv3 sin2 θ cos θ 2g

15. 7 s

21. (a)

17.

11. False

I + 2Mr 2 4Mr 2 − I

19. (a) µ < 1 (b) µ > 1

3ω 0R 4 µg

22. ω av = 4 rad/s, α av = − 6.0 rad- s −2

 αl 4 β l3  10. I =  +  3   4

v  16. ± cos −1   Rω 

lim can make the body move F → 0 20 N-m, 4 s 3

mm 1 2 is called the reduced mass of two masses. m1 + m2

(b)

3ω 20R 8 µg

10 $ 24. 25 rad- s −1 25. 7.29 × 10 −4 kg - m2 23. −  mRv  k  3  6

26. (a) 14.3 rad- s −1 = 2.27 rev - s −1 (b) E i = 39.9 J, E f = 181 J (c) 141.1 J 2m  27. (a) ω =  1 +  ω0  M  29.

(b)

1 2m  mω 20R 2  1 +   2 M 

28.

F each in opposite directions M + 3m

32. (a)

2u 3

(b)

30.

4F cos θ 3M + 8m

,

3MF cos θ 3M + 8m

,

MF cos θ 3M + 8m

4 mlω 3

31. (a)

2v 9L

(b)

1 9

2u 3

LEVEL 2 Objective Questions 1. 11. 21. 31. 41.

(d) (b) (c) (b) (a)

2. 12. 22. 32. 42.

(b) (c) (b) (b) (a)

3. 13. 23. 33.

(b) (b) (c) (c)

4. 14. 24. 34.

(d) (b) (d) (d)

5. 15. 25. 35.

(c) (d) (c) (a)

6. 16. 26. 36.

(a) (a) (a) (d)

7. 17. 27. 37.

(a) (b) (b) (b)

8. 18. 28. 38.

(b) (b) (d) (d)

9. 19. 29. 39.

(c) (d) (a) (c)

10. 20. 30. 40.

(c) (b) (c) (a)

More than One Correct Options 1. (a,c) 8. (b,d)

2. (all) 9. (a,b)

3. (a,b,d) 10. (a,c)

4. (a,d) 11. (a,b)

5. (a,c,d)

6. (all)

7. (a,c)

208 — Mechanics - II Comprehension Based Questions 1. (c)

2. (a)

3. (d)

Match the Columns

4. (b)

5. (d)

1. (a) → q

(b) → s

(c) → q

(d) → p

2. (a) → p

(b) → s

(c) → p

(d) → r

3. (a) → p,s

(b) → p,r

(c) → q,r

(d) → p,r

4. (a) → q

(b) → r

(c) → r

(d) → p

5. (a) → q

(b) → s

(c) → p

(d) → s

6. (a) → q,r

(b) → p

(c) → s

(d) → q

7. (a) → r

(b) → q

(c) → s

(d) → p

6. (c)

7. (c)

8. (d)

(b) 3g/4

(c) mg/4

9. (d)

10. (a)

Subjective Questions 1. In each case in clockwise direction 5. 0.75 ms −1, 1.98 ms −1 9. f =

6. (b)

1 Mg 3

4mg (2m + M )R

7.

8.

2 1 r  −  F , assuming f opposite to F 3 2 R

(d) f in same direction as F. (b)

3m1 F 3m1 + 8m2

12. (a) 1.43 ms −2

4.

F F , 2M 2

2v 0 R

(b) 7.15 rad-s −2

2F  1 (b) a =   (R + r ) (c) yes, if r is greater than R.  3mR  2

8F 4F , 3m1 + 8m2 3m1 + 8m2

11. (a)

m1 F

(between plank and cylinder)

3v (anticlockwise) 2 r

3m1 + 8m2

16. 1 m

17.

ω 0R ω 0 ; 6 6

18. S =

−1

21. (a) Sphere, 5.29 ms , disk 5.11 ms , hoop 4.43 ms (b) Sphere 4.2 N, disk 4.9 N, hoop 7.36 N 22. 2l

25. (a)

8 24. (a)  m 3 

g 3 l  7lω 2  4 mg 1 +   7  4g 

2

 3g    7 

(b)

(between cylinder and ground)

30. d1 =

3 mlω 2

(b) Fx = –

F , Fy = 4

r 2ω 20 (µ cos α − sin α ) 2 g (3µ cos α − sin α )

4µ  32. θ = tan−1    13 

2

, dmax =

33. 3.6 ms −2 ,

2l from the boy 3

38.

4 gt 11

81µg

19. v = 1.5 ms −1 20.

9 mv02 2 l

(c) Sphere, 1.51 s disk 1.56 s hoop 1.81 s

3 gl , (b)

2

+ (lω 2 )2

4 m 6 gl ( 2 + 1) 3 13 6g (c)  mg + mlω 2  ,  + lω 2   7   7  Ml 2 12d 2 + l 2

29. 2.0 ms −1,

3 mlω 2

2ω 20 r 2

−1

26. Linear momentum, angular momentum and kinetic energy, 28. (a)

14. 0.745 ms −1 (rightwards)

13. 2.25 s

(c) 1.43 N

−1

37. (b)

3. 0.612 m

mv02 2 mg mg sin θ, N = (17 cos θ − 10) + 7 7 (R − r )

10. (a) f =

15.

2. (a) 3g/2,

1.5 ms −1,

27. ω =

(d)

6g 7l

 cos θ 6v 0    l  1 + 3 cos 2 θ 

16.67 N

r 2 ω 20 (µ cos α − sin α ) 4 g sin α (3 µ cos α − sin α ) 34.

10 Mg, 4

F 21F , 26M 26M α

35.

6g 3 3g , 11r 11r

1 α = tan−1    3

36. 6.21 ms −2

13

Gravitation Chapter Contents 13.1

Introduction

13.2

Newton’s Law of Gravitation

13.3

Acceleration Due to Gravity

13.4

Gravitational Field and Field Strength

13.5

Gravitational Potential

13.6

Relation between Gravitational Field and Potential

13.7

Gravitational Potential Energy

13.8

Binding Energy

13.9

Motion of Satellites

13.10 Kepler’s Laws of Planetary Motion

210 — Mechanics - II

13.1 Introduction Why are planets, moon and the sun all nearly spherical? Why do some earth satellites circle the earth in 90 minutes, while the moon takes 27 days for the trip? And why don’t satellites fall back to earth? The study of gravitation provides the answers for these and many related questions. Gravitation is one of the four classes of interactions found in nature. These are : (i) the gravitational force (ii) the electromagnetic force (iii) the strong nuclear force (also called the hadronic forces) (iv) the weak nuclear forces. Although, of negligible importance in the interactions of elementary particles, gravity is of primary importance in the interactions of large objects. It is gravity that holds the universe together. In this chapter, we will learn the basic laws that govern gravitational interactions.

13.2 Newton's Law of Gravitation Along with his three laws of motion, Newton published the law of gravitation in 1687. According to him; “every particle of matter in the universe attracts every other particle with a force that is directly proportional to the product of the masses of the particles and inversely proportional to the square of the distance between them.” Thus, the magnitude of the gravitational force F between two particles m1 and m2 placed at a distance r is, mm mm F ∝ 1 2 2 or F = G 1 2 2 r r Here, G is a universal constant called gravitational constant whose magnitude is, G = 6.67 × 10 –11 N-m 2 /kg 2 The direction of the force F is along the line joining the two particles. Following three points are important regarding the gravitational force : (i) Unlike the electrostatic force, it is independent of the medium between the particles. (ii) It is conservative in nature. (iii) It expresses the force between two point masses (of negligible volume). However, for external points of spherical bodies the whole mass can be assumed to be concentrated at its centre of mass.

Gravity In Newton’s law of gravitation, gravitation is the force of attraction between any two bodies. If one of the bodies is earth then the gravitation is called ‘gravity’. Hence, gravity is the force by which earth attracts a body towards its centre. It is a special case of gravitation.

Chapter 13

Gravitation — 211

Extra Points to Remember ˜

Direct formula F =

Gm1m2

can be applied under following three conditions: r2 (a) To find force between two point masses m1

m2 r Fig. 13.1

(b) To find force between two spherical bodies m2

m1 r

Fig. 13.2

(c) To find force between a spherical body and a point mass. m2

m1 r

Fig. 13.3 ˜

To find force between a point mass and a rod single integration is required. In this case, we cannot assume whole mass of the rod at its centre to find force between them. Thus, r

c

m

M, L

Fig. 13.4

F≠

˜

GMm

r2 To find force between two rods double integration is required but normally, double integration is not asked in physics paper. M1 , L 1

M2 , L 2

Fig. 13.5 ˜

V

Two or more than two gravitational forces acting on a body can be added by vector addition method.

Example 13.1 Three point masses ‘m’ each are placed at the three vertices of an equilateral triangle of side ‘a’. Find net gravitational force on any one point mass. We are finding net force on the point mass kept at O. G ( m ) ( m ) Gm2 F= = 2 a2 a Since, the two forces are equal in magnitudes, therefore the resultant force will pass through the centre as shown in figure.

m

Solution

F net = F 2 + F 2 + 2( F ) ( F )cos 60°

F Fnet m

30° 30°

F

O

m

Fig. 13.6

= 3F =

3 Gm2 a2

Ans.

212 — Mechanics - II V

Example 13.2 Four particles each of mass ‘m’ are placed at the four vertices of a square of side ‘a’. Find net force on any one of the particle. Solution. We are finding net force on the particle at D.

G ( m ) ( m ) Gm = 2 = F (say) a2 a G ( m ) ( m ) 1 Gm2 F = = = 2 a2 2 ( 2a ) 2

FDC = FDA = FDB

B

Now, resultant of FDA and FDC is 2 F in the direction of DB. ∴

F net = 2 F +

F  1 =  2+  F 2  2

1 Gm2  =  2+   2 a 2 V

C

a

2

(towards DB)

a

A

√2 a F 2 F Fig. 13.7

F D

Ans.

Example 13.3 Six particles each of mass ‘m’ are placed at six vertices A, B, C, D, E and F of a regular hexagon of side ‘a’. A seventh particle of mass ‘M’ is kept at centre ‘O’ of the hexagon. (a) Find net force on ‘M’. (b) Find net force on ‘M’ if particle at A is removed. (c) Find net force on ‘M’ if particles at A and C are removed. GmM Solution (a) F A = FB = FC = FD = FE = FF = = F (say) a2 E

D FD FE

F

FF

O

FC

C

a FA

FB

A

a B

Fig. 13.8

So, net force will be zero, as three pairs of equal and opposite forces are acting on ‘M’ at O. (b) When particle at A is removed then F A is removed. So, there is no force to cancel FD . GMm Ans. F net = FD = 2 (towards D) ∴ a (c) When particles at A and C are removed then, F A and FC are removed. FB and FE are still cancelled. So, net force is the resultant of two forces FD and FF of equal magnitudes acting at 120°. So, the resultant will pass through the centre or towards E. Magnitude of this resultant is F net = F 2 + F 2 + 2 ( F ) ( F )cos 120° GMm = F = 2 (towards E) a

Ans.

Chapter 13 V

Gravitation — 213

Example 13.4 Five particles each of mass ‘m’ are kept at five vertices of a regular pentagon. A sixth particle of mass ‘M’ is kept at centre of the pentagon ‘O’. Distance between ‘M’ and ‘m’ is ‘a’. Find (a) net force on ‘M’ (b) magnitude of net force on ‘M’ if any one particle is removed from one of the vertices. GMm Solution (a) F A = FB = FC = FD = FE = = F (say) D a2 360° FD Angle between two successive force vectors is θ = = 72°. FE O FC 5 E a When these five force vectors are added as per polygon law of vector FB FA addition we get another closed regular polygon as shown below. FC

A

C

B

Fig. 13.9 FD

FB 72° FE

FA

Fig. 13.10

Therefore, net resultant force on ‘M’ is zero. Note This result is true ( Fnet = 0 ) for any number of particles, provided masses at vertices are equal and polygon is regular.

(b) When particle at A is removed, then F A will be removed. So, magnitude of net force will be F as shown below: FD

FC

FE

FB Fnet

Fig. 13.11

∴ V

F net = F =

GMm

Ans.

a2

Example 13.5 A mass m is at a distance a from one end of a uniform rod of length l and mass M. Find the gravitational force on the mass due to the rod. m a

l

Fig. 13.12

214 — Mechanics - II

Solution

∴ V

dF =

G ( dM ) m x

F =∫

2

x = (a + l )

x=a

=

M  G  ⋅ dx m  l 

dM dx

m

2

x GMm dF = a (l + a )

x Fig. 13.13

Ans.

Example 13.6 A uniform ring of mass m is lying at a distance 3 a from the centre of a sphere of mass M just over the sphere (where a is the radius of the ring as well as that of the sphere). Find the magnitude of gravitational force between them. a m √ 3a

a

M

Fig. 13.14

Solution

Net force on ring = ∫ =∫ =

as

whole ring

a

dF sin 60° dF

GM ( dm ) whole ring

( 2a ) 2

3a

dm 60° dF 2a

3 2

3 GMm 8a 2

M,a

∫whole ring dm = m

INTRODUCTORY EXERCISE

Fig. 13.15

13.1

1. Three uniform spheres each having a mass M and radius a are kept in such a way that each touches the other two. Find the magnitude of the gravitational force on any one of the spheres due to the other two.

2. Four particles having masses m, 2m, 3m and 4m are placed at the four corners of a square of edge a. Find the gravitational force acting on a particle of mass m placed at the centre.

3. Two particles of masses 1.0 kg and 2.0 kg are placed at a separation of 50 cm. Assuming that the only forces acting on the particles are their mutual gravitation, find the initial accelerations of the two particles.

4. Three particles A, B and C, each of mass m, are placed in a line with AB = BC = d . Find the gravitational force on a fourth particle P of same mass, placed at a distance d from the particle B on the perpendicular bisector of the line AC.

5. Spheres of the same material and same radius r are touching each other. Show that gravitational force between them is directly proportional to r 4.

Chapter 13

Gravitation — 215

13.3 Acceleration due to Gravity When a body is dropped from a certain height above the ground it begins to fall towards the earth under gravity. The acceleration produced in the body due to gravity is called the acceleration due to gravity. It is denoted by g . Its value close to the earth’s surface is 9.8 m/s 2 . Suppose that the mass of the earth is M, its radius is R, then the force of attraction acting on a body of mass m close to the surface of earth is GMm F= R2 According to Newton’s second law, the acceleration due to gravity F GM g= = 2 m R This expression is independent of m. If two bodies of different masses are allowed to fall freely, they will have the same acceleration, i.e. if they are allowed to fall from the same height, they will reach the earth simultaneously.

Variation in the Value of g The value of g varies from place to place on the surface of earth. It also varies as we go above or below the surface of earth. Thus, value of g depends on following factors :

Shape of the Earth The earth is not a perfect sphere. It is somewhat flat at the two poles. The equatorial radius is approximately 21 km more than the polar radius and since GM 1 g = 2 or g ∝ 2 R R The value of g is minimum at the equator and maximum at the poles. Height above the Surface of the Earth The force of gravity on an object of mass m at a height h above the surface of earth is, GMm F= ( R + h) 2 ∴ Acceleration due to gravity at this height will be, F GM g′ = = m ( R + h) 2 This can also be written as,

g′ =

or

g′ =

g h  1 +   R

2

h

R M

GM h  R 2 1 +   R

m

Fig. 13.16 2

as

GM R2

=g

216 — Mechanics - II Thus, g′ < g i.e. the value of acceleration due to gravity g goes on decreasing as we go above the surface of earth. Further, h  g ′ = g 1 +   R

−2

 2h  g ′ ≈ g 1 −   R

or

if

h 0, the spring force F = kx acts on the two masses in the directions shown in above figure. Thus, we can write d 2x1 …(ii) m1 = − kx dt 2 d 2x2 …(iii) = kx m2 dt 2 Multiplying Eq. (ii) by m2 and Eq. (iii) by m1 and subtracting the latter from the former, we have d 2x1 d 2x2 m1m2 − m1m2 = − (m2 + m1 ) kx 2 dt dt 2 d 2(x1 − x2) or m1m2 = − kx (m1 + m2) dt 2  m1m2  d 2 …(iv) or (x1 − x2) = − kx    m1 + m2 dt 2 Differentiating Eq. (i), twice with respect to time, we have d 2x d 2 = (x1 − x2) dt 2 dt 2

Chapter 14

Simple Harmonic Motion — 341

m1m2 = µ = reduced mass of the two blocks m1 + m2

Also,

Substituting these values in Eq. (iv), we have d 2x µ 2 = − kx dt d 2x d 2x1 d 2x2 = − = Relative acceleration) dt 2 dt 2 dt 2 This, is the standard differential equation of SHM. Time period of which is µa r = − kx

or

(Here, a r =

 x T = 2π   or T = 2π a r V

Example 30

µ k

Two particles move parallel to x-axis about the origin with the

A 3 from the origin on opposite sides but their velocities are found to be in the same direction. What is the phase difference between the two ? same amplitude and frequency. At a certain instant they are found at distance

Solution

Let equations of two SHM be

Give that and Which gives

x1 = A sin ωt x2 = A sin (ωt + φ ) A = A sin ωt 3 A − = A sin (ωt + φ ) 3 1 sin ωt = 3 1 sin (ωt + φ ) = − 3

K (i) K (ii)

K (iii) K (iv)

From Eq. (iv), sin ωt cos φ + cos ωt sin φ = − ⇒

1 3

1 1 1 cos φ + 1 − sin φ = − 3 9 3

Solving this equation, we get or ⇒

cos φ = − 1, φ=π

7 9 or

 7 cos −1    9

Differentiating Eqs. (i) and (ii), we obtain v1 = Aω cos ωt and v2 = Aω cos (ωt + φ ) If we put φ = π , we find v1 and v2 are of opposite signs. Hence, φ = π is not acceptable.  7 φ = cos −1   ∴  9

Ans.

342 — Mechanics - II V

Example 31 For the arrangement shown in figure, the spring is initially compressed by 3 cm. When the spring is released the block collides with the wall and rebounds to compress the spring again. m = 1 kg

k = 104 N/m

1 cm

(a) If the coefficient of restitution is

1 2

, find the maximum compression in the spring after

collision. (b) If the time starts at the instant when spring is released, find the minimum time after which the block becomes stationary. Solution

or Here,

(a) Velocity of the block just before collision, 1 1 1 mv02 + kx2 = kx02 2 2 2 k 2 v0 = (x0 − x2) m x0 = 0.03 m,

∴ After collision,

x = 0.01 m, k = 104 N/m, m = 1 kg v0 = 2 2 m/s 1 v = ev0 = 2 2 = 2 m/s 2

Maximum compression in the spring is 1 2 1 2 1 kxm = kx + mv2 2 2 2 m 2 or xm = x2 + v k = (0.01)2 +

(1) (2)2 m = 2.23 cm 104

Ans.

(b) In the case of spring-mass system, since the time period is independent of the amplitude of oscillation. Time = tAB + tBC + tCB + tBD =

T0  T0   1  T   1  T0 +   sin −1   +  0  sin −1   +  2π   3  2π   2.23 4 4 2.23 cm

A

D

3 cm

T0 = 2π

Here,

C B

1 cm

m k

Substituting the values, we get Total time =

m k

 −1 π + sin 

 1 −1  1     + sin    3  2.23 

Ans.

Chapter 14 V

Simple Harmonic Motion — 343

Example 32 A long uniform rod of length L and mass M is free to rotate in a horizontal plane about a vertical axis through its one end ‘ O ’. A spring of force constant k is connected between one end of the rod and PQ. When the rod is in equilibrium it is parallel to PQ.

L A

θ O

(a) What is the period of small oscillation that result when the rod P Q is rotated slightly and released? (b) What will be the maximum speed of the displaced end of the rod, if the amplitude of motion is θ 0? Solution

(a) Restoring torque about ‘O’ due to elastic force of the spring τ = − FL = − kyL τ = − kL2θ 1 d 2θ τ = Iα = ML2 2 3 dt 2 1 2 2 d θ ML = − kL θ 3 dt 2 d 2θ 3k =− θ M dt 2 ω=

⇒

(F = ky) (as y = Lθ)

3k M

T = 2π

M 3k

(b) In angular SHM, maximum angular velocity  dθ  = θ0 ω = θ0    dt  max

Ans.

3k M

 dθ  v=r    dt  So ,

V

 dθ  vmax = L   = Lθ 0  dt  max

3k M

Ans.

Example 33 A block with a mass of 2 kg hangs without vibrating at the end of a spring of spring constant 500 N/m, which is attached to the ceiling of an g elevator. The elevator is moving upwards with an acceleration . At time t = 0, the 3 acceleration suddenly ceases. (a) What is the angular frequency of oscillation of the block after the acceleration ceases? (b) By what amount is the spring stretched during the time when the elevator is accelerating? (c) What is the amplitude of oscillation and initial phase angle observed by a rider in the elevator in the equation, x = A sin(ωt + φ ) ? Take the upward direction to be positive. Take g = 10.0 m/s2.

344 — Mechanics - II Solution

(a) Angular frequency ω=

k m

or ω =

500 2

or ω = 15.81 rad/s (b) Equation of motion of the block (while elevator is accelerating) is,

Ans.

kx a=

g 3

mg

kx − mg = ma = m ∴

g 3

4mg 3k (4)(2)(10) = = 0.053 m (3)(500)

x=

or x = 5.3 cm (c) (i) In equilibrium when the elevator has zero acceleration, the equation of motion is,

Ans.

kx0

mg

or

kx0 = mg mg (2)(10) x0 = = k 500

= 0.04 m = 4 cm A = x − x0 = 5.3 – 4.0 = 1.3 cm (ii) At time t = 0, block is at x = − A. Therefore, substituting x = − A and t = 0 in equation, ∴ Amplitude

Ans.

x = +A

Mean position

x = –A

We get initial phase

x = A sin (ωt + φ ) 3π φ= 2

Ans.

Chapter 14 V

Simple Harmonic Motion — 345

Example 34 Calculate the angular frequency of the system shown in figure. Friction is absent everywhere and the threads, spring and pulleys are massless. Given that m A = mB = m.

A

k

B

θ

Solution

Let x0 be the extension in the spring in equilibrium. Then, equilibrium of A and B

give, T = kx0 + mg sin θ 2T = mg

and

…(i) …(ii)

Here, T is the tension in the string. Now, suppose A is further displaced by a distance x from its x mean position and v be its speed at this moment. Then, B lowers by and speed of B at this 2 v instant will be . Total energy of the system in this position will be, 2 2

E=

1 1 1  v k (x + x0 )2 + mAv2 + mB   + mA ghA − mB g hB  2 2 2 2

1 1 1 x k (x + x0 )2 + mv2 + mv2 + mgx sin θ − mg 2 2 8 2 1 5 x 2 2 or E = k(x + x0 ) + mv + mgx sin θ − mg 2 8 2 dE Since, E is constant, =0 dt dx 5  dv  dx mg  dx or 0 = k(x + x0 ) + mv   + mg (sin θ )   −    dt   dt  dt 4 2  dt  or

Substituting, and We get, Since,

E=

dx dv =v ⇒ =a dt dt mg kx0 + mg sin θ = 2 5 m a = − kx 4

[From Eqs. (i) and (ii)]

a∝−x

Motion is simple harmonic, time period of which is, x T = 2π   a = 2π ∴

ω=

5m 4k

2π 4k = T 5m

Ans.

346 — Mechanics - II V

Example 35 A solid sphere (radius = R) rolls without slipping in a cylindrical through (radius = 5R). Find the time period of small oscillations. 5R

R

Solution

For pure rolling to take place,

v = Rω ω′ = angular velocity of COM of sphere C about O v Rω ω = = = 4R 4R 4 dω′ 1 dω ∴ = dt 4 dt ω′ O θ

ω C v

or

where,

as, ∴

α 4 a α = for pure rolling R g sin θ a= I 1+ mR2 5 g sin θ = 7 2 I = mR2 5 5 g sin θ α′ = 28R α′ =

For small θ,sin θ ≈ θ , being restoring in nature, 5g α′ = − θ 28R ∴

θ T = 2π   α ′ = 2π

28R 5g

Ans.

Chapter 14 V

Simple Harmonic Motion — 347

Example 36 Consider the earth as a uniform sphere of mass M and radius R. Imagine a straight smooth tunnel made through the earth which connects any two points on its surface. Show that the motion of a particle of mass m along this tunnel under the action of gravitation would be simple harmonic. Hence, determine the time that a particle would take to go from one end to the other through the tunnel. Solution

Suppose at some instant, the particle is at radial distance r from centre of earth O. Since, the particle is constrained to move along the tunnel, we define its position as distance x from C. Hence, equation of motion of the particle is, ma x = Fx The gravitational force on mass m at distance r is, F =

GMmr R3

(towards O)

Therefore, Fx = − F sin θ =−

GMmr  x   R3  r 

r O

GMm =− ⋅x R3 Since, Fx ∝ − x, motion is simple harmonic in nature. Further, GMm ma x = − ⋅x R3 GM or ax = − 3 ⋅ x R

x

θ

x-axis

C

∴ Time period of oscillation is,  x T = 2π   a x = 2π

R3 GM

The time taken by particle to go from one end to the other is ∴

t=

T ⋅ 2

T 2

=π

R3 GM

Ans.

Exercises LEVEL 1 Assertion and Reason Directions : Choose the correct option. (a) If both Assertion and Reason are true and the Reason is correct explanation of the Assertion. (b) If both Assertion and Reason are true but Reason is not the correct explanation of Assertion. (c) If Assertion is true, but the Reason is false. (d) If Assertion is false but the Reason is true. 1. Assertion : In x = A cos ω t, x is the displacement measured from extreme position.

Reason : In the above equation x = A at time t = 0. 2. Assertion : A particle is under SHM along the x-axis. Its mean position is x = 2, amplitude is A = 2 and angular frequency ω. At t = 0, particle is at origin, then x-co-ordinate versus time equation of the particle will be x = − 2 cos ω t + 2. Reason : At t = 0, particle is at rest.

3. Assertion : A spring block system is kept over a smooth surface as shown in figure. If a constant horizontal force F is applied on the block it will start oscillating simple harmonically. k

Reason : Time period of oscillation is less than 2π

F m

m . k

4. Assertion : Time taken by a particle in SHM to move from x = A to x = taken by the particle to move from x =

3A is same as the time 2

3A A to x = . 2 2

Reason : Corresponding angles rotated in the reference circle are same in the given time intervals.

5. Assertion : Path of a particle in SHM is always a straight line. Reason : All straight line motions are not simple harmonic.

6. Assertion : In spring block system if length of spring and mass of block both are halved, then angular frequency of oscillations will remain unchanged. k Reason : Angular frequency is given by ω = m

7. Assertion : All small oscillations are simple harmonic in nature. Reason : Oscillations of spring block system are always simple harmonic whether amplitude is small or large.

Chapter 14

Simple Harmonic Motion — 349

8. Assertion : In x = A cos ω t, the dot product of acceleration and velocity is positive for time interval 0 < t
T2 > T3

(b) T3 > T2 > T1

(B)

(C)

(c) T2 > T1 > T3

(d) T2 > T3 > T1

360 — Mechanics - II 12. Three arrangements are shown in figure. m 2

m k

k

k m 2

m (B)

(A)

(C)

(a) A spring of mass m and stiffness k (b) A block of mass m attached to massless spring of stiffness k m m (c) A block of mass attached to a spring of mass and stiffness k 2 2

If T1, T2 and T3 represent the period of oscillation in the three cases respectively, then identify the correct relation. (a) T1 < T2 < T3 (c) T1 > T3 > T2

(b) T1 < T3 < T2 (d) T3 < T1 < T2

13. A block of mass M is kept on a smooth surface and touches the two springs as shown in the figure but not attached to the springs. Initially springs are in their natural length. Now, the block is shifted ( l0 / 2) from the given position in such a way that it compresses a spring and released. The time-period of oscillation of mass will be k

M

l0

π M 2 k 3π M (c) 2 k (a)

4k 2l0

M 5k M (d) π 2k (b) 2π

14. A particle moving on x-axis has potential energy U = 2 − 20x + 5x 2 joule along x-axis. The particle is released at x = − 3. The maximum value of x will be (x is in metre) (a) 5 m (c) 7 m

(b) 3 m (d) 8 m

15. A block of mass m, when attached to a uniform ideal spring with force constant k and free length L executes SHM. The spring is then cut in two pieces, one with free length n L and other with free length (1 − n ) L. The block is also divided in the same fraction. The smaller part of the block attached to longer part of the spring executes SHM with frequency f1. The bigger part of the block attached to smaller part of the spring executes SHM with frequency f2. The ratio f1 / f2 is (a) 1 (c)

1+ n n

n 1−n n (d) 1+ n

(b)

Simple Harmonic Motion — 361

Chapter 14

16. A body performs simple harmonic oscillations along the straight line ABCDE with C as the midpoint of AE. Its kinetic energies at B and D are each one fourth of its maximum value. If AE = 2R, the distance between B and D is A

(a)

3 R 2

(b)

B

C

R 2

D

(c)

E

(d)

3R

2R

17. In the given figure, two elastic rods P and Q are rigidly joined to end supports. A small mass m is moving with velocity v between the rods. All collisions are assumed to be elastic and the surface is given to be smooth. The time period of small mass m will be (A = area of cross section, Y = Young’s modulus, L = length of each rod) A/2, 2Y, L P

A, Y, L m

v

Q

L

2L mL (a) + 2π v AY 2L mL (c) +π v AY

2L + 2π v 2L (d) v (b)

2mL AY

18. A particle executes SHM of period 1.2 s and amplitude 8 cm. Find the time it takes to travel 3 cm from the positive extremity of its oscillation. [cos− 1( 5/ 8) = 0.9 rad]

(a) 0.28 s

(b) 0.32 s

(c) 0.17 s

(d) 0.42 s

19. A wire frame in the shape of an equilateral triangle is hinged at one vertex so that it can swing freely in a vertical plane, with the plane of the triangle always remaining vertical. The side of the frame is 1/ 3 m. The time period in seconds of small oscillations of the frame will be ( g = 10 m / s2 ) (a) π / 2 (c) π / 6

(b) π / 3 (d) π / 5

20. A particle moves along the x-axis according to x = A [1 + sin ω t ]. What distance does is travel in time interval from t = 0 to t = 2.5π/ ω ?

(a) 4 A

(b) 6 A

(c) 5 A

(d) 3 A

21. A small bob attached to a light inextensible thread of length l has a periodic time T when allowed to vibrate as a simple pendulum. The thread is now suspended from a fixed end O of a vertical rigid rod of length 3l/ 4. If now the pendulum performs periodic oscillations in this arrangement, the periodic time will be O 3l/4

l

A

(a) 3T /4

(b) 4T /5

(c) 2T /3

(d) 5 T /6

362 — Mechanics - II 22. A stone is swinging in a horizontal circle of diameter 0.8 m at 30 rev/min. A distant light causes a shadow of the stone on a nearly wall. The amplitude and period of the SHM for the shadow of the stone are (b) 0.2 m, 2s (d) 0.8 m, 2s

(a) 0.4 m, 4s (c) 0.4 m, 2s

23. Part of SHM is graphed in the figure. Here, y is displacement from mean position. The correct equation describing the SHM is y(cm) 2 O

0.6π

t(s)

0.3π

–2

π  10 (b) y = 2 sin  t−  3 2 π  (d) y = 2 cos  0.6 t +   2

(a) y = 4 cos (0.6 t )  π 10  (c) y = 2 sin  − t 2 3 

24. A particle performs SHM with a period T and amplitude a. The mean velocity of particle over the time interval during which it travels a distance a/ 2 from the extreme position is (a) 6 a /T (c) 3 a /T

(b) 2 a /T (d) a /2T

25. A man of mass 60 kg is standing on a platform executing SHM in the vertical plane. The displacement from the mean position varies as y = 0.5 sin ( 2πft ). The value of f, for which the man will feel weightlessness at the highest point, is ( y in metre)

(a) g/4π

(b) 4πg

(c)

2g 2π

(d) 2π 2 g

26. A particle performs SHM on a straight line with time period T and amplitude A. The average speed of the particle between two successive instants, when potential energy and kinetic energy become same is A T 2A (c) T

4 2A T 2 2A (d) T (b)

(a)

27. The time taken by a particle performing SHM to pass from point A to B where its velocities are same is 2 s. After another 2 s it returns to B. The ratio of distance OB to its amplitude (where O is the mean position) is (b) ( 2 − 1) : 1 (d) 1 : 2 2

(a) 1 : 2 (c) 1 : 2

28. A particle is executing SHM according to the equation x = A cos ω t. Average speed of the particle during the interval 0 ≤ t ≤ 3 Aω 2 3 Aω (c) π (a)

π is 6ω

3 Aω 4 3 Aω (d) (2 − 3 ) π (b)

Chapter 14

Simple Harmonic Motion — 363

More than One Correct Options 1. A simple pendulum with a bob of mass m is suspended from the roof of a car moving with horizontal acceleration a

(a) The string makes an angle of tan −1 (a / g ) with the vertical  a (b) The string makes an angle of sin −1   with the vertical  g (c) The tension in the string is m a 2 + g 2 g2 − a 2

(d) The tension in the string is m

2. A particle starts from a point P at a distance of A/ 2 from the mean position O and travels towards left as shown in the figure. If the time period of SHM, executed about O is T and amplitude A then the equation of the motion of particle is A/2 O

P A

π  2π (a) x = A sin  t+  T 6 π  2π (c) x = A cos  t+  T 6

5π   2π (b) x = A sin  t+  T 6 π  2π (d) x = A cos  t+  T 3

3. A spring has natural length 40 cm and spring constant 500 N/m. A block of mass 1 kg is attached at one end of the spring and other end of the spring is attached to a ceiling. The block is released from the position, where the spring has length 45 cm (a) (b) (c) (d)

the block will perform SHM of amplitude 5 cm the block will have maximum velocity 30 5 cm/s the block will have maximum acceleration 15 m/s 2 the minimum elastic potential energy of the spring will be zero

4. The displacement-time graph of a particle executing SHM is shown in figure. Which of the following statements is/are true?

y O

T/4

T/2

3T/4 T

t

(a) The velocity is maximum at t = T /2 (b) The acceleration is maximum at t = T (c) The force is zero at t = 3T /4 (d) The kinetic energy equals the total oscillation energy at t = T /2

5. For a particle executing SHM, x = displacement from mean position, v = velocity and a = acceleration at any instant, then

(a) v -x graph is a circle (c) a-x graph is a straight line

(b) v -x graph is an ellipse (d) a-x graph is a circle

364 — Mechanics - II 6. The acceleration of a particle is a = − 100x + 50. It is released from x = 2. Here, a and x are in SI units (a) (b) (c) (d)

the particle will perform SHM of amplitude 2 m the particle will perform SHM of amplitude 1.5 m the particle will perform SHM of time period 0.63 s the particle will have a maximum velocity of 15 m/s

7. Two particles are performing SHM in same phase. It means that (a) (b) (c) (d)

the two particles must have same distance from the mean position simultaneously two particles may have same distance from the mean position simultaneously the two particles must have maximum speed simultaneously the two particles may have maximum speed simultaneously

8. A particle moves along y-axis according to the equation y (in cm) = 3 sin 100πt + 8 sin 2 50πt − 6 (a) the particle performs SHM (b) the amplitude of the particle’s oscillation is 5 cm (c) the mean position of the particle is at y = − 2 cm (d) the particle does not perform SHM

Comprehension Based Questions Passage (Q Nos. 1 to 2) A 2 kg block hangs without vibrating at the bottom end of a spring with a force constant of 400 N/m. The top end of the spring is attached to the ceiling of an elevator car. The car is rising with an upward acceleration of 5 m/ s2 when the acceleration suddenly ceases at time t = 0 and the car moves upward with constant speed ( g = 10 m/ s2 )

1. What is the angular frequency of oscillation of the block after the acceleration ceases? (a) 10 2 rad/s (c) 20 2 rad/s

(b) 20 rad/s (d) 32 rad/s

2. The amplitude of the oscillation is (a) 7.5 cm (c) 2.5 cm

(b) 5 cm (d) 1 cm

Match the Columns 1. For the x -t equation of a particle in SHM along x-axis, match the following two columns. x = 2 + 2 cos ω t Column I

Column II

(a) Mean position

(p) x = 0

(b) Extreme position

(q) x = 2

(c) Maximum potential energy at

(r) x = 4

(d) Zero potential energy at

(s) Can’t tell

Chapter 14

Simple Harmonic Motion — 365

2. Potential energy of a particle at mean position is 4 J and at extreme position is 20 J. Given that amplitude of oscillation is A. Match the following two columns. Column I

Column II

(a) Potential energy at x = A 4 (c) Kinetic energy at x = 0 A (d) Kinetic energy at x = 2

A 2

(p) 18 J

(b) Kinetic energy at x =

(q) 16 J (r) 8 J (s) None

3. Acceleration-time graph of a particle in SHM is as shown in figure. Match the following two columns. a

t1

Column I (a) (b) (c) (d)

t

t2

Column II

Displacement of particle at t1 Displacement of particle at t2 Velocity of particle at t1 Velocity of particle at t2

(p) (q) (r) (s)

zero positive negative maximum

4. Mass of a particle is 2 kg. Its displacement-time equation in SHM is x = 2 sin (4π t )

(SI Units)

Match the following two columns for 1 second time interval. Column I (a) (b) (c) (d)

Column II

Speed becomes 30 m/s Velocity becomes + 10 m/s Kinetic energy becomes 400 J Acceleration becomes− 100 m/s 2

(p) (q) (r) (s)

two times four times one time None

5. x -t equation of a particle in SHM is, x = 4 + 6 sin πt. Match the following tables corresponding to time taken in moving from

Column I (a) x = 10 m to x = 4 m (b) x = 10 m to x = 7 m x = 7 m to x = 1 m (d) x = 10 m to x = − 2 m

(c)

Column II 1 second 3 (q) 1 second 2 (r) 1 second (p)

(s) None

366 — Mechanics - II Subjective Questions 1. A 1 kg block is executing simple harmonic motion of amplitude 0.1 m on a smooth horizontal surface under the restoring force of a spring of spring constant 100 N/m. A block of mass 3 kg is gently placed on it at the instant it passes through the mean position. Assuming that the two blocks move together. Find the frequency and the amplitude of the motion.

2. Two particles are in SHM along same line. Time period of each is T and amplitude is A. After how much time will they collide if at time t = 0. (a) first particle is at x1 = + towards positive x-axis and second particle is at x2 = −

A and moving 2

A

and moving towards negative x-axis, 2 (b) rest information are same as mentioned in part (a) except that particle first is also moving towards negative x-axis.

3. A particle that hangs from a spring oscillates with an angular frequency of 2 rad/s. The spring particle system is suspended from the ceiling of an elevator car and hangs motionless (relative to the elevator car) as the car descends at a constant speed of 1.5 m/s. The car then stops suddenly. (a) With what amplitude does the particle oscillate ? (b) What is the equation of motion for the particle ? (Choose upward as the positive direction)

4. A 2 kg mass is attached to a spring of force constant 600 N/m and rests on a smooth horizontal surface. A second mass of 1 kg slides along the surface toward the first at 6 m/s. (a) Find the amplitude of oscillation if the masses make a perfectly inelastic collision and remain together on the spring. What is the period of oscillation ? (b) Find the amplitude and period of oscillation if the collision is perfectly elastic. (c) For each case, write down the position x as a function of time t for the mass attached to the spring, assuming that the collision occurs at time t = 0. What is the impulse given to the 2 kg mass in each case?

5. A block of mass 4 kg hangs from a spring of force constant k = 400 N/m. The block is pulled down 15 cm below equilibrium and released. How long does it take the block to go from 12 cm below equilibrium (on the way up) to 9 cm above equilibrium?

6. A plank with a body of mass m placed on it starts moving straight up according to the law y = a(1 − cos ωt ), where y is the displacement from the initial position, ω = 11 rad/s. Find

(a) The time dependence of the force that the body exerts on the plank. (b) The minimum amplitude of oscillation of the plank at which the body starts falling behind the plank.

7. A particle of mass m free to move in the x-y plane is subjected to a force whose components are Fx = − kx and Fy = − ky, where k is a constant. The particle is released when t = 0 at the point (2, 3). Prove that the subsequent motion is simple harmonic along the straight line 2 y − 3x = 0.

8. Determine the natural frequency of vibration of the 100 N disk. Assume the disk does not slip on the inclined surface. /m

0N

20

30°

Chapter 14

Simple Harmonic Motion — 367

9. The disk has a weight of 100 N and rolls without slipping on the horizontal surface as it oscillates about its equilibrium position. If the disk is displaced, by rolling it counterclockwise 0.4 rad, determine the equation which describes its oscillatory motion when it is released. k = 1000 N/m

10. A solid uniform cylinder of mass m performs small oscillations due to the action of two springs of stiffness k each (figure). Find the period of these oscillations in the absence of sliding.

m

R

11. A block of mass m is attached to one end of a light inextensible string passing over a smooth light pulley B and under another smooth light pulley A as shown in the figure. The other end of the string is fixed to a ceiling. A and B are held by springs of spring constants k1 and k2. Find angular frequency of small oscillations of the system. k2 B A m k1

12. In the shown arrangement, both the springs are in their natural lengths. The coefficient of friction between m2 and m1 is µ.There is no friction between m1 and the surface. If the blocks are displaced slightly, they together perform simple harmonic motion. Obtain k2 m2 k1 m1

(a) Frequency of such oscillations. (b) The condition if the frictional force on block m2 is to act in the direction of its displacement from mean position. (c) If the condition obtained in (b) is met, what can be maximum amplitude of their oscillations?

368 — Mechanics - II 13. Two blocks A and B of masses m1 = 3 kg and m2 = 6 kg respectively

k

v0 A B are connected with each other by a spring of force constant k = 200 C N/m as shown in figure. Blocks are pulled away from each other by x0 = 3 cm and then released. When spring is in its natural length and blocks are moving towards each other, another block C of mass m = 3 kg moving with velocity v0 = 0.4 m/ s (towards right) collides with A and gets stuck to it. Neglecting friction, calculate

(a) velocities v1 and v2 of the blocks A and B respectively just before collision and their angular frequency. (b) velocity of centre of mass of the system, after collision, (c) amplitude of oscillations of combined body, (d) loss of energy during collision.

14. A rod of length l and mass m, pivoted at one end, is held by a spring at its mid-point and a spring at far end. The springs have spring constant k. Find the frequency of small oscillations about the equilibrium position.

15. In the arrangement shown in figure, pulleys are light and springs are ideal. k1 , k2 , k3 and k4 are force constants of the springs. Calculate period of small vertical oscillations of block of mass m. k2

k4

m k3

k1

16. A light pulley is suspended at the lower end of a spring of constant k1 , as shown in figure. An inextensible string passes over the pulley. At one end of string a mass m is suspended, the other end of the string is attached to another spring of constant k2. The other ends of both the springs are attached to rigid supports, as shown. Neglecting masses of springs and any friction, find the time period of small oscillations of mass m about equilibrium position.

k1

k2

m

Simple Harmonic Motion — 369

Chapter 14

17. Figure shows a solid uniform cylinder of radius R and mass M, which is free to rotate about a fixed horizontal axis O and passes through centre of the cylinder. One end of an ideal spring of force constant k is fixed and the other end is hinged to the cylinder at A. Distance OA is equal to R . An inextensible thread is wrapped round the cylinder and passes over a smooth, small 2 pulley. A block of equal mass M and having cross sectional area A is suspended from free end of the thread. The block is partially immersed in a non-viscous liquid of density ρ. k A O M,R M ρ

If in equilibrium, spring is horizontal and line OA is vertical, calculate frequency of small oscillations of the system.

18. Find the natural frequency of the system shown in figure. The pulleys are smooth and massless.

k

k

M

Answers Introductory Exercise 14.1 1 15 3. (a) 15.0 cm , 16 16 4. 0.101 m/s, 1.264 m/s 2 , 0.632 N

1. π

2.

(b) 0.726 s

(c) 1.38 Hz (d) 1.69 J

(e) 1.30 m/s

5. No

Introductory Exercise 14.2 2. ± 0.58 m/s, − 0.45 ms 2 , ± 0.60 m/s, zero

1. (a) 1.39 J (b) 1.1 s

3.

π 6

(d) Zero (e) 0.197 m/s 2

4. (a) 0.08 m (b) 1.57 rad/s (c) 1.97 N/m π π π 5. (a) sec (b) sec (c) sec 120 30 30

6. See the hints.

Introductory Exercise 14.3 π 1.   sec  2

2. 3.2 kg

3. 1 sec

16 9

5.

4. p

6. 11 %

7.

3 2π

Introductory Exercise 14.4 1. (a) 7.0 cm (b) 6.1 cm (c) 5.0 cm (d) 1.0 cm 2. (a) 2.6 unit (b) 1917unit (c) 6.0 × 10 5 unit

3. 2 A

4.

2π 3

Exercises LEVEL 1 Assertion and Reason 1. (d)

2. (b)

3. (c)

4. (a)

5. (d)

6. (d)

7. (d)

8. (a)

9. (a)

10. (d)

4. (c) 14. (b) 24. (b)

5. (d) 15. (b)

6. (b) 16. (a)

7. (b) 17. (a)

8. (d) 18. (a)

9. (c) 19. (b)

10. (a) 20. (d)

4. The clock will lose 10.37s 5. (a) 2.8 N/m

Objective Questions 1. (a) 11. (c) 21. (c)

2. (a) 12. (d) 22. (a)

3. (d) 13 (c) 23. (a)

Subjective Questions 1. T = 2 π

m k

2. 0.314 s

3. 0.78 s

7. T = 2 π

m k

 10  8.  T  9 

9. 6.0 N/m

10. 10.0 cm,

π rad 6

(b) 0.84 s

π 11. y = (0.1 m) sin  (4s−1)t +   4 

12. (a) 1 rad/s (b) Umean = 10 J, Kmean = 16 J, Uextreme = 26 J, K extreme = 0 (c) 4 m (d) x = 6 m and x = − 2 m 13. 4 s 15. 2 π

14. 1.8 s l  2  v 2  g +    r  

2 1/2

  

 v2  and inclined to the vertical at an angle θ = tan−1   away from the centre  rg 

16. 1.4 × 10 −3 s 17. (a) 0.28 s (b) 0.4%

18.

g 2r

19.(a)

1 2π

k ma (b) m k

π 20. θ =  rad cos [(40 πs −1)t ]  10 

Simple Harmonic Motion — 371

Chapter 14 22. 2 π

21. 7.2 m

24. (a) A2 = A1

l

M + m M , T = 2π M+m k

3R r2 + 2 2R 26. T = 2 π g m1 32. A = d m1 + m2 34. (a) 10 cm

l ( g + a)

23. (a) 2 π

qE g− m

(b) 2.5 J

F M , 2π k k

π sec 5

(c)

28. 2 2 s

(b)

(d) 20 cm

F2 2k

(c)

25.

k

29. (a) equal

l ( g 2 + a2 )1/2 mv0

k(M + m)

(b) equal

F2 2k

(e) 4.5 J

 2x 2  38. Parabola, y = A  1 − 2  A  

(c) Infinite (d) 2 π

M + m

(b) A2 = A1, T = 2 π

27. 1.4 × 10 5 g-cm2 33. (a)

l ( g − a)

(b) 2 π

35. 0.816

(f) 0.5 J

39. (a) –2.41 cm

37. 0.38 Hz

(b) 0.27 cm

LEVEL 2 Single Correct Option 1. (d) 11. (c) 21. (a)

2. (a) 12. (b) 22. (c)

3. (b) 13. (c) 23. (b)

4. (a) 14. (c) 24. (c)

5. (b) 15. (a) 25. (c)

6. (a) 16. (c) 26. (b)

7. (c) 17. (a) 27. (a)

8. (a) 18. (c) 28. (d)

9. (a) 19. (d)

10. (b) 20. (c)

More than One Correct Options 1. (a,c)

2. (b,d)

3. (b,c,d)

4. (b, c)

5. (b,c)

6. (b,c, d)

7. (b,c)

8. (a, b, c)

Comprehension Based Questions 1. (a)

2. (c)

Match the Columns 1. 2. 3. 4. 5.

(a) → (a) → (a) → (a) → (a) →

(b) → (b) → (b) → (b) → (b) →

q r r s q

(c) → (c) → (c) → (c) → (c) →

p,r s p q p

p,r q r s p

(d) → (d) → (d) → (d) → (d) →

s s r,s q r

Subjective Questions 19 11 3. (a) A = 0.75 m (b) x = − 0.75 sin 2t T (b) T 48 48 4. (a) 14.1 cm, 0.44 s (b) 23 cm, 0.36 s (c) x = ± (14.1cm) sin (10 2 t ), x = ± (23 cm) sin (10 3 t), 4 N-s, 8 N-s

1. 0.8 Hz, 0.05 m

5.

π s = 0.157 s 20

10. T =

π 2

3m k

2. (a)

6. (a) N = m( g + aω 2 cos ωt ) (b) 8.1 cm 11.

k1 k2 4m (k1 + k2 )

13. (a) 0.2 m/s, 0.1 m/s, 10 rad/s 1 14. 2π 17. f =

15k 4m 1 2π

6M

18.

1 2π

1 π

2k M

k1 + k2 m1 + m2

(b) 0.1 m/s (towards right)

1 1 | 1 m + + +  k3 k4   k1 k2

15. T = 4 π

k + 4 Aρg

12. (a)

8. 0.56 Hz 9. θ = 0.4 cos (16.16 t ) (b)

k1 m < 1 k2 m2

(c) 4.8 cm

16. T = 2 π

(c)

mk 1 2 − m2 k1

(d) 0.03 J

m (4k2 + k1) k1 k2

µ (m1 + m2 )m2 g

15

Elasticity Chapter Contents 15.1

Introduction

15.2

Elasticity

15.3

Stress and Strain

15.4

Hooke’s Law and Modulus of Elasticity

15.5

The Stress-Strain Curve

15.6

Potential Energy Stored in a Stretched Wire

15.7

Thermal Stresses of Strain

374 — Mechanics - II

15.1 Introduction The properties of material under the action of external deforming forces are very essential, for an engineer, to enable him, in designing him all types of structures and machines. Whenever a load is attached to a thin hanging wire it elongates and the load moves downwards (sometimes through a negligible distance). The amount by which the wire elongates depends upon the amount of load and the nature of wire material. Cohesive force, between the molecules of the hanging wire offer resistance against the deformation, and the force of resistance increases with the deformation. The process of deformation stops when the force of resistance is equal to the external force (i.e. the load attached). Sometimes the force of resistance offered by the molecules is less than the external force. In such a case, the deformation continues until the wire breaks. Thus, we may conclude that if some external deforming force is applied to a body it has two effects on it, namely : (i) deformation of the body, (ii) internal resistance (restoring) forces are developed.

15.2 Elasticity As we have already discussed that whenever a single force (or a system of forces) acts on a body it undergoes some deformation and the molecules offer some resistance to the deformation. When the external force is removed, the force of resistance also vanishes and the body returns back to its original shape. But it is only possible if the deformation is within a certain limit. Such a limit is called elastic limit. This property of materials of returning back to their original position is called the elasticity. A body is said to be perfectly elastic if it returns back completely to its original shape and size after removing the external force. If a body remains in the deformed state and does not even partially regain its original shape after the removal of the deforming forces, it is called a perfectly inelastic or plastic body. Quite often, when the external forces are removed, the body partially regains the original shape. Such bodies are partially elastic. If the force acting on the body is increased and the deformation exceeds the elastic limit, the body loses to some extent, its property of elasticity. In this case, the body will not return to its original shape and size even after removal of the external force. Some deformation is left permanently.

15.3 Stress and Strain Stress When an external force is applied to a body then at each cross-section of the body an internal restoring force is developed which tends to restore the body to its original state. The internal restoring force per unit area of cross-section of the deformed body is called stress. It is usually denoted by σ (sigma). Restoring force Thus, Stress (σ ) = Area Note In equilibrium, when further deformation stops, the restoring force is equal to the external force (or the suspended load from a hanging wire). Therefore, the stress is also sometime called the external deforming force per unit area.

Chapter 15

Elasticity — 375

Strain When the size or shape of a body is changed under an external force, the body is said to be strained. The change occurred in the unit size of the body is called strain. Usually, it is denoted by ε. Thus, ∆x ε= x Here, ∆x is the change (may be in length, volume etc.) and x the original value of the quantity in which change has occurred. For example, when the length of a suspended wire increases under an applied load, the value of strain is, ∆l ε= l

15.4 Hooke's Law and the Modulus of Elasticity According to Hooke’s law, “For small deformation, the stress in a body is proportional to the corresponding strain.” i.e. stress ∝ strain or

stress = ( E ) (strain)

stress is a constant called the modulus of elasticity. Now, depending upon the nature of strain deforming force applied on the body, stress, strain and hence modulus of elasticity are classified in following three types:

Here, E =

Young’s Modulus of Elasticity (Y ) When a wire is acted upon by two equal and opposite forces in the direction of its length,  ∆l  the length of the body is changed. The change in length per unit length   is called the  l  longitudinal strain and the restoring force (which is equal to the applied force in equilibrium) per unit area of cross section of the wire is called the longitudinal stress. For small change in the length of the wire, the ratio of the longitudinal stress to the corresponding strain is called the Young’s modulus of elasticity (Y ) of the wire. Thus, Y =

F/A ∆ l/ l

or

Y =

l M ∆l

Fig. 15.1

Fl A∆ l

Let there be a wire of length l and radius r. Its one end is clamped to a rigid support and a mass M is attached at the other end. Then, F = Mg

and

Substituting in above equation, we have Y =

Mgl ( πr 2 ) ∆l

A = πr 2

376 — Mechanics - II Bulk Modulus of Elasticity (B) When a uniform pressure (normal force) is applied all over the surface of a body, the volume of the body changes. The change in volume per unit volume of the body is called the ‘volume strain’ and the normal force acting per unit area of the surface (pressure) is called the normal stress or volume stress. For small strains, the ratio of the volume stress to the volume strain is called the ’bulk modulus’ of the material of the body. It is denoted by B. Then, B=

−p ∆V /V

or

−∆p ( ∆V /V )

V

Fig. 15.2

Here, negative sign implies that, when the pressure increases volume decreases and vice-versa.

Compressibility The reciprocal of the bulk modulus of the material of a body is called the ‘compressibility’ of that material. Thus, 1 Compressibility = B

Modulus of Rigidity ( η) When a body is acted upon by an external force tangential to a surface of the body, the opposite surface being kept fixed, it suffers a change in shape, its volume remaining unchanged. Then, the body is said to be sheared. The ratio of the displacement of a layer in the direction of the tangential force and the distance of the layer from the fixed surface is called the shearing strain and the tangential force acting per unit area of the surface is called the “shearing stress”. For small strain the ratio of the shearing stress to the shearing strain is called the “modulus of rigidity” of the material of the body. It is denoted by η. F /A Thus, η= KK ′/KN KK ′ Here, = tan θ ≈ θ KN F /A F or η= η= ∴ θ Aθ

K

F K′

L

θ

θ N

L′

M

Fig. 15.3

Chapter 15

Elasticity — 377

Extra Points to Remember ˜

If a spring is stretched or compressed by an amount ∆l, the restoring force produced in it is, Fs = k ∆l

…(i)

k = force constant of spring

Here,

Similarly, if a wire is stretched by an amount ∆ l, the restoring force produced in it is, YA F =   ∆l  l  F/ A ∆l / l

Y =

as,

…(ii)

Comparing Eqs. (i) and (ii), we can see that force constant of a wire is, k=

YA l

…(iii)

YA . So, all l formulae which we use in case of a spring can be applied to a wire also. From Eq. (iii), we may also conclude that force constant of a spring is inversely proportional to the length of the spring l or i.e. a wire is just like a spring of force constant

k∝

l , 2k 2 l, k

l , 2k 2

⇒

1 l

Fig. 15.4

i.e. if a spring is cut into two equal pieces its force constant is doubled. V

Example 15.1 Determine the elongation of the steel bar 1m long and 1.5 cm2 cross-sectional area when subjected to a pull of 1.5 × 104 N . (Take Y = 2.0 × 1011 N / m2 ) Solution

Y =

F/ A ∆l / l

∴

∆l =

Fl AY

Substituting the values, ∆l =

(1.5 × 104 )(1.0) (1.5 × 10–4 )(2.0 × 1011 )

= 0.5 × 10–3 m or V

∆l = 0.5 mm

Ans.

Example 15.2 The bulk modulus of water is 2.3 × 10 9 N/m 2 . (a) Find its compressibility in the units atm −1 . (b) How much pressure in atmospheres is needed to compress a sample of water by 0.1% ?

378 — Mechanics - II Solution

Here, B = 2.3 × 109 N/m 2 =

2.3 × 109 1.01 × 105

= 2.27 × 104 atm (a) Compressibility =

1 1 = = 4.4 × 10 −5 atm −1 B 2.27 × 104

Ans.

∆V = − 0.1% = − 0.001 V Required increase in pressure,

(b) Here,

 ∆V  ∆ p = B × −   V  = 2.27 × 10 4 × 0.001 = 22.7 atm

INTRODUCTORY EXERCISE

Ans.

15.1

1. Two wires A and B of same dimensions are stretched by same amount of force. Young’s modulus of A is twice that of B. Which wire will get more elongation?

2. A rod 100 cm long and of 2 cm × 2 cm cross-section is subjected to a pull of 1000 kg force. If the modulus of elasticity of the material is 2.0 × 106 kg/cm 2, determine the elongation of the rod.

3. A cast iron column has internal diameter of 200 mm. What should be the minimum external diameter so that it may carry a load of 1.6 MN without the stress exceeding 90 N/mm 2?

4. Find the dimensions of stress, strain and modulus of elasticity.

15.5 The Stress-Strain Curve

σ A plot of longitudinal stress (either tensile or compressive) (Fracture point) versus longitudingal strain for a typical solid is shown in figure. σE E (Elastic limit) σP The strain is directly proportional to the applied stress for values P (Proportional limit) of stress upto σP . In this linear region, the material returns to its original size when the stress is removed. Point P is known as the proportional limit of the solid. For stresses between σP and σE , O ε where point E is called the elastic limit, the material also returns The stress-strain curve for a typical solid Fig. 15.5 to its original size. However, notice that stress and strain are not proportional in this region. For deformations beyond the elastic limit, the material does not return to its original size when the stress is removed, it is permanently distorted. Finally, further stretching beyond the elastic limit leads to the eventual fracture of the solid. The proportionality constant for linear region or the slope of stress-strain curve in this curve is called the Young’s modulus of elasticity Y.

Chapter 15

Elasticity — 379

15.6 Potential Energy Stored in a Stretched Wire When a wire is stretched, work is done against the inter atomic forces. This work is stored in the wire in the form of elastic potential energy. Suppose on applying a force F on a wire of length l, the increase in length is ∆l. The area of cross-section of the wire is A. The potential energy stored in the wire should be, 1 U = k ( ∆l) 2 2 YA Here, k= l ∴

U =

1 YA ( ∆l) 2 2 l

Elastic potential energy per unit volume of the wire (also called energy density) is, 1 YA ( ∆l) 2 U 2 l or u = u= volume Al 1  ∆l   ∆l  1 or u =    Y ⋅  or u = (strain ) (Y × strain ) 2 l  l  2 u=

or

1 (strain) × (stress) 2

15.7 Thermal Stresses or Strains Whenever there is some increase or decrease in the temperature of the body, it causes the body to expand or contract. If the body is allowed to expand or contract freely, with the rise or fall of the temperature, no stresses are induced in the body. But if the deformation of the body is prevented, some stresses are induced in the body. Such stresses are called thermal stresses or temperature stresses. The corresponding strains are called thermal strains or temperature strains. Consider a rod AB fixed at two supports as shown in figure.

A

B

Fig. 15.6

l = length of rod A = area of cross-section of the rod Y = Young’s modulus of elasticity of the rod and α = thermal coefficient of linear expansion of the rod Let the temperature of the rod is increased by an amount t. The length of the rod would had increased by an amount ∆l, if it were not fixed at two supports. Here ∆l = lα t Let

380 — Mechanics - II But the rod is fixed at the supports. Hence a compressive strain will be produced in the rod. Because at the increased temperature, the natural length of the rod is l + ∆l, while being fixed at two supports its actual length is l. Hence, thermal strain ∆l ∆ l lα t ε= ≈ = = αt l + ∆l l l or

ε = αt

Therefore, thermal stress or or force on the supports,

σ = Yε σ = Yαt

(stress = Y × strain )

F = σA = YAαt This force F is in the direction shown below. F F

F F

Fig. 15.7 V

Example 15.3 A steel wire 4.0 m in length is stretched through 2.0 mm. The cross-sectional area of the wire is 2.0 mm 2 . If Young’s modulus of steel is 2.0 × 1011 N/m 2 . Find (a) the energy density of wire, (b) the elastic potential energy stored in the wire. Solution

Here,

l = 4.0 m,

∆l = 2 × 10−3 m, A = 2.0 × 10−6 m 2 , Y = 2.0 × 1011 N/m 2

(a) The energy density of stretched wire 1 U = × stress × strain 2 1 = × Y × (strain ) 2 2 =

 ( 2 × 10−3 ) 1  × 2.0 × 1011 ×  2 4  

2

= 0.25 × 105 = 2.5 × 104 J/m3 (b)

Elastic potential energy = energy density × volume = 2.5 × 104 × ( 2.0 × 10−6 ) × 4.0 J = 20 × 10−2 = 0.20 J

V

Ans.

Example 15.4 A rubber cord has a cross-sectional area 1 mm2 and total unstretched length 10.0 cm. It is stretched to 12.0 cm and then released to project a missile of mass 5.0 g. Taking Young’s modulus Y for rubber as 5.0 × 10 8 N / m2 . Calculate the velocity of projection.

Ans.

Chapter 15

Elasticity — 381

Solution Equivalent force constant of rubber cord

k=

YA (5.0 × 108 ) (1.0 × 10–6 ) = = 5.0 × 103 N/ m l (0.1)

Now, from conservation of mechanical energy, elastic potential energy of cord = kinetic energy of missile ∴

1 1 k ( ∆l ) 2 = mv 2 2 2

 k ⇒ ∴ v=  ∆l  m

 5.0 × 103   (12.0 – 10.0) × 10−2 = 20 m/s =  5.0 × 10−3   

Ans.

Following assumptions have been made in this problem : (i) k has been assumed constant, even though it depends on the length (l ). (ii) The whole of the elastic potential energy is converting into kinetic energy of missile.

Note

INTRODUCTORY EXERCISE

15.2

1. Find the dimensions of energy density. 2. (a) A wire 4 m long and 0.3 mm in diameter is stretched by a force of 100 N. If extension in the wire is 0.3 mm, calculate the potential energy stored in the wire. (b) Find the work done in stretching a wire of cross-section 1 mm 2 and length 2 m through 0.1mm. Young’s modulus for the material of wire is 2.0 × 1011 N/m 2.

Extra Points to Remember ˜

Modulus of elasticity E (whether, it is Y, B or η ) is given by stress E= strain Following conclusions can be made from the above expression : (i) E ∝ stress (for same strain), i.e. if we want the equal amount of strain in two different materials, the one which needs more stress is having more E. 1 (for same stress), i.e. if the same amount of stress is applied on two different materials, the (ii) E ∝ strain one having the less strain is having more E. Rather we can say that the material which offers more resistance to the external forces is having greater value of E. So, we can see that modulus of elasticity of steel is more than that of rubber or Esteel > Erubber ∆x (iii) E = stress for unit strain  = 1 or ∆x = x , i.e. suppose the length of a wire is 2 m, then the Young’s  x  modulus of elasticity(Y ) is the stress applied on the wire to stretch the wire by the same length of 2 m.

˜

The material which has smaller value of Y is more ductile, i.e. it offers less resistance in framing it into a wire. Similarly, the material having the smaller value of B is more malleable. Thus, for making wire we will more concentrate on Y.

382 — Mechanics - II ˜

˜

A solid will have all the three modulii of elasticity Y, B and η. But in case of a liquid or a gas only B can be defined, as a liquid or a gas cannot be framed into a wire or no shear force can be applied on them. For a liquid or a gas,  − dp  B=    dV/ V 

or

 ∆p  −   ∆V / V 

So, instead of p we are more interested in change in pressure dp or ∆p. ˜

In case of a gas, bulk modulus is process dependent and is given by, B = xp in the process pV x = constant For example, for x = 1, or pV = constant (isothermal process), B = p. i.e., isothermal bulk modulus of a gas (denoted by BT ) is equal to the pressure of the gas at that instant of time or BT = p Similarly, for x = γ =

Cp CV

γ

or pV = constant (adiabatic process), B = γ p.

i.e., adiabatic bulk modulus of a gas (denoted by Bs) is equal to γ times the pressure of the gas at that instant of time or Bs = γp In general for a gas, B ∝ p in all processes. Physically this can be explained as under: A

B

... . .... .. .. .. . ... . . ... . . .... . . . .. . p1 . .. . .. . . . . . . .. . . ... . ... . . .. .. ..

. . . . . . .. . . p2 . . . . . . .

Fig. 15.8

Suppose we have two containers A and B. Some gas is filled in both the containers. But the pressure in A is more than the pressure in B, i.e. p1 > p2 So, bulk modulus of A should be more than the bulk modulus of B, or B1 > B2 and this is quiet obvious, because it is more difficult to compress the gas in chamber A, i.e. it provides more resistance to the external forces. And as we have discussed earlier also, the modulus of elasticity is more for a material which offers more resistance to external forces. ˜

When a pressure is applied on a substance, its density is changed. The change in density can be calculated as under : mass ρ= (ρ = density ) volume 1 or (mass = constant) ρ∝ V ρ′ V V = = ρ V ′ V + dV or

 V  ρ′ = ρ    V + dV 

Chapter 15   V =ρ   V − (dp/ B)V 

Elasticity — 383 dp    as B = −   dV / V 

ρ dp B From this expression, we can see that ρ′ increases as pressure is increased (dp is positive) and vice-versa. We can also write the above equation as ρ′ =

1−

dp ρ′ = ρ 1−   B or

˜

−1

dp ρ′ ≈ ρ  1+   B

(ifdp y

(b) x = y

(c) x < y

(d) x = 2 y

Match the Columns 1. Match the following two columns. (dimension wise) Column I

Column II

(a) Stress

(p) coefficient of friction

(b) Strain

(q) relative density

(c) Modulus of elasticity

(r) energy density

(d) Force constant of a wire (s) None

2. A wire of length l, area of cross section A and Young’s modulus of elasticity Y is stretched by a longitudinal force F. The change in length is ∆l. Match the following two columns.

Note In column I, corresponding to every option, other factors remain constant. Column I

Column II

(a) F is increased

(p) ∆l will increase

(b) l is increased

(q) stress will increase

(c) A is increased

(r) ∆l will decrease

(d) Y is increased

(s) stress will decrease

3. In Column I, a uniform bar of uniform cross-sectional area under the application of forces is shown in the figure and in Column II, some effects/phenomena are given. Match the two columns. Column I (a)

F

(b)

F

F

(c)

Column II (p) Uniform stresses are developed in the rod. (q) Non-uniform stresses are developed in the rod.

F

Smooth F

(d)

F

Smooth

(r) Compressive stresses are developed in the rod. (s) Tensile stresses are developed in the rod.

Subjective Questions 1. A solid sphere of radius R made of a material of bulk modulus B is surrounded by a liquid in a cylindrical container. A massless piston of area A (the area of container is also A) floats on the surface of the liquid. When a mass M is placed on the piston to compress the liquid, fractional Mg . Find the value of α. change in radius of the sphere is α AB

Chapter 15

Elasticity — 401

2. A 0.1 kg mass is suspended from a wire of negligible mass. The length of the wire is 1 m and its cross-sectional area is 4.9 × 10−7 m 2. If the mass is pulled a little in the vertically downward direction and released, it performs SHM with angular frequency 140 rad s−1. If the Young’s modulus of the material of the wire is p × 109 Nm −2, find the value of p.

3. A wire having a length L and cross-sectional area A is suspended at one of its ends from a ceiling. Density and Young’s modulus of material of the wire are ρ and Y , respectively. Its strain ρ2g2 AL3 energy due to its own weight is . Find the value of α. αY

4. A wire of length 3 m, diameter 0.4 mm and Young’s modulus 8 × 1010 N / m 2 is suspended from a point and supports a heavy cylinder of volume 10−3 m3 at its lower end. Find the decrease in length when the metal cylinder is immersed in a liquid of density 800 kg / m3 .

5. A sphere of radius 10 cm and mass 25 kg is attached to the lower end of a steel wire of length 5 m and diameter 4 mm which is suspended from the ceiling of a room. The point of support is 521 cm above the floor. When the sphere is set swinging as a simple pendulum, its lowest point just grazes the floor. Calculate the velocity of the ball at its lowest position (Ysteel = 2 × 1011 N / m 2 ).

6. A uniform ring of radius R and made up of a wire of cross-sectional radius r is rotated about its axis with a frequency f. If density of the wire is ρ and Young’s modulus is Y. Find the fractional change in radius of the ring.

7. A 6 kg weight is fastened to the end of a steel wire of unstretched length 60 cm. It is whirled in a vertical circle and has an angular velocity of 2 rev/s at the bottom of the circle. The area of cross-section of the wire is 0.05 cm 2. Calculate the elongation of the wire when the weight is at the lowest point of the path. Young’s modulus of steel = 2 × 1011 N / m 2.

8. A homogeneous block with a mass m hangs on three vertical wires of equal length arranged symmetrically. Find the tension of the wires if the middle wire is of steel and the other two are of copper. All the wires have the same cross-section. Consider the modulus of elasticity of steel to be double than that of copper.

C

S

C

m

9. A uniform copper bar of density ρ, length L, cross-sectional area S and Young’s modulus Y is moving horizontally on a frictionless surface with constant acceleration a0. Find (a) the stress at the centre of the wire, (b) total elongation of the wire.

10. A 5 m long cylindrical steel wire with radius 2 × 10−3 m is suspended vertically from a rigid support and carries a bob of mass 100 kg at the other end. If the bob gets snapped, calculate the change in temperature of the wire ignoring radiation losses. (Take g = 10 m / s2) (For the steel wire, Young’s modulus = 21 . × 1011 N / m 2 ; Density = 7860 kg / m3 ; Specific heat = 420 J/ kg-°C).

Answers Introductory Exercise 15.1 1. Wire B

3. 250.2 mm 4. [ML−1 T −2 ], [M0 L0 T 0 ], [ML−1 T −2 ]

2. 0.0125 cm

Introductory Exercise 15.2 1. [ML−1 T −2 ]

2. (a) 0.015 J (b) 5.0 × 10 −4 J

Exercises LEVEL 1 Assertion and Reason 1. (a)

2. (c)

3. (d)

4. (b)

5. (a)

6. (d)

7. (c)

8. (c)

9. (a)

4. (c) 14. (d)

5. (b) 15. (a)

6. (a) 16. (c)

7. (b)

8. (a)

9. (b)

Objective Questions 1. (c) 11. (b)

2. (b) 12. (a)

3. (a) 13 (c)

10. (b)

Subjective Questions 1. 1169 . g/cm3

2. 1.91mm

5. Load = 14 kg, lower string

4. ∆l = 4.9 × 10 −6 m, U = 5.43 × 10 −5 J

3. 34.64 m/s 2

6. (a) 1 (b) Ratio =

13 20

7. 2.0 kg/m3

8. B = 176 . × 10 9 N/m2

LEVEL 2 Single Correct Option 1. (c)

2. (d)

3. (b)

4. (a)

5. (d)

6. (c)

7. (c)

8. (a)

9. (b)

10. (a)

More than One Correct Options 1. (a, d)

2. (a, c)

3. (c)

4. (a, c, d)

5. (a, b,c)

6. (a, c)

7. (c, d)

8. (a, b)

Comprehension Based Questions 1. (b)

2. (c)

3. (d)

4. (d)

5. (b)

6. (a)

Match the Columns 1.(a) → r

(b) → p, q

(c) → r

(d) → s

2.(a) → p, q

(b) → p

(c) → r, s

(d) → r

3.(a) → p, r

(b) → p, s

(c) → q, s

(d) → q, r

Subjective Questions 1. 3 8. T c =

2. 4

3. 6 4. ∆l = 2.34 × 10 −3 m

mg , T s = 2T c 4

9. (a)

5. v = 3123 m/s .

1 1 ρa0L2 Lρa0 (b) 2 2 Y

6.

10. 4.568 × 10 −3 °C

4 π 2 f 2 pR 2 Y

7. 3.8 × 10 −4 m

16

Fluid Mechanics Chapter Contents 16.1

Definition of a Fluid

16.2

Density of a Liquid

16.3

Pressure in a Fluid

16.4

Pressure Difference in Accelerating Fluids

16.5

Archimedes’ Principle

16.6

Flow of Fluids

16.7

Application based on Bernoulli’s Equation

16.8

Viscosity

16.9

Surface Tension

16.10 Capillary Rise or Fall

404 — Mechanics - II

16.1 Definition of a Fluid Fluid mechanics deals with the behaviour of fluids at rest and in motion. A fluid is a substance that deforms continuously under the application of a shear (tangential) stress no matter how small the shear stress may be. Thus, fluids comprise the liquid and gas (or vapour) phases of the physical forms in which matter exists. The distinction between a fluid and the solid state of matter is clear if you compare fluid and solid behaviour. A solid deforms when a shear stress is applied but it does not continue to increase with time. However, if a shear stress is applied to a fluid, the deformation continues to increase as long as the stress is applied. We may alternatively define a fluid as a substance that cannot sustain a shear stress when at rest. F

(a) Solid

Fig. 16.1

F

(b) Fluid

Behaviour of a solid and a fluid, under the action of a constant shear force

Ideal Fluid An ideal liquid is incompressible and non-viscous in nature. An incompressible liquid means the density of the liquid is constant, it is independent of the variations in pressure. A non-viscous liquid means that, parts of the liquid in contact do not exert any tangential force on each other. Thus, there is no friction between the adjacent layers of a liquid. The force by one part of the liquid on the other part is perpendicular to the surface of contact.

16.2 Density of a Liquid Density (ρ) of any substance is defined as the mass per unit volume or mass m or ρ= ρ= volume V

Relative Density (RD) In case of a liquid, sometimes an another term Relative Density (RD) is defined. It is the ratio of density of the substance to the density of water at 4°C. Hence, RD =

Density of substance Density of water at 4° C

RD is a pure ratio. So, it has no units. It is also sometimes referred as specific gravity. Density of water at 4°C in CGS is 1 g/cm 3 . Therefore, numerically the RD and density of substance (in CGS) are equal. In SI units, the density of water at 4°C is 1000 kg/m 3 .

Chapter 16

Fluid Mechanics — 405

Density of a Mixture of two or more Liquids Here, we have two cases: Suppose two liquids of densities ρ1 and ρ 2 having masses m1 and m2 are mixed together. Then, the density of the mixture will be ( m + m2 ) Total mass = 1 ρ= Total volume (V1 + V2 )

Case 1

=

If m1 = m2 , then

ρ=

( m1 + m2 )  m1 m2  +    ρ1 ρ 2  2ρ1ρ 2 ρ1 + ρ 2

If two liquids of densities ρ1 and ρ 2 having volumes V1 andV2 are mixed, then the density of the mixture is, m + m2 Total mass = 1 ρ= Total volume V1 + V2 Case 2

= If V1 = V2 , then

ρ=

ρ1V1 + ρ 2V2 V1 + V2 ρ1 + ρ 2 2

Effect of Temperature on Density As the temperature of a liquid is increased, the mass remains the same while the volume is increased 1  and hence, the density of the liquid decreases  as ρ ∝  . Thus,  V ρ′ V V V = = = ρ V ′ V + dV V + Vγ∆θ or Here, and

ρ′ 1 = ρ 1 + γ∆θ γ = thermal coefficient of volume expansion ∆θ = rise in temperature ρ 1 + γ∆θ

∴

ρ′ =

We can also write

ρ′ = ρ(1 + γ∆θ ) −1

or

ρ′ ≈ ρ(1 − γ∆θ )

(if γ∆θ F1 . Thus, hydraulic lift is a force multiplying device with a multiplication factor equal to the ratio of the areas of the two pistons. Dentist’s chairs, car lifts and jacks, many elevators and hydraulic brakes all use this principle.

Extra Points to Remember ˜

At same point on a fluid pressure is same in all directions. In the figure, p1 = p2 = p3 = p4

p4 p1

p3 p2

Fig. 16.8 ˜

Forces acting on a fluid in equilibrium have to be perpendicular to its surface. Because it cannot sustain the shear stress.

410 — Mechanics - II ˜

p0 In the same liquid pressure will be same at all points at the same level ρ1 (provided their speeds are same).

p1 ≠ p2 , p3 = p4 and p5 = p6

For example, in the figure ∴

ρ1h1 = ρ2 h2

˜

h∝

or

ρ2

2 h2 4 6

3 5

P0 + ρ1 gh1 = P0 + ρ2 gh2

or

h1

1

p3 = p4

Further,

p0

1 ρ

Fig. 16.9

Barometer It is a device used to measure atmospheric pressure. In principle, any liquid can be used to fill the barometer, but mercury is the substance of choice because its great density makes possible an instrument of reasonable size.

Vacuum (p = 0) h

p1 = p2 Here,

p1 = atmospheric pressure ( p0 )

and

p2 = 0 + ρgh = ρgh

1 2

ρ = density of mercury

Here, ∴

Fig. 16.10

p0 = ρgh

Thus, the mercury barometer reads the atmospheric pressure ( p0 ) directly from the height of the mercury column. For example, if the height of mercury in a barometer is 760 mm, then atmospheric pressure will be p0 = ρgh = (13.6 × 103 )(9.8)(0.760) = 1.01 × 105 N/m2 ˜

Manometer It is a device used to measure the pressure of a gas inside a container. The U-shaped tube often contains mercury. p1 = p2 p1 = pressure of the gas in the container ( p)

and

p2 = atmospheric pressure ( p0 ) + ρgh p = p0 + hρg

This can also be written as

˜

h 2

1

Here, ∴

p0

Hg

Fig. 16.11

p − p0 = gauge pressure = hρg

Here, ρ is the density of the liquid used in U-tube. Thus, by measuring h we can find absolute (or gauge) pressure in the vessel. Free body diagram of a liquid The free body diagram of the liquid (showing the vertical forces only) is shown in Fig. 16.12 (b). For the equilibrium of liquid, p0 A A ρ

h

(a)

W

(b)

pA

Fig. 16.12

net downward force = net upward force ∴

p0 A + W = pA here, W = ρghA ∴ p = p0 + ρgh

Chapter 16 V

Fluid Mechanics — 411

Example 16.2 For the arrangement shown in the figure, what is the density of oil ? C d = 12.3 mm Oil l = 135 mm A

B Water

Fig. 16.13

pB = p A p 0 + ρ w gl = p 0 + ρ oil ( l + d ) g ρ l (1000) (135) ρ oil = w = = 916 kg/m 3 l + d (135 + 12.3)

Solution

∴ ⇒ V

Ans.

Example 16.3 A U-tube of uniform cross-sectional area and open to the atmosphere is partially filled with mercury. Water is then poured into both arms. If the equilibrium configuration of the tube is as shown in figure with h 2 = 1.0 cm, determine the value of h 1 . h1

h2

Hg

Fig. 16.14

Solution

p1 = p 2

h1 + h2 + h

h h2

1

2

Fig. 16.15

∴ ∴

p 0 + ρ w g ( h1 + h2 + h ) = p 0 + ρ w gh + ρ Hg gh2 (ρ Hg − ρ w ) h2 h1 = ρw =

(13.6 − 1) (1.0) 1

= 12.6cm

Ans.

412 — Mechanics - II V

Example 16.4 A U-shaped tube open to the air at both ends contains some mercury. A quantity of water is carefully poured into the left arm of the U-shaped tube until the vertical height of the water column is 15.0 cm. (a) What is the gauge pressure at the water mercury interface ? (b) Calculate the vertical distance h from the top of the mercury in the right hand arm of the tube to the top of the water in the left-hand arm. Solution (a) Gauge pressure = ρ w ghw

h 15.0 cm

Water

Mercury

Fig. 16.16

= (103 ) ( 9.8) ( 0.15) = 1470 N/ m 2

Ans.

(b) Let us calculate pressure or two sides at the level of water and mercury interface. p 0 + ρ w g hw = p 0 + ρ Hg gh Hg ρ w hw = ρ Hg hHg ∴ (1) (15) = (13.6) (15 − h ) h = 13.9 cm V

Ans.

Example 16.5 For the system shown in figure, the cylinder on the left, at L, has a mass of 600 kg and a cross-sectional area of 800 cm2 . The piston on the right, at S, has cross-sectional area 25 cm2 and negligible weight. If the apparatus is filled with oil (ρ = 0.78 g/cm3 ), what is the force F required to hold the system in equilibrium? F S 8m L

600 kg

Fig. 16.17

Let A1 = area of cross-section on LHS and A 2 = area of cross-section on RHS Equating the pressure on two sides of the dotted line. Mg F Then, + ρgh = A2 A1

Solution

∴

(M = 600 kg)

 Mg  F = A2  − ρgh A  1    600 × 9.8 780 9.8 8 = ( 25 × 10− 4 )  − × ×  −4   800 × 10 = 31N

Ans.

Chapter 16

INTRODUCTORY EXERCISE

Fluid Mechanics — 413

16.1

1. Water and oil are poured into the two limbs of a U-tube containing mercury. The interfaces of the mercury and the liquids are at the same height in both limbs. h1 Determine the height of the water column h1 if that of the oil h2 = 20 cm. The density of the oil is 0.9.

h2

Fig. 16.18

2. The liquids shown in figure in the two arms are mercury (specific

gravity = 13.6) and water. If the difference of heights of the mercury columns is 2 cm, find the height h of the water column.

2 cm A

B

Fig. 16.19

3. The heights of mercury surfaces in the two arms of the manometer shown in Fig.16.20 are 2 cm and 8 cm. Atmospheric pressure is 1.01 × 105Nm −2. Find (a) the pressure of the gas in the cylinder and (b) the pressure of mercury at the bottom of the U tube.

gas

4. A cylinderical vessel containing a liquid is closed by a smooth piston of mass m as shown in the Fig. 16.21. The area of cross section of the piston is A. If the atmospheric pressure is p 0, find the pressure of the liquid just below the piston.

Fig. 16.20

Fig. 16.21

5. The area of cross section of the two arms of a hydraulic press are 1 cm 2

5N

F

2

and 10 cm respectively (Fig.16.22). A force of 5 N is applied on the water in the thinner arm. What force should be applied on the water in the thicker arm so that the water may remain in equilibrium? Fig. 16.22

6. The area of cross section of the wider tube shown in Fig. (16.23) is 900 cm 2. If the body standing on the position weighs 45 kg, find the difference in the levels of water in the two tubes.

Fig. 16.23

414 — Mechanics - II

16.4 Pressure Difference in Accelerating Fluids If fluids are at rest then pressure does not change in horizontal direction. It changes only in vertical direction. At a height difference ‘h’, difference in pressure (or change in pressure) is ∆p = ±ρ gh Pressure increases with depth. So, ∆p = +ρ gh in moving downwards and ∆p = −ρ gh in moving upwards. If fluids are accelerated then pressure changes in both horizontal and vertical directions.

In Horizontal Direction (i) Pressure decreases in the direction of acceleration. (ii) At a horizontal distance x change in pressure is ∆p = ±ρ ax

(a = horizontal acceleration)

Take ∆p = +ρ ax, while moving in opposite direction of acceleration. Because pressure increases in the opposite direction of acceleration and take ∆P = −ρ ax while moving in the direction of acceleration.

In Vertical Direction Instead of ‘g’ in the equation, ∆p = ±ρ gh take effective value of g or g e . Thus, ∆p = ±ρ g e h Here, g e = g if vertical acceleration of fluid is zero. g e = g + a if vertical acceleration of fluid ‘a’ is upwards and g e = g − a if vertical acceleration ‘a’ is downwards. In equation form, all above statements can be explained as below. Consider a liquid kept at rest in a beaker as shown in Fig. 16.24 (a). In this case, we know that pressure do not change in horizontal direction (x-direction), it decreases upwards along y-direction. So, we can write the equations, dp dp and …(i) =0 = − ρg dx dy y

y

ay ax

x

o

x

o

(a)

(b)

Fig. 16.24

Chapter 16

Fluid Mechanics — 415

But, suppose the beaker is accelerated and it has components of acceleration a x and a y in x and y directions respectively, then the pressure decreases along both x and y-directions. The above equations in that case become, dp = − ρa x dx dp = − ρ( g + a y ) dy

and

…(ii)

These equations can be derived as under : Consider a beaker filled with some liquid of density ρ accelerating upwards with an acceleration a y along positive y-direction. Let us draw the free body diagram of a small element of fluid of area A and length dy as shown in figure. y

(p + dp )A

p + dp ay

A

ay

A

dy

p

pA

x

Fig. 16.25

Equation of motion for this fluid element is, pA − W − ( p + dp) A = ( mass )( a y ) or −W − ( dp) A = ( Aρ dy)( a y ) dp or −( Aρg dy) − ( dp) A = ( Aρ dy)( a y ) or = − ρ( g + a y ) dy Similarly, if the beaker moves along positive x-direction with an acceleration a x , the equation of motion for the fluid element shown in figure is, y

ax p

dx o

(p + dp )A

pA A

p + dp

A x

ax

Fig. 16.26

or or

pA − ( p + dp) A = ( mass )( a x ) − ( dp) A = ( Aρ dx ) a x dp = − ρa x dx

416 — Mechanics - II Note (i)

dp = − ρ ax , ∆p = − ρ ax x = pressure difference in a horizontal distance x. dx Here, negative sign implies that pressure decreases with x or in the direction of acceleration.

(ii) In Fig. 16.26, pressure on left hand side of the fluid element should be more than the pressure on right hand side of this element. This is because this element is accelerated only due to this pressure difference. dp (iii) = − ρ (g + a y ) dy ⇒ ∆p = − ρ ( g + a y ) y or ∆p = − ρge h Here, negative sign implies that pressure decreases with ‘y’ or in moving upwards.

Free Surface of a Liquid Accelerated in Horizontal Direction Consider a liquid placed in a beaker which is accelerating horizontally with an acceleration ‘a’. Let A and B be two points in the liquid at a separation x in the same horizontal line. As we have seen in this case dp = − ρa dx or

dp = − ρa dx

y

h1

θ h2

x

Integrating this with proper limits, we get p A − pB = ρax

…(iii)

Further,

p A = p0 + ρgh1

and

pB = p0 + ρgh2

a

B

A

x

Fig. 16.27

Substituting in Eq. (iii), we get ρg ( h1 − h2 ) = ρax ∴ ∴

h1 − h2 a = = tan θ x g tan θ =

a g

Pressure Difference in Rotating Fluids In a rotating fluid (also accelerating) pressure increases in moving away from the rotational axis. At a distance ‘x’ from the rotational axis, pressure difference is ∆p = ±

ρω 2 x 2 2

ρω 2 x 2 in moving away from the rotational axis, as pressure increases in this direction 2 ρω 2 x 2 in moving towards the rotational axis. and take ∆p = − 2

Take ∆p = +

Chapter 16

Fluid Mechanics — 417

Proof Suppose that liquid of density ρ kept inside a tube of area of cross-section A is rotating with an angular velocity ‘ω’ as shown. ω dx p + dp

p x

Fig. 16.28

Consider a small element of length ‘dx’ at a distance x from the axis of rotation. Mass of this element is, dm = (density ) ( volume ) dm = (ρ Adx )

or

This element is rotating in a circle of radius ‘x’. So, this is accelerated towards centre with a centripetal acceleration (as a = R ω 2 )

a = xω2

To provide this acceleration, pressure on right hand side of the element should be more. ∴

( p + dp) A − ( p) A = ( dm) a = (ρ Adx ) ( xω 2 )

or

dp = (ρ x ω 2 ) dx

∴

∆p = ∫ (ρω 2 ) xdx

or

∆p =

x

0

V

ρω 2 x 2 2

Hence proved.

Example 16.6 A closed container shown in figure is filled with water ( ρ =103 kg / m3 ) A

B a = 2 m/s2

6m C

D 2m

Fig. 16.29

This is accelerated in horizontal direction with an acceleration, a = 2 m/ s 2 . Find (a) pC − pD and (b) pA − pD Solution (a) In horizontal direction, pressure decreases in the direction of acceleration. Thus, pC > p D or pC − p D = + ρ ax

418 — Mechanics - II Substituting the values, we have pC − p D = (103 )( 2)( 2) pC − p D = 4.0 × 103 N/ m 2

or

Ans.

(b) In vertical direction, pressure increases with depth. ∴ or

pC > p A p A − pC = − ρ gh

(g e = g)

= − (10 ) (10) ( 6) 3

= − 60 × 103 N/ m 2 Now,

p A − p D = ( p A − pC ) + ( pC − p D ) = ( − 60 × 103 ) + ( 4.0 × 103 ) = − 56 × 103 N/ m 2 = − 5.6 × 104 N/ m 2

Ans.

Note Container is closed. So, nowhere inside the container pressure is atmospheric pressure p0 . V

Example 16.7 A closed tube is filled with ω

A

B

C

Fig. 16.30

AB = 2 m BC = 4 cm water (ρ = 103 kg/ m3 ). It is rotating about an axis shown in figure with an angular velocity ω = 2 rad/ s. Find, pA − pC . Solution Pressure decreases in moving towards the axis of rotation and increases in moving  ρω 2 x 2   away from the axis  ∆p = ± 2   ∴

Here, and

p A > p B and p B < pC p A − pC = ( p A − p B ) + ( p B − pC )  ρω 2 x12   − ρω 2 x 22    + =  + 2   2   x1 = AB = 2 m x 2 = BC = 4 m

Chapter 16

Fluid Mechanics — 419

Substituting the values we have, p A − pC =

(103 ) ( 2) 2 ( 2) 2 (103 ) ( 2) 2 ( 4 ) 2 − 2 2

= − 2.4 × 103 N/ m 2 V

Ans.

Example 16.8 A liquid of density ρ is in a bucket that spins with angular velocity ‘ω’ as shown in figure. Prove that the free surface of the liquid has a parabolic shape. Find equation of this. ω

Fig. 16.31

Solution In the figure shown, suppose the coordinates of point M are (x, y) with respect to the coordinate axes. y

M Q

x

N

Fig. 16.32

Then, MN = y, NQ = x Points M and Q are open to atmosphere. ∴ p M = p Q = atmospheric pressure p 0 Now, p M − p N = − ρ gy ⇒

p 0 − p N = − ρ gy pN

∴

…(i)

ρω 2 x 2 − pQ = 2

p N − p0 =

ρω 2 x 2 2

…(ii)

420 — Mechanics - II Adding Eqs. (i) and (ii), we have 0 = − ρ gy + y=

ρω 2 x 2 2

ω 2x2 2g

This is the required equation of free surface of the liquid and we can see that this is an equation of a parabola.

INTRODUCTORY EXERCISE

16.2

1. In example 16.6, which point has the maximum pressure and which has the minimum pressure. 2. A cubical closed vessel of side 5 m filled with a liquid is accelerated with an acceleration a. Find the value of a so that pressure at mid point M of AC is equal to pressure at N. N a

A

C M Fig. 16.33

3. Water (ρ = 103 kg/m 3 ) is filled in tube AB as shown in figure. ω = 10 rad/s. Tube is open at end A. Atmospheric pressure is p 0 = 105 N/m 2. Find absolute pressure at end B. ω A

B

2m Fig. 16.34

16.5 Archimedes’ Principle If a heavy object is immersed in water, it seems to weightless than when it is in air. This is because the water exerts an upward force called buoyant force. It is equal to the weight of the fluid displaced by the body. A body wholly or partially submerged in a fluid is buoyed up by a force equal to the weight of the displaced fluid. This result is known as Archimedes’ principle. Thus, the magnitude of buoyant force ( F ) is given by, F = Vi ρ L g Here, Vi = immersed volume of solid, ρ L = density of liquid and g = acceleration due to gravity

Chapter 16

Fluid Mechanics — 421

Proof Consider an arbitrarily shaped body of volume V placed in a container filled with a fluid of density ρL . The body is shown completely immersed, but complete immersion is not essential to the proof. To begin with, imagine the situation before the body was immersed. The region now occupied by the body was filled with fluid, whose weight wasVρ L g. Because the fluid as a whole was in hydrostatic equilibrium, the net upwards force (due to difference in pressure at different depths) on the fluid in that region was equal to the weight of the fluid occupying that region. Now, consider what happens when the body has displaced the fluid. The ρL pressure at every point on the surface of the body is unchanged from the value at the same location when the body was not present. This is because the pressure at V any point depends only on the depth of that point below the fluid surface. Hence, the net force exerted by the surrounding fluid on the body is exactly the same as that exerted on the region before the body was present. But we know the latter to beVρ L g, the weight of the displaced fluid. Hence, this must also be the buoyant Fig. 16.35 force exerted on the body. Archimedes’ principle is thus proved.

Law of Floatation Consider an object of volume V and density ρ S floating in a liquid of density ρ L . Let Vi be the volume of object immersed in the liquid. For equilibrium of object, Weight = Upthrust ∴ Vρ S g = Vi ρ L g Vi ρ S …(i) = ∴ V ρL

Vi ρL

Fig. 16.36

This is the fraction of volume immersed in liquid. ρ V Percentage of volume immersed in liquid = i × 100 = S × 100 V ρL Three possibilities may now arise: (i) If ρ S < ρ L , only fraction of body will be immersed in the liquid. This fraction will be given by the above equation. (ii) If ρ S = ρ L , the whole of the rigid body will be immersed in the liquid. Hence, the body remains floating in the liquid wherever it is left. (iii) If ρ S > ρ L , the body will sink.

Buoyant Force in Accelerating Fluids Suppose a body is dipped inside a liquid of density ρ L placed in an elevator moving with an acceleration a. The buoyant force F in this case becomes, F = Vρ L g eff Here,

g eff = | g − a |

(V = immersed volume of solid or Vi )

422 — Mechanics - II For example, if the lift has an upward acceleration a, the value of g eff is g + a and if it has a downward acceleration a, the g eff is g − a. In a freely falling lift g eff is zero (as a = g) and hence, net buoyant force is zero. This is why, in a freely falling vessel filled with some liquid, the air bubbles do not rise up (which otherwise move up due to buoyant force). The above result can be derived as follows. Suppose a body is dipped inside a liquid of density ρ L in an elevator moving up with an acceleration a. As was done earlier also, replace the body into the liquid by the same liquid of equal volume. The replaced liquid is at rest with respect to the elevator. Thus, this replaced liquid is also moving up with an acceleration a together with the rest of the liquid. The forces acting on the replaced liquid are, (i) the buoyant force F and (ii) the weight mg of the substituted liquid. From Newton’s second law, F − mg = ma or F = m( g + a ) Here, m = Vρ L ∴ F = Vρ L ( g + a ) = Vρ L g eff where g eff = g + a

Extra Points to Remember ˜

In weight of a solid, take total volume of solid, density of solid and g. In upthrust (or buoyant force) take immersed volume of solid, density of liquid and geff . Thus, W = Vρ s g F = Vi ρL geff

˜

V

Upthrust force also makes a pair of equal and opposite forces. On solid it is upwards and on liquid, it is downwards.

Example 16.9 Density of ice is 900 kg/m3 . A piece of ice is floating in water of density 1000 kg / m3 . Find the fraction of volume of the piece of ice outside the water. Let V be the total volume and Vi the volume of ice piece immersed in water. For equilibrium of ice piece, Solution

∴

weight = upthrust Vρ i g = Vi ρ w g

Here, and

ρ i = density of ice = 900 kg/m 3 ρ w = density of water = 1000 kg/m 3

Substituting in above equation, we get Vi 900 = = 0.9 V 1000 i.e. the fraction of volume outside the water, f = 1 − 0.9 = 0.1

Ans.

Chapter 16 V

Example 16.10

A metallic sphere floats in an immiscible mixture of water 4 ( ρ w = 10 kg / m ) and a liquid ( ρ L = 13.5 × 103 kg / m3 ) such that its th volume is 5 1 in water and th volume in the liquid. Find the density of metal. 5 Solution Total upthrust = weight of metal sphere 4  1  3 5 ∴  V  (10 ) g +  V  (13.5 × 10 ) g = (V ) (ρ metal ) ( g ) 5  5  3

∴ V

Fluid Mechanics — 423

3

ρ metal = 3.5 × 103 kg / m 3

Ans.

Example 16.11 A block of mass 1 kg and density 0.8 g/cm3 is held stationary with the help of a string as shown in figure. The tank is accelerating vertically upwards with an acceleration a = 1.0 m /s2 . Find

a

Fig. 16.37

(a) the tension in the string, (b) if the string is now cut find the acceleration of block. (Take g = 10 m/ s 2 and density of water = 103 kg/m3 ). (a) Free body diagram of the block is shown in Fig. 16.38. In the figure, Solution

F a

w+T

Fig. 16.38

F = upthrust force = Vρ ω ( g + a )  mass of block  =  ρω (g + a )  density of block   1  =  (1000)(10 + 1) = 13.75 N  800 w = mg = 10 N

424 — Mechanics - II Equation of motion of the block is, ∴

F − T − w = ma 13.75 − T − 10 = 1 × 1

∴

T = 2.75 N

Ans.

(b) When the string is cut, T = 0 F −w m 13.75 − 10 = 1

∴

a=

= 3.75 m/s 2 V

Ans.

Example 16.12 The tension in a string holding a solid block below the surface of a liquid (of density greater than that of solid) as shown in figure is T 0 when the system is at rest. What will be the tension in the string if the system has an upward acceleration a ?

a

Fig. 16.39

Solution

Let m be the mass of block.

Initially for the equilibrium of block, F = T0 + mg

…(i)

Here, F is the upthrust on the block. F′

F

a

T0 + mg (a)

T + mg (b)

Fig. 16.40

When the lift is accelerated upwards, g eff becomes g + a instead of g. Hence,  g + a F′ = F    g 

…(ii)

Chapter 16

Fluid Mechanics — 425

From Newton’s second law, F ′ − T − mg = ma Solving Eqs. (i), (ii) and (iii), we get

…(iii)

 a T = T0 1 +   g

INTRODUCTORY EXERCISE

Ans.

16.3

1. A block of material has a density ρ1 and floats three-fourth submerged in a liquid of unknown density. Show that the density ρ 2 of the unknown liquid is given by ρ 2 =

4 ρ1. 3

2. A block of wood weighing 71.2 N and of specific gravity 0.75 is tied by a string to the bottom of a tank of water in order to have the block totally immersed. What is the tension in the string?

3. A beaker when partly filled with water has total mass 20.00 g. If a piece of metal with density 3.00 g/cm 3 and volume 1.00 cm 3 is suspended by a thin string, so that it is submerged in the water but does not rest on the bottom of the beaker, how much does the beaker then appear to weigh if it is resting on a scale?

4. A small block of wood of density0.4 × 103 kg/m 3 is submerged in water at a depth of 2.9 m. Find (a) the acceleration of the block towards the surface when the block is released and (b) the time for the block to reach the surface. Ignore viscosity.

16.6 Flow of Fluids Steady Flow If the velocity of fluid particles at any point does not vary with time, the flow is said to be steady. Steady flow is also called streamlined or laminar flow. The velocity at different points may be different. Hence, in the figure, v 1 = constant, v 2 = constant, v 3 = constant but v 1 ≠ v 2 ≠ v 3 v3 v1 v2

Fig 16.41

Principle of Continuity It states that, when an incompressible and non-viscous liquid flows in a stream lined motion through a tube of non-uniform cross-section, then the product of the area of cross section and the velocity of flow is same at every point in the tube. 1 Thus, or Av = constant or v ∝ A1 v1 = A2 v 2 A This is basically the law of conservation of mass in fluid dynamics.

Q P

v2

v1

A2

A1

Fig. 16.42

426 — Mechanics - II Proof Let us consider two cross sections P and Q of area A1 and A2 of a tube through which a fluid is flowing. Let v1 and v 2 be the speeds at these two cross sections. Then, being an incompressible fluid, mass of fluid going through P in a time interval ∆t = mass of fluid passing through Q in the same interval of time ∆t. A1 A2 v2

v1

Q P

Fig. 16.43

∴ or

A1 v1ρ∆t = A2 v 2ρ ∆t A1 v1 = A2 v 2

Proved.

Therefore, the velocity of the liquid is smaller in the wider parts of the tube and larger in the narrower parts. as or A2 < A1 v 2 > v1 Note The product Av is the volume flow rate

dV , the rate at which volume crosses a section of the tube. dt

Hence, dV = volume flow rate = Av dt

The mass flow rate is the mass flow per unit time through a cross-section. This is equal to density (ρ) dV times the volume flow rate . dt We can generalize the continuity equation for the case in which the fluid is not incompressible. If ρ1 and ρ 2 are the densities at sections 1 and 2 then, ρ1 A1 v1 = ρ 2 A2 v 2 So, this is the continuity equation for a compressible fluid.

Bernoulli’s Equation Bernoulli’s equation relates the pressure, flow speed and height for flow of an ideal (incompressible and non-viscous) fluid. The pressure of a fluid depends on height as in the static situation and it also depends on the speed of flow. A2 p2 v2

A1 v1 h1

b b′ h2

p 1 a′ a

Fig. 16.44

Chapter 16

Fluid Mechanics — 427

The dependence of pressure on speed can be understood from the continuity equation. When an incompressible fluid flows along a tube with varying cross section, its speed must change, and so, an element of fluid must have an acceleration. If the tube is horizontal, the force that causes this acceleration has to be applied by the surrounding fluid. This means that the pressure must be different in regions of different cross section. When a horizontal flow tube narrows and a fluid element speeds up, it must be moving towards a region of lower pressure in order to have a net forward force to accelerate it. If the elevation also changes, this causes additional pressure difference. To derive Bernoulli’s equation, we apply the work-energy theorem to the fluid in a section of the fluid element. Consider the element of fluid that at some initial time lies between two cross sections a and b. The speeds at the lower and upper ends are v1 and v 2 . In a small time interval, the fluid that is initially at a moves to a′ a distance aa ′ = ds1 = v1 dt and the fluid that is initially at b moves to b′ a distance bb′ = ds2 = v 2 dt. The cross-section areas at the two ends are A1 and A2 as shown. The fluid is incompressible, hence, by the continuity equation, the volume of fluid dV passing through any cross-section during time dt is the same. dV = A1 ds1 = A2 ds2

That is,

Work Done on the Fluid Element Let us calculate the work done on this fluid element during time interval dt. The pressure at the two ends are p1 and p2 , the force on the cross section at a is p1 A1 and the force at b is p2 A2 . The net work done dW on the element by the surrounding fluid during this displacement is, dW = p1 A1 ds1 − p2 A2 ds2 = ( p1 − p2 ) dV

…(i)

The second term is negative, because the force at b opposes the displacement of the fluid. This work dW is due to forces other than the conservative force of gravity, so it equals the change in total mechanical energy (kinetic plus potential). The mechanical energy for the fluid between sections a and b does not change.

Change in Potential Energy At the beginning of dt the potential energy for the mass between a and a′ is dmgh1 = ρdVgh1 . At the end of dt the potential energy for the mass between b and b′ is dmgh2 = ρdVgh2 . The net change in potential energy dU during dt is, dU = ρ( dV ) g ( h2 − h1 )

…(ii)

Change in Kinetic Energy At the beginning of dt the fluid between a and a′ has volume A1 ds1 , mass ρA1 ds1 and kinetic energy 1 1 ρ( A1 ds1 ) v12 . At the end of dt the fluid between b and b′ has kinetic energy ρ( A2 ds2 ) v 22 . The net 2 2 change in kinetic energy dK during time dt is, 1 …(iii) dK = ρ( dV )( v 22 − v12 ) 2

428 — Mechanics - II Combining Eqs. (i), (ii) and (iii) in the energy equation, dW = dK + dU We obtain, 1 ( p1 − p2 ) dV = ρdV ( v 22 − v12 ) + ρ( dV ) g ( h2 − h1 ) 2 1 p1 − p2 = ρ( v 22 − v12 ) + ρg ( h2 − h1 ) 2

or

…(iv)

This is Bernoulli’s equation. It states that the work done on a unit volume of fluid by the surrounding fluid is equal to the sum of the changes in kinetic and potential energies per unit volume that occur during the flow. We can also express Eq. (iv) in a more convenient form as, 1 1 p1 + ρgh1 + ρv12 = p2 + ρgh2 + ρv 22 2 2

Bernoulli’s equation

The subscripts 1 and 2 refer to any two points along the flow tube, so we can also write 1 p + ρgh + ρv 2 = constant 2

Extra Points to Remember ˜

1 In Bernoulli's equation, there are three terms; p, ρv 2 and ρgh. Under following three cases, this equation 2 reduces to a two term Bernoulli. Case 1 If all points are open to atmosphere then pressure at every point may be assumed to be constant (= p0 ) and the Bernoulli equation can be written as, 1 2 ρv + ρ gh = constant 2 At greater heights ‘h’, speed ‘v’ will be less as ρ and g are constants. Case 2 If the liquid is passing through a pipe of uniform cross-section, then from continuity equation ( Av =constant ), speed v is same at all points. Therefore, the Bernoulli equation becomes, p + ρ gh = constant or

p1 + ρ gh1 = p2 + ρ gh2

or

p1 − p2 = ρ g (h2 − h1 )

This is the pressure relation we have already derived for a fluid at rest or pressure decreases with height of liquid and increases with depth of liquid. Case 3 If a liquid is flowing in a horizontal pipe, then height ‘h’ of the liquid at every point may be assumed to be constant. So, the two term Bernoulli becomes, 1 p + ρ v 2 = constant 2 From this equation, we may conclude that pressure decreases at a point where speed increases. V

Example 16.13 Water is flowing through a horizontal tube of non-uniform cross section. At a place, the radius of the tube is 1.0 cm and the velocity of water is 2 m/s. What will be the velocity of water where the radius of the pipe is 2.0 cm?

Chapter 16 Solution

Fluid Mechanics — 429

Using equation of continuity, A1 v1 = A 2 v 2 A  v 2 =  1  v1  A2   πr 2  r  v 2 =  12  v1 =  1  v1  r2   πr2  2

or

Substituting the values, we get 2

 1.0 × 10−2   ( 2) v2 =   2.0 × 10−2  v 2 = 0.5 m/s

or V

Ans.

Example 16.14 Calculate the rate of flow of glycerine of density 1.25 × 103 kg /m3 through the conical section of a pipe, if the radii of its ends are 0.1 m and 0.04 m and the pressure drop across its length is 10 N / m2 . Solution

From continuity equation, A1 v1 = A 2 v 2 v1 A 2 πr22  r2  = = =  v 2 A1 πr12  r1 

or

2

2

4  0.04  =  =  0.1  25

…(i)

From Bernoulli’s equation, A1, p1 A2, p2 v1

v2

Fig. 16.45

p1 + or

2( p1 − p 2 ) 1 2 1 ρv1 = p 2 + ρv 22 or v 22 − v12 = 2 2 ρ

v 22 − v12 =

2 × 10 1.25 × 103

= 1.6 × 10−2 m 2/s 2

…(ii)

Solving Eqs. (i) and (ii), we get v 2 ≈ 0.128 m/s ∴ Rate of volume flow through the tube Q = A 2 v 2 = ( πr22 )v 2 = π ( 0.04) 2 ( 0.128) = 6.43 × 10−4 m 3/s

Ans.

430 — Mechanics - II V

Example 16.15 Water is flowing smoothly through a closed-pipe system. At one point the speed of the water is 3.0 m/s, while at another point 1.0 m higher the speed is 4.0 m/s. If the pressure is 20 kPa at the lower point, what is the pressure at the upper point ? What would the pressure at the upper point be if the water were to stop flowing and the pressure at the lower point were 18 kPa ? 1 2 1 ρv1 + ρgh1 = p 2 + ρv 22 + ρgh2 2 2 1 1 ( 20 × 103 ) + × 103 × ( 3) 2 + 0 = p 2 + × 103 × ( 4 ) 2 + 103 × 10 × 1 2 2

Solution (i)

∴

p1 +

p 2 = 6.5 × 103 N/ m 2 = 6.5 kPa

Ans.

(ii) Again applying the same equation, we have (18 × 103 ) + 0 + 0 = p 2 + 0 + (103 ) (10) (1) ⇒

p 2 = 8 × 103 N/ m 2 = 8 kPa

INTRODUCTORY EXERCISE

Ans.

16.4

1. Water flows through a tube shown in figure. The areas of cross section at A and B are 1cm 2 and 0.5 cm 2 respectively. The height difference between A and B is 5 cm. If the speed of water at A is 10 cms −1, find (a) the speed at B and (b) the difference in pressures at A and B.

A

2. Water flows through a horizontal tube of variable cross section as

B

Fig. 16.46

A B

shown in figure. The area of cross section at A and B are 4 mm 2 and 2

2 mm respectively. If 1 cc of water enters per second through A, find (a) the speed of water at A, (b) the speed of water at B and (c) the pressure difference p A − p B .

Fig. 16.47

3. Water from a tap emerges vertically downwards with an initial speed of 1.0 m/s. The

cross-sectional area of tap is 10−4 m 2. Assume that the pressure is constant throughout the stream of water and that the flow is steady, the cross-sectional area of stream 0.15 m below the tap is (JEE 1998)

(a) 5.0 × 10−4 m 2

(b) 1.0 × 10−4 m 2

(c) 5.0 × 10−5 m 2

(d) 2 .0 × 10−5 m 2

4. A horizontal pipeline carries water in a streamline flow. At a point along the pipe, where the cross-sectional area is 10 cm 2, the water velocity is 1 ms −1 and the pressure is 2000 Pa. The

pressure of water at another point where the cross-sectional area is 5 cm 2, is ...…Pa. (Density of water = 103 kg-m 3) (JEE 1994)

Chapter 16

Fluid Mechanics — 431

16.7 Application Based on Bernaulli’s Equation Atomizer or Spray Gun Fig. 16.48 shows the essential parts of spray gun. When the piston is pressed, the air rushes out of the horizontal tube B decreasing the pressure to p2 which is less than the pressure p1 in the container. As a result, the liquid rises up in the vertical tube A. When it collides with the high speed air in tube B, it breaks up into a fine spray. Filter pumps, bunsen burner and sprayers used for perfumes or to spray insecticides work on the same principle. Spray

B p2

p1

A Liquid Container

Fig. 16.48

Principle of Lifting of an Aircraft We all know that airplanes fly. This is a result of a lift force acting on the wings of the aircraft. Lift can be generated when an asymmetric object moves through a fluid. The asymmetry of the object requires the fluid particles to travel different distances along different paths as they pass around the object. The molecules will either speed up or slowdown in order to remain even with the fluid molecules on the other side. For example, in Fig. 16.49, the air particles passing along the top of the body must travel a greater distance than those travelling below. LIFT High velocity

Low velocity

Low pressure

High pressure

Fig. 16.49. The principle of lift of an aircraft

In order to arrive at the other side of the foil at the same time, these particles must travel more quickly than those travelling below the foil. The difference in the velocities of the fluid molecules results in a pressure difference according to the Bernoulli principle: high velocities are associated with relatively low pressure regions, while low velocities are associated with relatively high pressure regions. This pressure difference produces a force in upward direction.

432 — Mechanics - II Magnus Effect and Spinning of a Ball Spinning objects can also generate force. When an object spins in a fluid medium, the fluid boundary layer spins along with the object. This results in a difference in velocity on two sides. A pressure difference is created from one side of the object to the other and the object curves towards the area of low pressure. This influence of rotation of an object on its flight path is termed the magnus effect. This magnus force changes the path of a rotating cricket ball. This is called spinning of ball. Path of motion

Relative low pressure Relative high velocity flow

Relative high pressure Relative low velocity flow

Fig. 16.50. The magnus effect

Blowing Off the Roof during Wind Storm During certain wind storm or cyclone, the roofs of some houses are blown off without damaging the other parts of the house. The high wind blowing over the roof creates a low pressure p2 in accordance with Bernoulli's principle. The pressure p1 below the roof is equal to the atmospheric pressure which is larger than p2 . The difference of pressure ( p1 − p2 ) causes an upward thrust and the roof is lifted up. Once the roof is lifted up, it is blown off with the wind.

Venturimeter Figure shows a venturimeter used to measure flow speed in a pipe of non-uniform cross-section. We apply Bernoulli’s equation to the wide and narrow parts of the pipe, with h1 = h2 1 1 p1 + ρv12 = p2 + ρv 22 2 2 A1 v1 From the continuity equation v 2 = A2

h p1

p2 v1

v2 A2

A1

Fig. 16.51

Substituting and rearranging, we get 1 p1 − p2 = ρv12 2

 A12   − 1 2  A2 

…(i)

Because A1 is greater than A2 , v 2 is greater than v1 and hence the pressure p2 is less than p1 . A net force to the right accelerates the fluid as it enters the narrow part of the tube (called throat) and a net force to the left slows as it leaves.

Fluid Mechanics — 433

Chapter 16

The pressure difference is also equal to ρgh, where h is the difference in liquid level in the two tubes. Substituting in Eq. (i), we get v1 =

2gh 2

 A1    −1  A2 

Note The discharge or volume flow rate can be obtained as, dV = A1v 1 = A1 dt

2 gh 2

 A1    −1  A2 

Speed of Efflux Suppose, the surface of a liquid in a tank is at a height h from the orifice O on its sides, through which the liquid issues out with velocity v. The speed of the liquid coming out is called the speed of efflux. If the dimensions of the tank be sufficiently large, the velocity of the liquid at its surface may be taken to be zero. Applying Bernoulli’s equation at the surface and just outside the orifice.

h v

H

O

v

H

O

H–h

R Fig. 16.52

1 2 ρv + ρgh + p = constant 2 with h = 0 at the orifice, we have 1 ρgh + p0 = ρv 2 + p0 or v = 2gh 2 Evangelista Torricelli show that this velocity is the same as the liquid will attain in falling freely through the vertical height ( h) from the surface to the orifice. This is known as Torricelli’s theorem and may be stated as, “The velocity of efflux of a liquid issuing out of an orifice is the same as it would attain if allowed to fall freely through the vertical height between the liquid surface and orifice. ”

Range (R) Let us find the range R on the ground. Considering the vertical motion of the liquid, ( H − h) =

1 2 gt 2

or

t=

2( H − h) g

434 — Mechanics - II Now, considering the horizontal motion, R = vt or

 2( H − h)   R = ( 2gh )  g  

or

R = 2 h( H − h)

From the expression of R, following conclusions can be drawn, h

H –h

O

H

v

H –h h

Fig. 16.53

(i) R h = R H − h

as

R h = 2 h( H − h)

This is shown in Fig. 16.53. H and (ii) R is maximum at h = 2 Proof

R max = H . R 2 = 4( Hh − h 2 )

dR 2 =0 dh H That is, R is maximum at h = 2 For R to be maximum,

and

R H − h = 2 ( H − h) h

and

or

R max = 2

H − 2h = 0 or

h=

H 2

H  H H −  = H 2  2

Hence Proved.

Time Taken to Empty a Tank We are here interested in finding the time required to empty a tank if a hole is made at the bottom of the tank. Consider a tank filled with a liquid of density ρ upto a height H. A small hole of area of cross section a is made at the bottom of the tank. The area of cross-section of the tank is A. Let at some instant of time the level of liquid in the tank is y. Velocity of efflux at this instant of time would be, v = 2gy  dV  Now, at this instant volume of liquid coming out of the hole per second is  1  .  dt 

Chapter 16

Fluid Mechanics — 435

 dV  Volume of liquid coming down in the tank per second is  2  .  dt  dV1 dV2 = dt dt

y a

∴

 av = A  − 

∴

 dy  a 2gy = A  −   dt  t

or

∫0 dt = − a

∴

t=

∴

t= V

v

dy   dt 

A

Fig. 16.54

0

∫ 2g H

2A a 2g A a

A

y −1/ 2 dy

…(i)

H

[ y ]0

2H g

Example 16.16 Water flows through a horizontal tube as shown in figure . If the difference of heights of water column in the vertical tubes is 2 cm and the areas of cross-section at A and B are 4cm2 and 2 cm2 respectively. Find the rate of flow of water across any section.

A

B

Fig. 16.55

Solution Applying Bernoulli’s equation at A and B

or

1 1 p A + ρv 2A + ρgh A = p B + ρv B2 + ρghB 2 2 1 2 1 2 ρv B − ρv A = ρg ( h A − h B ) 2 2

or

v B2 − v 2A = 2g ( h A − hB )

or

v B2

( h A = hB )

= 2 × 10 × 0.02

or ∴ Solving Eqs. (i) and (ii), we get

−

v 2A

= 0.4 m 2 /s 2

v A A A = vB A B 4v A = 2v B v B = 2v A

…(i)

…(ii)

v A = 0.363 m/s Volume flow rate = v A A A = ( 0.365) ( 4 × 10− 4 ) = 1.46 × 10− 4 m 3 /s = 146 cm 3 / s

Ans.

436 — Mechanics - II V

Example 16.17 Water flows through the tube as shown in figure. The areas of cross-section of the wide and the narrow portions of the tube are 5 cm2 and 2 cm2 respectively. The rate of flow of water through the tube is 500 cm3 / s. Find the difference of mercury levels in the U-tube. 2 1

Fig. 16.56

Solution Applying continuity equation

A1 v1 = A 2 v 2 = rate of flow of water 5v1 = 2v 2 = 500 cm 3 / s ∴

v1 = 100 cm/s = 1.0 m/s v 2 = 250 cm/ s = 2.5 m/s Now, applying Bernoulli’s equation at 1 and 2 1 1 p1 + ρ w v12 + ρ w gh1 = p 2 + ρ w v 22 + ρ w gh2 2 2 1 ρ w ( v 22 − v12 ) = p1 − p 2 = ρ Hg ghHg ⇒ 2 1 or ρ w ( v 22 − v12 ) = ρ Hg g hHg 2

∴

hHg = =

( h1 = h2 )

ρ w ( v 22 − v12 ) 2 ρ Hg g 103 ( 6.25 − 1) 2 × 13.6 × 103 × 9.8

= 0.0196 m = 1.96cm V

Ans

Example 16.18 A tank is filled with a liquid upto a height H. A small hole is made at the bottom of this tank. Let t1 be the time taken to empty first half of t the tank and t2 the time taken to empty rest half of the tank. Then find 1 . t2 Solution

Substituting the proper limits in Eq. (i), derived in the theory, we have t1

∫0

dt = −

A a

H /2

∫ 2g H

y −1/ 2 dy

Chapter 16 t1 =

or

2A

H

a 2g

[ y ]H / 2

or

t1 =

2A   H − a 2g 

or

t1 =

A a

Similarly,

t2

∫0

dt = − t2 =

or

A a

Fluid Mechanics — 437

H  2

H ( 2 − 1) g A

0

a 2g

∫H / 2

…(ii)

y −1/ 2 dy

H g

…(iii)

From Eqs. (ii) and (iii), we get t1 = 2−1 t2 t1 = 0.414 t2

or

Ans.

Note From the above example we can see that t 1 < t 2 . This is because initially the pressure is high and the liquid comes out with greater speed.

INTRODUCTORY EXERCISE

16.5

1. There is a small hole at the bottom of tank filled with water. If total pressure at the bottom is 3 atm (1 atm = 105 Nm −2), then find the velocity of water flowing from hole.

2. Liquid is filled in a container upto a height of H. A small hole is made at the bottom of the tank. Time taken to empty from H to

H H is t 0. Find the time taken to empty the tank from to zero. 3 3

3. Water is filled in a cylindrical container to a height of 3 m. The ratio of the cross-sectional area of the orifice and the beaker is 0.1. The square of the speed of the liquid coming out from the orifice is ( g = 10 m /s 2 ) (JEE 2005)

3m 52.5 cm

Fig. 16.57 2

(a) 50 m /s

2

(c) 51m 2 /s 2

(b) 50.5 m 2 /s 2 (d) 52 m 2 /s 2

438 — Mechanics - II

16.8 Viscosity Viscosity is internal friction in a fluid. Viscous forces oppose the motion of one portion of a fluid relative to the other. F

v

y

x

Fig. 16.58

The simplest example of viscous flow is motion of a fluid between two parallel plates. The bottom plate is stationary and the top plate moves with constant velocity v. The fluid in contact with each surface has same velocity at that surface. The flow speeds of intermediate layers of fluid increase uniformly from bottom to top, as shown by arrows. So, the fluid layers slide smoothly over one another. According to Newton, the frictional force F (or viscous force) between two layers depends upon the following factors, (i) Force F is directly proportional to the area ( A ) of the layers in contact, i.e. F ∝A  dv  (ii) Force F is directly proportional to the velocity gradient   between the layers. Combining  dy  these two, we have F ∝A

dv dy

or

F = − ηA

dv dy

Here, η is constant of proportionality and is called coefficient of viscosity. Its value depends on the nature of the fluid. The negative sign in the above equation shows that the direction of viscous force F is opposite to the direction of relative velocity of the layer. The SI unit of η is N-s/m 2 . It is also called decapoise or pascal second. Thus, 1 decapoise = 1 N-s /m 2 = 1 Pa-s = 10 poise The SI unit of viscosity is sometimes referred to as the poiseuille (symbol Pl). Thus,

1 Pl =1 N-s/ m 2

Dimensions of η are [ML–1T –1 ]. Coefficient of viscosity of water at 10°C is η = 1.3 × 10 −3 N-s/m 2 . Experiments show that coefficient of viscosity of a liquid decreases as its temperature rises.

Chapter 16

Fluid Mechanics — 439

Stoke’s Law and Terminal Velocity When an object moves through a fluid, it experiences a viscous force which acts in opposite direction of its velocity. The mathematics of the viscous force for an irregular object is difficult, we will consider here only the case of a small sphere moving through a fluid. The formula for the viscous force on a sphere was first derived by the English physicist G. Stokes in 1843. According to him, a spherical object of radius r moving at velocity v experiences a viscous force given by (η = coefficient of viscosity) F = 6πηrv This law is called Stoke’s law.

Terminal Velocity (v T ) Consider a small sphere falling from rest through a large column of viscous fluid. The forces acting on the sphere are, (i) Weight w of the sphere acting vertically downwards (ii) Upthrust Ft acting vertically upwards (iii) Viscous force Fv acting vertically upwards, i.e. in a direction opposite to velocity of the sphere. Fv = 0

Initially,

(as v = 0)

Ft + Fv

v

w

Fig. 16.59

w > Ft

and

and the sphere accelerates downwards. As the velocity of the sphere increases, Fv increases. Eventually a stage in reached when w = Ft + Fv

…(i)

After this net force on the sphere is zero and it moves downwards with a constant velocity called terminal velocity ( vT ) . v

vT

t

O

Fig. 16.60

Substituting proper values in Eq. (i) we have, 4 3 4 πr ρg = πr 3σg + 6πηrvT 3 3 Here, ρ = density of sphere, σ = density of fluid and From Eq. (ii), we get

η = coefficient of viscosity of fluid vT =

2 r 2 (ρ − σ ) g 9 η

Figure shows the variation of the velocity v of the sphere with time.

…(ii)

440 — Mechanics - II Extra Points to Remember ˜

˜

Terminal velocity vT ∝ r 2 If the fluid is air, then its density σ is negligible compared to density of sphere. So, in that case upthrust will be zero and terminal velocity will be vT =

˜

2 r 2ρ g 9 η

In this case, weight is equal to the viscous force when terminal velocity is attained. If density of fluid is greater than density of sphere (σ >ρ) then terminal velocity comes out to be negative. Ft vT

W + FV

Fig. 16.61

So, in this case terminal velocity is upwards. In the beginning upthrust is greater than the weight. Viscous force in this case will be downwards. vT =

2 r 2 (σ − ρ)g 9 η

When terminal velocity is attained,

Ft = W + Fv This is the reason why, air bubbles rise up in water. V

Example 16.19 A plate of area 2 m2 is made to move horizontally with a speed of 2 m/s by applying a horizontal tangential force over the free surface of a liquid. If the depth of the liquid is 1 m and the liquid in contact with the bed is stationary. Coefficient of viscosity of liquid is 0.01 poise. Find the tangential force needed to move the plate. Solution

Velocity gradient =

m/s ∆v 2 − 0 = =2 ∆y 1 − 0 m v = 2 m/s F

1m

Fig. 16.62

From, Newton’s law of viscous force, | F | = ηA

∆v ∆y

= ( 0.01 × 10−1 )( 2)( 2) = 4 × 10−3 N. So, to keep the plate moving, a force of 4 × 10−3 N must be applied.

Ans.

Chapter 16 V

Fluid Mechanics — 441

Example 16.20 Two spherical raindrops of equal size are falling vertically through air with a terminal velocity of 1 m/s. What would be the terminal speed if these two drops were to coalesce to form a large spherical drop? vT ∝ r 2

Solution

…(i)

Let r be the radius of small rain drops and R the radius of large drop. Equating the volumes, we have 4 4  πR 3 = 2  πr 3    3 3 ∴

R = ( 2)1/ 3 . r or

R = ( 2)1/ 3 r

vT ′  R  =   = ( 2) 2/ 3  r vT 2

∴ ∴

v T ′ = ( 2) 2/ 3 v T = ( 2) 2/ 3 (1.0) m/s = 1.587 m/s

V

Ans.

Example 16.21 With what terminal velocity will an air bubble 0.8 mm in diameter rise in a liquid of viscosity 0.15 N-s/m2 and specific gravity 0.9 ? Density of air is 1.293 kg / m3 . Solution

The terminal velocity of the bubble is given by, 2 r 2 (ρ − σ )g vT = 9 η

Here,

r = 0.4 × 10−3 m, σ = 0.9 × 103 kg / m 3 , ρ = 1.293 kg / m 3 , η = 0.15 N-s/m 2

and

g = 9.8 m/s 2

Substituting the values, we have vT =

2 ( 0.4 × 10−3 ) 2 (1.293 − 0.9 × 103 ) × 9.8 × 9 0.15

= − 0.0021 m/s or

v T = − 0.21cm

Ans.

Note Here, negative sign implies that the bubble will rise up. V

Example 16.22 A spherical ball of radius 3.0 × 10 −4 m and density 104 kg/m3 falls freely under gravity through a distance h before entering a tank of water. If after entering the water the velocity of the ball does not change, find h. Viscosity of water is 9.8 × 10 −6 N - s/ m2 . Solution

Before entering the water the velocity of ball is 2gh. If after entering the water this

velocity does not change then this value should be equal to the terminal velocity. Therefore, 2gh =

2 r 2 (ρ − σ )g 9 η

442 — Mechanics - II  2 r 2 (ρ − σ )g    9 η   h= 2g

∴

2

=

2 r 4 (ρ − σ ) 2 g × 81 η2

=

2 ( 3 × 10−4 ) 4 (104 − 103 ) 2 × 9.8 × 81 ( 9.8 × 10−6 ) 2

= 165 . × 103 m V

Ans.

Example 16.23 A small sphere falls from rest in a viscous liquid. Due to friction, heat is produced. Find the relation between the rate of production of heat and the radius of the sphere at terminal velocity. (JEE 2004) Solution Terminal velocity v T =

2r 2 g (ρ S − ρ L ) 9η

and viscous force F = 6πηrv T Rate of production of heat (power) : as viscous force is the only dissipative force. Hence, dQ = Fv T = ( 6πηrv T )( v T ) dt = 6πηrv T2  2 r2 g  = 6πηr (ρ S − ρ L ) 9 η  =

2

8πg 2 dQ (ρ S − ρ L ) 2 r 5 or ∝ r5 27 η dt

INTRODUCTORY EXERCISE

Ans.

16.6

1. A typical riverborne silt particle has a radius of 20 µm and a density of 2 × 103 kg/m 3. The viscosity of water is 1.0 mPl. Find the terminal speed with which such a particle will settle to the bottom of a motionless volume of water.

2. Two equal drops of water are falling through air with a steady velocity v. If the drops coalesced, what will be the new velocity ?

3. A large wooden plate of area 10 m 2 floating on the surface of a river is made to move horizontally with a speed of 2 m/s by applying a tangential force. If the river is 1 m deep and the water in contact with the bed is stationary, find the tangential force needed to keep the plate moving. Coefficient of viscosity of water at the temperature of the river = 10−2 poise.

4. The velocity of water in a river is 18 km/h near the surface. If the river is 5 m deep, find the shearing stress between the horizontal layers of water. The coefficient of viscosity of water = 10−2 poise.

Chapter 16

Fluid Mechanics — 443

16.9 Surface Tension A needle can made to float on a water surface if it is placed there carefully. The forces that support the needle are not buoyant forces but are due to surface tension. The surface of a liquid behaves like a membrane under tension. The molecules of the liquid exert attractive forces on each other. There is zero net force on a molecule inside the volume of the liquid.But a surface molecule has a net force towards inside of the liquid. Thus, the liquid tends to minimize its surface area, just as a stretched membrane does.

Fig. 16.63

Freely falling raindrops are spherical because a sphere has a smaller surface area for a given volume than any other shape. Hence, the surface tension can be defined as the property of a liquid at rest by virtue of which its free surface behaves like a stretched membrane under tension and tries to occupy as small area as possible. Let an imaginary line AB be drawn in any direction in a liquid surface. The surface on B F either side of this line exerts a pulling force on the surface on the other side. This force F is at right angles to the line AB. The magnitude of this force per unit length of AB is A taken as a measure of the surface tension of the liquid. Thus, if F be the total force acting on either side of the line AB of length L, then the surface tension is given by, Fig. 16.64 F T= L Hence, the surface tension of a liquid is defined as the force per unit length in the plane of the liquid surface, acting at right angles on either side of an imaginary line drawn on that surface.

Examples of Surface Tension Example 1 Take a ring of wire and dip it in a soap solution. When the ring is taken out, a soap film is formed. Place a loop of thread gently on the soap film. Now, prick a hole inside the loop. The thread is radially pulled by the film surface outside and it takes a circular shape.

Fig. 16.65

Before the pricking, there were surfaces both inside and outside the thread loop. Surfaces on both sides pull it equally and the net force is zero. Once the surface inside was punctured, the outside surface pulled the thread to take the circular shape so that area outside the loop becomes minimum (because for given perimeter area of circle is maximum). Reason

444 — Mechanics - II A piece of wire is bent into a U-shape and a second piece of wire slides on the arms of the U. When the apparatus is dipped into a soap solution and removed, a liquid film is formed. The film exerts a surface tension force on the slider and if the frame is kept in a horizontal position, the slider quickly slides towards the closing arm of the frame. If the frame is kept vertical, one can have some weight to keep it in equilibrium. This shows that the soap surface in contact with the slider pulls it parallel to the surface.

Example 2

w

Fig. 16.66

Example 3 Needle supported on water surface Take a greased needle of steel on a piece of blotting paper and place it gently over the water surface. Blotting paper soaks water and soon sinks down but the needle keeps floating. The floating needle causes a little depression. The forces F and F due to surface tension of the curved surface are inclined as shown in Fig.16.67. The vertical components of these two forces support the weight of the needle. F

F

Water W

Fig. 16.67

Example 4 Small mercury droplets are spherical and larger ones tend to flatten. Reason Small mercury droplets are spherical because the

forces of surface tension tend to reduce their area to a minimum value and a sphere has minimum surface area for a Fig. 16.68 given volume. Larger drops of mercury are flattened due to the large gravitational force acting on them. Here the shape is such that the sum of the gravitational potential energy and the surface potential energy must be minimum. Hence the centre of gravity moves down as low as possible. This explains flattening of the larger drops. Example 5 The hair of a painting brush cling together when taken out of water.

Fig. 16.69

Reason

This is because the water films formed on them tend to contract to minimum area.

Note The surface tension of a particular liquid usually decreases as temperature increases. To wash clothing thoroughly, water must be forced through the tiny spaces between the fibers. This requires increasing the surface area of the water ,which is difficult to do because of surface tension. Hence, hot water and soapy water is better for washing.

Chapter 16

Fluid Mechanics — 445

Surface Energy When the surface area of a liquid is increased, the molecules from the interior rise to the surface. This requires work against force of attraction of the molecules just below the surface. This work is stored in the form of potential energy. Thus, the molecules in the surface have some additional energy due to their position. This additional energy per unit area of the surface is called ‘surface energy’. The surface energy is related to the surface tension as discussed below. dx

2Tl

F

Fig. 16.70

Let a liquid film be formed on a wire frame and a straight wire of length l can slide on this wire frame as shown in figure. The film has two surfaces and both the surfaces are in contact with the sliding wire and hence, exert forces of surface tension on it. If T be the surface tension of the solution, each surface will pull the wire parallel to itself with a force Tl. Thus, net force on the wire due to both the surfaces is 2Tl. One has to apply an external force F equal and opposite to it to keep the wire in equilibrium. Thus, F = 2Tl Now, suppose the wire is moved through a small distance dx, the work done by the force is, dW = F dx = (2Tl) dx But (2l) ( dx ) is the total increase in area of both the surfaces of the film. Let it be dA. Then,

or

dW T= dA

dW = T dA ∆W or ∆A

Thus, the surface tension T can also be defined as the work done in increasing the surface area by unity. Further, since there is no change in kinetic energy, the work done by the external force is stored as the potential energy of the new surface. dU ∆U or (as dW = dU ) T= ∴ dA ∆A Thus, the surface tension of a liquid is equal to the surface energy per unit surface area.

Excess Pressure inside a Bubble or Liquid Drop Surface tension causes a pressure difference between the inside and outside of a soap bubble or a liquid drop.

Excess Pressure Inside Soap Bubble A soap bubble consists of two spherical surface films with a thin layer of liquid between them. Because of surface tension, the film tend to contract in an attempt to minimize their surface area. But

446 — Mechanics - II as the bubble contracts, it compresses the inside air, eventually increasing the interior pressure to a level that prevents further contraction.

O R

Fig. 16.71

We can derive an expression for the excess pressure inside a bubble in terms of its radius R and the surface tension T of the liquid. Each half of the soap bubble is in equilibrium. The lower half is shown in figure. The forces at the flat circular surface where this half joins the upper half are Surface tension force = (2T )(2πR )

Force due to pressure difference = (∆p) πR 2

Fig. 16.72

(a) The upward force of surface tension. The total surface tension force for each surface (inner and outer) is T (2πR ), for a total of (2T )(2πR ). (b) downward force due to pressure difference. The magnitude of this force is ( ∆p)( πR 2 ). In equilibrium these two forces have equal magnitude. ∴ or

(2T )(2πR ) = (∆P )(πR 2 ) 4T ∆p = R

Note Suppose, the pressure inside the air bubble is P, then p − p0 =

4T R

P0 P R

Fig. 16.73

Excess Pressure Inside a Liquid Drop A liquid drop has only one surface film. Hence, the surface tension force is T (2πR ), half that for a soap bubble. Thus, in equilibrium, T (2πR ) = ∆p( πR 2 ) or

∆p =

2T R

Chapter 16

Fluid Mechanics — 447

Note (i) If we have an air bubble inside a liquid, a single surface is formed. There is air on the concave side and liquid on the convex side. The pressure in the concave p2 side (that is in the air) is greater than the pressure in the convex side (that is in p1 2T the liquid) by an amount . R 2T p2 − p1 = ∴ Fig. 16.74 R The above expression has been written by assuming P1 to be constant from all sides of the bubble. For small size bubbles this can be assumed. (ii) From the above discussion, we can make a general statement. The pressure on the concave side of a 2T spherical liquid surface is greater than the convex side by . R

Extra Points to Remember ˜

Radius of curvature of a curve : To describe the shape of a curved surface or interface, it is necessary to know the radii of curvature to a curve at some point. Consider the curve AB as shown in figure: y A

P

R B x

O

Fig. 16.75

Let P be a point on this curve. The radius of curvature R of AB at P is defined as the radius of the circle which is tangent to the curve at point P. ˜

Principal radii of curvature of a surface : The spherical and cylindrical surfaces are rather simple cases for mathematical treatment. In many other cases however, the shapes are more complicated. Let us now consider a curved surface. At each point on a given surface, two radii of curvature (which are denoted by r1 and r2 ) are required to describe the shape. If we want to determine these radii at any point (say P), the normal to the surface at this point is drawn and a plane is constructed through the surface containing the normal. This will intersect the surface in a plane curve. The radius of curvature of the curve at point P is denoted by r1. An infinite number of such planes can be constructed each of which intersects the surface at P. For each of these planes, a radius of curvature can be obtained. If we construct a second plane through the surface, containing the normal and perpendicular to the first plane, the second line of intersection and hence the second radius of curvature at point P (i.e., r2 ) is obtained. These two radii define the curvature at P completely. It can be shown that (1/ r1 + 1/ r2 ) called mean radius of curvature of the surface is constant, which is independent of the choice of the planes. An infinite set of such pairs of radii is possible. For standardization, the first plane is rotated around the normal until the radius of curvature in that plane reaches minimum. The other radius of curvature is therefore maximum. These are the principal radii of curvature (denoted by R1 and R 2 ).

˜

Young-Laplace Equation : There exists a difference in pressure across a curved surface which is a consequence of surface tension. The pressure is greater on the concave side. The Laplace equation relates the pressure difference to the shape of the Young surface.

448 — Mechanics - II This difference in pressure is given by  1 1 ∆p = T  +   R1 R 2  The simplified forms of spherical, cylindrical and planar surfaces are given below For a spherical surface: 2T R1 = R 2 = R (say), therefore ∆p = R For a cylindrical surface: R1 = R and R 2 = ∞, therefore ∆p =

T R

For a planer surface:

˜

˜

R1 = R 2 = ∞, therefore ∆p= 0 As we know that molecules of a liquid reach to its surface after struggling with the net inward force acting on them from other liquid molecules. So, we can say that the surface molecules have some extra energy compared to inner molecules and every system has a tendency to keep its energy minimum. So, a liquid also has a tendency to keep its energy minimum by putting least number of molecules on the surface or by making its surface area minimum. This property of a liquid is called surface tension. Surface tension is a property of a liquid. It does not depend on the surface area. A1 A2

(i)

A2 > A1

(ii)

Fig. 16.76

For example, both containers have the same liquid under same conditions. A2 > A1 but surface tension will be same in both cases. V

Example 16.24 How much work will be done in increasing the diameter of a soap bubble from 2 cm to 5 cm ? Surface tension of soap solution is 3.0 × 10 −2 N / m . Solution

Soap bubble has two surfaces. Hence, W = T ∆A

Here,

∆W   T =   ∆A 

∆A = 2 [ 4π {( 2.5 × 10−2 ) 2 − (1.0 × 10−2 ) 2 }] = 1.32 × 10−2 m 2

∴

W = ( 3.0 × 10−2 )(1.32 × 10−2 ) J = 3.96 × 10–4 J

V

Ans.

Example 16.25 Calculate the energy released when 1000 small water drops each of same radius 10 −7 m coalesce to form one large drop. The surface tension of water is 7.0 × 10 −2 N/m.

Chapter 16

Fluid Mechanics — 449

Let r be the radius of smaller drops and R of bigger one. Equating the initial and final volumes, we have 4 4  πR 3 = (1000)  πr 3  3  3

Solution

or

R = 10r = (10)(10−7 ) m

or

R = 10−6 m

Further, the water drops have only one free surface. Therefore, ∆A = 4πR 2 − (1000)( 4πr 2 ) = 4π [(10−6 ) 2 − (103 )(10−7 ) 2 ] = − 36π(10−12 ) m 2 Here, negative sign implies that surface area is decreasing. Hence, energy released in the process. U = T | ∆A | = ( 7 × 10−2 )( 36π × 10−12 ) J = 7.9 × 10−12 J V

Ans.

Example 16.26 What should be the pressure inside a small air bubble of 0.1 mm radius situated just below the water surface. Surface tension of water = 7.2 × 10 −2 N/m and atmospheric pressure = 1.013 × 105 N/m2 . Solution

Surface tension of water T = 7.2 × 10−2 N/m

Radius of air bubble R = 0.1 mm = 10–4 m The excess pressure inside the air bubble is given by, 2T p 2 − p1 = R ∴ Pressure inside the air bubble, 2T p 2 = p1 + R Substituting the values, we have p 2 = (1.013 × 105 ) +

( 2 × 7.2 × 10−2 )

= 1.027 × 105 N/ m 2 V

10−4 Ans.

Example 16.27 Two separate air bubbles (radii 0.004 m and 0.002 m) formed of the same liquid (surface tension 0.07 N/m) come together to form a double bubble. Find the radius and the sense of curvature of the internal film surface common to both the bubbles. 4T 4T Solution p1 = p 0 + ⇒ p2 = p0 + r1 r2 ∴

r2 < r1 p 2 > p1

450 — Mechanics - II i.e. pressure inside the smaller bubble will be more. The excess pressure r2 p

p2 p1

r1

Fig. 16.77

 r − r2  p = p 2 − p1 = 4T  1   r1 r2 

…(i)

This excess pressure acts from concave to convex side, the interface will be concave towards smaller bubble and convex towards larger bubble. Let R be the radius of interface then, 4T …(ii) p= R From Eqs. (i) and (ii), we get rr ( 0.004 )(0.002) R= 12 = r1 − r2 (0.004 – 0.002) = 0.004 m V

Ans

Example 16.28 Under isothermal condition two soap bubbles of radii r1 and r2 coalesce to form a single bubble of radius r. The external pressure is p0 . Find the surface tension of the soap in terms of the given parameters. Solution

As mass of the air is conserved,

∴

n1 + n 2 = n p1V1 pV pV + 2 2 = RT1 RT2 RT

∴

(as pV = nRT )

As temperature is constant, p p1 r1

p2 +

= r2

r

Fig. 16.78

∴ ∴

T1 = T2 = T p1V1 + p 2V2 = pV  4S   4 3  4S   4 3   4S   4 3    p0 +   πr1  +  p 0 +   πr2  =  p 0 +   πr       r1   3 r2   3 r  3 

Chapter 16

Fluid Mechanics — 451

Solving, this we get S =

p 0 ( r 3 − r13 − r23 )

Ans.

4( r12 + r22 − r 2 )

To avoid confusion with the temperature, surface tension here is represented by S.

Note V

Example 16.29 Assume that a drop of liquid evaporates by decrease in its surface energy, so that its temperature remains unchanged. What should be the minimum radius of the drop for this to be possible? The surface tension is T , density of liquid is ρ and L is its latent heat of vaporization (2013 Main) (a)

ρL T

T ρL

(b)

(c)

T ρL

(d)

2T ρL

Solution Decrease in surface energy = heat required in vaporization.

∴

T ( dS ) = L( dm )

∴

T ( 2) ( 4πr ) dr = L( 4πr 2 dr ) ρ

∴

r=

2T ρL

The correct options is (d).

INTRODUCTORY EXERCISE

16.7

1. A mercury drop of radius 1 cm is sprayed into 106 droplets of equal size. Calculate the increase in surface energy if surface tension of mercury is 35 × 10−3 N/m.

2. A water film is made between two straight parallel wires of length 10 cm each and at a distance of 0.5 cm from each other. If the distance between the wires is increased by 1 mm, how much work will be done? Surface tension of water = 7.2 × 10−2 N/m

3. A soap bubble of radius R has been formed at normal temperature and pressure under isothermal conditions. Compute the work done. The surface tension of soap solution is T.

4. A small air bubble of radius ‘r’ is at a depth ‘h’ below the water surface (density of water = ρ). Surface tension of water is T, atmospheric pressure is p 0. Find pressure inside the air bubble for the condition r 90° less than 90° (also called acute angle). However, those liquids which don't wet Fig. 16.79 the walls of the container (say in case of mercury and glass) have meniscus convex upwards and their value of angle of contact is greater than 90° (also called obtuse angle). The angle of contact of mercury with glass is about 140°, whereas the angle of contact of water with glass is about 8°. But, for pure water, the angle of contact θ with glass is taken as 0°.

Shape of Liquid Meniscus When the adhesive force ( P ) between solid and liquid molecules is more than the cohesive force (Q ) between liquid-liquid molecules (as with water and glass), shape of the meniscus is concave and the angle of contact θ is less than 90°. In this case, the liquid wets or adheres to the solid surface. The resultant ( R ) of P and Q passes through the solid. Glass P R

Q Water Fig. 16.80

Chapter 16

Fluid Mechanics — 453

On the other hand when P < Q (as with glass and mercury), shape of the meniscus is convex and the angle of contact θ > 90°. The resultant ( R ) of P and Q in this case passes through the liquid. Glass

P R

Q Mercury

Fig. 16.81

Let us now see why the liquid surface bends near the contact with a solid. A liquid in equilibrium can not sustain tangential stress. The resultant force on any small part of the surface layer must be perpendicular to the surface at that point. Basically three forces are acting on a small part of the liquid surface near its contact with solid. These forces are, (i) P, attraction due to the molecule of the solid surface near it (ii) Q, attraction due to liquid molecules near this part and (iii) The weight w of the part considered. We have considered very small part, so weight of that part can be ignored for better understanding. As we have seen in the last figures, to make the resultant ( R ) of P and Q perpendicular to the liquid surface the surface becomes curved (convex or concave).

Capillarity Surface tension causes elevation or depression of the liquid in a narrow tube. This effect is called capillarity. When a glass capillary tube (A tube of very small diameter is called a capillary tube) open at both ends is dipped vertically in water the water in the tube will rise above the level of water in the vessel as shown in figure (a). In case of mercury, the liquid is depressed in the tube below the level of mercury in the vessel as shown in figure (b).

(a)

(b)

Fig. 16.82

When the contact angle is less than 90° the liquid rises in the tube. For a non-wetting liquid angle of contact is greater than 90° and the surface is depressed, pulled down by the surface tension forces.

Explanation When a capillary tube is dipped in water, the water meniscus inside the tube is concave. The pressure 2T just below the meniscus is less than the pressure just above it by , where T is the surface tension of R water and R is the radius of curvature of the meniscus. The pressure on the surface of water is p0 , the atmospheric pressure. The pressure just below the plane surface of water outside the tube is also

454 — Mechanics - II 2T .We know that pressure at all points in R 2T the same level of water must be the same. Therefore, to make up the deficiency of pressure below R the meniscus water begins to flow from outside into the tube. The rising of water in the capillary stops 2T at a certain height h. In this position, the pressure of water column of height h becomes equal to , R i.e. 2T hρg = R 2T r or h= θ Rρg R p0 , but that just below the meniscus inside the tube is p0 −

If r is the radius of the capillary tube and θ the angle of contact, then r R= cos θ

θ

Fig. 16.83

2T cos θ h= rρg

∴

Alternative Proof for the Formula of Capillary Rise As we have already seen, when the contact angle is less than 90°, the total surface tension force just balances the extra weight of the liquid in the tube. The water meniscus in the tube is along a circle of circumference 2πr which is in contact with the glass. Due to the surface tension of water, a force equal to T per unit length acts at all points of the circle. If the angle of contact is θ, then this force is directed inward at an angle θ from the wall of the tube. T cos θ T cos θ T θ T sin θ h

T cos θ (2πr )

θ T θ

θ

T sin θ

Fig. 16.84

In accordance with Newton’s third law, the tube exerts an equal and opposite force T per unit length on the circumference of the water meniscus. This force which is directed outward, can be resolved into two components T cos θ per unit length acting vertically upward and T sin θ per unit length acting horizontally outward. Considering the entire circumference 2πr, for each horizontal component T sin θ there is an equal and opposite component and the two neutralise each other. The vertical components being in the same direction are added up to give a total upward force (2πr )(T cos θ ). It is this force which supports the weight of the water column so raised.

Chapter 16

Fluid Mechanics — 455

Thus, (T cos θ )(2πr ) = Weight of the liquid column. = ( πr 2ρgh) ∴

h=

…(i)

2T cos θ rρg

The result has following notable features : (i) If the contact angle θ is greater than 90°, the term cos θ is negative and hence, h is negative. The expression then gives the depression of the liquid in the tube. r

r Fig. 16.85

(ii) The correction due to weight of the liquid contained in the meniscus can be made for contact angle θ = 0° . The meniscus is then hemispherical. The volume of the shaded part is

h

Fig. 16.86

V = ( πr 2 )( r ) −

1 4 3 1 3  πr  = πr  3 2 3

1 3 πr ρg. 3 Therefore, we can write Eq. (i) as, The weight of this shaded part is

(T cos 0° )(2πr ) = πr 2ρgh + h=

or

1 3 πr ρg 3

2T r − rρg 3

(iii) Suppose a capillary tube is held vertically in a liquid which has a concave meniscus, then capillary rise is given by, h= or

hR =

2T cos θ 2T = rρg Rρg 2T ρg

 r   as R =  cos θ   …(ii)

456 — Mechanics - II When the length of the tube is greater than h, the liquid rises in the tube, so as to satisfy the above relation. But if the length of the tube is insufficient (i.e. less than h) say h′ , the liquid does not emerge in the form of a fountain from the upper end (because it will violate the law of conservation of energy) but the angle made by the liquid surface and hence, the R changes in such a way that the force 2πrT cos θ equals the weight of the liquid raised. Thus, 2πrT cos θ ′ = πr 2ρgh′ h′ =

2T cos θ ′ rρg

or

h′ =

2T R ′ ρg

or

h′ R ′ =

2T ρg

hR = h′ R ′ =

From Eqs. (ii) and (iii), we have

…(iii) 2T ρg

Practical Applications of Capillarity 1. The oil in a lamp rises in the wick by capillary action. 2. The tip of nib of a pen is split up, to make a narrow capillary so that the ink rises upto the tin or nib continuously. 3. If one end of the towel dips into a bucket of water and the other end hangs over the bucket, the towel soon becomes wet throughout, due to capillary action. 4. Ink is absorbed by the blotter due to capillary action. V

Example 16.30 A capillary tube whose inside radius is 0.5 mm is dipped in water having surface tension 7.0 × 10 −2 N/m. To what height is the water raised above the normal water level? Angle of contact of water with glass is 0°. Density of water is 103 kg/m3 and g = 9.8 m/s2 . Solution

h=

2T cos θ rρg

Substituting the proper values, we have h=

( 2)( 7.0 × 10−2 ) cos 0° ( 0.5 × 10−3 )(103 )( 9.8)

= 2.86 × 10–2 m = 2.86 cm V

Ans.

Example 16.31 A glass tube of radius 0.4 mm is dipped vertically in water. Find upto what height the water will rise in the capillary ? If the tube in inclined at an angle of 60° with the vertical, how much length of the capillary is occupied by water ? Surface tension of water = 7.0 × 10 −2 N/m, density of water = 103 kg/m3 .

Chapter 16 Solution

Fluid Mechanics — 457

For glass-water, angle of contact θ = 0° . 60°

h

l

Fig. 16.87

2T cos θ h= rρg

Now,

=

( 2)( 7.0 × 10−2 ) cos 0° ( 0.4 × 10−3 )(103 )( 9.8)

= 3.57 × 10−2 m = 3.57cm l=

V

h 3.57 = = 7.14 cm 1 cos 60° 2

Ans. Ans.

Example 16.32 Mercury has an angle of contact of 120° with glass. A narrow tube of radius 1.0 mm made of this glass is dipped in a trough containing mercury. By what amount does the mercury dip down in the tube relative to the liquid surface outside. Surface tension of mercury at the temperature of the experiment is 0.5 N/m and density of mercury is 13.6 × 103 kg / m3 . (Take g = 9.8 m/s2 ). Solution

h=

2T cos θ rρg

Substituting the values, we get

h=

2 × 0.5 × cos 120° 10

−3

× 13.6 × 103 × 9.8

= − 3.75 × 10−3 m or Note

h = − 3.75 mm

Ans.

Here, negative sign implies that mercury suffers capillary depression. V

Example 16.33 If a 5 cm long capillary tube with 0.1 mm internal diameter open at both ends is slightly dipped in water having surface tension 75 dyne cm −1 , state whether (a) water will rise half way in the capillary, (b) Water will rise up to the upper end of capillary and (c) water will overflow out of the upper end of capillary? Explain your answer. Solution Given that surface tension of water, T = 75 dyne/cm 01 . Radius r = mm = 0.05 mm = 0.005 cm 2 Density ρ = 1gm/cm 3 , angle of contact, θ = 0°

458 — Mechanics - II Let h be the height to which water rises in the capillary tube. Then 2T cos θ 2 × 75 × cos 0° cm h= = 0.005 × 1 × 981 rρg = 30.58 cm But length of capillary tube is h′ = 5 cm h′ (a) Because h > therefore the first possibility does not exist. 2 (b) Because the tube is of insufficient length therefore the water will rise upto the upper end of the tube. (c) The water will not overflow of the upper end of the capillary. It will rise only up to the upper end of the capillary. The liquid meniscus will adjust its radius of curvature R′ in such a way that   2T = constant Q hR = ρ g  

R ′ h′ = Rh

where, R is the radius of curvature that the liquid meniscus would possess if the capillary tube were of sufficient length. ∴

R′ =

Rh rh 0.005 × 30.58 = = = 0.0306 cm h′ h′ 5

INTRODUCTORY EXERCISE

  r r = = r Q R = cos θ cos 0 °  

16.8

1. Water rises in a capillary tube to a height of 2.0 cm. In another capillary tube whose radius is one third of it, how much the water will rise?

2. Water rises up in a glass capillary upto a height of 9.0 cm, while mercury falls down by 3.4 cm in the same capillary. Assume angles of contact for water glass and mercury glass 0° and 135° respectively. Determine the ratio of surface tension of mercury and water (cos 135° = − 0.71).

3. A tube of insufficient length is immersed in water (surface tension = 0.7 N/m) with 1 cm of it projecting vertically upwards outside the water. What is the radius of meniscus ? Given, radius of tube = 1 mm.

4. A capillary tube is dipped in a liquid. Let pressures at points A, B and C be p A, p B and pC respectively, then B A C

Fig. 16.88

(a) p A = p B = pC

(b) p A = p B < pC

(c) p A = pC < p B

(d) p A = pC > p B

Chapter 16

Fluid Mechanics — 459

Final Touch Points 1. Application on Surface Tension (i) When the detergent materials are added to liquids, the angle of contact decreases and hence the wettability increases. On the other hand, when water proofing material is added to a fabric it increases the angle of contact, making the fabric water-repellant. (ii) Surface tension of all lubricating oils and paints is kept low so that they spread over a large area. (iii) Oil spreads over the surface of water because the surface tension of oil is less than the surface tension of cold water.

2. Effect of Temperature and Impurities on Surface Tension The surface tension of a liquid decreases with the rise in temperature and vice-versa. Surface tension becomes zero at a critical temperature. It is for this reason that hot soup tastes better. Surface tension of a liquid changes appreciably with addition of impurities. For example, surface tension of water increases with addition of highly soluble substances like Nacl, ZnSO4 etc. On the other hand surface tension of water gets reduced with addition of sparingly soluble substances like phenol, soap etc.

3. Laminar and Turbulent Flow, Reynolds Number When a liquid flowing in a pipe is observed carefully, it will be seen that the pattern of flow becomes more disturbed as the velocity of flow increases. Perhaps this phenomenon is more commonly seen in a river or stream. When the flow is slow the pattern is smooth, but when the flow is fast, eddies develop and swirl in all directions. At the low velocities, flow is calm. This is called “laminar flow”. In a series of experiments, Reynolds showed this by injecting a thin stream of dye into the fluid and finding that it ran in a smooth stream in the direction of the flow at low speeds. As the velocity of flow increased, he found that the smooth line of dye was broken up, at high velocities, the dye was rapidly mixed into the disturbed flow of the surrounding fluid. This is called “turbulent flow”.

Turbulent

Laminar

After many experiments Reynolds saw that the expression ρud η where, ρ = density, u = mean velocity, d = diameter and η = viscosity would help in predicting the change in flow type. If the value is less than about 2000 then flow is laminar, if greater than 4000 then turbulent and in between these two in the transition zone. This value is known as the Reynolds number, Re. Re =

ρud η

Laminar flow: Re < 2000, Transitional flow: 2000 < Re < 4000, Turbulent flow: Re > 4000 SI Units of Reynolds Number ρ = kg/ m 3, u = m /s , d = m , η = Ns /m 2 = kg/ ms Re =

ρud (kg/m 3 )(m /s )(m ) = =1 η (kg/ms )

i.e. it has no units. A quantity that has no units is known as a non-dimensional (or dimensionless) quantity. Thus, the Reynolds number, Re is a non-dimensional number.

Solved Examples TYPED PROBLEMS Type 1. Based on Law of Floatation. Concept Whenever a block floats in a liquid (density of block should be less than density of liquid), there is one single equation. Weight of solid = Upthrust on solid

1

V ρs g = V i ρl g

∴

Here, V is total volume of solid and V i is immersed volume of solid.

⇒

Vi ρ = fi = s V ρl

∴

(i)

2

Here, f1 is immersed fraction of volume of solid. 1 Further, in figure (i) immersed volume is less and in figure (ii) immersed volume is more. Upthrust on immersed volume in figure (i) is equal to weight of block-1 and upthrust on immersed volume in (ii) figure (ii) is equal to weight of both the blocks 1 and 2. In other words, we can also say that upthrust on extra immersed volume in figure (ii) is equal to extra weight. In equation form, we can write as m1g = (V i )1 ρl g and or Note V

( m1 + m2 ) g = (V i )2ρl g

m2g = [(V i )2 − (V i )1 ]ρl g or ( ∆m ) g or ∆w = ( ∆V i )ρl g If fluid is accelerated, then in the expression of upthrust g is replaced be geff .

Example 1 A block of wood floats in a bucket of water placed in a lift. Will the block sink more or less if the lift starts accelerating up? Solution Under nromal conditions, fraction of volume immersed under floating condition is f1 =

ρs ρl

…(i)

If the lift starts accelerating up, then upthrust − weight = ma ∴ Solving this equation, we get

Vi ρl ( g + a ) − Vρ s g = Vρ sa Vi ρ = f2 = s V ρl

f1 = f2 So, the block neither sinks less nor more.

…(ii)

Chapter 16

Fluid Mechanics — 461

Note From this example, we can make a general concept that by the acceleration of container, fraction (or percentage of volume immersed ) does not change. V

Example 2 A raft of wood (density = 600kg / m3 ) of mass 120 kg floats in water. How much weight can be put on the raft to make it just sink? Solution Weight of raft + external weight = upthrust on 100% volume of raft ∴ ∴

 120 3 (120 + m) g =   × 10 × g  600 m = 80 kg

Ans.  m 120 3  = m .  ρ 600 

Note Here immersed fraction used in upthrust is the total volume of solid  =

Type 2. Level Problems. Concept Let us first note down the following four results: (i) If ice is floating in water, then after melting of ice the level ‘h’ remains unchanged. h (ii) In floating to floating condition level ‘h’ remains unchanged. For example Suppose some wooden blocks are floating in water kept inside a boat. If these wooden blocks are thrown into the water, then level will remain unchanged. This is because under both the conditions wooden blocks are floating. (iii) In the floating to sink condition, level ‘h’ decreases. For example In the above case if wooden blocks are replaced by stone pieces then level will fall. Because, initially stone pieces were floating (with the help of boat) but eventually those pieces will sink. (iv) A solid is floating in a liquid of density ρ1 . After sometime, the solid melts and density (of solid) after melting (in liquid state) is suppose ρ 2 , then there are following three cases : Case 1 If ρ 2 = ρ1 , then level remains unchanged. Note that this is also result (i). Case 2 If ρ 2 < ρ1 , then level will increase. Case 3 If ρ 2 > ρ1 , then level will decrease. V

Example 3 A piece of ice is floating in a glass vessel filled with water. Then prove that level of water in the vessel remains unchanged after melting of ice .

Note This is result (i). Solution Let m be the mass of ice piece floating in water. In equilibrium, weight of ice piece = upthrust or mg = Viρw g m or Vi = ρw

…(i)

462 — Mechanics - II Here, Vi is the volume of ice piece immersed in water. When the ice melts, let V be the volume of water formed by m mass of ice. Then, m V = ρw

…(ii)

From Eqs. (i) and (ii), we see that Vi = V Hence, the level will not change. V

Ans.

Example 4 A piece of ice having a stone frozen in it floats in a glass vessel filled with water. How will the level of water in the vessel change when the ice melts?

Note This is result (iii). Solution Let, m1 = mass of ice, m2 = mass of stone ρS = density of stone and ρw = density of water In equilibrium, when the piece of ice floats in water, or ∴

weight of (ice + stone) = upthrust (m1 + m2) g = Viρw g m m Vi = 1 + 2 ρw ρw

…(i)

Here, Vi = Volume of ice immersed When the ice melts, m1 mass of ice converts into water and stone of mass m2 is completely submerged. Volume of water formed by m1 mass of ice, m V1 = 1 ρw Volume of stone (which is also equal to the volume of water displaced) m V2 = 2 ρS ρS > ρw Therefore, V1 + V 2 < Vi or the level of water will decrease. Since,

V

Ans.

Example 5 A solid floats in a liquid of different material. Carry out an analysis to see whether the level of liquid in the container will rise or fall when the solid melts.

Note This is result (iv). Solution Let

M = Mass of the floating solid ρ 2 = density of liquid formed by the melting of the solid ρ1 = density of the liquid in which the solid is floating

The mass of liquid displaced by the solid is M. Hence, the volume of liquid displaced is

M . ρ1

Chapter 16 When the solid melts, the volume occupied by it is will rise or fall according as M > or ρ2


σ ) fill half the circle. Find the angle that the radius passing through the interface makes with the vertical.

r

O θ

Chapter 16

Fluid Mechanics — 467

Solution hAB = r − r cos (90° − θ ) = r − r sin θ E

A 90°–θ

θ

θ D 90°–θ C

B

hBC = r − r cos θ hCD = r sin (90° − θ ) = r cos θ hDE = r sin θ Writing pressure equation between points A and E we have pA + (r − r sin θ ) ρg − (r − r cos θ ) ρg − (r cos θ ) (σ ) g − (r sin θ ) σ g = pE But pA = pE = pgas Solving this equation, we get  ρ − σ tan θ =    ρ − σ

Type 5. Based on concept of syphon. Concept v1 ≈ v2 = 0 p1 = p2 = p6 = p0

In the figure shown,

4 h2 3

1

5

2

h1 6 v

Note Point 3 is just above point 2. v 2 = 0 but v 3 = v If area of cross-section of pipe is uniform, then from continuity equation, v3 = v4 = v5 = v6 = v (say ) Applying Bernoulli’s equation at 1 (or 2), 3, 4, 5 and 6, we have 1 1 p0 + 0 + 0 = p3 + ρv 2 + 0 = p4 + ρv 2 + ρgh2 2 2

= p5 +

1 2 1 ρv + 0 = p0 + ρv 2 – ρgh1 2 2

Ans.

468 — Mechanics - II From this equation, following conclusions can be made: (i) p1 = p2 = p6 = p0 (ii) p3 = p5 < p0 (iii) v1 = v 2 = 0 (vi) v3 = v 4 = v5 = v 6 = v (v) v = 2gh1 so, h1 should be greater than zero (vi) p4 = p0 − ρ g ( h1 + h 2 ) From the last equation we can see that p4 decreases as ( h1 + h 2 ) increases. Minimum value of pressure is at 4 or p4 and this minimum value is zero and this will occur at, 0 = p0 − ρ g ( h1 + h2 ) max

Thus, ( h1 + h 2 ) max

p = 0 and simultaneously h1 > 0 ρg

Thus, syphon will work when h1 > 0 and ( h1 + h 2 )
> area of pipe v1 ≈ 0, also So,

p1 = p2 = atmospheric pressure v2 = 2 g (h1 − h2) = 2 × 9.8 × 7 = 11.7 m/s

Ans.

Chapter 16

Fluid Mechanics — 469

(b) The minimum pressure in the bend will be at A. Therefore, applying Bernoulli’s equation between (1) and (A) 1 1 p1 + ρv12 + ρgh1 = pA + ρvA2 + ρghA 2 2 Again, v1 ≈ 0 and from continuity equation vA = v2 1 or pA = p1 + ρg (h1 − hA ) − ρv22 2 Therefore, substituting the values, we have 1 pA = (1.01 × 105 ) + (1000)(9.8)(−1) − × (1000)(11.7)2 2 Ans. = 2.27 × 104 N/m2

Type 6. Based on pressure force and its torque. Concept p=

F A

⇒ F = pA

Let us call this force as the pressure force. Now, this force is calculated on a surface. If surface is horizontal then pressure is uniform at all points. So, F = pA can be applied directly. For calculation of torque, point of application of force is required. In the above case, point of application of force may be assumed at geometrical centre of the surface. If the surface is vertical or inclined, pressure is non-uniform (it increases with depth) so pressure force and its torque can be obtained by integration. After finding force and torque by integration, we can also find point of application of this force by the relation. r⊥ =

V

τ F

(as τ = F × r1)

Example 14 A liquid of density ρ is filled upto a height of ‘h’ in a container as shown in figure. Base of the container is a square of side L. Ignoring the atmospheric pressure find A

D

B

C

h

(a) pressure force F1 on its base. (b) torque of force F1 about an axis passing through C and perpendicular to plane of paper. (c) pressure force F2 on the vertical side wall DC. (d) torque of force F2 about the same axis mentioned in part (b). (e) point of application of force F2.

470 — Mechanics - II Solution (a) Base is horizontal. So, we can directly apply F1 = pA = (ρ gh ) (L ) (L ) = ρ ghL2

Ans. L (b) Point of application of F1 is at geometrical centre of base or at a perpendicular distance 2 from the axis mentioned in the question.  L ∴ τF1 = F1 × r⊥ = (ρ ghL2)    2 1 Ans. = ρ gh L3 2 (c) and (d) : Side wall is vertical. Pressure is non-uniform. So, force and torque both will be obtained by integration.

A

D x dF2

p dx

L p p

(h – x) B

dx

C

Pressure at depth x, p = ρ gx Area of small element shown in figure is dA = L (dx) ∴ dF2 = ( p)(dA ) = (ρ gx)(L dx) Perpendicular distance of this small force dF2 from the axis mentioned in the question is (h − x). Therefore, small torque of force dF2 is d τ = (dF2) (h − x) = (ρ gx)(h − x) (L dx) x =h

Now,

F2 = ∫

and

τF2 = ∫ dτ

dF2

x =0 x =h x =0

Substituting the values and then integrating, we get ρ gLh 2 F2 = 2 ρ gLh3 and τF2 = 6 (e) For point of application of F2 , we can apply τF r⊥ = 2 F2 =

(ρ gLh3 / 6) h = (ρ gh 2L / 2) 3

(from C)

Chapter 16

Fluid Mechanics — 471

Note that point of application of F2 is below the centre, as pressure is not uniform. It is increasing with depth. F1 and F2 and their points of application are shown in figure below. A

D 2h/3

B

L/2

L/2

F2 =

h/3 C

ρgLh2 2

F1 = ρghL2

Note

O

In the figure shown, torque of hydrostatic force about point O, the centre of a semicylindrical (or hemispherical) gate is zero as the hydrostatic force at all points passes through point O.

Type 7. Based on the concept of pressure force and upthrust. Concept (i) Pressure in a fluid (gas or a liquid) increases with depth. At a height (or depth) difference of h change in pressure is ∆p = ± ρ gh. Here, ρ is the density of fluid. Now, density of air is almost negligible, so for small height differences it is almost constant ( = p0 ). Therefore, value of atmospheric pressure is almost same everywhere. But this is not the case with a liquid, whose density is not negligible. (ii) If a solid is floating in a liquid then net pressure force (including p0 A) on this solid is upwards. This force is called upthrust and this is numerically equal to V i ρ l g and in equilibrium this is equal to weight of the solid. F1 A

h F2

Density of liquid = r

A plank of area of cross-section Ais floating in a liquid of density ρas shown in figure. Net upward pressure force = F2 − F1 = p2 A − p1 A

472 — Mechanics - II = ( p0 + ρ gh ) − p0 A = ( hA) ρ g (hA = immersed volume) = Vi ρ g = upthrust = weight of plank in equilibrium (iii) In some cases when a plank floats in two or more than two liquids then net force by a liquid on a plank is zero but immersed volume of plank on this liquid contributes in the upthrust. F1 A F2

F2 r1

F3

F3

F4

F4

h1 h2

r2

F5

A plank of area of cross-section A is floating in two liquids. h1 height of the plank is immersed in first liquid and h 2 height in second liquid. Atmospheric pressures is constant at small height differences. So F1 and F2 forces shown in the figure can be obtained directly ( = pressure × area). Inside the liquids pressure increases with depth. So, F3 and F4 will be obtained by integration as they are acting on vertical surfaces. F5 is acting on a horizontal surface so this force can also be obtained directly. In the above figure Net force on the plank by atmosphere = F1 = p0 A Net force on the plank by liquid -1 = 0 Net force on the plank by liquid - 2 = F5 = pA = ( p0 + ρ1 gh1 + ρ 2 gh 2 ) A Net upthrust pressure force on the plank = F5 − F1 = ( h1 A) ρ1 g + ( h 2 A) ρ 2 g (V i = immersed volume) = (V i )1 ρ1 g + (V i ) 2 ρ 2 g (U = upthrust) = U 1 +U 2 = U total = Weight of plank in equilibrium. In this case, we can see that net force on plank by liquid-1 is zero, but immersed volume of plank in liquid-1 contributes in the upthrust ( =U 1 ).

Chapter 16 V

Fluid Mechanics — 473

Example 15 A uniform solid cylinder of density 0.8 g / cm3 floats in equilibrium in a combination of two non-mixing liquids A and B with its axis vertical. The densities of the liquids A and B are 0.7 g / cm3 and 1.2 g / cm3 , respectively. The height of liquid A is h A = 1.2 cm. The length of the part of the cylinder immersed (JEE 2002) in liquid B is h B = 0.8 cm. Air

h

A

hA

B

hB

(a) Find the total force exerted by liquid A on the cylinder. (b) Find h, the length of the part of the cylinder in air. (c) The cylinder is depressed in such a way that its top surface is just below the upper surface of liquid A and is then released. Find the acceleration of the cylinder immediately after it is released. Solution (a) Liquid A is applying the hydrostatic force on cylinder uniformly from all the sides. So, net force is zero. Air

h

A

hA

B

hB

(b) In equilibrium Weight of cylinder = Net upthrust on the cylinder Let s be the area of cross-section of the cylinder, then weight = (s) (h + hA + hB ) ρ cylinder g and upthrust on the cylinder = upthrust due to liquid A + upthrust due to liquid B = shAρ A g + shBρB g Equating these two, or Substituting,

s (h + hA + hB ) ρ cylinder g = sg (hAρ A + hBρB ) (h + hA + hB ) ρ cylinder = hAρ A + hBρB

hA = 1.2 cm, hB = 0.8 cm and ρA = 0.7 g/cm3 ρB = 1.2 g/cm3 and ρ cylinder = 0.8 g/cm3 In the above equation, we get h = 0.25 cm

Ans.

474 — Mechanics - II (c) ∴

Net upward force = extra upthrust = shρB g Force Net acceleration a = mass of cylinder

or

a=

shρB g s(h + hA + hB ) ρ cylinder

or

a=

hρB g (h + hA + hB ) ρ cylinder

Substituting the values of h, hA hB , ρB and ρ cylinder , g we get, a= 6

(upwards)

Type 8. Based on surface tension force. Concept Surface tension is given by T=

F l

∴Surface tension force, F = Tl This force acts only on a line on free surface of the liquid. This force has a tendency to decrease the free surface of the liquid. Following are given some figures for the direction of this force. F

F

F

F F F

V

Example 16 Water is filled up to a height h in a beaker of radius R as shown in the figure. The density of water is ρ, the surface tension of water is T and the atmospheric pressure is p0 . Consider a vertical section ABCD of the water column through a diameter of the beaker. The force on water on one side of this section by water on the other side of this section has magnitude (JEE 2007) (a) |2 p0Rh + πR 2ρgh − 2 RT| (b) |2 p0Rh + Rρgh 2 − 2 RT| (c) | p0πR 2 + Rρgh 2 − 2RT| (d)| p0πR 2 + Rρgh 2 + 2 RT|

2R B A C

D

h

Chapter 16

Fluid Mechanics — 475

Solution Force from right hand side liquid on left hand side liquid. (i) Due to surface tension force = 2RT (towards right) (ii) Due to liquid pressure force

2R p

x=h

=

∫

( p0 + ρgx)(2R ⋅ dx)

At depth 'x'

dx

x=0

= (2 p0Rh + Rρgh 2) ∴ Net force is |2 p0Rh + Rρgh − 2RT | ∴ The correct option is (b).

(towards left)

2

V

Example 17 On heating water, bubbles beings formed at the bottom of the vessel detach and rise. Take the bubbles to be spheres of radius R and making a circular contact of radius r with the bottom of the vessel. If r I B as more mass is concentrated towards B.

=

1 2 1  ml   3 J  Iω =     2 2  3   ml  3J 2 2m

2

ω

Rotational Mechanics — 563

Chapter 12

24. a = (µg cos θ ) + (g sin θ )

20. No rotational motion. τ 21. α = net I  l mg   + mgl  2 = (ml 2 / 3)

= 0.5 × 10 × 0.8 + 10 × 0.6 = 10 m/s2 (µ mg cos θ ) R 2 µg cos θ α= = 1 R mR 2 2 2 × 0.5 × 10 × 0.8 = = 20 rad /s2 0.4

α mg

α

mg

4.5 g N l acoin = lα = 4.5 g a mg − N = macoin ∴ N = mg − macoin = − 3.5 g mg Since, N can't be negative. So, N = 0 and coin can not remain in contact with rod. 22. Let F = the applied force and f = force of friction (leftwards) Net horizontal force = 0 ...(i) ∴ F cos θ = f Net torque about centre = 0 ∴ F ⋅r = f R f r Substituting = in Eq. (i) we have, R R r cos θ = R

a

=

r2 R2  r2  θ = sin −1  1 − 2   R  

sin θ = 1 −

or ∴

23. a =

mg − T m

...(i)

α=

τ = I

T ⋅ l − mg ⋅

T

mg

T

mg

37°

...(iii) Ans. a

Ans.

25. Angular momentum will remain conserved at the point of impact P. and just after impact it starts rotating about point P. ω

P

Li = Lf R 1  v  3 ∴ Mv0  R −  + MR 2  0  = MR 2 ⋅ω   R 2 4 2 5v0 ∴ ω= 6R 5v Ans. ∴ vCOM = ωR = 0 6 µ mg 26. a = = µg m u 2

...(ii)

µ mg cos θ

Pure rolling will start when, v = Rω or at = R (ω 0 − αt ) ∴ 10 t = 0.4 (54 − 20t ) Solving this equation, we get t = 1.2s

α

l 2

(ml 2 / 3) 1  3mg  or α= 3T −  ml  2  a = lα Solving these three equations we get, 3g a= 8 α

mg sin θ

ω a fmax = µmg

(µ mg ) R 5 µ g ⇒ α= = 2R 2/ 5 mR 2 Pure rolling will start when v = Rω u   − at = R (αt − ω ) 2  u u   5 µg or t−   − µ gt = R  2   2R R 3u Solving we get, t = 7 µg u ω= R

Ans.

564 — Mechanics - II 27. Viscous liquid has only translational kinetic energy ∴ KE = translational + rotational kinetic energy of hollow sphere + translational kinetic energy of liquid. 2 1 1 2 1  v = mv 2 + × mR 2   + mv 2  R 2 2 3 2 4 2 Ans. = mv 3 g sin θ g sin θ 28. a = = 2 1 + (1/ 2) 1 + I / mR 2g sin θ = 3 3 ∴ g sin θ = a 2 mg sin θ m (3/ 2 a) Now, f = = 2 1+ 2 1 + mR / I 1 Ans. = ma 2 29. Only two forces are acting on rod, normal reaction (vertically upwards) and weight (vertically downwards). Since, both forces are vertical centre of mass falls in vertical direction (downwards). Q α O

C

N 30°

30. aP = (OP ) α l   12 g  =  cos 30°   2   7 l 3 3g Ans. 7 31. Angle between acceleration and velocity is 45°. =

Rω = v

2v

45°

α = Rω2

v

F M FL / 2 6F α= = 2 ML /12 ML

32. a =

F

a

α

B

A

L 2F α −a= 2 M v sin θ v sin θ 33. ω = = r b/sin θ aB =

θ

mg r

ap P

l l sin 30° = 2 4 τ α= I l  (mg )  sin 30° 2  = 2 2  ml ml  +   16   12

OC =

12 g 7 l aC = (OC ) α  l   12 g  3 =    = g  4  7 l  7 3 mg Now, mg − N = maC = 7 4 mg N = ∴ 7

b

θ

ac

(downwards)

v v sin θ

v sin 2 θ (as v and bare constants) ∝ sin 2 θ b θ is decreasing. Therefore, ω will decrease. 34. Decrease in gravitational potential energy = increase in rotational kinetic energy =

(aboutO)

ω

r

T

=

C

∴ Ans.

Mg

2Mg 3

L 1  ML2  2 3g =   ⋅ω ∴ ω 2 = 2 2 3  L r=

L 2L / 3 2L + = 3 2 3

Chapter 12 T − 2Mg  2M   2M  2 =  a =  (r ω )  3  C  3  3 2Mg  2Mg   2L  3g  T = + ∴       3   3  L 3 = 2 Mg

Ans.

before and just after impact about point O we have, ω

JH

Just after impact

So, the disc will return to its initial position for all values of µ. x 38. x = 30 tan θ

C

m

1 mR 2ω 0 4 = + ve or anticlockwise During slip, friction acts about bottommost point. So, its torque is v ω zero or angular momentum about bottommost point should also v = Rω remain anticlockwise when pure rolling starts. So, figure should be as shown below. =

35. From conservation of angular momentum, just O

v

dθ  dX  2 ∴   = (30 sec θ ) ⋅  dt  dt

J J

or

∴

∴

Li = Lf L  mL2  mv =   ⋅ω 2  3 

3v 2L 1 3 vC = ⋅ ω = v 2 4 Impulse J has changed the momentum of particle from mv to O. Hence, J = mv For rod 3 J H + J = mvC = mv 4 3 3 J H = mv − J = mv − mv ∴ 4 4 mv mv or | J H | = Ans. =− 4 4 36. Li = Lf ∴

ω=

(I + mR 2 ) ω − mvR I 37. About bottommost point, ω′ =

ω0

Ans.

ω R v0 = 0 4

Angular momentum, (anticlockwise) L = IC ω 0 − mv0R 1  R 2 = mR ω 0 − mω 0  R  4 2

30 m

vcar = (30 sec2 θ ) ω vcar ω= 30 sec2 θ 40 = = 1 rad /s (30) sec2 30°

θ

39. Two forces normal reaction and weight are the only forces acting on the rod during motion. Both forces are vertical. So centre of mass will fall downwards in a vertical line. y C P d x

C

θ O

l cos θ − d cos θ 2 x cos θ = l   − d 2 

X P = x (say ) = or

(I + mR 2 ) ω = (mvR ) + I ω′ ∴

Rotational Mechanics — 565

...(i)

Similarly, yP = y = d sin θ y ...(ii) sin θ = ∴ d Squaring and adding Eqs. (i) and (ii), we get X2 y2 + =1 2 d2 l  − d   2  l This is an equation of a circle for d = . For any 4 other value of ' d ' it is equation of ellipse.

566 — Mechanics - II  3 v2 = mgh = mg   4 g

40. Instantaneous axis of rotation will pass through O. C O

A

1 2 mr 2 So it is either solid cylinder or disc. 2  ml 2  l  2. (a) I x = I y = 2  + m    2  12  2 2 = ml 3 Solving we get I =

vc

ω

∴ v0 = 0 41. Decrease in gravitational potential energy = increase in rotational kinetic energy about O l 1  ml 2  2 mg =  ∴ ω 2 2 3  3g l vA = rω = lω = 3gl

y

ω=

or

Ans.

z

µ mg 2 42. a = = µg = g m 7

x r

α

a µmg

P

4 2 ml 3 (c) I = I z + (4 m) r2

(b) I z = I x + I y =

v + Rω

P

µ mgR 2 mR 2 5 5 µg = 2R 5 g = 7 R Pure rolling will start when, vP = v + R ω = v0 or at + R (αt ) = v0 2  5 ∴  gt + gt = v0 7  7 v ∴ t= 0 g 1 S = at 2 2 α=

2

4 2  l  ml + (4 m)    2 3 10 2 = ml 3 ml 2 5 (d) I = 2 + ml 2 = ml 2 3 3 3. O → instantaneous axis of rotation =

P

O

1  2   v0  v2  g   = 0 2 7   g 7g

ω

v = rω or v ∝ r More the value of r from O more is the speed of point P.

2

=

v

r

Ans.

4. v0 + Rω 0 = v

...(i)

v0 − Rω 0 = 3v

...(ii)

More than One Correct Options 1.

KR I = KT mr2 ∴

 mr2  KT =   KTotal  I + mr2 

or

 I + mr2   1 2 KTotal =    mv    mr2   2

v0 ω0

Solving these two equations, we get v0 = 2v v and ω0 = − R

Chapter 12 5. x 2 + y2 = l 2 = (0.25) m2

∴

yω = 4 xω = 3

...(i) ...(ii) O

3 m/s

ω

l c

y

x

∴

4 m/s

Kf =

∴

ω = 10 rad /s l OC = = 0.25m 2 vCM = (OC ) ω = 2.5 m/s 1 1 2 KTotal = mvcm + I cm ω 2 2 2 1 1  2 × 0.25 2 = × 2 × (2.5)2 +   (10) 2 2  12  25 = J 3 6. Net acceleration of any point on the rim is vector sum of a, Rω 2 and Rα with a = R α Rω2

A

a C

Rω2 B

a

a + Rα

Rω2

Rα

aA = Rω 2 → vertically upwards If a = Rω 2 , aB is vertically downwards and so on.

7. C 1 is centre of mass of rod C 2 is centre of mass of both ⇒

l 3 l 6

m

v

Pi = Pf

1 1  v (3m)   + IC2 ω 2  3 2 2 2

(0.25) ω 2 = 25

Rα

v ...(i) 3 Li = Lf about C 2 we have, l mv = IC2 ω 6 2 2  2ml 2   l  l  =   + 2m   + m     3  6  12   2 v 1 2 ω= ⇒ K i = mv 5 l 2 mv = 3mv0 ⇒ v0 =

2

Squaring and adding, we get (x 2 + y2 ) ω 2 = 25 or

Rotational Mechanics — 567

C1 C2 ω

1 1 5   2 v mv 2 +  ml 2     5 l 6 2  12 1 = mv 2 5 3 ∴ Loss of kinetic energy = K i − K f = mv 2 10 R 8. I = mK 2  where, K = = radius of gyration   2 =

2

 R 1 = m   = mR 2  2 4 J (h – R) ω

h

Ball will roll purely if v = Rω  J (h − R )  J   ∴   =R 2  m  (1/ 4 ) mR  Solving this equation, we get 5R h= 4 Further, if ball is struck at centre of mass, there will be no rotation only translation. 9. Friction always acts upwards. If the ball moves upwards, it becomes a case of retardation with pure rolling. tan α tan α 2 µ min = = = tan α mR 2 1 + 3 5 1+ 2 I f

v0

v

v

ω α a α

568 — Mechanics - II 10. I 1 ω 1 = I 2 ω 2 ∴

decrease in potential energy of rod = increase in rotational kinetic energy of rod + translational kinetic energy of block.

I ω2 = 1 ⋅ ω1 I2

v = vP sin θ = (lω ) sin θ 2 l l   1  ml 2 ∴ mg  − sin θ =  ω 2 2  2 3 

(1/ 2 mR 2 ) ⋅ω (1/ 2 mR 2 + mR 2 ) ω = 3 Pi = mvi = m (2ωR ) = 2 mωR ω  1 Pf = mv f = m  R = mωR 3  3

But

=

+ From here we get, mg (1 − sin θ ) lω 2 = an = m / 3 + M sin 2 θ

ω 3

ω

Pf Pi

Impulse J = | Pf − Pi | (asθ = 90°)

= Pi2 + Pf2 37 mω R 3 11. Velocity component along AB = 0 =

y A v θ

v0 B

∴ or at

x

l   mg cos θ − Nl sin θ    τ 2 at = lα = l   = l   2 I0 ml / 3     3 3N sin θ ...(ii) = g cos θ − 2 m For block ...(iii) N = Ma = M [ at sin θ − an cos θ ] Now putting values of at and an from Eqs. (i) and (ii) in Eqs. (iii) and then putting N = 0 and θ = 30° in the equation we get, M 4 Ans. = m 3 M 4 2. = m 3 4 M = m 3 60°

ωl 2

lω 30°

M v

ω

ωl 2v =v ∴ ω= 2 l Decrease in potential energy of rod = increase in rotational kinetic energy of rod + translational kinetic energy of block l l  1 ∴ mg  − sin 30° = 2 2  2

Comprehension Based Questions 1. At angle θ shown in figure : a N θ P N an mg θ at Oθ ω

...(i)

ωl cos 60° =

v0 cos θ = v sin θ v = v0 cot θ = f1 (θ ) 4 θ = 37° , v = v0 3 Relative velocity ⊥ to AB ω= l v0 sin θ + v cos θ = = f2 (θ ) l (v ) (3/ 5) + (4 / 3 v0 ) (4 / 5) 5v0 = 0 = l 3l

⇐

1 M (ωl sin θ )2 2

V

Solving this equation, we get 3gl v= 4

2

2

 ml   2v       3  l  1  4 m 2 +   (v ) 2 3 

Ans.

Chapter 12 l 2

3. an = ω 2 =

l  4 v2   2  l2 

Rotational Mechanics — 569

∴

f =

F 1 = ma 3 2

N=0

30°

f

at

7. About point P net torque of F and f is zero. So,

mg

α

angular momentum is conserved.

 2  3gl  3 =    = g  l   16  8

P

 mg l l  τ l  at = α =   =  2 2 I 2   3 3g = 8 ∴

F

l   cos 30°  2   2 ml / 3  

f=

=

∴

3g 4

 3g  = −   $j  4

1  3 mv 2 =   (mgh)  5 2

6gh 5 2 2 v sin sin 37° 9. H = 2g v=

=

Ans.

(6gh/ 5) (3/ 5)2 27 h = 2g 125 2u2 sin 37° cos 37° g (2) (6gh/ 5) (3/ 5) (4 / 5) = g 144 = h 125

Ans.

10. R = x =

C 60° 30° 60° 30°

y

x

3√3 g 8

Match the Columns

mg (− $j) + FHinge = maC

1. (a) t =

Substituting the values we get  mg  FHinge =   $j  4 

Ans.

5. Bottommost point where friction actually acts is at 6. α =

KT =

∴

3 3 g  3 + g cos 30° + g cos 60° (− $j) 8  8 

rest.

F 3

KR 2 = KT 3

3 3 g  3 4. aC =  cos 60° − g cos 30° $i 8  8 

3g 8

R/2

8. KTotal after falling a height h is mgh

a = an2 + at2

Now,

a

F

an

F ⋅R 2F = 3 3 mR 2 mR 2 2F a = Rα = 3m 2F F − f = ma = 3

F

2S and a

g sin θ 1 + I / mR 2

I of hollow sphere is maximum. So, a is minimum and t is maximum. (b) KTotal = mgh for all KR 2 (c) = for solid sphere KT 5 2 for hollow sphere 3 1 = for disc 2 It is maximum for hollow sphere. =

α

a=

Ans.

570 — Mechanics - II But

KT is maximum for solid sphere. KR

τ 6. a = Rα = R   I

   20 × 1  F − 20 = (1)   10  1 × 10 × 12  2 

2. In all cases, J v= m J × r⊥ ω= IE

(J = linear impulse)

Here, r⊥ is from centre E. v is rightwards. So, for pure rolling ω should be clockwise. Hence, J should be applied at A. If it is below A, angular velocity is anticlockwise and it will cause forward slip.

3. In case of pure roll (upwards or downwards) required value of friction acts in upward direction. In case of slip (forward or backward) maximum friction will act in backward or forward direction. 4. (a) K = I / m =

m (2a)2 2 = a 3 3

(b) K =

m (2a)2 a = 12 3

(c) K =

ma2 a = 3 3

(d) K =

ma2 a = 12 12

KTotal = mgh K (say ) KR 2 = KT 5 2 2 K R = K = mgh ∴ 7 7 h (b) KTotal = mg 2 KR 2 = KT 5

∴

α fmax = µmg = 20 N

Solving this equation, we get F = 60 N So, when F = 60 N, friction reaches its maximum value or slipping will start. F becomes 60 N at 6 s.  τ (d) a = R α = R   I   F − 10  10 × 1  or = (1)   1 10 2  × 10 × 1  2  F = 30 N and F becomes 30 N at 3 s.

a

F α

5. (a)

∴ KT =

a

F

10 N

Subjective Questions 1. Torque about bottommost point in each case is clockwise. τ 2. α = = I

 l mg    2 ml 2 3

5 h 5 mgh  mg  = 7 2 14

=

3 g 2 l

(c) At 3 KR =

h 1 2  mg  = mgh 7 2 7

From 3 to 4, K R will become constant. (d) KT = KTotal − K R 1 6 = mgh − mgh = mgh 7 7

T C

A α

mg

B

Chapter 12 3g 2 l 3g (b) aC = α = 2 4 mg − T 3g mg − T or (c) aC = = m 4 m mg T = ∴ 4 3 3. mgh = K R + KT = mv 2 4 Here h = s sin θ 3 3v 2 ∴ gs sin θ = v 2 or s = 4 4 g sin θ (a) aB = lα =

=

Ans.

vF = v 2 + (Rω )2 + 2v (Rω ) cos (90 + 30)  1 = 0.5625 + 5.0625 + 2 × 0.75 × 2.25 ×  –   2

Ans.

Ans.

Rotational Mechanics — 571

= 1.98 m/s Mg – T 6. a = M

Ans. …(i)

TR 2T …(ii) = 1 MR 2 MR 2 …(iii) a = Rα Solving these three equations, we get 2g Mg and T = Ans. a= 3 3 7. From conservation of mechanical energy, decrease in gravitational PE = increase in rotational KE 1  1  or mg (R ) = MR 2 + mR 2  ω 2  2   2  α=

3 × (2.0)2 = 0.612 m 1 4 × 9.8 × 2

Ans.

4. I = MR 2 For pure rolling to take place. a = Rα

or

ω=

4 mg (2m + M )R

Ans.

8. Initially there is forward slipping. Therefore,

F

friction is backwards and maximum.

f

F– f f  f .R  =R  =  MR 2  M M F ∴ f = 2 F– f F and a= = M 2M 5. v + Rω = 1.5

v0

or

and Rω – v = 3.0 From Eq. (i) and (ii), we have

ω0

Ans. …(i) …(ii)

Rω – v ω v

µ mg

Ans.

Let velocity becomes zero in time t1 and angular velocity becomes zero in time t2. Then, 0 = v0 − at1 v v ...(i) or t1 = 0 = 0 a µg and or

⇒ v + Rω

Here,

Rω = 2.25 m/s and v = – 0.75 m/s Thus, velocity of point C is 0.75 m/s (towards left). ∴ Rω 30° 30° F v

0 = ω 0 − αt2 ω t2 = 0 α µmgR 2µg α= = 1 R mR 2 2 ωR t2 = 0 2µg

Disk will return back when ω 0R v0 t2 > t1 or > 2µg µg 2v or ω0 > 0 R

…(ii)

Ans.

572 — Mechanics - II 9. h = (R − r)(1 − cos θ )

…(i)

q v

or

Ans.

F– f m Substituting value of f from part (a), we get 2F Ans. a= (R + r) 3mR 2 F R (c) a > if Ans. (R + r) > 1 or r > 3R m 2 (d) In this case force of friction is in forward direction. Ans. (b) Acceleration a =

h

QIn case of pure rolling 5 KT = K 7 1 1 5 2 ∴ mv = mv02 − mgh 2 2 7 10 2 2 ∴ v = v0 − gh 7 Equation of motion at angle θ is, N − mg cos θ =

…(ii)

∴

forces at different contacts. a1

mv 2 (R − r)

m2 f1

Substituting value of h from Eq. (i)  m  N = mg cos θ +    R − r   2 10 g (R − r)(1 − cos θ )  v0 − 7   mg mv02 (17 cos θ − 10) + 7 (R − r)

Ans.

Force of friction, mg sin θ (for pure rolling to take place) f = mr2 1+ I mg sin θ 2 2  =  I = mr  5   5 1+ 2 2 Ans. = mg sin θ 7 10. (a) For pure rolling to take place, F

f

11. We can choose any arbitrary directions of frictional F

m  2 10  N = mg cos θ + gh  v0 − (R − r)  7 

=

 Fr + fR  F– f =R   1 m  mR 2    2 Solving this equation, we get 2  1 r f =  –  F 3  2 R a = Rα

Kinetic energy at angle θ is, 7 1  K =  mv02 − mgh  5 2

In the final answer, the negative value will show the opposite directions. Let f1 = friction between plank and cylinder f2 = friction between cylinder and ground a1 = acceleration of plank a2 = acceleration of centre of mass of cylinder andα = angular acceleration of cylinder about is COM. Directions of f1 and f2 are as shown here f1 a2

m1 f2

Since, there is no slipping anywhere …(i) ∴ a1 = 2a2 (Acceleration of plank = acceleration of top point of cylinder) F − f1 …(ii) a1 = m2 f + f2 …(iii) a2 = 1 m1 ( f − f2 ) R α= 1 I

Chapter 12 (I = moment of inertia of cylinder about COM)

Solving, we get a=

a1 = 2a2 a2

( f1 − f2 ) R 1 m1 R 2 2 2( f1 − f2 ) α= m1 R 2( f1 − f2 ) a2 = Rα = m1

…(iv) …(v)

(Acceleration of bottommost point of cylinder = 0)

Rα

T = 1.43 N Tr T and α= = = 7.15 rad /s2 I Mr 13. Initially the cylinder will slip on the plank, therefore kinetic friction will act between the cylinder and the plank. µmg ac = – = – µg m µmg µg ap = + =+ 2m 2 (µmg ) (R ) 2µg αc = + =+ R (mR 2 / 2) For pure rolling,

a2

(a) Solving Eqs. (i), (ii), (iii) and (v), we get 8F a1 = 3m1 + 8m2 4F and a2 = 3m1 + 8m2 3m1F (b) f1 = 3m1 + 8m2 m1F f2 = 3m1 + 8m2 Since, all quantities are positive, they are correctly shown in figures.

Note Above calculations have been done at t = 0

when ω = 0. 12. If α be the angular acceleration of the hoop and a be the acceleration of its centre, acceleration of m would be αr + a. N

µmg µmg mg FBD of cylinder

∴ ∴ ∴

T

Here, Tr = Iα [where I = moment of inertia of the hoop about the horizontal axis passing through its centre] Also, T = Ma and mg – T = m [ a + αr ]

2mg + N FBD of plank

vp = vc – R ω c µg  2µg  t = v0 – µgt – (R )   (t )  R  2 v0 7 t= = =2s 3.5µg 3.5 × 01 . × 10 1 1  µg  × (µg ) (t 2 ) –   (t 2 ) 2 2 2 1 = (7 × 2) – (0.1) (10) (4 ) 2 1  0.1 × 10 –   (4 ) = 11 m 2 2 

sc – sp = v0t –

 µg  vc – vP = (v0 – µgt ) –   (t )  2 0.1 × 10 × 2 = 7 – 0.1 × 10 × 2 – = 4 m/s 2 Hence, the remaining distance (12 – 11 = 1 m ) is travelled in a time, 1.0 t′ = = 0.25 s 4 Total time = 2 + 0.25 = 2.25 s ∴ Also,

mg FBD of block of mass m

N′

N

T

Mg FBD of hoop

mg 2 = = 1.43 m/s2 [ M + 2m ] 1.4

Hence,

α=

∴

Rotational Mechanics — 573

574 — Mechanics - II v

14. In rolling without sliding on a stationary ground, work done by friction is zero. Hence work done by the applied force = change in kinetic energy 1 ∴ (30)(0.25) = × 9 × v 2 2 1 v2  1 1 + 2  × 6 × v 2 + × × 6 × r2 × 2  2 2 r  2 7.5 = 13.5v 2

or ∴

ω

⇒ ω=

∴

v = ωR =

and

v = 0.745 m/s

v v0

v0 − rω 0 = 2v and v0 + rω 0 = − v Solving Eqs. (i) and (ii), we have 3 v ω0 = − 2 r

…(i) …(ii)

Hence, the angular velocity of disc is anticlockwise.

16. Let x be the distance of centre

F1

point C of rod from D. Then,

0.2m

F2 – F1 = ma or F1 = 3 N Further, τ c = 0

or

µ mg

µ mg

2v

∴

Ans.

ω0

ω0

Let pure rolling starts after time ‘t’. Then ωr − v = v ∴ ωr = 2v (ω 0 − αt )r = 2(a1t ) Substituting the values, 2 ω 0r t= 9 µg

3 v 2 r Ans.

A D

F2

x C

F2 x = F1 (0.2 + x ) B 5x = F1 (0.2 + x ) 5x = 3 (0.2 + x ) x = 0.3 m Length of rod = 2 (x + 0.2) = 1.0 m Ans.

17. Li = Lf

ω 0R 6

a1 = linear acceleration of sphere (towards right), a2 = linear acceleration of plank (towards left) and α = angular retardation of sphere µmg µmgr 5 µg a1 = a2 = = µg ⇒ α = = 2 2 2 r m mr 5

Ans.

velocity of the disc as shown in figure then,

∴ or ∴

Ans.

18. Let,

15. Let v0 be the linear velocity and ω 0 the angular

∴

ω0 6

ω

∴

1 2ω 20r2 (a2 ) t 2 = 2 81µg

s=

Ans.

19. Between A and B, there is forward slipping. Therefore, friction will be maximum and backwards (rightwards). At point B where v = Rω, ball starts rolling without slipping and force of friction becomes zero. ω0

(about bottommost point)

v0

Iω 0 = 2 [ Iω + mRv ] 1 1  2 2  mR  ω 0 = 2  mR ω + mR (ωR ) 2  2 

ω

ω0 0.7 v0

v

ω0 B

A

Rotational Mechanics — 575

Chapter 12

1 3 × 9.8 × mg sin θ 2 = 4.2 N f = =  1  1 + mR 2 / I 1+    0.4 

From conservation of angular momentum between points A and B about bottommost point (because torque of friction about this point is zero) LA = LB m(0.7v0 )R − Iω 0 = mvR + Iω v v Substituting ω 0 = 0 ,ω = R R 2 and I = mR 2, we get 5 3  3 v= v0 =   (7)m/s = 1.5 m/s  14  14

∴

4 2 4 2 1 2  ml + ml +  ml + m (l 3 )2  3 3 3  2 = 6 ml =

Ans.

…(i) …(ii)

If ω is the angular velocity in the second position, then using conservation of mechanical energy, we have C

A

2l

l √3

l

l

A

C

2×4 = 1.51s 3.5

Similarly, the values for disk and hoop can be obtained. 22. I A = I AB + I AC + I BC

20. Let J be the linear impulse applied at B and ω the angular speed of rod. J = mv0 2  l  ml J  = .ω  2 12 Solving these two equations, 6v ω= 0 l Linear speed of D (mid-point of CB) relative to C,  l 3 v = ω   = v0  4 2

2s = a

t=

ω

ωA B

2 l √3 3

v0 O B

D

∴ Force exerted by upper half B J on the lower half,  m 2   v  2 F=  l    4 3 Substituting v = v0, we get 2 9 mv02 Ans. F= 2 l I 2 21. = = 0.4 for sphere 2 5 mR 1 = = 0.5 for disc and = 1 for hoop 2 2 s= =4m sin 30° For sphere 1 9.8 × g sin θ 2 = 3.5 m/s a= = I 1 + 0.4 1+ mR 2 v = 2as = 2 × 3.5 × 4 = 5.29 m/s ∴

hi = +

For COM

hf = −

and

C

3l 3

2 3 l 3

 l 3 1  − 2l 3  3mg   = (6ml 2 )ω 2 + 3mg   3 2    3  ω=

or

g 3 l

Now, velocity of C at this instant is 2lω or 2 gl 3 and maximum. Ans. 23. (i) C is the centre of mass of the rod. Let ω be the angular speed of rod about point O at angle θ. From conservation of mechanical energy, ω, α Ο θ

O θ0

θ c

Fr c

Mg sin θ + Ft

c′

A A′

Mg cos θ

576 — Mechanics - II Mg

L 1  ML2  2 (cos θ − cos θ 0 ) =  ω 2 2 3  3g (cos θ − cos θ 0 ) L  L Fr − Mg cos θ = M   ω 2  2 ω2 =

∴ Now,

…(i) …(ii)

From Eqs. (i) and (ii), we get 1 Fr = Mg (5 cos θ − 3 cos θ 0 ) Hence proved. 2 (ii) Angular acceleration of rod at this instant, L Mg sin θ τ 2 α= = I ML2 3 3 g sin θ = 2 L Tangential acceleration of COM,  L 3 at = (α )   = g sin θ …(iii)  2 4 Now, Ft + Mg sin θ = Mat …(iv) From Eqs. (iii) and (iv), we get 1 Hence proved. Ft = − Mg sin θ 4 Here negative sign implies that direction of Ft is opposite to the component Mg sin θ. 24. (a) From conservation of mechanical energy. A

ω′

ω 3l

60°

C B

J=

l A

l O ω, α

C

2l B

mg

Let V be the vertical reaction (upwards) at axis, then 3mg mg − V = maC = 7 4 …(i) ∴ V = mg 7 If H be the horizontal reaction (towards CO) at axis, then …(ii) H = mlω 2 ∴ Total reaction at axis,

1 (3m)(g )(2l ) = Iω 2 2 1  (3m)(4 l )2  2 2 2 =   ω = 8ml ω 2 3  ∴

8 3g ml 3 l 8 Ans. or J = m 3gl 3 (b) Let ω′ be the angular speed in opposite direction. Again applying conservation of mechanical energy, 1 (3m)(g )(l ) = I (ω′ )2 = 8ml 2 (ω′ )2 2 1 3g ∴ ω′ = l 2 2 Now, applying, angular impulse = change in angular momentum 3g  1 2 1 ∴ J (3l ) = I (ω + ω′ ) = (16ml ) 1 +   2 l 2 4 Ans. ∴ J = m 6gl ( 2 + 1) 3 mgl 3 g 25. α = = ⋅ 7 l m(4 l )2 2 + ml 12 3 (downwards) (aC )V = lα = g ∴ 7 ∴

ω=

1 2

3g l

Applying, angular impulse = change in angular momentum J (3l ) = Iω  1 3g  or 3 Jl = (16ml 2 )   2 l 

N = H2 + V2 = (b)

4 mg 1 + 7

 7lω 2     4g 

2

Ans.

aC = (aC )V2 + (lω 2 )2 2

 3g  =   + (lω 2 )2  7

Ans.

(c) Let ω′ be the angular speed of the rod when it becomes vertical for the first time. Then from conservation of mechanical energy, 1 I (ω′ 2 − ω 2 ) = mgl 2

Rotational Mechanics — 577

Chapter 12 2mgl I 2mgl 2 =ω + 7 2 ml 3 6 g = ω2 + 7 l Acceleration of centre of mass at this instance will be, 6g Ans. aC = lω′ 2 = lω 2 + 7 Let V be the reaction (upwards) at axis at this instant, then, 6mg V − mg = maC = mlω 2 + 7 13 2 Ans. ∴ V = mg + mlω 7 (d) From conservation of mechanical energy, 1 mgl = Iω 2min 2 2mgl 2mgl ω min = = ∴ 7 2 I ml 3 6 g Ans. = 7 l ∴

ω′ 2 = ω 2 +

26. Linear momentum, angular momentum and kinetic energy are conserved in the process. From conservation of linear momentum, Mv′ = mv m or v′ = v M Conservation of angular momentum gives,  Ml 2  mvd =  ω  12  or

 12mvd  ω=   Ml 2 

…(i)

27. Let v = linear velocity of rod after impact (upwards), v ω J θ

v0

θ A After impact

A At the time of impact

ω = angular velocity of rod and J = linear impulse at A during impact J = ∆ P = Pf − Pi J = mv − (− mv0 ) ∴ J = m(v + v0 ) Angular impulse = ∆L ml 2 l  ∴ J  cos θ = Iω = ω 2  12 Then,

lω 2

…(i) …(ii)

v θ

θ

Collision is elastic (e = 1) ∴ Relative speed of approach = Relative speed of separation at point of impact l …(iii) v0 = v + ω cos θ 2 Solving above equations, we get 6v0 cos θ Ans. ω= l (1 + 3 cos2 θ )

28. (a) The distance of centre of mass (COM) of the system about point A will be A

…(ii)

Collision is elastic. Hence, e=1 or relative speed of approach = relative speed of separation ∴ v = v ′ + dω Substituting the values, we have 12mvd 2 m v= v+ M Ml 2 Solving it, we get Ml 2 Ans. m= 12d 2 + l 2

y

ω,α l

l

√3/2l

x

COM F

B

l

C

l 3 Therefore, the magnitude of horizontal force exerted by the hinge on the body is F = centripetal force or F = (3m)rω 2 r=

578 — Mechanics - II  l  F = (3m)  ω 2  3 F = 3 mlω 2

or or

(b) Angular acceleration of system about point A is τ α= A IA  3  (F )  l  2  = 2ml 2 3F = 4 ml Now, acceleration of COM along x-axis is F  l   3 F ax = rα =     or ax =  3   4 ml  4m Now, let Fx be the force applied by the hinge along x-axis. Then, Fx + F = (3m)ax  F  or Fx + F = (3m)    4 m 3 F or Fx + F = F or Fx = − 4 4 Further if Fy be the force applied by the hinge along y-axis. Then, Fy = centripetal force or Fy = 3 mlω 2

29. From conservation of linear momentum ω

N − mg =

∴ N = mg +

mvr2 = (0.5)(10) R−r (0.5)(3.5)2 Ans. = 16.67 N 0.525 ⇒ µmg cos α > mg sin α

+

30. Given µ > tan α

a = (µg cos α – g sin α ) (µmg cos α ) r 2µg cos α α= = 1 2 r mr 2 Slipping will stop when, v = rω or at = r (ω 0 – αt )   rω 0 rω 0 t= ∴ =  a + rα  3µg cos α – g sin α  d1 = =

1 2 at 2

  1 rω 0 (µg cos α – g sin α )    3µg cos α – g sin α  2

=

r2ω 20 (µ cos α – sin α ) 2g (3µ cos α – sin α )2

Ans.

nα

i

s mg

ω0 µ

c mg

α os

α

…(i) mv1 = Mv2 ∴ Velocity of cylinder axis relative to block …(ii) v r = v1 + v2 Applying conservation of mechanical energy, 1 1 1 mg (R − r) = mv12 + Iω 2 + Mv22 …(iii) 2 2 2 1 2 v Here, I = mr and ω= r 2 r Solving the above equations with given data, we get v1 = 2.0 m/s and Ans. v2 = 1.5 m/s

2

  rω 0 v = at = (µg cos α – g sin α )    3µg cos α – g sin α  rω (µ cos α – sin α ) = 0 (3µ cos α – sin α )

v1

v2

mvr2 R−r

Further,

α

a

α

Once slipping is stopped, retardation in cylinder, g sin α g sin α 2 a′ = = = g sin α I 1 3 1+ 1+ 2 2 mr

Rotational Mechanics — 579

Chapter 12 d2 =

v2 3r2ω 20 (µ cos α – sin α )2 = 2a′ (3µ cos α – sin α )2 (4 g sin α )

∴

dmax = d1 + d2

 3 (µ cos α – sin α )   1 + 2 sin α   2 2 r ω 0 (µ cos α – sin α ) Ans. = 4 g sin α (3µ cos α – sin α ) =

r2ω 20

(µ cos α – sin α ) 2g (3µ cos α – sin α )2

Once slipping was stopped, pure rolling continues if tan α µ> mr2 1+ I tan α tan α or or µ > µ > 1+ 2 3

32. Figure (a) and (b) ω : Decrease in gravitational potential energy = increase in rotational kinetic energy l 1 ∴ mg sin θ = I 0ω 2 4 2 2 1  ml 2  l  =  + m   ω 2  4 2  12  ∴

 24 g sin θ  7l  

ω=

Note

and already in the question it is given that 2 µ > tan α .That’s why we have taken a′ = g sin α . 3 31. Point A is momentarily at rest.

l mg cos θ τ 4 α= = 2 I  ml 2  l  + m     4  12 

α:

12 g cos θ 7l ΣFy = may or mg cos θ − N = mat =

3l/4

l/4

C O

C

l/4

aC A

θ

l mg cos θ 3 g cos θ 2 α= = 2 2 l ml 3 l 3 ∴ aC = α = g cos θ 2 4 Now µN = max or µN = maC sin θ 3 …(i) or µN = mg sin θ cos θ 4 Further, mg − N = may or N = mg − maC cos θ 3 or …(ii) N = mg − mg cos2 θ 4 Dividing Eq. (i) by Eqs. (ii), we have 3 sin θ cos θ 3 sin θ cos θ 4 µ= = 2 3 1 − cos2 θ 4 − 3 cos θ 4 3 sin θ cos θ Hence proved. = 1 + 3 sin 2 θ

…(i)

(a) ω,α θ O

mg (b) y

µN

x

N

θ

an O at mg (c)

…(ii)

580 — Mechanics - II N = mg cos θ − mat l = mg cos θ − m α 4 Substituting value of α from Eq. (ii), we get 4 …(iii) N = mg cos θ 7 or

Rod begins to slip when µN − mg sin θ = man 4 l µmg cos θ − mg sin θ = m ω 2 7 4

or

( f1 – f2 ) R 1 MR 2 2 2( f1 – f2 ) or α= MR For no slipping condition, a2 + Rα = – a1 and a2 – Rα = a3 α=

M

Substitution value of ω from Eq. (i), we get 4µ tan θ = 13 −1  4µ  ∴ θ = tan    13 

f1 a2

M, R

…(iv)

…(v) …(vi) a1

F f1

α

f2

Ans.

2F

2M

f2

33. Writing equations of motion, we get 5g – T2 = 5a2 (T2 – T1 ) R2 = α2 1 M 2R22 2 T1 + f = M 1a1 (T1 – f ) R1 = α1 1 M 1R12 2 a1 = R1α 1 a1 + R1α 1 = R2α 2 R2α 2 = a2 α1

T1 a1

M1

…(ii) …(iii) …(iv)

…(v) …(vi) …(vii) α2

T1 M2

R1

a3

…(i)

T2

R2

a2

We have six unknowns, f1 , f2 , a1 , a2 , a3 and α. Solving the above six equations, we get 21F F and a2 = Ans. a1 = 26M 26M 35. Angular velocity From conservation of mechanical energy, decrease in gravitational PE = increase in rotational KE or mgr sin 60° + mg (2r sin 60° )  1  3mr2 =  + m (2r)2  ω 2 2 2  3 3mgr 11 2 2 ∴ = mr ω 2 4 6g 3 11r Angular acceleration τ mgr cos 60° + mg (2r cos 60° ) α= = I  3mr2 2  2 + m (2r)    3mgr 2 = 3g = 11 2 11r mr 2 36. a B = a 0 + a B / 0 ∴

f T2

5g

We have seven unknowns,T1 , T2 , a1 , a2, α 1 , α 2 and f solving above equations, we get 4 Ans. a2 = g = 3.6 m/s2 11 4 gt Ans. v = a2t = 11 34. Equations of motion are, …(i) F + f1 = Ma1 …(ii) f1 + f2 = Ma2 …(iii) 2F – f2 = 2Ma3

ω=

Ans.

Ans.

Here, a B / 0 has two components at (tangential acceleration) and an (normal acceleration)

Rotational Mechanics — 581

Chapter 12 at = rα = (0.3)(5) = 1.5 m/s2 an = rω 2 = (0.3)(4 )2 = 4.8 m/s2 and

a0 = 2 m/s2

∴

aB = (Σax )2 + (Σay )2 =

60°

(2 + 4.8 cos 45° − 1.5 cos 45° )2

O

ω, α

+ (4.8 sin 45° + 1.5 sin 45° )2

= 6.21 m/s2  M   l BC =      M + m  2 and

O C

From conservation of linear momentum, (M + m) v = mv0  m  or v=  v0  M + m

ω

v

B

Fx

α

…(i)

Mg

1  Ml 2  2 l   ω = Mg 2 3  4

momentum about

ω=

∴

3 g ⋅ 2 l

…(ii)

 l Fx = M   ω 2 = M  2

2

Ml 2  l   +  4 12

 l   3g  3     = Mg  2  2l  4

 l Mg − Fy = M (α )    2

2  m   l2  +M    ω  M + m  4   

Putting

…(i)

Fy

 m   l OC =      M + m  2

From conservation of angular point C we have, mv0 ( BC ) = Iω   M 2 mMv0l or = m   2 (M + m)   M + m 

α=

A

37. C is the COM of (M + m)

l 2 =3 g 2 l Ml 2 3

(Mg )

Fy = Mg −

mv0 =v M +m

3 Mg Mg = 4 4

F

Fy

from Eq. (i), we have v l  4m + M  = ω 6  M + m  v , from C ω (towards O) will be at rest. Hence, distance of point P from boy at B will be BP = BC + x  M   l l  4m + M  =    +  M + m  2 6  M + m 

α

Now, a point (say P) at a distance x =

2l Ans. 3 38. Let ω be the angular velocity and α the angular acceleration of rod in horizontal position. Then

Fx

∴

F = Fx2 + Fy2 10 Mg 4 Fy (M g / 4 ) 1 tan α = = = Fx (3M g / 4) 3 =

=

∴

 1 α = tan −1    3

Ans.

Ans.

13. Gravitation INTRODUCTORY EXERCISE

5. m1 = m2 = ( volume)(density)

13.1

4  =  πr3 ρ 3 

1. Fnet = F 2 + F 2 + 2FF cos 60° Fnet F

F

2F

Gm1m2 r2 4  4  G  πr3  πr3 ρ 2 3  3  = r2 or F ∝ r4

∴

2F 30°

30°

60°

= 3F = 3

3 GM 2 4 a2

GMM = (2a)2 3m

Fnet

4F

F

3F

F

2F F

m

2m

G. m. m r2 2 2 Gm2 4 2 Gm2 = = (a/ 2 )2 a2 F Gm1m2 Gm2 3. a1 = = = 2 m1 r2 r =2 2

(6.67 × 10−11 ) (2) (0.5)2

= 5.3 × 10−10 m/s2 Gm1 r2 = 2(F1 cos 45° ) + F2 a2 =

Similarly,

4. Fnet

F1

P

m

13.2

GM M or g ∝ 2 R2 R Mass and radius both are two times. Therefore, value of g is half. g 2. (a) g′ = 2 h  1 +   R g At h = R , g′ = 4 d  (d) g′ = g 1 −   R R g At d = , g′ = 2 2 g h  3. = g 1 −  2  R h  1 +    R Solving this equation, we get  5 − 1 h= R  2 

45° d

4. ∆g = g′ − g = − Rω 2 cos2 θ

m

2

  2π 2 = − (6.37 × 106 )   cos 45°  24 × 3600

F1

F2 A

Hence proved.

1. g =

4m

=

2r

INTRODUCTORY EXERCISE

2. Fnet = 2 (2F )

r

F=

2d

d

B m

= − 0.0168 m/s2

m C

g′ = 0.64 g =

5. Fnet

= 2 F1 + F2 =

( 2 ) Gmm Gmm + ( 2d )2 d2

 2 + 1 Gm2 =  2  d2 

(Along PB)

∴

g 2

h  1 +   R h 5 1+ = R 4 R h = = 1600 km 4

Gravitation — 583

Chapter 13 6. g′ = ∴ or

− Gm 5Gm  =−  a  a

2. V = 5 

3 g = g − Rω 2 cos2 0° 5 2 Rω 2 = g 5 2g 2 × 9.8 ω= = 5R 5 × 6400 × 103

For E Five vectors of equal magnitudes, when added as per polygon law of vector addition make a closed regular pentagon. Hence, net E is zero. Gm Gm 3. V = 4  −  = − 4  a  a

= 7.8 × 10−4 rad /s

7. At equator, g′ = g − Rω 2

For E

  2π = (9.8)2 − (6400 × 103 )    24 × 3600 = 9.766 m/s2 Now, mg = 1000 1000 ∴ m= g  g′  w′ = mg′ = (1000)    g (1000) (9.766) = = 997 N (9.8) 0 = g − Rω g 9.8 ω= = R 6400 × 1000 2

∴

E

E

E

E net

Gm a2 4. Vp = due to particle + due to Enet = E =

shell

8. g′ = g − Rω 2 ∴

E

2

= 1.237 × 10−3 rad /s and T =

2π ω

By putting the value of ω, we get T ≈ 84.6 min g 9. = g − Rω 2 h  1 +   R By putting the values of g , h, R and ω, we get h ≈ 10 km

INTRODUCTORY EXERCISE

13.3

m

m

E 30° 30°

13.4

 ∂V $ ∂V $ ∂V $  i+ j+ k ∂y ∂z   ∂X 2

2 2  ∂V  ∂V   − ∂V  2. | E | =  −  + −   +  ∂x   ∂z   ∂y 

∂V = 10 J/kg-m is given ∂x No information is given about ∂V ∂V and − − ∂y ∂z | E | ≥ 10 N/kg 

O

E

Enet = E 2 + E 2 + 2(E )(E )cos 60° = 3E =

m,R

1. E = − 

So,

Gm E= 2 a 3Gm a2

INTRODUCTORY EXERCISE

P R/2

−

Gm Gm 2Gm 1. V = − − =− a a a

Enet

m

Gm Gm =− − R /2 R 3Gm =− R F 4 $i 5. E = = = (200 $i ) N/kg m 20 × 10−3

3. F = mE = m  − 

∂V  $  ∂V  $   i + −  j  ∂y   ∂x 

= − (10 i$ + 10 $j) N or | F | = 10 2 N 4. F = mE = (2$i + 3$j) N/kg We can check that path given (therefore displacement) is perpendicular to force. ∴ W =0

584 — Mechanics - II INTRODUCTORY EXERCISE

13.5

1 2

6. Apply mv 2 =

1. K i + U i = K f + U f v

2v 10kg

∴ 0−

INTRODUCTORY EXERCISE

20kg

Gm1m2 1 1 = × 10 × (2v )2 + × 20 v 2 ri 2 2

Gm1m2 30

v=

 1 1  −   rf ri 

6.67 × 10−11 × 10 × 20  1 1 −    0.5 1 30

=

= 2.1 × 10−5 m/s 2v = 4.2 × 10−5 m/s 4 (4 − 1) 2. Total pairs = = 6. Four pairs are at a 2 distance of a and two pairs at distance of 2a. mgh 3. ∆U = h 1+ R Put h=R mgh 4. (i) ∆U = h 1+ R Putting h = nR, we get and

1  2GM  m  2  R  m = (gR 2 ) R = mgR =

2. ve =

1 = mv 2 2 2ngR v= n+1

∴

5. h =

v2 2g − (v 2 / R ) =

(104 )2 (104 )2 2 × 9.8 − 6.4 × 106

= 2.51 × 107 m = 2.51 × 104 km

(as GM = gR 2 )

2GM R

M R 3. (a) Total mechanical energy = E0 − 2E0 = − E0. or

ve ∝

Since, it is negative, it will not escape to infinity. (b) Ei = E f ⇒ E0 − 2E0 = 0 + U ⇒U = − E0

INTRODUCTORY EXERCISE

13.7

1. Delhi does not lie on equator. 2. T ∝ r3/ 2  T2   r2    =   π   r1 

3/ 2

∴

r  T2 =  2   r1 

3/ 2

or

 n  ∆U =   mgR  n + 1  n  (ii) ∆U =   mgR  n + 1

13.6

1 1. KE = mve2 2

Gm1m2 − rf ∴

mgh h 1+ R

T1

 2 × 104  =   104 

3/ 2

(28) = 56 2 h

3. r1 = R + h1 = R + R = 2R r2 = R + h2 = R + 3R = 4 R 1 KE and PE ∝ r K 1 U 1 r2 2 ∴ = = = K 2 U 2 r1 1 GM GM = r R+h GMm GMm (b) KE = = 2r 2 (R + h) GMm GMm (c) PE = − =− r (R + h) 2π 3/ 2 2π (d) T = r = (R + h)3/ 2 GM GM

4. (a) v =

Chapter 13 5. (a) W = Energy of satellite in first orbit − energy of satellite on the surface of earth GMm  GMm =− − −  2 (2R )  R 

Gravitation — 585

(b) W = energy of satellite in second orbit – energy in first orbit  GMm − GMm  =− − 2 (3R )  2 (2R ) 

3 GMm 4 R 3 3 = mgR = × 2 × 103 × 10 × 6.4 × 106 4 4 = 9.6 × 1010 J

1 GMm 1 = mgR 12 R 12 1 = × 2 × 103 × 10 × 6.4 × 106 12 = 1.07 × 1010 J

=

=

Exercises LEVEL 1

9. Geostationary satellites lies about equator and

Assertion and Reason G m1m2 r If r decreases,U also decreases. GM 2. VC = − 1.5 R GM VS = − R

1. U = −

GM R2 C → centre, S → surface. 3. If a mass m is displaced from centre along the line AB, force on it is away from the centre. So, it is in unstable equilibrium position at centre. So, potential energy and hence the potential at centre is maximum.  ∂V $ ∂V $ ∂V $  4. E = −  i+ j+ k ∂y ∂z   ∂x EC = 0

and

ES =

∂V = 0 does not mean field strength is zero. ∂X ∂V ∂V and . Because it also depends on ∂y ∂z

5.

6. 7. 8.

u2 or h ∝ u2 cannot be applied in this case. h= 2g Because, for higher values of v, acceleration due to gravity g does not remain constant. Force on planet is towards centre of sum. Hence, torque is zero only about centre of sun. Polar satellites don’t have equatorial plane. Earth’s gravity is utilised in providing the necessary centripetal force. But weight is felt due to moon’s weight is felt due to moon’s gravity.

Moscow does not lies over equator. GMm GMm GMm 10. V = − ,K = ,E=− r 2r 2r GM and V = r

11. By changing the radius, moment of inertia will

change. Hence, angular speed ω will change. But ω has no effect on the value of g on pole.

Single Correct Option 2. Angular momentum is conserved only about centre of sun. 3. g′ = g − Rω 2 cos2 φ At φ = 90°, g′ = g, independent of ω.

5. V is negative. V r

6. Comparing with  q

∫ E ⋅ dS =  ε 0  We have, ⇒ ⇒

1 ≡G 4 π ε0 1 ≡ 4 πG ε0 q≡m

8. E = 0 inside a sheet. Therefore, gravitational force on m is zero. 9. T ∝ r3/ 2

586 — Mechanics - II 10. Area velocity remains constant. If area is half time taken is half. 1 11. T ∝ g T2 = T1

g1 g2

1.05 T = T

or

g

h  g 1 −   R Solving this equation we get, h = 64 km g M GM 12. g = 2 ⇒ ∴ = 2 G R R 1 13. E ∝ 2 r E2  r1  =  E1  r2 

ω 2 2π /T2 T1 = = ω 1 2π /T1 T2

 3R + R  =   R 

or

gpole 2

Ans.

6

 2π  (ω 1 ) 6 +   = 2π  23 

3/ 2

=8

∴ T2 = 8T1 15. gequator = g − Rω 2 and gpole − Rω 2 ∴

24 × 60 = 17 84.6

21. ω 1 t + ω 2t = 2π

2

or 3/ 2

Ans

20. At 84.6 min, g at equator becomes zero.

=

14. T ∝ r3/ 2 T2  r2  =  T1  r1 

R2

g ∝ ρR g1 = g2 ρ 1 R1 = ρ 2 R2 ρ1 R  1 R2 = ⋅ R1 =   R =  2 ρ2 2

or

h = 9 Re

or

4  G  π R 3ρ 3 

2

 Re  1 =  100  Re + h

or

GM 19. g = 2 = R or or ∴

g = g − Rω 2 2 g ω 2R 9.8 = 2 × 6.4× 106

16. U = mV or U ∝V U is half, it means gravitational potential is half. GM 1  3 GM  − = − (R + h) 2  2 R  R Solving we get, Ans. h= 3 mgh 17. ∆U = h 1+ R GM 18. T = r T is independent of R, the radius of earth.

3π 2

 π or ω 1 =   rad /h  4

23. Only due to M 1. 24. Time taken in one complete oscillations is 84.6 min. 25. W = ∆U = U f − Vi  GMM   GMM  = 3 −  − 3 −    2l  l 

or

= 8.75 × 10−4 rad /s

6ω 1 =

3 GM 2 2 l 26. Change in kinetic energy = change in potential energy 1 GMm  GMm mv 2 = − − − ∴  2 R+R  R  =

2GM / R GM = R 2 ve = 2 where, ve = escape velocity v=

27. U = ∴

− GMm r (− GM ) =

Ur m

GM (Ur / m) U = = 4 mr (2r)2 4 r2 U F = mE = 4r E=−

Gravitation — 587

Chapter 13 Subjective Questions 1. Potential at the surface of sphere, GM R (6.67 × 10−11 )(20) =− J/kg 1 −9 = − 1.334 × 10 J/kg

V =−

i.e. 1.334 × 10−9 J work is obtained to bring a mass of 1 kg from infinity to the surface of sphere. Hence, the same amount of work will have to be done to take the particle away from the surface of sphere. Thus, W = 1.334 × 10−9 J GM 2. g = 2 R dg − 2GM ⇒ = dR R3 dg − 2GM 1 ⇒ . = h R R2 dg  h =−2  ⇒  R g

3. Let vr be their velocity of approach. From conservation of energy, Increase in kinetic energy = decrease in gravitational potential energy 1 2 Gm1m2 …(i) or µvr = 2 r Here, µ = reduced mass m1m2 = m1 + m2

5. (a) ve = 2 vo . Given v = 1.5 vo If vo < v < ve, satellite will move in elliptical orbit with initial position as the perigee position. (b) v = 2vo > 2 vo or

ve

Hence, the satellite will escape to infinity.

6. Present angular momentum of earth 2 MR 2ω 5 New angular momentum because of change in radius L1 = Iω =

2

2  R M   ω′  2 5 If external torque is zero then angular momentum must be conserved L1 = L2 2 1 2 2 MR ω = × MR 2ω′ 5 4 5 i.e. ω′ = 4 ω 1 1 T ′ = T = × 24 = 6 h 4 4 7. Particle lies between two shells. Therefore, net force in exerted only by the inner shell. Gm1m 4 Gm1m F= = ∴ 2 2 ( R  R1 + R2  1 + R2 )    2  L2 =

8. L = πR ∴

R=

L π

=∫

π

⇒ F = ∫ dF sin θ 0

π GdM . m 0

R2

sin θ

Substituting in Eq. (i), we get vr =

2G (m1 + m2 ) r

dM

dF

dF

T = 2π

or

dM

θ

4. Close to earth,

or

dθ

R = 2π g

R GM / R 2

4π R GM 4 π 2R 3 = 4  G  πR 3ρ 3 

T2 =

ρT 2 =

m

=∫

2 3

3π = a universal constant G

π

M  G  dθ m π 

0

L2

(π 2 ) sin θ

2πG mM L2 9. K i + U i = K f + U f =

∴

1 GMm 1 m (2 gR )2 − = mv 2 + 0 2 R 2

588 — Mechanics - II Gm G (2πa3ρ 0 ) = 2 (2a) 4 a2 πGρ 0a = 2 13. By conservation of momentum, their speeds are same. Using the energy equations we have Ki + U i = K f + U f Gmm 1  Gmm 0− = 2  mv 2 − 2  ri rf

2GM R Putting GM = gR 2, we get

∴

v 2 = 4 gR −

E=

v 2 = 4 gR − 2gR v = 2gR

∴

GM G (4 M ) 10. (a) 2 = (12 R − r)2 r P 12R – r Q

r

M, R

∴ 4M, 2R

Solving this equations, we get r = 4 R vQ vP = 2  GM   − G (4 m)  ∴ −  =  r   12 R − r  − G (4 m) GM = − 2R 10 R Solving this equation, we get r = 7.65 R and 1.49 R 3a 11. (a) At distance which lies between solid sphere 2 and shell field is only due to solid sphere. GM 4GM (towards centre) E= = 2 (3a/ 2) 9a2 5a (b) At distance which lies outside the shell. So, 2 field strength is due to both G (M + M ) 8 GM (towards centre) ∴ E= = (5a/ 2)2 25a2

 1 1 v = Gm  −   rf ri  Gm ri

=

6.67 × 10−11 × 1030 1010 = 8.16 × 104 m/s

=

= 81.6 km/s

Using Eq. (i), we have  1 1 v = Gm  −   rf ri   1 1  = 6.67 × 10−11 × 1030  − 10  5  2 × 10 10  = 1.8 × 107 m/s = 1.8 × 104 km/s

14. Using the equation ∴ r

∴

dV

Ki + U i = K f + U f 1 GmM 2 mv − =0+ 0 2 2R GM v= R

Ans. v R

15. (a) W = U f − U i = U B − U A

 ρ a =  0  (4 πr2 )dr = (4π aρ 0 ) (rdr)  r  a

∴ Mass of whole sphere = ∫ (4 πaρ 0 ) rdr 0

m = (2π a3ρ 0 )

Ans.

rf = 2R

12. Volume of small strip dV = (4 πr2 ) dr mass of this strip dm = (ρ ) (dV )

…(i)

(V = gravitational potential) = mVB − mVA = m (VB − VA )  100 400 100 400  = (1) − − + + (6.67 × 10−11 )  2 8 8 2  = 7.5 × 10−9 J

Gravitation — 589

Chapter 13 (b) First let us find the point (say) cbetween A and B where field strength due to 400 kg and 100 kg masses is zero. Let its distance from 400 kg is r. G (400) G (100) Then, = r2 (10 − r)2 20 10 r= m and (10 − r) = m 3 3 So, we have to move the body only from A to C. After that 100 kg mass will pull 1 kg mass by its own. ∴Minimum kinetic energy = ∆U = UC − U A = m (VC − VA )  400 100 100 400  (6.67 × 10−11 ) = (1) − − + + 8 2   20/ 3 10/ 3 = 8.17 × 10−9 J

16. (a) F =

GM . M GM 2 = (2R )2 4 R2

M

R COM R M

18.

2π 3/ 2 r = 1yr GM G (M ) (M ) Mv 2 = (r / 2) r2 GM 2r 2π (r / 2) πr T = = v GM 2r v=

∴

 πr3/ 2 1  = yr  as 2  GM  GMm GMm GMmh 19. W 1 = ∆U = − + = (R + h) R R (R + h)

=

v

(b) Immediately after collision velocity of combined mass = 0 Path is straight line as shown only potential energy is there. G (2m) M v=− r 2GMm =− r

or or

2m

1 1  GM  mv 2 = m   2 2  r 

2

1  GM  GMm m = 2  R + h 2 (R + h) W1 > W2 h 1 R or h > > R 2 2 6370 h> km 2 h > 3185 km

If

 GMm − GMm = 2 −  =  2r  r

M

M

 1 = 2   yr = 0.71 yr  2

v

(c) W = E f − Ei = [U f − U i ] + [ K f − K i ] 2   GM   1   −GMM   = 0 −      + 0 − 2 × × M   2R    2  4R      2 GM = 4R 17. (a) E = 2 (energy of one satellite)

r/2 r/2

 π  3/ 2 = 2  r  GM 

W2 =

GM 2 MV 2 (b) = R 4 R2 GM ∴ v= 4R 4 πR 3/ 2 2π (R ) 2πR T = = = v GM GM / 4 R

...(i)

LEVEL 2 Single Correction Option 1. Just before collision, v0 =

GM = r

GM GM = R+R 2R

From conservation of linear momentum, m m mv0 = × 0 + v 2 2 2GM ∴ v = 2v0 = R Increase in mechanical energy = K f − Ki 1 m 2 1 = v − mv02 2 2 2 1 1 2 = m (2v0 ) − mv02 4 2

590 — Mechanics - II 1 1  GM  1 GMm mv02 = m   = 2 2  2R  4 R 1 GM   = mgR  as g = 2   4 R  =

π

0

y Enet dE dM

dE θ

=∫

θ

π G ⋅ dM

dθ

0

GM  F  y  as a = E =   R3   m 4  G  πR 3ρ 3  4 = y = π Gρy 3 R3 6. Earth rotates from west to east. Net velocity of train = (ω R − θ )2

∴

 ω (ωR − 2v ) v 2  Ans. = mg 1 − − g Rg  

7. Gravitational field at any point inside the cavity is uniform (both in magnitude as well as direction). So, let us find its value at centre of cavity. (at centre of cavity) ER = ET − EC R → Remaining, T → total, C → cavity So, (as EC = 0) ER = ET  GM  =  3  a or ER ∝ a  R 

Mv 2 G. MM  GMM = 2 +  a2  (2R )2 R

or

v2 2 GM GM = + R ( 2R )2 4 R2

∴

GM 2 2 + 1 v= R 4

8. Potential (and hence potential energy) at centre is 3 times the value on surface. So, required kinetic 2 3 3 energy is times or required speed is times. 2 2

9. F1 R

a = √2 R

C F2

∫

dV = − ∫ Edr

10. a = 2R cos 30° = 3 R

F1

45° P

m (ωR − θ )2 R m (ωR − v )2 N = mg − R  ω 2R 2 + v 2 − 2 vRω  = mg 1 −  gR  

Now, mg − N =

sin θ

R2 M  G  dθ π π  sin θ =∫ 0 R2 2GM 2 GπM = = πR 2 l2 l = πR l R= ∴ π 3. Net force on P towards centre F = 2 F1 + F2 ∴

So,

5. a = E = 

2. Enet = ∫ dE sin θ

dM

GM GM  GM  (1 − k 2 ) =  as gR =  r r  R  R r= 1− k2

or

a

v

4. K i + U i = K f + U f 1 GMm GMm m (kve )2 − =0− 2 R r 1 2 GM GM k (2gR ) − =− ∴ 2 R r

F

F R C 30° a F

Fnet F

Fnet = 3 Fis towards centre C ∴

3F=

MV 2 R

Chapter 13 GMM MV 2 = R a2 3GM 2 MV 2 = R ( 3R )2

or

From conservation of angular momentum about centre of earth

3

or

v=

or

11. T ∝ R 2

3

⇒ ∴ T

2/ 3

 m n 

GM 3R

1 GMm mv 2 − 2 =0+ 0 2 2R v=

2 2GM R

14. W = − ∆U = U i − U f =−

=

 (1) (2) (1) (3) (2) (3)    1 + 1 + 2    (1 × 2) (1 × 3) (2 × 3)  − + +  1 1   1

Gmm a

6 Gm2  1 1 −   a 2

15. K i + U i = K f + U f 1 3 GMm GMm mv 2 − =0− 2 2 R R Solving, we get GM v= R 2GM 16. v1 = nve = n 2gR = n R ∴

v1

v2

From mechanical energy conservative, we have 1  m n 2 

2

…(ii)

17. Let v is the velocity at centre then. Ki + U i = K f + U f GMm 1 3 GMm or 0− = mv 2 − R 2 2 R GM v= ∴ R From conservation of linear momentum. 2mv − mv = (2m + m) v v 1 GM v′ = = ∴ 3 3 R Again applying energy conservation, Ki + U i = K f + U f

2GM  GMm 1 GMm ...(i) − = mv22 − R 2 R  R R+ 2

1 1 (3m)  2 3

2

GM  3 GM (3m) −  R  2 R GM  = 0 − 3m (1.5 R 2 − 0.5 r2  R 3  Solving this equations, we get R r= 3 R So, is the amplitude of oscillations. 3 18. ve = 2gR ∴

or ve ∝ gR dU = − 6r2 dr mv 2 Now = 6r2 r 1 2 ∴ mv = 3r3 2 E = K + U = 3r3 + 2r3 = 5r3

19. F = −

At R 2

 R   R = mv2  R + 2  

n = 0.6

1 ∝ R or T ∝ (1/ R 3 ) 2

= (K A + U A ) − (K α + U α ) = (K A + mVA ) − (0 + 0) 1 ∴ − 5.5 = × 1 × (3)2 + (1) VA 2 or VA = − 10 J/kg 13. K i + U i = K f + U f

∴

2GM R

Solving these two equations we get,

12. W = ∆E = EA − Eα

∴

Gravitation — 591

r = 5m, E = 625 J

20. E on the surface of earth, F 10 = m 1 = 10 N /kg 1 E∝ 2 r

E1 =

Ans.

592 — Mechanics - II 2

 R  E2  r1  4 =  =  =  3R / 2 E1  r2  9

∴

E2 =

21. g =

So, just over the point masses, E = ∞. Hence, in moving from one point mass to other point mass, E first decreases and then increases. V due to a point mass is Gm V =− r As r → 0, V → − ∞ So, just over the point mass, V is − ∞. Hence, in moving from one point mass to other point mass, V first increases and then decreases. 5. Inside a shell, V = constant and E = 0.

2

∴

4 40 Ef = N/kg 9 9  40 F2 = m E2 = (200)   = 889 N  9

GM R2

⇒ ∴ GM = gR 2

v=

GM = r

GM = R+x

gR 2 R+x

22. C → cavity, T → Total, R → remaining F1 =

GMm (2R )2

…(i)

FR = F2 = FT − FC = F1 − FC M G  m  8 GMm − F2 = (2R )2 (3R / 2)2

or

14 GMm 72R 2 From Eqs. (i) and (ii) we get, F2 7 = F1 9 F2 =

or

…(ii)

Between A and B, Enet = 0, Vnet = constant because these points µ inside both shells. Between B and C E of m ≠ 0, E of 2m = 0 V of m ≠ constant, V of 2m = constant. Beyond C E and V due to both shells are neither zero nor constant. Gm 6. E = 2 r 4m 4E

ve = 2gR =

=

2E r

0−

2m

E′ net = 2 2 E = (asρ is same)

4  2G  πR 3ρ 3  R

v=

2GM R

4. E due to point mass is E = As r → 0, E → ∝

Gm r2

2 2 Gm r2

E′net = (9E )2 + E 2 = 10 E

2GM R

GMm 1 3 GMm = mv 2 − 2R 2 2 R

or

2E

2E

E

m

10 Gm r2

=

E′net > Enet 4m

3m 4E

So, ve or v ∝ R (as ρ is same). 3. K i + U i = K f + U f ∴

Enet

3E

More than One Correct Options 4  G  π R 3ρ 3  GM 1. g = 2 = 2 R R or g ∝R

3m

Enet

3E 3E r 2m

E 2E m

Potential in both cases is G Vnet = − (m + 2m + 3m + 4 m) r 10 Gm =− r

E

Chapter 13 7. At highest point velocity of particle-1will becomes zero. But velocity of particle-2 is non zero. 8. At maximum distance (at A) kinetic energy is minimum. But angular momentum about centre of sun always remains constant. GMm 9. E1 = − R GMm U2 = − (r = R + R = 2R ) 2R GMm K2 = 4R GMm and E2 = − 4R GM 10. v = r or v ∝ GM 2π 3/ 2 T = r GM 1 or T ∝ GM

Match the Columns 1. Inside the shell, V , U , K , and v remains constant. F , E and a are zero. From A to B, kinetic energy increase and potential energy decreases. 2. Resultant of five field strength vectors of equal magnitude acting at same angles is zero. 5 GMm (b) V = − r (c) If E5 is removed then resultant of rest four vectors is equal and opposite of E5. E4

(d) V = −

3.

GM  y xy y x =  = y3  2  y3 2 2 y GM xy x (d) |V | = = = 2y 2y 2 mgh 4. (a) W = ∆U = h 1+ R 1 Put h = R , W = mgR 2 GMm GMm 1 (b) KE = = = mgR (asGM = gR 2 ) 2r 4R 4 GMm (c) E = − 2r GMm GMm 1 E1 = − =− = − mgR 2 (2R ) 4R 4 GMm 1 E2 = − = − mgR 2 (3R ) 6 1 E1 ~ E = mgR 12 mgh (d) KE = ∆U = h 1+ R 1 Putting h = R , KE = mgR 2 (c) E =

Subjective Questions 1. W = ∆U = U f − U i =3

| Enet | = | E5 | = 4Gm r

GM = x or GM = xy y GM xy x (a) E = = = 2 2 4y (2 y) 4y

=

E2

2. h =

E1

∴

y3  GM  1.5 y2 − 0.5  3 4 y 

 xy   11 2 11 =  3  y= x y 8  8

E3

Enet = 0

E5

(b) |V | =

Gm r2

Gravitation — 593

∴

 − Gmm   Gmm  −3 − a   2a  

3 Gmm 3Gm2 = 2a 2a

Ans.

u2 2ge u = 2geh

…(i)

For the asked planet this u should be equal to the escape velocity from its surface. ∴ 2geh = 2gp Rp or

geh = gp Rp GM p GM e .h = . Rp 2 Re Rp2

594 — Mechanics - II

or

 4 3  πRe  ρh 3  Re2

=

 4 3  πRp  ρRp 3 

=

Rp2

    1 G Mm   = − 1 2 d2  R   81 −     2d   

RP = Reh

or

= (6.41 × 106 )(1.5) = 31 . × 103 m

Ans.

v 3. (a) vo = e 2

G Mm G Mm − d2 8(d − R/2)2

6. Let m1 be the mass of the core and m2 the mass of outer shell.

2GM GM R ∴ = r 2 ∴ r = 2R or h = r − R = R or height = radius of earth. (b) Increase in kinetic energy = decrease in potential energy 1 2 mgh mv = ∴ h 2 1+ R 2hg ∴ v= h 1+ R Substituting the values we have, 2 × 9.81 × 6400 × 103 R 1+ R = 7924 m/s ≈ 7.92 km/s

v=

B A R

2R

gA = gB Gm1 G (m1 + m2 ) = (2R )2 R2

Then ∴

4 m1 = (m1 + m2 ) 4 4 3  or 4  πR ρ 1  = πR 3 ⋅ ρ 1 3  3 4  4 +  π (2R )3 − πR 3  ρ 2 3  3

∴ Ans.

4. (i) At point A, field strength due to shell will be zero. Net field is only due to metal sphere. Distance between centre of metal sphere and point A is 4R. G (m) Gm EA = = ∴ (4 R )2 16R 2 (ii) At pointB , net field is due to both, due to shell and due to metal sphere. Gm Gm ∴ EB = + (5R )2 (6R )2 61 Gm Ans. = 900 R 2 R 5. Radius of hollow sphere is , so mass in this hollow 2 M portion would had been, . 8 Now, net force on m due to whole sphere = force due to remaining mass + force due to cavity mass. ∴ Force due to remaining mass = force due to whole sphere − force due to cavity mass

(given)

4ρ 1 = ρ 1 + 7ρ 2 ρ1 7 = ρ2 3

∴

Ans.

7. Total mechanical energy of a satellite in an elliptical orbit of semi major axis ‘a’ is −

GMm . 2a

E = K +U GMm 1 2 GMm − = mv − 2a 2 r  2 1 2 Hence Proved. v = GM −  r a 

∴ or

8. dF = force on a small mass ‘ dm’ of the ring by the sphere. dm dF

θ

a √3 a

dm dF

θ 2a

M

Gravitation — 595

Chapter 13 Net force on ring = Σ(dF sin θ ) or ∫ dF sin θ =Σ

GM (dm) 3 × = 2 (2a)2

10. Let mass of the ball be m. u=0

3GM Σ(dm) 8a2

A

But Σ(dm) = m, the mass of whole ring. Net force =

∴

B

3GMm 8a2

Ans.

v

9. Let there are two stars 1 and 2 as shown below. 16 M

M C1

r1

P

r2

C2

a 2a 2

1

Let P is a point between C 1 and C 2 , where gravitational field strength is zero. Or at P field strength due to star 1 is equal and opposite to the field strength due to star 2. Hence, GM G (16M ) r or 2 = 4 = r1 r12 r22 also ∴

r1 + r2 = 10a  4  r2 =   (10a) = 8a  4 + 1

and r1 = 2a Now, the body of mass m is projected from the surface of larger star towards the smaller one. Between C 2 and P it is attracted towards 2 and between C 1 and P it will be attracted towards 1. Therefore, the body should be projected to just cross point P because beyond that the particle is attracted towards the smaller star itself. 1 2 From conservation of mechanical energy mvmin 2 = Potential energy of the body at P – Potential energy at the surface of the larger star.  GMm 16GMm  1 2 ∴ mvmin = − − r1 r2  2   GMm 16GMm  − − − 2a   10a − 2a  GMm 16GMm   GMm 8GMm  = − − − − −  2a 8a   8a a  1  45 GMm 2 or mvmin =   8 a 2 ∴

vmin =

3 5  GM    2  a 

Ans.

1 mv 2 = m(VA − VB ) 2 GM    GM  = m − −  −15 .   R    R GMm = 2R GM ∴ v= R Velocity of ball just after collision, 1 GM v′ = ev = 2 R Let r be the distance from the centre upto where the ball reaches after collision. Then, 1 mv′ 2 = m [V (r) − V (centre)] 2  3GM GM  3R 2 r2   1 GMm or =m − 3  −  8 R 2 R  2  2R 2 1 3 3 r or = − + 8 2 2 2R 2 R r2 1 or r = ∴ = 2 2 4 R ∴ The desired distance, R R Ans. s = R + + = 2R 2 2 11. Applying conservation of mechanical energy, Increase in kinetic energy = decrease in gravitational potential energy 1 or m0v 2 = UB − UA = m0 (VB − VA ) 2 …(i) ∴ v = 2(VB − VA ) Potential at A VA = potential due to complete sphere – potential due to cavity 1.5 GM  Gm  =− − −  R/2  R 2Gm 1.5 GM = − R R

596 — Mechanics - II ∴ ∴

3

4  R πρR 3 π  ρ= 3  2 6 4 and M = πR 3ρ 3 Substituting the values, we get  G  πρR 3 5 VA =  − 2πρR 3  = − πGρR 2 R  3 3  Here,

m=

Potential at B 2 GM   R   1.5Gm VB = − 3 1.5R 2 − 0.5    +  2 R/2 R   11 GM 3Gm =− + 8 R R  4 G  πρR 3 11 =  − ⋅ πρR 3  = − πGρR 2 3 R  2 6  1 VB − VA = πGρR 2 ∴ 3 So, from Eq. (i) 2 v= πGρR 2 3

(c) vr = constant 14. (a)

r1

r2

m2

Solving these two equations we get,  m2   m1  r1 =   d or r2 =   d  m1 + m2   m1 + m2 

Ans.

Ans.

(b) Apply conservation of linear momentum and conservation of mechanical energy. 13. (a) Mean radius of planet, r +r m2 = 1 2 = 1.4 × 108 km 2 Now, T ∝ r3/ 2

The centripetal force is provided by gravitational force, Gm1m2 m1r1ω 2 = m2r2ω 2 = d2 Solving these equations, we get G (m1 + m2 ) ω= d3 ∴

T =

d3 G (m1 + m2 )

2π = 2π ω

Ans.

1 I ω2 K1 2 1 I m r2 (b) = = 1 = 1 12 K 2 1 I ω 2 I 2 m2r2 2 2 2

2

m  r  m  m  m =  1   1  =  1   2  = 2 Ans.  m2   r2   m2   m1  m1 (c)

L1 I 1ω I 1 m2 = = = L2 I 2ω I 2 m1 = (m1r12 + m2r22 ) ω  m m2d 2 m2m12d 2  = 1 2 2 + ω (m1 + m2 )2   (m1 + m2 ) = µωd 2

3/ 2

T2 = 2 (1.4)3/ 2 Ans.

(b) For m2 , point P is perigee position. So, speed at this point is greater than orbital speed for circular orbit.

Ans.

(d) L = L1 + L2 = (I 1 + I 2 ) ω

∴ Time period of m2 :

= 3.31 yr

…(i) …(ii)

d

Solving, we get

or

COM

m1

displacement of the particle is x0 − x, or (3.0 − x ) m. Centre of mass will not move. Hence, (5.4 × 109 )x = (6 × 108 )(3 − x )

 1.4 × 108  T2 = T1    108 

r1 + r2 = d m1r1 = m2r2

12. (a) Let x be the displacement of ring. Then

x = 0.3 m

U m2 = U m1 Em2 > Em1 K m2 > K m1

where, (e)

m1m2 = reduced mass m1 + m2 1 1 K = (I 1 + I 2 ) ω 2 = µω 2d 2 2 2

Ans.

µ=

Ans.

14. Simple Harmonic Motion INTRODUCTORY EXERCISE

14.1

INTRODUCTORY EXERCISE

1. a = − 4 x Comparing with a = − ω 2x, we get

2

ω = 2 rad/s 2π T = = (π )sec ω

Now,

=

2

1 2 1  A kx = k   2 2  4 1  1 2 =  kA   16  2 1 = E 16 1 15 fraction is potential energy and Hence, 16 16 fraction is kinetic energy. 3. (a) A = initial displacement from mean position

Ans.

π  π t (b) 0.5 = 1.5 sin  1 +   4 6  From here find t. π  πt Then, − 0.75 = 1.5 sin  2 +   4 6

1 2

U =

= 130 . m/s

1  π × 2.0 ×   × (15 . )2  4 2

= 1.39 J

2. E = kA 2

= 15 cm m 2.0 (b) T = 2π = 2π = 0.726 s k 150 1 (c) F = = 1.38 Hz T 1 (d) E = kA 2 2 1 = × 150 × (0.15)2= 1.69 J 2 (e) vmax = ωA  2π   2π  = A = . )  (015 T   0.726

14.2

1 1 2 1. (a) E = mvmax = (m) ω 2 A 2 2 2

From here find t2. Now, t1 ~ t2 is the required time. 2. ω = 3 rad /s A = 0.2 m At x = 5 cm v=±ω

A2 − x2

= ± 3.0 (0.2)2 − (0.05)2 = ± 0.58 m/s

Ans.

a = − ω x = − (3) (0.05) 2

Ans.

2

= − 0.45 m/s2

Ans.

At x = 0 v = ± ωA = ± (3.0) (0.2) = ± 0.6 m /s

Ans.

Ans.

a = − ω x = − (3) (0) 2

2

=0 Ans.

4. vmax = ωA = 2π fA

Ans.

3.

= (2π ) (2) (8 × 10− 3 ) = 0.101 m/s

Ans.

–A

amax = ω 2 A = (2πf )2 A = 4 × π 2 × 4 × 8 × 10− 3 = 1.264 m/s2 Fmax = mamax = (0.5) (1264 . ) = 0.632 N

Ans.

Ans.

5. No, as acceleration in SHM a = − ω 2x is variable.

A/2 x 60° +A 30° A

sin

x = A sin (ωt + 30° ) δ = 30° 2π 2π 4. (b) ω = = = 1.57 rad /s T 4 ∴

Ans. Ans.

598 — Mechanics - II (c)

k m ∴ k = m ω 2 = (0.8) (1.57)2 = 1.97 N/m ω=

Ans.

(d) At t = 1 s, v = slope of x -t graph = 0 (e) At t = 1 s, x is maximum (= + A ). Therefore. magnitude of acceleration is also maximum. a = maximum = ω 2 A = (157 . )2 (0.08) = 0.197 m/s

5.

2

Ans.

π  X = 5 sin  20t +   3 dx π  v= = 100 cos  20 t +   dt 3 dv π  a= = − 2000 sin  20t +   dt 3

Comparing with a = − ω 2x ⇒ω = 4 rad/s T =

2π  π  =   sec ω  2

M 2. 2 = 2π k M +4 3 = 2π k Solving two equations, we get M = 3.2 kg

…(ii)

ω=

p

M k 5T M +m = 2π 3 k Solving these two equations, we get m 16 = M 9 6. T ∝ l T = 2π

T′ l′ 121 . l = = = 1.1 T l l ∴ T ′ = 1.1 T Hence, percentage increase in T is 11%.

7. ω 2x = ω A 2 − x 2 =

A2 − x2 = 2πf x

A2 − x2 2πx Substitute, A = 2cm and x = 1 cm

∴

f =

INTRODUCTORY EXERCISE 14.4 1. In such situation, amplitudes are added by vector method. AR = (4 )2 + (3)2 + 2 (4 ) (3) cos φ …(i)

Now, we can substitute different values of φ given in different parts in the question and can find the value of AR .

1. a = − 16x

Now,

∴

= 25 + 24 cos φ

14.3

…(i)

ω2 = p

5.

(a) v = 0 for the first time when, π π 20t + = 3 2 π Ans. t= s ∴ 120 (b) a = 0 for the first time, when π 20 t + = π 3 π Ans. t= s ∴ 30 (c) When a = 0 for the first time, its speed will be maximum. π Ans. t= s 30 6. Simple harmonic with mean at x = 10, amplitude 4 and extreme positions at x = 6 and x = 14. At t = 0, it starts from x = 6. Here, x is coordinate, not displacement from mean position.

INTRODUCTORY EXERCISE

(0.3 + 01 .) k 0.1 T = 2π k Solving these two equations we get, T =1s 4. Comparing with a = − ω 2x 2 = 2π

3.

2. AR = (4 )2 + (3)2 + 2 (4 ) (3)cos 60° 3

…(i) …(ii)

AR 60° φ

4

= 6.1units 3 sin 60° tan φ = = 0.472 4 + 3 cos 60°

Chapter 14 ∴ ∴

φ = 25.3° x = 61 . sin (100 π t + 25.3° ) (a) At t = 0, x = 61 . sin 25.3° = 2.6 unit (b) vmax = ωA = (100 π ) (6.1) = 1917 unit (c) amax = ω 2 A

Simple Harmonic Motion — 599

3. Resultant of 1 and 3 is also A in the direction of 2. A

A 3

60°

2 60° 1

A

AR = 2 A

∴

4. A = A + A + 2 A. A cos φ 2

2

Solving we get, 1 cos φ = − 2

= (100 π )2 (6.1) = 6.0 × 105 units

or

φ = 120°

or

2π 3

Exercises LEVEL 1

7. F = − kx. For all displacements. From the mean

Assertion and Reason 1. x is measured from the mean position. 2. y = (− A cos ωt ) at the particle starts from − A.

position, whether they are small or large. π π T 8. = = . In the given time particle 2ω 2(2π /T ) 4 moves from x = A to x = 0.

So, displacement from the mean position will be − A cos ω t. x 2

.

X=0 At t = 0

X=2 Mean position

x = y+ 2 = − 2 cos ω t + 2 3. By applying a constant force on spring-block system mean position is changed but time period remains unchanged. θ π /6 T 4. t1 = t2 = = = ω 2π /T 12 ∴

t1

O 30° A 2

x 1 T  =   = 2 = constant for a given SHM.  2π  a ω

10. If a particle P rotates in a circle with constant

y

30°

2

9.

t2 3A 2

angle speed ω and we draw a perpendicular on and diameter. This perpendicular cuts the diameter at point Q then motion of Q is simple harmonic. But motion of P is circular.

Single Correct Option 1. In equation x = A cos ω t, putting x =

A we get 2

π 3 π  2π  ∴  t = T  3 T or t= 6 2. Kinetic energy and potential energy in SHM oscillate with double frequency. ωt =

3. v = ω A 2 − x 2 5. In angular SHM, path is not straight line. 6. k ∝ ∴

1 l

⇒

ω= ω∝

k m 1 ml

It m and l both are halved then ω will become 2 times.

v2 x2 + = A2 ω 2 (1)2 Hence, v − x graph is an ellipse. l 4. l in the equation T = 2π is measured from g centre of mass and centre of mass in both cases remains at same location. ∴

600 — Mechanics - II 5. At t = 1 s

1 2 kx 2 i.e.U versus x 2 graph is a straight line passing through origin.

13. U =

π φ1 = + φ 2 2π φ2 = +φ 3

and

14. T = 2π

π ∆φ = φ 2 − φ 1 = 6

∴

6. v = ω A 2 − x 2 3 ωA A = ωA A 2 − x 2 or x = 2 2 7. All other equation can be converted into the from y = a sin (ωt ± φ ) or a cos (ωt ± θ ). 8. Phase difference between cos πt and sin πt is 90°. 4

17.

(θ 1 = θ 2 + 2π )

2π t= ω1 − ω2 2π = (2π /T1 ) − (2π /T2 ) T T = 1 2 T2 − T1 4 × 4.2 = = 84 4.2 − 4

Number of vibration of X in this time are, t 84 N1 = = = 21 T1 4 mg 10. mg = kx ⇒ ∴ k = x M+m (M + m) x T = 2π = 2π R mg

11. ω 1 A1 = ω 2 A2 k2 / m k1 / m

1 3 12. I = mR 2 + mR 2 = mR 2 2 2 I T = 2π mgl l=R

kx0 = mg sin θ mg sin θ x0 = k = amplitude of oscillations

=

d 2x = − π 2x dt 2 d 2x = − ω 2x dt 2 ω=π 2πf = π 1 f = Hz 2

Comparing with

9. ω 1t = ω 2t + 2π

Here

F ∝−x Hence, F is − cos ωt graph. 16. In equilibrium, let x0 is extension in spring then ∴

AR = 4 2 units

A1 ω 2 = = A2 ω 1

15. x = A cos ωt

AR

4

∴

l=R

Put

∴

∴

1 1 1  g +   l R

k2 k1

We have, ∴

18. f =

1 2π

k µ µ = reduced mass M m = M +m

where,

19. T ∝

1 ∝ g

1 M / R2

R M R and M both are doubled. So, T will become 2 times. or

T ∝

T−1 = 2T = 2 2 s

∴

(as T = 2 s)

20. x3 can be written as 5 y 37°

3

15

x3 = 15 sin (ωt − π / 2)

x

Simple Harmonic Motion — 601

Chapter 14 A = 3 i$ + 5 cos 37° $j + 5 sin 37° i$ − 15 $j = 7$i + 12$j | A | = (7)2 + (12)2 =13.89

∴

37°

…(i)

(b)

= 3.53 rad /s 24. Let x0 is the compression in equilibrium. Then,

Ans.

Subjective Question 1. Resultant of k and k is 2k. Then, resultant of 2k and 2k is k. Now

T = 2π

m k

∆t =

∴

0.5 4 × 10− 2

kx0 = ma ma 1 × 2 x0 = = k 100 = 0.02 m Amplitude = x0 = 0.02 m

∆T t T1 But T′ ≈ T 1 ∴ ∆t = α ∆θ (t ) 2 1 = × 0.000012 × 20 × 24 × 3600 2 Ans. = 10.37 s At higher temperature, length of pendulum clock will be more. So, time period will be more and it will lose the time. 5. (a) kx = mg ⇒

A2 − x2

x = a

Ans.

1 2

x = 0.6 A = 6cm ωA (b) = ω A2 − x2 2 3 ∴ x= A 2 x 23. T = 2π a 2π = T

2.7 180

4. ∆T = Tα ∆θ

∴

ω=

m = 2π k = 0.78 s

T = 2π

This is equation of a straight line passing through origin. 22. (a) Kinetic energy is 0.64 times, it means speed is 0.8 times

∴

F 9 = = 180 N/m x 0.05 W 27 m= = = 2.7 kg g 10

x

= (a sin ωt + 0.8 b sin ωt ) y = S 2 sin 37° = 0.6 b sin ωt y sin ω t = 0.6 b Substituting this value in Eq. (i), we have  a + 0.8 b x=  y  0.6 b 

0.8 ωA = ω

Ans.

k=

∴

y

S1

m 0.2 π = 2π = s k 80 10 = 0.314 s

3. F = kx

21. X = S 1 + S 2 cos 37° S2

2. T = 2π

mg (20 × 10− 3 ) (10) = x 7 × 10− 2 20 = N/m 7 m T = 2π k k=

= 2π

Ans.

50 × 10− 3 (20/ 7)

Ans. = 0.84 s 1 6. k ∝ l Length is halved, so value of force constant of each part will become 2k. Now net force constant of 2k and 2k as shown in figure will becomes 4k. m T = 2π k 1 or T ∝ k k has becomes 4 times. Therefore, T will remain half.

602 — Mechanics - II 8. ge = g −

Upthrust mass =g− T ′ = 2π

V (δ /10) g 9 = g Vδ 10 l = 2π ge

10  l    9  g

 10  = T  9

9. For small values of x, the term x 2 can be neglected. ∴ Comparing with

F =−6x F = − kx, we get k = 6 N/ m

Ans.

2π 2π = = (8π ) rad /s T 0.25 x = A sin (ωt + φ ) dx U = = ωA cos (ωt + φ ) dt At t = 0 x = A sin φ …(i) ∴ 5 = A sin φ U = ω A cos φ = (8π ) A cos φ …(ii) ∴ 218 = (8π A ) cos φ Solving Eqs. (i) and (ii) we get, π or 30° A = 10 cm and φ = 6 11. KE at mean position 1 1 2 K = m vmax = m ω2A2 2 2  2K  1 ω= ∴   m A

10. ω =

 2 × 10− 3 × 8  1  = = 4 rad /s   0.1 0.1  

∴ (x − 2) = ± 4 Hence, x = 6 m and x = − 2 m are the extreme positions. (a) K max = E − U 0 = 16J 1 ∴ mω 2 A 2 = 16 2 1 or × 2 × ω 2 × (4 )2 = 16 2 or ω = 1 rad /s (b) At mean position E = 26J U = U 0 = 10J ∴ K = 16J At extreme position K =0 ∴ U = E = 26 J g 13. g′ = 2 h  1 +   R Putting

h=R g g′ = 4 T = 2π = 4π

x=2

x=6 A

= 10 J At extreme position U = Total mechanical energy = 26 J = 10 + (x − 2)2

l g

= 4 l = 4 10 .

(as g = π 2)

=4s

Ans.

14. In such situation, upthrust also behaves like a spring force of force constant = ρAg

k

Now, y = A sin (ωt + π / 4 ) is the required equation. 12. U 0 = minimum potential energy at mean position x = –2

l (g / 4 )

A

ρ

k net in the given situation is (k + ρAg ) m T = 2π ∴ k + ρAg = 2π = 1.8 s

10 100 + 1000 × 20 × 10− 4 × 10 Ans.

Chapter 14  v2

Here, I = moment of inertia of the ring about point of suspension = mr2 + mr2 = 2mr2

2

15. ge =   + g 2  r

and l = distance of point of suspension from centre of gravity = r

θ

∴

∴ ∴ or ∴ ∴

k = m

v 2 )2

√( r

m

mg

T = 2π

g ge

⇒

θ = tan − 1

= 2π

or

4 × 103 = 200 rad/s 01 . X = 10 sin (200 t ) 6 = 10 sin 200 t1  3 200 t1 = sin − 1   = 0.646 rad  5

t2 = 4.62 ms

Ans.

17. (a) For small displacement x, the term x can be neglected. ∴ F = − 100x Comparing with F = − kx we have, k = 100 N/m m T = 2π k 0.2 = 0.28 s 100 | ∆F | = 10x 2 = 10 (0.04 )2 = 0.016 N

Ans.

18. It is a physical pendulum, the time period of which T = 2π

I mgl

Ans.

1 k 2π m (Frequency is independent of g in spring) (b) Extension in spring in equilibrium mg initial = k Extension in spring in equilibrium in m (g + a) accelerating lift = k m (g + a) mg ma Ans. ∴ Amplitude = − = k k k 2π 2π 20. ω = = = (40π ) rad /s T 0.05 Since, the body starts from the extreme position, we can write θ = θ 0 cosωt  π or Ans. θ =   cos 40πt  10 2π 2π π = = rad /s T 16 8 If at t = 0, particle passes through its mean position (x = A sin ωt ) with maximum speed its v - t equation can be written as v = ωA cos ωt Substituting the given values, we have  π  π 2 =   A cos   (2)  8  8

21. ω =

| F | = 100 x = 100 (0.04 ) = 4 N | ∆F | ∴ % error = × 100 = 0.4% |F | is,

2π T g ω= 2r

19. (a) Frequency =

t1 = 3.23 ms 8 = 10 sin 200 t2  4 200 t2 = sin − 1   = 0.925 rad  5

= 2π

2r g

ω=

 v2     rg 

∆t = 1.4 ms = 1.4 × 10− 3 s

2mr2 mgr

∴ Angular frequency

+ g2

2

(b)

T = 2π

mv 2 = pseudo force r

θ

16. ω =

Simple Harmonic Motion — 603

∴

16 2 m π = 7.2 m

A=

Ans.

604 — Mechanics - II Substituting the proper values, we have M A2 = A1 M +m

22. The two forces acting on the bob are shown in figure Fe = qE

Ans.

Note E2 < E1 , as some energy is lost into heating up the block and putty. M +m Ans. k (b) When the putty drops on the block, the block is instantaneously at rest. All the mechanical energy is stored in the spring as potential energy. Again the momentum in horizontal direction is conserved during the process. But now it is zero just before and after putty is dropped. So, in this case, adding the extra mass of the putty has no effect on the mechanical energy, i.e. 1 E2 = E1 = kA12 2 and the amplitude is still A1. Thus, A2 = A1 M +m and Ans. T2 = 2π k Further, T2 = 2π

w = mg

w − Fe m mg − qE geff = m qE =g− m l T = 2π geff

geff in this case will be or

∴

= 2π

l g−

qE m

Ans.

23. (a) ge = g + a (b) ge = g − a (c) ge = 0

25. Let v is speed of combined mass just after

(d) ge = g + a 2

2

24. (a) Before the lump of putty is dropped the total mechanical energy of the block and spring is 1 E1 = kA12. 2 Since, the block is at the equilibrium position, U = 0, and the energy is purely kinetic. Let v1 be the speed of the block at the equilibrium position, we have 1 1 E1 = Mv12 = kA12 2 2 ∴

v1 =

k A1 M

During the process momentum of the system in horizontal direction is conserved. Let v2 be the speed of the combined mass, then (M + m)v2 = Mv1 M v2 = v1 ∴ M +m Now, let A2 be the amplitude afterwards. Then, 1 1 E2 = kA22 = (M + m)v22 2 2

collision. Then, from conservation of linear momentum, we have (M + m) v = mv0 mv0 ∴ v= M +m This is maximum speed of combined mass at mean position ∴ v = ωA or ∴

 m v0   k    =   A  M + m  M + m A=

26. Mass per unit area =

mv0 k (M + m)

m = σ (say) π (R 2 − r2 )

Whole mass m1 = (πR 2 )σ  R2  = 2  m  R − r2 

Chapter 14 Mass of cavity m2 = (πr2 ) σ

Simple Harmonic Motion — 605

At a depth h below the surface, h  g′ = g 1 −   R

 r2  = 2  m  R − r2 

m [ 3R 4 − r4 − 2r2R 2 ] 2 (R 2 − r2 )

=

m (3R 2 + r2 ) 2 T = 2π

Here,

l=R

I mgl

or

T′ ∝

R R−h

1 R−h

Hence proved.

R R −R/2

=2 2s

Ans.

29. (a) Distance travelled in first 4 s = OP + PO = A + A = 2 A Distance travelled in next 4 s = OQ + QO = A + A = 2 A Two distances are equal.

3R r2 + 2 2R g

Q

O

T = 2π

(b) Distance travelled in first 2 s = OP = A and distance travelled in next 2 s = PO = A Again the two distances are same.

l g

3R r2 + 2 2R 3R as r → 0 l= 2 and l = 2R as r → R l I 27. 2π = 2π g mgl ′ l of pendulum =

30. (a) u = ω A 2 − x12

… (i)

v = ω A 2 − x22 Solving Eqs. (i) and (ii), we get the result, A or a = (b) u1 = ω

I ml ′

∴

l=

or

I = (m) (l ) (l′ ) = (200) (20) (35) = 1.4 × 105 g - cm 2

A 2 − x12

…(iv) u12 − u23 2π = T x22 − x12

given by 1 g

…(iii)

Solving Eqs. (iii) and (iv), we find ω= ∴

…(ii)

v 2x12 − u2x22 v 2 − u2

u2 = ω A 2 − x22

28. At earth’s surface, the value of time period is or T ∝

P

At t = 0

Comparing with

L g

=

R R−h

TR / 2 = 2

…(i)

I 3R r2 = + mR 2 2R Substituting in Eq. (i) we have,

T = 2π

h  1 −   R

T′ = T

∴

T = 2π

1

∴

Further,

Now,

g g′

=

  r2  + 2  mR 2   R − r2   =

T′ = T

∴

3 1  I = m1R 2 − m r2 + m2R 2  2 2  2 2 2 1  r  3 R  =  2 2  mR 2 −   2  mr2 2 2 R − r  2  R − r 

T = 2π

x22 − x12 u12 − u22

606 — Mechanics - II 31. At equilibrium position, kx0 = mg

M k

T = 2π 2

(b) E =

Natural length x0

(c) Kinetic energy at mean position F2 =E= 2k 34. (a) F = kx0 F 10 Ans. ∴ x0 = = = 0.1 m k 100 1 1 (b) E = mv 2 + kx02 2 2 1 1 = × 1 × (2)2 + × 100 × (0.1)2 = 2.5 J Ans. 2 2 M (c) T = 2π k 1 π Ans. = 2π = sec 100 5 (d) From P to Q

Equilibrium position (h = 0) y Displaced position

In displaced position, 1 U = k (x0 + y)2 − mg y 2 1 2 1 2 = kx0 + ky + kx0 y − mgy 2 2 Substituting kx0 = mg we get, 1 1 U = kx02 + ky2 2 2 1 2 1 2 = kx0 + ky + kx0 y − mgy 2 2 1 2 But k x0 = constant say U 0 2 1 ∴ U = U 0 + ky2 2 32. While returning to equilibrium position, 1 1 k d 2 = (m1 + m2 ) v 2 2 2  k1  ∴ v=  d  m1 + m2 

x0 F

θ

Now, after mean position m2 is detached from m1 and keeps on moving with this constant velocity v towards right. Block m1starts SHM with spring and this v becomes its maximum velocity at mean position. ∴ v = ωA    k  k ∴   d=  A  m1 + m2   m1  ∴

 m1  A=  d  m1 + m2 

Ans.

33. (a) In equilibrium, let x0 is the elongation then, F = kx0 F ∴ x0 = k This x0 is the amplitude F ∴ A = x0 = k

1 2 1  F F2 kx0 = k   = 2 2  k  2k

Natural length

2 m/s P

A

Work done by applied force = change in mechanical energy. ∴ FA = EQ − EP 1 ∴ (10) A = k ( A + x0 )2 − 2.5 2 1 = × 100 ( A + 0.1)2 − 2.5 2 Solving this equation, we get Ans. A = 0.2 m 1 (e)U Q = k ( A + x0 )2 2 1 = × 100 (0.2 + 0.1)2 2 Ans. = 4.5 J (f) P = mean position 0.1 m Q

P

A = 0.2 m

0.1 m R

A = 0.2 m

S

Chapter 14 Q , S = extreme position R = natural length 1 U S = k (0.1)2 2 1 = × 100 × (0.1)2 = 0.5 J 2 Due to work done by the applied fore, F, answer are different. I 35. TA = 2π mgl = 2π

(md 2 / 3)  2  =   2π mg (d / 2)  3 

and

TB = 2π

∴

TA = TB

d g

d g

2 = 0.816 3

Ans.

36. In displaced position, 1 2 1 2 1 kx + mv + I ω 2 2 2 2 1 2 I = mR 2 v ω= R 1 3 E = kx 2 + mv 2 2 4 = constant dE =0 dt 1  dx  3 dv 0 = k   (2x ) + m (2v )   2 dt 4 dt dx dv = v and =a dt dt  2k  F = (ma) = −   x  3 E=

Putting and we get Since, E ∴ or Putting, we get,

Since, F ∝ − x motion is simple harmonic 2k ke = − 3 m 3m Hence Proved. T = 2π = 2π ke 2k

37. In equilibrium ∴

k x0 = mg mg 0.5 × 10 x0 = = k 20 = 0.25 m

Simple Harmonic Motion — 607

When displaced downwards by x from the mean position, total mechanical energy, 1 1 1 E = − mgx + mv 2 + Iω 2 + (x + x0 )2 2 2 2 v 2 Substituting I = 0.6 MR and ω = R We have, 2

1 1  v mv 2 + (0.6 MR 2 )    R 2 2 1 + k (x + x0 )2 2 Since E = constant dE =0 ∴ dt dv   dx  1  or 0 = − mg   + m  2v ⋅   dt  2  dt  dv 1 dx + (0.6 M ) + (v ) + k [ 2 (x + x0 )] dt 2 dt dx dv Putting = v , kx0 = mg and =a dt dt We have 0 = (ma + 0.6 Ma) + kx   k or a=−  x  m + 0.6 M  Since, a ∝ − x motion is SHM. 1 a f = 2π x E = − mgx +

=

1 2π

k m + 0.6 M

=

1 2π

20 0.5 × 0.6 × 5

= 0.38 Hz

38. x = A sin ωt

…(i)

y = A sin (2ωt + π / 2) = A cos 2ωt = A (1 − 2 sin 2 ωt ) From Eq. (i), x sin ωt = A  2x 2  ∴ y = A 1 −  A 

39. (a) x = x1 + x2 At

Ans.

t = 0.0125 s

608 — Mechanics - II x1 = − 2 cm

2π 2π π = = rad /s T 16 8 At t = 0, particle crosses the means position. Hence its velocity is maximum. So, velocity as a function of time can be written as v = vmax cos ωt or v = ωA cos ωt  π  π 1 =   A cos   (2) ∴  8  8

5. ω =

and x2 = − 1 cm ∴ x = x1 + x2 = − 2.41 cm (b) At t = 0.025 s. x1 = 2cm and x2 = − 1.73cm ∴ x = x1 + x2 = 0.27cm

 π  =  A  8 2

LEVEL 2 ∴

Single Correct Option 1.

1 2 kA = 1.0 2

or

2. vmax

8 2 m π

6. ge = g 2 + a2 = (10)2 + (4 )2

1 k (0.4 )2 = 1.0 2 25 N/m k = 12.5 or 2 m 2 4π T = 2π = 2π = s 25 k 5 2  K = ωA =  a  m

= 10.77 m/s2 T = 2π

= 1.90 s dv = v0 ω cos ωt dt v sin ωt = v0 a cos ωt = v0 ω a=

vmax

 K ∆P = 2m vmax = 2m  a  m ∆P ∆P 2ma k / m 2aK = = = ∆t T /2 π π m/ k

3. A1 = 40 units 10c

A2 = 10 √c 2 + 1

l 1 = 2π ge 10.77

7. If v = v0 sin ωt, then

vmax

F=

A=

∴

 v  v 2 =  − 0  a2 + v02  ω

Hence, v 2 versus a2 equation is a straight line with positive intercept and negative slope. l 8. T = 2π g

A1 = A2 we get,

l

c = 15

4. T =

1 1 = = 0.4 s F 2.5

3T 4 At the given time, particle will be in its extreme position if at t = 0 it crosses the mean position. t = 0.3 s =

…(ii)

Squaring and adding these equations, we get v a2 1 = 22 + v0 v0 ω

10

Putting

…(i)

1000 g 20 = 50 cm = 0.5 cm l = 0.25 m 0.25 T = 2π =1s 9.8

2l = ⇒ Now

Chapter 14 9. T = 2π

m k

Here

m = (ρ 0 ) (a3 )

and

K = ρ l A g = ρa2g

∴

T = 2π

10. T = 2π

ρ 0a ρg

ml 2 + md 2 12 mgd

= 2π

l 2 + 12d 2 12 d

= 2π

 l2  + d   12d 

T is minimum when first derivation of the quantity inside the root with respect to d is zero. l2 − + 1= 0 12 d 2 d 1 or = l 12

11. In solved example, we have shown that

and

T1 = 2π

m k

T2 = 2π

4m k

T3 = 2π

m 4k

m + ms / 3 (in general) k Here ms = mass of spring and m = mass of block 13. Half the oscillation is completed with one spring and the other half with other spring. Hence, M M 2π 2π T1 T2 k k 4 T = + = + 2 2 2 2 M 1  3π M =π 1+ = k  2  2 k dU 14. F = − = 20 − 10x dx F =0

12. T = 2π

at x=0 So, from x = − 3 to x = 2, amplitude in 5. Hence, other extreme position will be x = 2+ 5 = 7 1 15. k ∝ Length of spring

I mgd

= 2π

Simple Harmonic Motion — 609

Length

Force constant

L

k

nL

k /n

(1 − n)L

k /(1 − n) m1 = nm m2 = (1 − n) m

and

1 2π f1 = f2 1 2π

16.

k /n (1 − n) m k / (1 − n) nm

=1

1 1 1  k ( A 2 − x 2 ) =  kA 2  2 4 2 ∴

x=

3 A 2

3 R 2 or BD = 2 (CD ) or 2 (CB ) = 3R YA 17. K Q = L (2Y ) ( A / 2) YA Kp = = L L YA ∴ KP = KQ = K = L Half oscillation is completed with P and half with Q. But their value of K is same. Hence, we can say that in one oscillation one time period is completed with spring. ∴

CD = CB =

 2L T =   + 2π  v

m 2L = + 2π K v

mL AY

18. X = A cos ωt x

3 cm x=5 t=t

5 = 8 cos ωt

x=8 t=0

610 — Mechanics - II ωt = cos− 1 (5/ 8) = 0.9 0.9 0.9 t= = ω (2π /T ) 0.9T 0.9 × 1.2 = = 2π 2π = 0.17 s

∴

∴ Ans.

3a 2 2 a OC = l = (OP ) = 3 3  ma2  ma2 I = I0 = 2   + m (OP )2 + 12  3 

19. OP = a sin 60° =

23. T = 0.6π 2π 2π = T 0.6 π 10 = rad /s 3 Particle starts from y = − A = − 2 cm ∴ y = − A cos ωt = A sin (ωt − π / 2)  10  = 2 sin  t − π / 2  3 

∴

2 3 1 3 ma2 + ma2 + ma2 = ma2 3 4 12 2 I O T = 2π (3m) gl =

= 2π

(3/2m a2 )  a (3m) g    3 3 a 2 g

= 2π

1 m and g = 10 m/s2 3 π We get, T = s 5 2.5 π 2.5 π 20. t = = ω (2π /T )

Diameter = 0.4 m 2 30 1 1 f = = rps or Hz 60 2 2 1 T = = 2 s. f

22. A = radius =

a

ω=

24. Mean velocity =

displacement time

C

T/6

60° P

x=0

=

Putting a =

a/2

(a/ 2) 3a = (T / 6) T

25. Maximum acceleration = g Ans.

T  = (2.5)    2 T  =5   4 Particle starts from extreme position and in every T time it travels a distance A. So, in time 4 t = 5 (T / 4 ) it will travel a distance 5 A. 21. Half of the oscillation is completed with length l and rest half with l/4. T T ∴ Time period = 1 + 2 2 2 1 l l/4  = 2π + 2π 2 g g  3 l 3 = 2π = T 4  g  4

∴

ω2A = g

∴

(2πf )2 (0.5) = g

∴

26.

f2 =

2g (2π )2

f =

2g 2π

1 2 1 kX = k ( A 2 − X 2 ) 2 2 T/8 –A/ 2

T/8

x=0

A/ 2

2A / 2 = 2 A

∴

X =±

Average speed = =

A 2

d 2A = t T /8 + T / 8 4 2 A T

Simple Harmonic Motion — 611

Chapter 14 27. V1 = V2

T = Fnet v2

v1

B

C 1s

tAB = 2 s tOB = 1 s tBCB = 2 s BC = 1 s tOB = tBC A OB = 2 OB 1 = A 2

∴ ∴ ∴ or

60°

π 6ω

(2 − 3 ) A / 2 3ωA = (2 − 3 ) (π / 6ω ) π

More than One Correct Options

θ

Q A 2

cos ω = 2π T

sin

=

1. tan θ =

P

A

∴ ωt = (π / 6) According to the equation, X = A cos ωt X = A at t = 0 3A at ωt = π /6 and X = 2 ∴ Distance travelled 3 =A− A 2 A = (2 − 3 ) 2 distance Average speed = time

ma a = mg g

5π   2π = A sin  t+  T 6 x = A cos (ωt + 60° ) π  2π = A cos  t+  T 3

or

3. k x0 = mg mg 1 × 10 = k 500 = 0.02 m = 2 cm So, equilibrium is obtained after an extension of 2 cm of at a length of 42 cm. But it is released from a length of 45 cm. ∴ A = 3cm = 0.03 m k (b) vmax = ωA = A m  500  =  (0.03) = 0.3 5 m/s  1  ∴

x0 =

= 30 5 cm/s (c) amax = ω 2 A k  500 2 =   ( A) =   (0.03) = 15 m/s  m  1 

T

θ

Pseudo force = ma Fnet

mg

∴

2. x = A sin (ωt + 150° )

2s

∴

28. t =

A

O

= m a2 + g 2

 a θ = tan − 1    g

(d) Mean position is at 42 cm length and amplitude is 3 cm. Hence block oscillates between 45 cm length and 39 cm. Natural length 40 cm lies in between these two, where elastic potential energy = 0. 4. v or KE = 0 at y = ± A v or KE = maximum at y = 0 F or a is maximum at y = ± A F or a is zero at y = 0

612 — Mechanics - II ma 2 × 5 = m K 400 = 0.025 m = 2.5cm A = x0 = 2.5 cm

5. v = ω A 2 − x 2

x0 = 2

2

v x + 2 = A2 2 ω (1) v - x graph is an ellipse. a = − ω 2x

∴ i.e.

∴

Match the Columns 1. xmin = 2 − 2 = 0→ extreme position

i.e. a - x graph is a straight line passing through origin with negative slope. 6. a = 0 at x = 0.5 m and particle is released from x = 2m Hence, A = 2 − 0.5 = 1.5 m 2 ω = 100 ∴

ω = 10 rad /s 2π 2π T = = ω 10 = 0.63 s vmax = ωA = (10) (1.5) = 15m/s 7. Two particles shown in figure are in same phase, although they have distances from the mean position at t = 0 –A1

+A1 x=0

+A2 +A2/2

x=0

8. The given equation can be written as y = 3 sin 100 πt + (4 − 4 cos 100 πt ) − 6 = 3 sin 100 πt + 4 sin (100 πt + π / 2) − 2 or y = 5 sin (100 π − 53° ) − 2 53°

15  1 2 15 × 16 = 15 J  kA  =  16 16  2 1 (c) KE = k ( A 2 − 0) 2 1 = kA 2 = 16 J 2 1  2 A2 (d) KE = k  A −  2  4 =

+A1/2

–A2

xmin = 2 + 2 = 4 → extreme position x + xmin Mean position = min =2 2 Maximum potential energy is at extreme positions. 1 2. U 0 + KA 2 = U max 2 1 1 2 4 + kA 2 = − 20 or kA = 16J 2 2 1 (a) U = U 0 + kX 2 2 2 1  A =U0 + k   2  2 16 =4+ =8J 4 1  A2  (b) KE = k  A 2 −  2  116

=

3

4

ymax = 5 − 2 = 3 ymin = − 5 − 2 = − 7 y + ymin Mean position = max = − 2cm 2

3  1 2 3  kA  = (16 J) = 12 J  4 4 2

3. (a) x ∝ − α At time t1 acceleration is positive. Hence, X will be negative. (b) At time t2 a = 0 ∴ x = 0 (c) At time t1, x is negative and a is increasing. So, the particle is moving toward extreme position. a

Comprehension Based Questions 1. ω =

k = m

2. Kx0 = ma

400 = 10 2 rad /s 2

–A

x=0

v x

∴ Velocity is negative.

+A

Chapter 14 (d) At mean position, velocity is maximum. Just after few seconds acceleration becomes positive. So displacement will becomes negative. Hence, the velocity should be negative. 4. vmax = ωA = (4 π )(2) = 25.12 m/s 2π 2π T = = ω 4π = 0.5 s Hence the given time t = 1 s = 2T 1 2 Maximum KE = m vmax 2 1 = × 2 × (25.12)2 2 = 631 J Given kinetic energy of 400 J is less than this maximum kinetic energy. So in one time period kinetic energy becomes 400 J four times. In time 2T it becomes 400 J eight times. (d) | amax | = ω 2 A

Simple Harmonic Motion — 613

Subjective Questions k 1 100 2.5 Ans. = = = 0.8 Hz m 2π 4 π 1 100 5 with 1 kg mass, f0 = = Hz 2π 1 π Further, from conservation of linear momentum (at mean position) 1 ωA = ω 0 A0 4 1 or fA = f0 A0 4 f A (5/π )(0.1) or A= 0 0 − 4f (4 )(2.5/π )

1. f =

1 2π

= 0.05 m

Ans.

2. (a) θ 0 = ω 1t + ω 2t 1

2 π/4

π/3 θ0

= (4 π )2 (2) = 315.5 m/s2 Given acceleration is less than this. So, it becomes two times in one time period and four times in time 2T . 5. xmax = 4 + 6 = 10m xmin = 4− 6 = − 2m x + xmin Mean position = max = 4m 2 2π 2π T = = =2s ω π T 1 (a) x = 10m to x = 4 m t = = s 4 2 T 1 (b) x = 10m to x = 7m, t = = s 6 3 A/2 –2 m

1m

A/2 4m

2π T π π 19 and θ0 = π + + = π 3 4 12 19π 2π 2π = t+ t ∴ 12 T T 19 or t= T 48 19 5π (b) θ = 2π − θ 0 = 2π − π = 12 12 ω1 = ω2 =

but

A/2 7m

1

2 θ

10m

X

XN

6m

T  1 (c) x = 7m to x , t = 2   = s  12 3 T (d) x = 10m to x = − 2m, t = = 1 s 2

Two particles will collide when line XX ′ becomes the line of bisector of angle θ.

614 — Mechanics - II ∴ Any one of the particles (say-2) has rotated an angle ωt = π /4 + θ/2 2π π 5π 11π or t= + = T 4 24 24 11T t= ∴ 48 3. 1.5 = ωA = 2 A ⇒ ∴ A = 0.75 m Further, the particle will start its journey from its mean position in downward direction (− ve direction). k 600 4. (a) ω = = = 10 2 rad/s m 2+1 From conservation of linear momentum (at mean position) velocity of combined mass will be 2 m/s. This is the maximum velocity of combined mass. ∴ v = ωA v 2 or A= = m ω 10 2 = 0.141 m = 14.1 cm 2π 2π T = = = 0.44 s ω 10 2 (b) If the collision is elastic, 600 ω′ = = 10 3 rad/s 2 In elastic collision,  m − m1   2m1  v2′ =  2  v2 +   v1  m2 + m1   m1 + m2   2 × 1 =  (6) = 4 m/s  1 + 2 v2′ = ω′ A′ A′ =

or

N = m (g + aω 2 cos ωt )

 d y (b)  2  = aω 2 or  dt  max 2

∴

a=

Ans.

aω 2 = g

g 980 = ω 2 (11)2

= 8.1 cm

Ans.

7. F = − kxi$ − ky$j

P rr

(x,y)

(0,0)

v2′ 4 = ω′ 10 3 = 0.23 m = 23 cm 2π 2π T′ = = = 0.36 s ω′ 10 3 (c) In both cases journey is started from mean position ∴ x = ± A sin ωt will be the displacement-time equation. For impulse we can apply the equation. Impulse = change in linear momemtum. or

or

d2 y = a ω 2 cos ωt dt 2 d2 y N − mg = m . 2 dt N = mg + maω 2 cos ωt

(As v2 = 0)

This is maximum velocity. ∴

k 400 = = 10 rad/s m 4 Let t1 be the time from x = 0 to x = 12 cm and t2 the time from x = 0 to x = 9 cm. Then, 12 = 15 sin (10t1 ) or t1 = 0.093 s 9 = 15 sin (10t2 ) or t2 = 0.064 s Total time = t1 + t2 = 0.157 s ∴ 6. (a) y = a(1 − cos ωt )

5. ω =

at F=0 (0, 0) When it is displaced to a point P whose position vector is r = xi$ + y$j Force on it is F = − k (x$i + y$j) = − kr Since, F ∝ − r, motion is simple harmonic. At t = 0 particle is at (2, 3)

(2,3) y x

Chapter 14 y 3 or 2 y − 3x = 0 = x 2 i.e. the particle will oscillate simple harmonically along this line. 8. In equilibrium, mg sin θ = kx0 …(i) When displaced by x, 1 1 1 E = mv 2 + Iω 2 + k (x + x0 )2 − mgx sin θ 2 2 2 Since, E = constant dE =0 dt dx  dv   dω  0 = mv   + Iω   + k (x + x0 )  dt   dt  dt dx − mg sin θ dt Substituting, v 1 dv = a, ω = , I = mR 2 R 2 dt dω a dx =α = , =v dt R dt and kx0 = mg sin θ We get, 3ma = − 2kx 1  a 1 2k ∴ = f = 2π  x  2π 3m Substituting the values, 1 2 × 200 f = 2π 3 × 100 9.8 = 0.56 Hz

Simple Harmonic Motion — 615

We have ω= ∴ or

8 × 1000 = 16.16 rad/s 3 × 100 9.8

θ = θ 0 cos ωt θ = 0.4 cos (16.16t )

Ans.

10. Similar to Q.9 11. Let F be the restoring force (extra tension) on block m when displaced by x from its equilibrium position. x = 2x1 + 2x2  2F 2F  =2 + k 2   k1  k + k2 = 4F  1   k 1k 2  k 1k 2 x 4 (k 1 + k 2 ) k 1k 2 a=– x 4 m (k 1 + k 2 )

F =–

or ∴

ω=

12. (a) f =

1 2π

k 1k 2 4 m (k 1 + k 2 )

k eff 1 = total mass 2π

Ans. k1 + k2 m1 + m2

(b) Suppose the system is displaced towards left by a distance x. Restoring force on m1 : f

Ans.

k1x

f

m1

9. In the displaced position, 1 1 1 mv 2 + Iω 2 + k (2x )2 2 2 2 1 v 2 and ω = I = mR 2 R 3 ∴ E = mv 2 + 2k x 2 4 E = constant dE ∴ =0 dt 3 dv dx or mv + 4 kx =0 2 dt dt dx dv Substituting, = v and =a dt dt 8k a=− .x 3m Comparing with, a = − ω 2x

8k = 3m

m2

k2x

(towards right) F = m1ω 2x  k1 + k2  = m1   x  m1 + m2 

E=

Friction f on it will be towards right if, k 1x < F  k + k2  or k 1 x < m1  1  x  m1 + m2  k 1 m1 or < k 2 m2

Ans.

 k + k2  (c) k 1 Am + µm2g = m1  1  Am  m1 + m2   m k + m1k 2  Am  1 1 − k 1 = µ m2 g  m1 + m2  or

Am =

µ (m1 + m2 ) m2g m1k 2 − m2k 1

Ans.

616 — Mechanics - II 13. (a)

1 2 1 2 kx0 = µvr 2 2

14. Restoring torque 5  l  l τ = − (klθ )l −  k θ = − kl 2θ  2 2 4

6×3 Here, µ = reduced mass = 6+ 3 = 2 kg  200  k x0 =  ∴ vr =  (3 × 10−2 ) µ  2  = 0.3 m/s = 2v + v v = 0.1 m/s v1 = 0.2 m/s v2 = 0.1 m/s

∴ ∴ and A

2v

(b) vcm

Ans.

 2F   2F   2F   2F  x = 2  + 2  + 2  + 2   k4   k3   k2   k2 

200 2 or

Ans.

(c) After collision velocity of combined blocks (A + C ) (3 × 0.2) + (3)(0.4) v0 = = 0.3 m/s 3+ 3 and velocity of block B is v2 = 0.1m/s A

position by a distance x, let F be the restoring (extra tension) force produced in the string. By this extra tension further elongation in the springs are 2F 2F 2F 2F and respectively. , , k1 k2 k3 k4 Then,

= 0.1 m/s (towards right)

C

1 α 1 15k = 2π θ 2π 4 m

15. When the mass m is displaced from its mean

= 10 rad/s m1v1 + m2v2 + m3v3 = m1 + m2 + m3 (3)(0.2) – (6)(0.1) + 3(0.4) = 3+ 6+ 3

0.3 m/s

 ml 2  5    α = −  kl 2 θ 4   3  f =

B

v

Angular frequency k ω= = µ

Now,

0.1 m/s B

The spring will compress till velocity of all the blocks become equal to the centre of mass. Applying conservation of mechanical energy, 1 1 1 (3 + 3)(0.3)2 + (6)(0.1)2 = 2 2 2 1 (3 + 3 + 6)(0.1)2 + kA 2 2 Solving this we get, A = 0.048 m or Ans. A = 4.8 cm 1 1 (d) ∆E = (3)(0.4)2 + (3)(0.2)2 2 2 1 − (3 + 3)(0.3)2 2 Ans. = 0.24 + 0.06 – 0.27 = 0.03 J

4 4 4 4 + +  =−x F +  k1 k2 k3 k4

Here negative sign shows the restoring nature of force. x a=− 4 4 4 4 m + + +   k1 k2 k3 k4 x T = 2π   a = 4π

 1 1 1 1 m + + +   k1 k2 k3 k4

Ans.

16. Let F be the extra tension in the string, when the block is displaced x from its mean position. Extension in spring-2 is F x2 = k2 Extension is spring-1 is 2F k1 4F F x = 2x1 + x2 = + k1 k2 x1 =

Extra tension F will become restoring force for the block. Therefore, above equation can be written as,       1  x =  k 1k 2  x F =−  4 k2 + k1  4 + 1    k1 k2 

Chapter 14 or

ke =

k 1k 2 4 k2 + k1

T = 2π = 2π

Ans.

17. In equilibrium, …(i) T + F = Mg When the block is further depressed by x, weight Mg remains unchanged, upthrust F increases by ρAxg and let ∆T be the increase in tension. T+F M

k  + ρAg  4  or α= θ 3 M 2 Here negative sign has been used for restoring nature of torque. –

m ke m(4 k 2 + k 1 ) k 1k 2

Simple Harmonic Motion — 617

∴

f =

1 2π

=

1 2π

α θ k + 4ρAg 6M

Ans.

18. If the mass M is displaced by x from its mean position each spring further stretches by 2x. 2kx 2kx 2kx 2kx

F = Upthrust

Mg

If a is the acceleration of block then, …(ii) ∆T + ρAxg = Ma Restoring torque on the cylinder,  kx R   kxR  τ= – ∆T R = – (Ma – ρAxg ) R  2 2   4   kR 2θ  1 MR 2α =  – (MRα – ρAgRθ ) R  2 4   2   3 kR or MR 2α =  + ρAgR 2  θ 2  4 

M

Net restoring force F = − 8kx ∴ M ⋅ a = − 8 kx 1  a f = 2π  x  1 8k 2π M 1 2k = π M =

Ans.

15 Elasticity INTRODUCTORY EXERCISE

15.1

Fl 1 or ∆l ∝ 1. ∆l = AY Y Fl 1000 × 100 2. ∆l = = AY 4 × 2 × 106

R= =

15.2

1. Energy density = energy per unit volume 1 (stress) (strain) (volume) 2 1  F   ∆l  or U =     ( Al ) 2  A  l  1 = F ⋅ ∆l 2 1 = (100)(0.3 × 10−3 ) 2 Ans. = 0.015 J (b) Work done = Potential energy stored 1 = k (∆l )2 2 1  YA  YA   =   (∆l )2  as k =   2 l  l 

2. (a) Energy stored U =

= 0.0125 cm F F σ= = A π (R 2 − r2 )

3.

INTRODUCTORY EXERCISE

F + r2 σπ . × 106 16 . )2 + (01 . 90 × 106 × 314

= 0.1251 m = 1251 . mm Diameter = 2R = 250.2 mm Force 4. Stress = Area ∆l Strain = l Stress Modulus of elasticity = Strain

Substituting the values, we have W =

1 (2.0 × 1011 )(10–6 ) (0.1 × 10–3 )2 2 (2)

= 5.0 × 10−4 J

Ans.

Exercises LEVEL 1

8. Case I F = ma ⇒ a =

Assertion and Reason 2. At higher pressure, it is difficult to press the gas

a

F

more. So, bulk modulus is high.

3. If length is doubled, ∆l will also becomes two times and Y will remain same.

4. Bulk modulus is related to volume change and volume change is possible in all three states. Young’s modulus is related to length change, which is possible only in solids. 6. Stress ∝ strain only in proportionality limit. 7. BT = P ⇒

BS = γ p

∴

BT 1 = BS γ

F m

x

m xa L m F x  T = F − x ⋅ = F 1 −   L m L Case II F − mg = ma′ F − T = m′ a =

F x a′

mg

…(i)

Elasticity — 619

Chapter 15 F −g m F − T − m′ g = m′ a′ F  F − T − m′ g = m′  − g m  a′ =

5. B = −

=

Radius of the wire is r = 2 × 10−3 m Increase in length is ∆l = 0.031 ×10−3 m

T m′g

g = 3.1 × π ms−2 Young’s modulus is (mg )l Fl = Y= A ∆l πr2 (∆l ) …(ii)

∴

produced in steel is more. 1 Also, Work done = Stress × Strain 2 1 Volume = Y × ε 2 2

Single Correct Option 1.

B=−

∆p ∆V / V

∆V = 0 for an incompressible liquid. B=∞ 2. Young’s modulus of elasticity is a materials property. F 3. σ max = max A ∴ Fmax = (σ max ) A , ∴

which is independent of length of wire.

4. T = 0 is free fall.

4 × 3.1 × π × 20 π (2 × 10−3 )2 × 0.031 × 10−3

l, r

You can appreciate the extension of rod in first case, by comparing it with a case of many identical blocks connected by ideal springs.

Extension in rod occurs due to force acting at any point on the rod. In certain cases, when net force acts at the centre of rod like weight, extension may not occur like the given case (II). 9. Ysteel > Ycopper . So, for the same strain, stress to be

Y=

= 2 × 1012 Nm −2

Tension at a point on the rod (of length L) at a distance x from point of application of force is same in both cases. [from Eqs. (i) and (ii)] Hence, weight has no effect on tension in case (II). Note

Ans.

6. Length of the wire is l = 20 m

a′ m′ x

⇒

(1.2 × 105 )(1 × 10− 3 ) 0.3 × 10− 3

= 4 × 105 N / m 2

F

F F − T = m′ ⋅ m m F x F −T = x⋅ = F L m L x  T = F 1 −   L

∆p ∆V /V

m

7. Volume of wire is V = L × πr2 = 1 × π (10−3 )2 = π × 10–6 m 3 Area of square cross-section = (2 × 10−3 )2 = 4 × 10−6 m 2 Length of new wire =

Volume π × 10−6 π = m = Area 4 × 10−6 4

Initially extension is x =

FL F  1  =   AY Y  π × 10−6 

F = (π × 10−6 ) x Y FL′ F L′ Finally extension is x′ = = ⋅ A′ Y Y A′

∴

x′ = (π × 10−6 ) x ⋅

π /4 π2 = x −6 16 4 × 10

8. In the figure, the reciprocal of slope of stress-strain (x and y-axes) curve, upto proportionality limit, gives Young’s modulus. The measure of ductility is obtained as the length of the stress-strain curve between yield point and ultimate load.

620 — Mechanics - II 9. ρ′ =

ρ dp  ≈ ρ 1 +  dp  B 1− B

13. ∆l =

ρ (dp) ∆ρ = ρ′ − ρ = B ∆ρ × 100 = 0.1 ρ ∆ρ = 0.001 ρ

∴ From Eq. (i),

dp

or

∆p = B

∴ …(i)

∆ρ ρ

= (2 × 109 ) (0.001) = 2 × 106 N/ m 2 1 1 YA (∆l )2 F (∆l ) = ⋅ 2 2 l A (QIn both the cases ∆ l is same) U ∝ l 2 U 2 A2 l1  r2  l1 = ⋅ =   ⋅ = (4 )(2) U 1 A1 l2  r1  l2

10. W = U = ∴

∴

U 2 = 8 U 1 = 8(2) = 16 J 1 1 11. U = σ ε = (Y ε ) ε 2 2 1 2 ⇒ U = Y ε is similar to x = ky2 2 Which is a parabola passing through origin and symmetric about x-axis. – ∆p ⋅ Vi 12. Change in volume, ∆V = B Hence, density at depth of about 11 km is ρ × Vi ρ0 B Mass = = 0 = Volume V – ∆pvi ( B – ∆p) i B ρ0 ρ0 = = ∆p 1 × 108 1– 1– B 2 × 109 ρ0 ρ0 ρ = = ⇒ ρ= 0 1 0.95 0.95 1– 20 ⇒ ρ 0 = 0.95ρ ρ − ρ0 ∴ % change in density = × 100 ρ0  1  −1  0.95  = × 100 1      ≈5%

∴

FL FL = AY (πr2 )Y L r2 ∆l1 L/R 2 = =2 ∆l2 2L/(2R )2 ∆l ∝

14. The net pressure at a depth of 1 km in the ocean is p = ρgh + patm = (103 ) (9.8) (103 ) + 105 = 99 × 105 Pa = 9.9 × 106 Pa This pressure acts uniformly on all sides of the balloon (which is in equilibrium) and the restoring forces with in the balloon are equal to external forces. So, normal stress will be same as external pressure. 15. Let T be the tension in the rope. Then, 2T = 10 kN ⇒ T = 5000 N Longitudinal stress in the rope is T 5000 N σ= = 3 = 5 N mm −2 A 10 mm 2 Stress Extension in the rope = ×L Y 5 = 3 × (600 + 900) = 7.5 mm 10 7.5 ∴ Downward deflection of the load = = 3.75 mm 2 16. Change in length of element is a x

θ

dx

3.14 kg b

F dx d (∆L) = AY Cross-sectional area of the element is 2

b−a   A=π a+ x L   b− a where, a = x is radius of the wire at the L location of element.

Chapter 15 F dx 2 b−a   x Y π a+  L  Total change in length of wire is 1 F L dx ∆L = ∫ 2 0 πY b−a   x a +  L  b−a Let a+ x=t L b−a L dx = dt ⇒ dx = dt ⇒ L b−a When x = 0, t = a and when x = L, t = b F b −2 L t ⋅ dt ∴ ∆L = πY ∫ a b−a FL  1 1 = (−1)  −   b a πY (b − a)

4. Refer solved example 1

d (∆L) =

∆l = =

1 1 YA k (∆l )2 = (∆l )2 2 2 l 1 Y × π r2 = × (∆l )2 2 l (2 × 1011 ) (3.14 ) (6 × 10− 3 )2 (4.9 × 10− 6 )2 = 2×5 = 5.43 × 10− 5 J

5. σ max ∴

= 10−3 m ⇒ ∴ ∆L = 10−3 m

or

or

Ans.

4 Fl π (∆l ) Y 4 × 400 × 3 3.14 × 0.2 × 10− 2 × 2.1 × 1011

=

= 1.91 × 10− 3 m = 1.91 mm m (g + a) 1 3. = σ max A 3 σ max A a= −g ∴ 3m =

m = 18 kg

8 × 108 =

(m1 + m′ ) g 0.003 × 10− 4 (10 + m′ ) × 10 0.003 × 10− 4

or m′ = 14 kg So, answer is 14 kg and lower string will break earlier. F 6. (a) σ = A F and A both are same. Hence, the ratio is 1. Stress 1 (b) Strain = ∝ Y Y Ycopper 13 (Strain)steel Ans. ∴ = = (Strain )copper Ysteel 20 ρ dp  ≈ ρ 1 +  dp  B 1− B ρ (dp) ∆ρ = ρ′ − ρ = ∴ B ρ (ρgh) = B (1030)2 × 9.8 × 400 = 2 × 109

7. ρ′ = Ans.

(3 × 108 ) (4 × 10− 4 ) − 9.8 3 × 900

= 34.64 m/s2

(10 + 20 + m) × 10 0.006 × 10− 4

σ max on lower string =

1. The changed density, ρ′ =

d=

Ans.

(m + m2 + m) g on upper string = 1 0.006 × 10− 4 8 × 108 =

Subjective Questions

∴

gl 2 ρ (9.8) (5)2 (8000) = 2Y 2 × 2 × 1011

U =

(3.14) (9.8) (10) (3.14) (2 ×1011 ) (5 × 10−4 ) (9.8 × 10−4 )

ρ dp 1− B Substituting the value, we have 11.4 or ρ′ =1169 ρ′ = . g / cm 3 2.0 × 108 1– 8.0 × 109 Fl Fl 2. ∆l = = AY (πd 2 / 4 ) Y

mgl ρ (π r2l ) gl = 2 AY 2 (π r2 ) Y

= 4.9 × 10− 6 m

FL (b − a) FL = ⋅ = πY (b − a) ab πY ab =

Elasticity — 621

Ans.

= 2.0 kg/m 3

Ans.

622 — Mechanics - II 8. B =

∆p ρgh = ∆V /V (∆V /V ) =

3. T = ml ω 2 T mlω 2 = σ max = A A σ max A ω= ml

3

(10 ) ( 9.8) (180) (0.1/100)

= 1.76 × 109 N / m 2

Ans.

∴

=

LEVEL 2 Single Correct Option 1. Extension in the first case is L, A, Y

w

FL wL = AY AY Extension in the second case is l=

L — , A, Y 2

w

L — , A, Y 2

w

w (L / 2) w (L / 2) wL + = AY AY AY It is clear that l' = l l′ =

Note

In the above problem, even if lengths of wire are unequal on two sides of the pulley, the elongation will still be l.

= 4 rad/s

Stress Hence,

⇒

g  3 mg = 2 2 F F σ= ⇒ A= A σ F A min = σ max 3 m⋅ g 2 π dmin = 2 4 σ 3 mg × 4 2 dmin = 2 σπ dmin =

6 mg ⋅ π σ

Ans.

M L  g =  sρ g 2  2 T 1 Ans. σ = = ρgL ∴ S 2 dU 5. F = − = − slope of U - r graph dr At B, slope = 0 ∴ F =0 Hence, atoms are in equilibrium at B. From A to B or even before A slope is negative. Hence, force is positive and positive force mean repulsion. Tl 6. …(i) l1 − l = 1 AY Tl …(ii) l2 − l = 2 AY Dividing Eq. (i) by Eq. (ii) and then simplifying, we get lT −lT Ans. l= 2 1 1 2 T1 − T2 Fl Fl 7. ∆l = = (V = Volume) AY (V / l ) Y T =

4.

Fl 2 VY ∆l ∝ l 2 =

2. Force on wire = Load carried  = m (g + a) = m  g + 

4.8 × 107 × 10− 6 10 × 0.3

∴

(as F , V and Y = constants)

8. ∆l = lα ∆θ ∆l = (α∆θ ) l Stress = Y × Strain = Yα∆θ Energy stored per unit volume 1 = × stress × strain 2 1 = × Y × (α ∆θ )2 2 1 = × 1011 × (12 × 10−6 × 20)2 2

∴

Strain =

= 2880 J / m 3

Ans.

Chapter 15 9. ∆l due to temperature rise

∴

= 2mm But 1 mm is allowed (1000 mm to 1001 mm) ∴ ∆le = 1 mm Stress = Y × Strain  1 mm  = (1011 )    1000 mm  = 108 N / m 2

10.

i.e. F versus ∆x graph is a straight line of slope

∴ Ans.

a F0

 m  F   F  T = mx a =   x  0  =  0  x  l   m  l  L Tdx a ∆l = ∫ 0 SY T 2 x  F0  l =  ×  lsY  2 ∆l F Ans. strain = = 0 l 2SY

More than One Correct Options 1 2 U -axis. At x = 0.2 mm, U = 0.2 J (from the figure) 1 −4 2 0.2 = k (2 × 10 ) ∴ 2 ⇒ k = 107 Nm −1 YA k= L A k 107 …(i) ⇒ = = = 5 × 10−5 L Y 2 × 1011

1. U = k x 2. It is a parabola symmetric about

…(ii)

On solving Eqs. (i) and (ii), we get A = 10−4 m 2 and L = 2m

2. W F = elastic potential energy stored in the wire 1 2 1 = 2

=

k (∆l )2

(Slope)B > (Slope)A  YA   YA    >   L B  L A

or ( A ) B > ( A )A They are of same material. Hence, YB = YA as elastic potential energy. Mgl l= AY 1 2 1  YA  2 U = Kl =  l 2 2 L  From Eqs. (i) and (ii), we can prove that 1 U = Mgl 2 mg 5. When, TQ = 3 mg 4 mg and TP = mg + = 3 3 T Stress, σ= π r2 σ P TP  rQ  = ⋅  σ Q TQ  rP 

2

If rP = rQ , σ P = 4 σ Q , then P breaks. If rP < 2rQ , σ P > σ Q , then P breaks. If rP = 2rQ , σ A = σ B , then either P or Q may break. πd 2 6. Area of cross-section is A = 4 F L ∆L = ⋅ πd 2 / 4 Y ⇒ ∴

2

 YA  2 YAl   (l ) =  L 2L

YA . L

4. Half of energy is lost in heat and rest half is stored

F a= 0 m

AL = Volume = 200 × 10−6 m 3

Fl AY  AY  F =  ∆x  L 

∆x =

3.

= l α ∆θ = (1000) (10−4 ) (20)

Elasticity — 623

⇒

1 d2 ∆LB (2d )2 = ∆LA (d )2 ∆L ∝

∆ LB = 4 ∆LA

…(i) …(ii)

624 — Mechanics - II ∆L 4F = L πd 2Y 1 Strain ∝ 2 ⇒ d ∴ (Strain)B = 4 (Strain)A 7. If stress in steel = stress in brass, TS T then = B A S AB

4 σB 3 Now, F = σ S AS + σ B AB or 5000 = σ S × 16 + σ B × 20 From Eqs. (i) and (ii), we get σ B = 120.9 kg cm −2

⇒

Strain =

AS TS 10−3 1 = = = −3 TB AB 2 × 10 2 Steel

and

…(i) …(ii)

σ S = 161.2 kg cm −2

Comprehension Based Questions …(i)

1. Initially, ∆L = =

Brass TB

TS

σS =

FL AY

10 × 1 10−3 × 2 × 105

= 0.05 m = 5 cm

P

2. Force constant of string is Force Elongation F = FL / AY

k= x

2–x

System is in equilibrium. So, taking moments about P TS⋅ x = TB (2 − x ) TS 2 − x …(ii) ⇒ = TB x From Eqs. (i) and (ii), we get x = 1.33 m Stress Strain = Y If strain in steel = strain in brass, TS / AS TB / AB then = YS YB ∴

TS A Y (10−3 ) (2 × 1011 ) =1 = S S = TB AB YB (2 × 10−3 ) (1011 )

or

YA 1 × 105 × 10−3 = L 1 = 200 Nm −1

k=

Initial elastic potential energy of string 1 = k (0.05)2 2 1 = (200) (25 × 10−4 ) 2 = 25 × 10−2 J …(iii)

F′

From Eqs. (ii) and (iii), we get x = 1m

8. Area of steel rod, A S = 16 cm 2 Area of two brass rods, AB = 2 × 10 = 20 cm 2 F = 5000 kg σ S = Stress in steel and σ B = Stress in brass Decrease in length of steel rod = Decrease in length of brass rod σS σ ⋅ LS = B ⋅ LB YS YB YS LB ⇒ σS = ⋅ ⋅ σB YB LS or

 2 × 106   20 σS =     σB  106   30

10N 10N 10N

Let after force F = 10 N is applied extra elongation is x, then 1 0.25 + 30x = (200) (x 2 + 25 × 10−4 + 0.1x ) 2 0.25 + 30x = 100x 2 + 0.25 + 10x ∴ 100x 2 = 20x 20 1 ⇒ x= = m = 0.2 m 100 5 = 20 cm ∴ x max = 20 + 5 = 25cm

Chapter 15 3. When the displacement of the pulley is 25 cm, the

3. Consider an element as shown in the figure.

string gets loosened on both the sides. So, point A moves down by 50 cm. 4. Young’s modulus is a material constant. 5. On crossing the yield region, the material will experience the breaking stress and further elongation causes reduction in stress and breaking of the wire. 6. Stress-strain curve flattens on crossing elastic region.

Match the Columns 1. Stress, modulus of elasticity, energy density and pressure have SI units J/m 3 or N/m 2. Strain, coefficient of friction and relative density are dimensionless. Force constant have SI unit N/m. Fl F 2. ∆l = and stress = AY A 3. For A and B, the rod is in equilibrium and hence internal restoring force developed per unit area across any cross-section is same and thus stress is uniform. But in C and D, the case is opposite.

Subjective Questions Mg A ∆p Bulk modulus is B = (∆V /V ) ∆V ∆p Mg ∴ = = V B AB Also, the volume of the sphere is ∆V 3∆R 4 = V = πR 3 ⇒ V R 3 ∆R 1 ∆V , or = ⋅ R 3 V Using Eq. (i), we get ∆R Mg = ⇒ ∴ α=3 R 3 AB 2. Force constant of the wire is F F YA k= = = ∆L FL / AY L

1. Increase in pressure is ∆p =

…(i)

k YA YA = ⇒ ∴ = 140 m Lm Lm Y × 4.9 × 10−7 = 140 × 140 1 × 0.1 140 × 140 Y= = 4 × 109 49 × 10−7

x dx

⇒

p=4

xAρg

Force xA ρg = = xρg Area A Now, elastic potential energy stored in the wire is 1 dU = (Stress) (Strain) (Volume) 2 1 (Stress)2 = ⋅ (Volume) 2 Y 1 (x ρg )2 1 ρ 2g 2 A 2 dU = ⋅ A dx = ⋅ x dx 2 Y 2 Y 1 ρ 2g 2 A L 2 Total elastic potential energy = ⋅ x dx 2 Y ∫0 2 2 3 ρ g AL = 6Y Fl 4. ∆l = ∆Y πd 2 Here, F = Upthrust = Vi ρ l g ⇒ A = 4 4Vi ρ l gl ∴ ∆l = π d2Y Stress in the element =

=

(4 ) (10− 3 ) (800) (9.8) (3) (314 . ) (0.4 × 10−3 )2 (8 × 1010 )

= 2.34 × 10− 3 m

Ans.

5. ∆l = 521 − 500 − 20 = 1 cm = 0.01 m T

500 cm v

ω= ⇒

Elasticity — 625

20 cm ∆l

mg

mv 2 R   v2 v2 T = m g +  = m g +  R l  

T − mg = ∴

626 — Mechanics - II  v2 m g +  l l 

Tl = AY (πd 2 / 4 )Y mgl + mv 2 = (πd 2 / 4 ) Y

∴

πd ∆ lY − gl 4m

or

∆l =

v=

∴

8. 2TC + TS = mg ∆lC = ∆ls TC lC T l = S S AC YC AS YS

∴

TS = 2TC (as YS = 2 YC , lS = lC , AS = AC ) …(ii) TS = 2TC

2

Solving these two equations we get, mg TC = 4 TS = 2TC

(3.14)(4 × 10− 3 )2 (0.01)(2 × 1011 ) − 9.8 × 5 4 × 25

=

≈ 31 m / s

Ans.

6. 2T sin dθ = (dm) Rω

2

9.

T

dm

M L  (a) T =   a0 =  Sρ a0  2 2 

dθ dθ T

T

T Lρa0 = S 2 (b) Tx = M x a0 = xρSa0 L T dX L xρSa 0 ∆l = ∫ x =∫ dX 0 SY 0 SY ρa L2 = 0 2Y Fl 10. ∆l = AY 1 U = K (∆l 2 ) 2 1  YA  2 =   (∆l ) 2 l  ∴

For small angles, sin dθ ≈ dθ 2T (dθ ) = (2R dθ ) (πr2 ) (ρ ) (R ) (2πf ) T = (4 π 3 f 2R 2r2ρ ) Tl Now ∆l = AY ∆l T T = = l AY πr2Y l = 2πR ∆l = 2π (∆R ) ∆l ∆R T ∴ = = 2 l R πr Y 4 π 3 f 2R 2 r2 ρ = πr 2 Y

∴

∴

T − mg = mlω 2 = ml (2πf )2

∴

T = mg + 4 π 2mlf 2 = 6 × 9.8 + 4 π 2 (6)(0.6) (2)2 Now,

= 628 N Tl ∆l = AY =

(628) (0.6) (0.05 × 10− 4 ) (2 × 1011 )

= 3.8 × 10−4 m

and

F

T

L/2

dθ

dθ

7.

…(i)

Ans.

Stress =

=

1  YA   Fl      2  l   AY 

=

F 2l = mS∆θ 2 AY

Ans.

Ans.

2

F 2l 2 AYms (Mg )2 l = 2(πr2 ) Y (l πr2 )ρs

∆θ =

=

M 2g 2 2π 2r4Yρs

=

(100 × 9.8)2 2π (2 × 10 ) (2.1 × 1011 ) (7860) (420) 2

−3 4

≈ 4.568 × 10− 3 ° C

Ans.

16. Fluid Mechanics INTRODUCTORY EXERCISE

16.1

p0 + ρ 1gh1 = p0 + ρ 2gh2 ρh h1 = 2 2 ρ1 20 × 0.9 = 1 = 18 cm

1. ∴

INTRODUCTORY EXERCISE

1. In vertical direction pressure increases with depth. In horizontal direction pressure decreases in the direction of acceleration. h 2. pN + ρ gh − ρa   = pM  2 Ans.

2. Suppose the atmospheric pressure = p0. Pressure at A = p0 + h(1000 kgm −3 )g Pressure at B = p0 + (0.02m )(13600 kgm −3 )g These pressures are equal as A and B which are at the same horizontal level. Thus, h = (0.02m ) × 13.6 = 0.27m = 27cm 3. (a) p = p0 + ρ Hg g∆h (b) p = p0 + ρ Hg ghRHS Here, ∆h = (8 − 2) = 6 cm = 6 × 10−2 m −2

and hRHS = 8 cm = 8 × 10

16.2

pN = pM a = 2g ρω 2x 2 ρω 2x 2 3. pB = pA + = p0 + 2 2 But ∴

INTRODUCTORY EXERCISE

16.3

1. Weight = upthrust

3  Vρ 1 g =  V  ρ 2 g 4  ρ2 =

∴

4 ρ1 3

2.

m

4. Let the pressure of the liquid just below the piston be p. The forces acting on the piston are: (a) its weight, mg (downward) (b) force due to the air above it, p0 A (downward) (c) force due to the liquid below it, pA (upward) It the piston is in equilibrium, pA = p0 A + mg mg or p = p0 + A 5. In equilibrium, the pressure at the two surfaces should be equal as they lie in the same horizontal level. If the atmospheric pressure is p and a force F is applied to maintain the equilibrium, the pressure are 5N F and p0 + respectively. p0 + 1 cm 2 10 cm 2 This gives F = 50 N

6. ∆p1 = ∆p2

∴

F = (∆h) ρg A F 45g = ∆h = Aρg (900 × 10−4 )(103 )g = 0.5 m = 50 cm

∴

T + weight = upthrust T = upthrust − weight  1  = (71.2)   − 71.2  0.75 = 23.7 N

Ans.

3. Reading = weight of water + magnitude of upthrust on piece of metal (acting downwards) = (0.02 × 9.8) + (10− 6 ) (103 ) (9.8) Ans. = 0.206 N upthrust − weight 4. (a) a = mass (V ) (1000) (9.8) − (V ) (0.4 × 10− 3 ) (9.8) = (V ) (0.4 × 103 ) = 14.7 m/s2 2s (b) t = a 2 × 2.9 = 14.7 = 0.63 s

Ans.

Ans.

628 — Mechanics - II INTRODUCTORY EXERCISE

16.4

A   10 v2 =  1  v1 =   (1) = 2 m/s  5  A2 

∴

1. (a) Apply continuity equation

Applying Bernoulli’s theorem at 1 and 2 1 1 p2 + ρv22 = p1 + ρv12 2 2 1 ∴ p2 = p1 + ρ (v12 − v22 ) 2 1   =  2000 + × 103 (1 − 4 )   2

(b) Apply Bernoulli's equation dV 2. (a) = Av dt dV (b) = Av dt (c) Apply Bernoulli's equation 3. From conservation of energy v22 = v12 + 2gh

…(i)

[can also be found by applying Bernoulli’s theorem] From continuity equation A1 v1 = A2 v2 v1 A1

∴

p2 = 500 Pa

INTRODUCTORY EXERCISE

h

A2

2. t =

A  v2 =  1  v1  A2 

…(ii)

A22 =

∴

A2 =

=

2 (3 atm − 1 atm ) ρ

=

2 × 2 × 105 = 20 m/s 103

A 2H a g

(to empty the complete tank)

Now, tH→ 0 = t

H→

Substituting value of v2 from Eq. (ii) in Eq. (i), A12 2 . v1 = v12 + 2gh A22 or

2∆p ρ

v=

∴

v2

A12 v12 2 v1 + 2gh A1v1

16.5

1 1. ∆p = ρv 2 2

Given,

A a

H 3

+ tH 3

→0

2H A 2 (H /3) = t0 + g a g

From here find, A 2 (H /3) a g

3. Applying continuity equation at 1 and 2, we have

v12 + 2gh

A1v1 = A2v2

Substituting the given values, we get (10−4 )(1.0) A2 = (1.0)2 + 2 (10) (0.15)

…(i) 1

h

A2 = 5.0 × 10−5 m 2

4. From continuity equation,

2

A1v1 = A2v2 1

2 v1

v2

Further applying Bernoulli’s equation at these two points, we have 1 1 p0 + ρgh + ρv12 = p0 + 0 + ρv22 …(ii) 2 2

Fluid Mechanics — 629

Chapter 16 Solving Eqs. (i) and (ii), we have v22 =

2gh A2 1 − 22 A1

Substituting the values, we have 2 × 10 × 2.475 = 50 m 2/s2 v22 = 1 − (0.1)2

INTRODUCTORY EXERCISE

16.6

2 r2 (ρ − σ ) g 1. Using, V = 9 η η = 1.0 mPi = 1.0 × 10−3 Pi =

2 (20 × 10−6 )2 × (2 × 103 − 103 ) × 9.8 9 1.0 × 10−3

= 0.871 mm/s 2. Q Volume of bigger drop = n × volume of smaller drop 4 4 3 π R = 2 × πr 3 3 3 R = 21/ 3 r Terminal velocity ∝ r2 ∴ v′ = 22/ 3 v ∆V 3. F = η A ∆x = 10−3 × 10 × 2 × 10−1

2. W = T (∆A ) = T (2) (l ) (∆d ) = 7.2 × 10−2 × 2 × 0.10 × 10−3 = 1.44 × 10−5 J

3. W = T (∆A ) Soap bubble has two free surfaces. ∴ W = T (8πR 2 ) 2T 4. p − ph = r 2T ∴ p = ph + ph p r 2T = p0 + hρg + r

INTRODUCTORY EXERCISE

∴ ∴

h2 = (2.0) (3)

4 4 1. πR 3 = 106  πr3 3  3 ⇒

r = (10−2 ) R = 10−2 cm Ai = 4 πR 2 A f = (106 ) (4 πr2 )

∴

∆A = A f − Ai ∆U = (T ) (∆A )

 r2 1  =   r1 3

= 6.0 cm

= 0.02 N ∆V F = ηA ∆x F ∆V = shearing stress = η A ∆x 5 = 10−3 × = 10−3 N/m 2 5

INTRODUCTORY EXERCISE

2T cos θ = constant ρg hr or h2 = 1 1 h1r1 = h2r2 r2 hr =

Substituting the values, we get

= 2 × 10−2

4. Q

16.8

2T cos θ 1. h = rρg

2. Q

2T cos θ h= rρg h1ρ 1rg 2 cos θ 1 h ρ rg T2 = 2 2 2 cos θ 2 T2 h2 ρ 2 cos θ 1 = ⋅ ⋅ T1 h1 ρ 1 cos θ 2 T1 =

16.7

= 7.23 2T 3. h = ( R = radius of meniscus at the top) Rρg 2T 2 × 0.07 ∴ R= = hρg (10−2 ) (103 ) (10) = 1.4 × 10−3 m = 1.4 mm

4. pA = pC = ( pB + ρgh)

Ans.

Exercise LEVEL 1

11. B will have a tendency to keep its area as low as possible.

Assertion and Reason F is always perpendicular A to the surface or it is always specific. Hence, it is not a vector. 3. p1 = p0 + ρgh

1. Direction of force in p =

5. 6. 9. 10. 11.

p2 = p0 + ρg (2h) ≠ 2 p1 Area of cross-section is different. So, heights are different. In pressure height is more important. Speed will also depend on h. On moon, atmospheric pressure is zero. Hence barometer height is zero. Pressure at P is less. As liquid is flowing at P while liquid is at rest at Q. Force of buoyancy, U = Vρ air g Since, ρ air is negligible. Hence, U is negligible.

Single Correct Option 1. v = constant ∴ or  η

a=0 Fnet = 0 −1

 ML

T

− 1

0 2 −1 2.   =   = [M L T ] −3  ρ   ML 

9. n = n1 + n2 pV pV pV = 1 1+ 2 2 RT RT RT or ∴

pV = p1V1 + p2V2  4T   4 3  4T   4 3    πr  =    πr1   r  3   r1   3   4T   4  +    πr23    r2  3

∴

r = r12 + r22 = 5 cm

10. Drag force + upthrust = weight ∴ or

ku + F = mg mg − F u= k

A B

12. m ∝ r3 and terminal velocity ∝ r2 mass is eight times. So, radius is two times and terminal velocity will be four times. 4 4 14. N  πr3 = πR 3  3 3 ∴

 1 r=R  N

1/ 3

W = σ (∆A ) = σ ( A f − Ai ) = σ [ N (4 πr2 ) − 4 πR 2 ] 2/ 3    1 = 4 πσ NR 3   − R2 N   2 1/ 3 = 4 σπR (N − 1) 4T 15. ( p2 − p1 ) = R where R = radius of common surface 4T 4T 4T − = ∴ R2 R1 R 1 1 1 or = − R R2 R1

Ans.

16. Pressure (and hence the level of liquid) will keep on decreasing in the direction of motion of liquid. 2 r2 (ρ − σ ) g 17. vt = 9 η Ignoring the density of air σ we have, 2 r2ρg vT = 9 η 9η vT r= ∴ 2ρg =

9 × 1.8 × 10− 5 × 0.3 2 × 103 × 9.8

= 0.49 × 10− 4 m ≈ 0.05 mm

Ans.

Chapter 16

Fluid Mechanics — 631

18. If angle of contact in 90°, then liquid will neither

25. F1 will decrease and F2 will increase. So f1 may or

rise nor fall. 19. vT ∝ r2

may not be greater than f2. Total weight to system in both conditions will remain same. Hence, f1 + f2 = F1 + F2 1 2 26. ∆p = ρv 2

Radius or diameter is half. So, uniform or terminal 1 speed is th. 4 20. hρg = 105 Pa (given) ∴

h  px =  h −  ρg = 0.8 hρg  5

∴

v=

2 × 0.5 × 105 103

= 10 m/s

= 0.8 × 105 Pa U

21. Fraction of volume immersed ρ f = s ρl f1 ρ l 2 = f2 ρ l 1

2∆p = ρ

1 r∝ ρl

or

 f  or ρ l2 =  1  ρ l 1  f2 

27.

(V − V / 3) = ρ l1 (V / 3)

Mg U a0

= 2ρ l 1

22. Let ρ = density of sugar solution > ρw m = mass of ice. In floating condition weight = upthrust ∴ mg = VDρg m or VD = ρ

(M – m)g

…(i)

When ice melts, m mass of ice becomes m mass of water. Volume of this water formed. m …(ii) VF = ρw Since, ρ w < ρ , VF > VD . Hence, level will increase. 23. Upthrust = Viρ l ge Value of ge will decrease. So upthrust will decrease.

h

h

24. 1

2

…(i) Mg − U = Ma0 …(ii) U = (M − m) g = (M − m) a0 Solving these two equations we get, 2Ma0 m= g + a0 1 28. Diameter of left hand side is times. So area will be 5 1 times. 25 F (Mg ) ∆pLHS = ∆pRHS ⇒ = ( A / 25) A Mg ∴ F= 25 Equating the volumes Displacement on LHS = 5 times Displacement on RHS ∴ S = (5) (0.5) = 2.5 cm W = FS  Mg  −2 (500) =  ∴  (2.5 × 10 )  25  ∴

∴ ⇒

p1 = p2 p0 + ρ w gh = p0 + ρgh ρ = ρw

(with g = 10 m / s2) M = 5 × 104 kg Vd + Vd2 29. 4 = 1 2V or …(i) d1 + d2 = 8

632 — Mechanics - II 3=

m+ m or (m / d1 ) + (m / d2 )

2d1d2 =3 d1 + d2

…(ii)

Solving Eqs. (i) and (ii), we get d1 = 6 and d2 = 2 30. Total weight = total upthrust ( A × 0.5 × 900) g + (100) g = A × 0.5 × 1000 × g Ans. ∴ A = 2 m2

31. Change in weight = upthrust. ∴(38.2 − 36.2) g = (Vgold + Vcavity ) ρ w g 1g   38.2  ∴ 2=  + Vcavity  as ρ w =   19.3   cm 3  ∴

Vcavity = 0.02 cm 3

Ans.

2T cos θ 1 or h ∝ rρg r Hence, h will become 2 times. h=

38. In steady state, Volume flow rate entering the vessel = volume flow rate leaving the vessel Q2 ∴ Q = av = a 2gh or h = 2ga2 =

Now, p1 +

of Level 1. dV = av = a 2gH dt This is independent of ρ liquid . 34. (ΣQ )inflow = (ΣQ )outflow

∴

⇒

∴(4 × 10− 6 ) + Q + (4 × 10− 6 )

4  4 8  πr3 = πR 3 or 3  3

+ (6 × 10− 4 ) Q = 13 × 10− 6 m 3 /s 3  Mg =  a3 ρg 4   4M  a=   3ρ 

∴

36. h =

1/ 3

2T cos θ rρg

2T1 cos 0° rρ 1g 2T cos 135° − 3.42 = 2 rρ 2 g From these two equations, we get T1 10. (cos 135° ) = T2 (− 3.42) (cos 0° ) ρ 2 10 =

(10) (− 1/ 2 ) (1) = (− 3.42) (1) (13.6) = 1: 6.5 37. Area is halved, means radius of tube is made 1 times. 2

R = 2r

v T ∝ (radius)2

35. Weight = Upthrust ∴

1 2 1 ρv1 = p2 + ρv22 2 2 1 p2 = p1 + ρ (v12 − v22 ) 2 1 = (8000) + × 1000 (4 − 16) 2 Ans. = 2000 Pa

40. Equating the volume, we have

= (8 × 10− 6 ) + (2 × 10− 6 ) + (5 × 10− 6 ) ∴

Ans.

39. Area at other point is half. So, speed will be double.

32. See the hint of Q-No. 23 (a) of subjective questions 33. v = 2gH

(102 )2 = 5 cm (2 × 1000) (1)2

…(i) …(ii)

Radius has become two times.Therefore, terminal velocity will become 4 times. 41. Under normal conditions, upward surface tension force supports the weight of liquid of height h in the capillary. In artificial satellite effective weight will be zero, but upward surface tension force will be there. Hence, liquid rises upto the top. 42. Pressure inside a soap bubble is given by, 4T P = P0 + r So, pressure inside a smaller soap bubble will be more and air will flow from smaller drop to bigger drop. 2T cos θ 1 43. h = or h∝ Rρg R Now,

M = (πR 2h) ρ

or

M ∝ R 2h

or M ∝R Radius is doubled, so mass in the capillary tube will also become two times. 4T rhρg 44. ρgh = ⇒ T = r 4

Chapter 16 45. hρg = =

2T r

⇒ h=

Fluid Mechanics — 633

2T rρg

2 × 70 = 28 cm 0.005 × 1 × 1000

46. N  πa3 =   4 3

4 3 b3 πb ⇒ N = 3 3 a

Ai = N (4 πa2 ) =

4 πb a

3

A f = 4 πb2

Since, weight = upthrust Apparent weight = 0 3. F = pA (ρgh) B ∴

b  ∆A = Ai − A f = 4 πb2  − 1 a  b  W = T (∆A ) = 4 πb  − 1 T a 

F ∝h 1

4. If volume is doubled, then radius becomes (2) 3 times. W = T (8 πR 2 )

2

1 1 4  = mv 2 =  πb3ρ v 2  2 2 3 Solving we get v= 2T 47. p = + p0 r =

6T  1 1  −  ρ  a b

2 × 0.7 + 105 0.14 × 10−3

= 1.01 × 105 N/m 2

48. From two surfaces,

or

W ∝. R 2

Therefore

W ′ = (2) 3 W .

2

5. Concept is same as in Q.No-2 of same exercise. 6. Relative density of metal in this case is given by weight in air change in weight in water 0.096 = = 3.84 0.096 − 0.071

RD =

∴

Density of metal = 3.84 ρ w = 3840 kg/ m 3

Ans.

7. Weight = upthrust F

∴ (25 + 5) g = (V + 2) ρ w g Putting ρ w = 1 g/ cm 3 V = 28 cm 3

We get,

Ans.

8. Weight of both (block + woman) S.T force

F = 2 (Tl ) = 2 (T ) (2 πr) = 2 (T ) (2 π ) (1.5) = (6 πT ) dyne 2T 49. ∆p = hρg + r 2 × 70 = 2 × 1 × 980 + 0.05 = 4760 dyne/cm 2

Subjective Questions 1. In floating condition, weight of liquid displaced = weight of solid 2. Weight = Vρ w g Upthrust = Vρ eg = Vρ w g

∴ ∴

= upthrust on 100% volume of block 50 g + V (850) g = V (1000) g Ans. V = 0.33 m 3

9. Weight of ice block + weight of metal piece = upthrust on 100% volume of ice cube. Let a = side of ice cube. Then, [(a3 ) × 900 × g + 0.5 g ] = (a3 ) (1000) (g ) ∴ or

a3 = (5 × 10− 3 )m 3 a ≈ 0.17 m or 17cm

10. Fraction of volume immersed is given by ρs ρl ρs ρs 0.6 = = ρw 1 f =

In first case

Ans.

634 — Mechanics - II or ρ s = 0.6 g/cm 3 = density of wood In second case

ρ s 0.6 = ρl ρe 0.6 ρl = 0.85 = 0.705 g/cm 3

0.8 = or

3 400 − 0 = v2 400 − 100

∴

v2 = 2.25 m/s

v2 = 0 at r2 = R

Ans.

volume. ∴ (∆m) g = (πr2∆h) ρ w g

17. (a) Q =

∆m = πr ∆h ρ w = (π ) (0.8)2 (3) (1)

= 6.92 × 105 N / m 2

= 7 × 10− 2 m 3 /s Ans.

Ans.

ρ s hw = ρ w hs 10 = = 0.8 12.5 = relative density of mercury 14. ( p0 + ρ w hw g ) − ( p0 + ρ s hs g ) = ρ Hg hHg g ∴

hHg =

ρ wρ w − ρ s hs ρ Hg

(1) (25) − (0.8) (27.5) 13.6 = 0.221 cm

(b) Q1 = Q2 ∴

13. p0 + ρ w hw g = p0 + ρ s hs g

∴

η  Q2 =  1  Q1  η2 

or

 1.005  −2 =  (7 × 10 )  0.469 Ans.

15. (i) Absolute pressure in part (a) = 76 cm of Hg + 20 cmof Hg = 96 cm of Hg Gauge pressure in part (a) = 20 cm of Hg Absolute pressure in part (b) = 76 cm of Hg − 18 cm of Hg = 58 cm of Hg Gauge pressure in part (b) = − 18 cm of Hg. (ii) 13.6 cm of water is equivalent to 1 cm of Hg. So the new level difference will become 19 cm. p1 − p2 2 (R − r2 ) 4ηL or (a)

v1 R 2 − r12 = v2 R 2 − r22

v ∝ R 2 − r2

[( p1 − p2 ) R 4 ] i = [( p1 − p2 ) R 4 ]f

Since, diameter or radius has decreased to half. Therefore gauge pressure should become 16 times. or ( p1 − p2 ) f = 16 × 1400 Ans. = 2.24 × 104 Pa 1 (c) Q ∝ η Q1 η2 ∴ = Q2 η1

=

16. v =

π  R 4   p1 − p2      8  η  L 

(0.04 )4 1400  π =  ×  8  1.005 × 10− 1 0.2

2

= 6.03 g F 3000 × 9.8 12. p = = A 425 × 10− 4

Ans.

 R 2 − r2  (b) v2 =  2 22  v1  R − r1 

11. Extra weight = extra upthrust on extra immersed

∴

or

= 0.15 m 3 /s

18. (a) Relative density of metal =

weight in air 10 = =5 change in weight in water 2

∴ Density of metal = 5ρ w = 5000 kg/ m 3 mass Now, volume = density 10 × 10− 3 5000 = 2 × 10− 6 m 3 =

(b) Change in weight = upthrust on 100% volume of solid or ∆w = Vs ρ l g ∴ ∆w ∝ ρ l

Ans.

Chapter 16 ∆wl ρ = l ∆ww ρ w

∴

∴

 ∆wl   1.5 ρl =   ρ w =   (1000)  2  ∆ww 

or

= 750 kg/ m

∴

19. Let (x ) mm is in water and (60 − x ) mm is in mercury. Total upthrust = weight

or

= (0.06)2 (x × 10− 3 ) (103 ) g + (0.06)2 (60 − x ) (10− 3 ) (13.6 × 103 ) g Solving this equation, we get x = 28 mm and 60 − x = 32 mm 20. T + U = 60 + w …(i) l 2

U

= force from right hand side kx = ∆pA (∆p) A x= k (30 + 101) × 103 × 0.5 × 10− 4 = 60 Ans. = 0.109 m or10.9 cm (∆p) A k 30 × 103 × 0.5 × 10− 4 = 60 = 0.025 m = 2.5 cm

x=

(b)

y

w

dy

dF r⊥ = h – y

Σ (Moments) about = O  l  l = U   + T (l ) = W   ∴  4  2

O

…(ii) U + 4T = 2W …(iii) w = 120 N Solving these three equations we get T = 20 N and U = 160 N V  V  Now, 160   ρ w g =   (1000) (10)  2  2 V = 32 × 10− 3 m 3

Ans.

21. (a) Upthrust − Weight − T = ma T = Upthrust − Weight − ma  2  =  (1000) (10 + 2) − 20 − 4  500 = 48 − 20 − 4 = 24N

ρglh2 2 (b) Perpendicular distance of about force from point O is r⊥ = h − y ∴ dτ = (dF ) r⊥ = (ρgly) (h − y) dy ∴

or

∴

Ans.

23. (a) dF = ( py ) (dA ) = (ρgy) ( ldy)

60 N

∴

Ans.

∆

l 4 O

l 4

T

Acceleration w.r.t. tank. = 14 − 2 = 12 m / s2

22. (a) Force from left hand side

3

= (0.06)3 (7.7 × 10− 3 ) (g )

Fluid Mechanics — 635

Ans.

(b) Downward force T suddenly becomes zero. Therefore upthrust − weight 48 − 20 a= = m 2 = 14 m / s2

h

F = ∫ dF = 0

h

= total torque = ∫ dτ 0

After substituting the values we get, ρglh3 Ans. τ= 6 (c) F × r⊥ = τ τ ∴ r⊥ = F Substituting the values we get, h Ans. r⊥ = 3 24. PQ is original level of mercury. Equating the pressures at the level of M from left hand side and right hand side.

636 — Mechanics - II 3A

30 cm

30 cm

A ⇒

P M

x 3x

∴

4 ηL vmax R2 (4 × 4 × 10−3 ) (10−3 ) (0.66 × 10−3 ) = (2 × 10−6 )2

( p1 − p2 ) =

Q

= 2.64 × 103 N/m 2 = hρg 2.64 × 103 h= ρg

p0 + ρ w g hw = p0 + ρ Hg ghHg ρ w hw = ρ Hg hHg

or

∴ (1) (30 + 3x ) = (13.6) (4 x ) Solving we get, x = 0.58 cm

=

Ans.

= 0.0195 m of Hg ≈ 19.5 mm of Hg

H

25.

2.64 × 103 m of Hg 13.6 × 103 × 9.81

30.

h

(0.11–h)

R

P

R = 2 Hh ∴

H = F

26.

h

R2 4h

1 2

Ans.

v

p− ∴

1 2 1 ρv1 = p2 + ρv22 2 2 F 1 2   p0 +  + 0 = p0 + ρv  A 2 p1 +

∴

27.

v=

2F ρA

For water θ = 0° Ans.

1 ρ air v 2 = ρ w ghw 2 2ρ w g hw v= ρ air 2 × 103 × 10 × 10− 2 1.3 = 12.4 m/s

∴ or

Ans.

=

2Mg + 2gh ρA

29.

vmax =

( p1 − p2 ) R 2 4ηL

2T + P0 r

p0 (0.11) = p (0.11 − h) 2T   p0 (0.11) =  p0 +  (0.11 − h)  r

(1.01 × 105 (0.11) = [1.01 × 105 2 × 5.06 × 10−2   (0.11 = h) 10−5 

Solving this equation we get, h = 0.01 m or 1 cm If seal is broken, then water will rise in the capillary. 31. W = T (∆A )

2 × 20 × 10 + 2 × 10 × 0.5 0.5 × 103

= 3.28 m/s

P=

or

+

2

v=

R=r

Substituting the values, we get

Mg 1 28.   + ρgh = ρv 2 ∴

∴

As temperature is constant. p1V1 = p2V2

=

 A

2T = p0 R 2T P= + P0 R r R= , cos θ

Ans.

= T (2ld ) = (72 × 10−3 ) (2) (0.1) (10−3 ) = 14.4 × 10−6 J

Ans.

Fluid Mechanics — 637

Chapter 16 32. Let the pressures in the wide and narrow capillaries of radii r1 and r2 respectively be p1 and p2. Then pressure just below the meniscus in the wide and narrow tubes respectively are  2T   p1 −   r1 

and

 2T   p2 −   r2  [excess pressure =

p1 h1

A B

h

C D

2T ] r

p2

h2

h1 − h2 = h

Difference in these pressures  2T   2T   = hρg  p1 −  −  p2 −  r1   r2  ∴ True pressure difference = p1 − p2  1 1 = hρg + 2T  −   r1 r2  = 0.2 × 103 × 9.8 + 2 × 72 × 10−3

where, r1 and r2 are the radii of the two limbs But pA = pC 2T 2T ∴ pB + = pD + r1 r2

where, h is the difference in water levels in the two limbs 2T  1 1  Now, h=  −  ρg  r1 r2  T = 0.07 Nm −1, ρ = 1000 kgm −3 3 3 3 m r1 = mm = cm = 2 20 20 × 100

Given that

  1 1 −  −3 −4  144 . × 10 7 . 2 × 10   = 186 . × 10 = 1860 N / m 3

 1 1 pD − pB − 2T  −   r1 r2 

or

= 15 . × 10−3m, r2 = 3 × 10−3 m

2

33. The surface tension of the liquid is rhρg 2 (0.025cm)(3.0cm)(1.5gm / cm 3 )(980cm / s2 ) = 2 = 55 dyne/cm Hence, excess pressure inside a spherical bubble 4T 4 × 55dyne/cm p= = R (0.5cm ) T =

= 440 dyne / cm 2

34. Suppose pressure at the points, A , B , C and D be pA, pB , pC and pD respectively. The pressure on the concave side of the liquid 2T surface is greater than that on the other side by . R An angle of contact θ is given to be 0°, hence R cos 0° = r or R = r 2T 2T and pC = pD + ∴ pA = pB + r1 r2

∴

h=

2 × 0.07  1 1  −  m 1000 × 9.8  15 . × 10−3 3 × 10−3 

= 4.76 × 10−3 m = 4.76mm

35. The total pressure inside the bubble at depth h1 is (p is atmospheric pressure) = ( p + h1ρg ) +

2T = p1 r1

and the total pressure inside the bubble at depth h2 is 2T = ( p + h2ρg ) + = p2 r2 Now, according to Boyle’s law p1V1 = p2V2 4 3 4 where, V1 = πr1 and V2 = πr23 3 3 Hence, we get  2T  4 3 ( p + h1ρg ) + r  3 πr1  1   2T  4 3 = ( p + h2ρg ) + πr2 r2  3 

638 — Mechanics - II  2T  3  2T  3 or ( p + h1ρg ) + r1 = ( p + h2ρg ) + r r2 r   1  2  Given that h1 = 100 cm, r1 = 01 . mm = 0.01 cm, mm = 0.0126 cm, T = 567 dyne/cm, r2 = 0126 . p = 76 cm of mercury. Substituting all the values, we get h2 = 9.48 cm. 36. Let n be the number of little droplets. Since, volume will remain constant, hence volume of n little droplets = volume of single drop 4 4 ∴ n × πr 3 = πR 3 3 3 or nr3 = R 3

LEVEL 2 Single Correct Option 1. When ice melts into water its volume decreases. Hence, over all level should decrease. Now suppose m is the mass of ice, V1 is volume immersed in water and V2 the volume immersed in oil. Then in floating condition, weight = upthrust ∴ mg = V1 ρ w g + V2 ρ 0 g m − V2 ρ 0 V1 = ∴ ρw m V2 ρ 0 …(i) = − ρw ρw

Decrease in surface area = n × 4 πr2 − 4 πR 2 or ∆A = 4 π[ nr2 − R 2 ]  nr3   R3  = 4π  − R2 = 4 π  − R2 r r     1 31 = 4 πR −  r R  Energy evolved W = T × decrease in surface area 1 1  = T × 4 πR 3 −  r R  W 4 πTR 3  1 1  Heat produced, Q = = − J J  r R  But Q = msdθ where, m is the mass of big drop, s is the specific heat of water and dθ is the rise in temperature. 4 πTR 3  1 1  ∴ − = Volume of big drop J  r R  × density of water × sp. heat of water × dθ 4 3 4 πTR 3  1 1  or πR × 1 × 1 × dθ =  −  J  r R 3 3T  1 1  or dθ = − J  r R 

37. 1 h

2

Applying pressure equations between 1 and 2, 2T p0 + ρgh − = p0 r 2T h= ⇒ rρg

When ice melts, m mass of ice converts into m mass of water. Volume of water so formed is m …(ii) V3 = ρw From Eqs. (i) and (ii), V3 > V1 ∴ Interface level will rise. 2. ge = 0 so, pressure inside and pressure outside the hole will be same. T P

U C

3. l /4 O

l/2

l /4 M w1 = 240 N

w2 = 120 N

or

…(i) T + U = w1 + w2 = 360 N V 3 U = ρ w g = (V / 2) (10 ) (10) 2 …(ii) U = 0.5 × 104 V

(Σ Moments) about M = 0  l  3l   l ∴ 120   + T   = 240    4  4  4 240 − 120 or T = 3 = 40 N = 4g Now from Eqs. (i) and (ii), we get V = 6.4 × 10− 2 m 3 V 4. Volume flow rate = av = t 120 × 10− 3 V v1 = = ∴ A1t 5 × 10− 4 × 2 × 60

Ans.

Fluid Mechanics — 639

Chapter 16 A1v1 = A2v2 1 A2 is th of A1. 5 Hence, v2 is five times of v1 or 10 m/s. 2h 2×1 t0 = = g 10 Now,

= 0.447 s R = v2t0 = 4.47 m y

5.

θ

z dz

h p

θ

z = y sec θ dz = dy sec θ p = (ρgy) dA = (b) dz = (b) dy sec θ dF = pdA = (ρgb sec θ ) y dy h 1 ∴ F = ∫ dF = ρbh2g sec θ 0 2 6. Velocity of body just before touching the lake surface is, v = 2gh Retardation in the lake, upthrust − weight a= mass Vρg − Vρg  σ − ρ = =  g  ρ  Vρ Maximum depth dmax =

v2 hρ = 2a σ − ρ

7. ρgh = ρaL aL ∴ h= g R 8. Radius of meniscus = =r cos θ 2σ ∆p due to spherical surface = r 2σ cos θ = R pB = patm 9. a1v1 = a2v2 ∴

(L2 ) 2gh = (πR 2 ) 2g (4 h) R=

L 2π

10. Let h = final height of liquid.

h1

⇒ h2

h

Equating the volumes we have, Ah1 + Ah2 = A (2h) h + h2 h= 1 ∴ 2 W mg = − ∆U = U i − U f h  h   =  Ah1 ρg 1  +  Ah2 ρg 2   2  2    h + h2   h1 + h2   − 2 A  1  ρ  2   4    Simplifying this expression we get the result. 11. ∆p = ρ 0il g h0il + ρ w ghw = (600)(10)(10 × 10− 2 ) + (1000 × 10 × 2 × 10− 2 ) = 800 N / m 2 1 12. ∆p = ρv 2 2 ∴

v=

2∆p = ρ

2 × 5 × 105 1000

= 31.5 m/s

13. From continuity equation, vA = v B = v0 ∴

pA + ρgh = pB + 0

…(i) ∴ pB − pA = ρgh Now, let us make pressure equation from manometer. pA + ρg (h + H ) − ρ Hg gh = p3 Putting pB − pA = ρgh we get h = 0

14. F =

∆p  ∆m  =  (∆v ) ∆t  ∆t   ∆V  =ρ   (2v )  ∆t  = ρ ( Av ) (2v ) = 2ρAv 2

15. ∆pA = ρgh or ρal 16. At every instant half of the length remains up the surface and half below the surface of liquid. Since, D >> d , over all level will remain unchanged.

640 — Mechanics - II Let at t = 0, length of candle is 10 cm.

1 1 (2ρ ) (g ) h2 = (3ρ ) (g ) R 2 2 2 3 h= R 2

∴

Time (s)

Length of candle (cm)

Length about the surface (cm)

Length below the surface (cm)

0 1

10

5

5

8

4

4

∴

Water

22. 17.

1m

dF

dx

(x – 0.5)

Net clockwise torque of applied force = F (0.5) =

20 cm3

ρ 1gh1 = ρ 2gh2 ∴ (4 ) (g ) (x ) = (1) (g ) (60) or x = 15 cm Total volume of liquid = (20 + 15) cm 3 F 2

Equating the two torques we get, ρg F= 6 18. See the typed example 12. ρ − ρ2 1 tan θ = 1 = tan 30° = ρ1 + ρ2 3 Solving this equation we get, ρ1 3+1 = ρ2 3 −1 ∆p  ∆m  19. F = =  (∆v ) ∆t  ∆t   ∆V  =ρ   (v + v2 ) = ρV (v1 + v2 )  ∆t  1 2 r (ρ − σ ) g 9 η 2 × (0.003)2 (1260 × 2 − 1260) × 10 = 9 × 1.26 = 0.02 m / s d 10 × 10− 2 t= = = 5 s. vT 0.02

vT =

60 cm3

x cm3

Net torque about hinge = 0 dF = (ρgx ) (1⋅ dx ) dτ = (ρgx ) (x − 0.5) dx ∴ Net anticlockwise torque 1  ρg  = ∫ dτ =   0  12 

20.

60 cm (x) cm

x

2

21. Torque of hydrostatic force about centre of sphere is already zero as hydrostatic force passes through the centre. ∴ FLHS = FRHS For finding force refer Q-No. 23 (a) subjective questions of section Level 1.

= 35 cm 3

23. Viscous force = mg sin θ ∴

v 3 = mg sin 37° = mg t 5 3 3 3ρagt = (a ρ ) g ⇒ ∴ η = 5 5v η (a2 )

24. Reading = weight of bucket of water + magnitude of upthrust on block = (10g ) +

1  7.2    ρwg 2  7.2 ρ w 

= 10.5 g = 10.5 kg

25. Sum of all three terms are different at three points A , B and C. a=

26.

upthrust − weight (upwards) mass

Vσg − Vρg  σ  (upwards) =  − 1 g ρ  Vρ 2H + H 27. hTop = = 1.5H 2 Since, this point lies in the tank. So hole should be made at this point. 28. Viscous force F ∝ (area) =

So let

F = kA F0 = k ( A1 + A2 )

…(i)

and

T = k A1

…(ii)

Chapter 16

Fluid Mechanics — 641

32. p1 = p2

Dividing Eq. (ii) by Eq.(i) we get, T A1 = F0 A1 + A2 F0 A1 T = ∴ A1 + A2

1

2

h

29.

25 cm original level of water 2

25 cm 1

∴

water

p1 = p2 ∴

ρ 0h0 = ρ w hw

or

(0.8) (h + 50) = (1) (50)

or

h = 12.5 cm

tension force supports the weight of liquid of height ‘h’. S.T. Force h

30. dθ F

A dθ dθ B

2T 2T + ρgh = p0 − r1 r2 ρ g hr1 r2 T = 2 (r2 − r1 )

33. When the capillary is inside the liquid, the surface

p0 + ρ 0gh0 = p0 + ρ w g hw

or

p0 −

dθ F

F = Tension Surface tension force is radially outwards. On AB : 2F sin d θ = Surface tension force from two films from two sides For small angle, sin d θ ≈ d θ ∴ 2F d θ = 2 (Tl ) = 2 (T ) (2Rd θ ) ∴ F = 2 TR 31. Surface tension force on liquid is F U downwards. But on the disc it is upwards. In the figure, F = Surface tension force = Tl cos θ = T (2 πr) cos θ U = upthrust = ω W = weight of disc For equilibrium of disc, W W = U + F = ω + 2 πTr cos θ

When the capillary is taken out from the liquid similar type of surface tension force acts at the bottom also, as shown in second figure. Hence, now it can support weight of a liquid of height 2h. 34. Let L be the width of plates (perpendicular to paper inwards). Surface tension force in upward direction = weight of liquid of height h ∴ Tl cos θ = Vd g (V = volume) S.T. Force

h

x

or ∴

T (2L) cos θ = (Lxh) dg 2T cos θ h= xdg

642 — Mechanics - II 35. Let R1 and R2 are the radii of soap bubbles before and after collapsing. Then given that, 4  4 V = 2  π R13 − π R23 3  3 S = 2 (8 πR12 ) − 8 πR22

...(i) ...(ii)

PV 1 1 = PV 2 2   4T   4 4T   4 3 3 ∴  P0 +   πR2    2 × πR1  =  P0 +      R1  3 R2   3 ...(iii) Solving these three equation we get the desired result. 36. piVi = pf V f 3 4 σ   4 3  4σ   4   r  ∴  p1 +   πr  =  p2 +   π      r  3 r / 2  3   2

Solving this equation we get, 24 σ p2 = 8 p1 + r 37. W = 2π (r1 + r2 ) T ∴

T =

7.48 × 10 × 10−3 W = 2π (r1 + r2 ) 2π × 17 × 10−2

= 70 × 10−3 N/m

More than One Correct Options 1. Velocity gradient =

∆v 2m / s = = 2 s− 1 ∆h 1 m

∆v = (10− 3 ) (10) (2) ∆h = 0.02 N 2. For contact angle θ = 90° , liquid neither rises nor F =ηA

falls.

3. Restoring force = − (ρAg ) x or F ∝ − x This is just like a spring-block system of force constant K = ρAg 4. From continuity equation Av = constant v2 > v2 as A2 < A1 From Bernoulli’s equation, 1 p + ρv 2 = constant (as h = constant) 2 p2 < p1 as v2 > v2 5. Fraction of volume immersed, ρ f = s ρl

This fraction is independent of atmospheric pressure. With increase in temperature ρ s and ρ l both will decrease. 2hB 6. U = 2ghT : t = , R = 2 hT hB g Here, hT = distance of hole from top surface of liquid and hB = distance of hole from bottom surface 7. Pressure increase with depth in vertical direction and in horizontal direction it increases in opposite direction of acceleration based on this concept pressure is maximum at point D and minimum at B. 8. In air, a1 = g (downwards) upthrust − weight In liquid, a2 = mass (V ) (2ρ ) (g ) − (V ) (ρ ) (g ) = (Vρ ) = g (upwards) ∴ a1 ≠ a2 9. Initially dV1 …(i) = v1a1 = ( 2gh ) (π ) (2R )2 dt dV2 …(ii) = v2a2 = ( 2g (16h)) (π ) (R )2 dt From Eqs. (i) and (ii), we can see that dV1 dV2 = dt dt After some time v1 and v2 both will decrease, but decrease in the value of v1 is more dominating. So, dV1 dV2 a1v1 or < a2 v2 or dt dt 10. Fraction of volume immersed is given by ρ f = s ρl ρ s and ρ l are same. Hence : f1 = f2 = f3 Base area in third case is uniform. Hence h3 is minimum.

Comprehension Based Questions 2. x3 = 2 hTop × h bottom = 2 3a × a = 2 3 a

3. x1 = 2 a × 3a = 2 3 a x2 = 2 2a × 2a = 2 4 a ∴

x3 = 2 3a x 1 = x3 < x2

Chapter 16 4. Weight = upthrust 3L A  L A  A ∴  L Dg = × × 2d × g + × ×d×g  4 5  5 4 5 5d ∴ D= 4 5. pA = p0 A + weight of two liquids + weight of cylinder H H = p0 A +   ( A ) (d ) (g ) +   ( A ) (2d ) (g )  2  2  A   5d  + L    g  5  4  ∴

p = p0 +

(L + 6H ) dg 4

6. Applying Bernoulli’s equation just inside and just outside the hole, 1 H H  p0 +   (d )g +  − h (2d )(g ) = (2d ) v 2 + p0  2  2  2 ∴

7. t =

v= 2h g

g (3H − 4 h) 2

⇒ ∴ x = vt = (3H − 4 h) h

Match the Columns 1. v = 2ghT and R = 2 hT hB Here, hT = height of hole from top surface and hB = height of hole from bottom 2. (a) and (b) In floating condition upthrust = weight By increasing temperature of density liquid, weight remains unchanged. Hence, upthrust is unchanged. (c) When density of solid is increased (with constant volume) mass and hence weight of solid will increase. So, upthrust will increase. 3. Ball will oscillate simple harmonically. The mean position in at depth A

B

ρ = AB or BC α From A to B : ρ > ρ l , weight > upthrust At B , ρ = ρ l , weight = upthrust From B to C ,ρ e > ρ , upthrust > weight. From A to C → upthrust will increase and gravitational potential energy will decrease. From C to A, upthrust will decrease and gravitational potential energy will increase. From A to B, speed will increase. From B to C, speed will decrease. F 4. Surface tension = e ∆v Viscous force f = η A ` ∆y Energy Energy density = Volume Volume Volume flow rate = Time 5. F2 − F1 = upthrust Amplitude =

= weight of cylinder in equilibrium or floating condition. From liquid-1, horizontal pair of forces cancel out. So, net force = 0

Subjective Questions 1. w = Weight, F = Upthrust F

ρ = αh

or

hmax =

h=

θ

2ρ = AC α

A θ

w

0.5 m

D O

OB 0.5 sec θ = = 0.25 sec θ 2 2 About point O , clockwise moment of w = anticlockwise moment of F. L  ∴ w  sin θ = F (OA sin θ ) = F (0.25 sin θ ⋅ sec θ ) 2  OA =

∴ ρ = AB α

C B

Given, C

Fluid Mechanics — 643

or ∴

L =1m F (0.5 sec θ )( A )(1.0)g cos θ = = 2w 2(1)( A )((0.5)g cos2 θ =

1 1 , cos θ = 2 2

θ = 45°

Ans.

644 — Mechanics - II 2. Fraction of volume immersed before putting the ρ 800 new weight = block = = 0.8 ρ water 1000

i.e. 20% of 3 cm or 0.6 cm is above water. Let w is the new weight, then spring will be compressed by 0.6 cm. ∴ w + weight of block = Upthrust on whole volume of block + spring force or

∴

w = (3 × 10−2 )3 × 1000 × 10 + 50 × −2

∴

Oscillations are simple harmonic in nature. Rewriting Eq. (i) 6mgx ma = − 5h0 6g or a=− x 5h0

−2 3

(0.6 × 10 ) − (3 × 10 ) × 800 × 10 Ans. w = 0.354 N

3. Net force on the block at a height h from the bottom is

f =

1 a 2π x

f =

1 2π

6g 5h0

Ans.

4. (a) v = 2gh = 2 × 10 × 5 = 10 m/s (b) From conservation of energy, v′ 2 = v 2 + 2gH

x h=

= 100 + 2 × 10 × 5 = 200

h0 2

(c) (upwards) Fnet = upthrust – weight    m   3h =  ρ 0  4 −  g − mg 5  h0   ρ0  2  h Fnet = 0 at h = 0 2 h0 is the equilibrium position of the block. 2 h For h > 0 , weight > upthrust 2 h i.e. net force is downwards and for h < 0 2 weight < upthrust i.e. net force is upwards. For upward displacement x from mean position, net downward force is So, h =

     m   h0  3(h + x )  F = −   ρ 0 4 − g − mg   h =  5 h 2   0  ρ0    2   6mg …(i) ∴ F =− x 5h0 h (because at h = 0 upthrust and weight are equal) 2 Since F ∝−x

v′ = 14.1 m/s 2A  H t=  H − 2 a 2g   2 2 × π × (1) = −4 [ 5 − 2.5 ] 10 2 × 10 = 9200 s

5. Writing equation of motion for the block T – mg sin 30° = ma For the sphere Weight – Buoyant force – T = ma mg or mg – – T = ma 2 Solving, we get a=0

6. w = (0.25)Vρ Hg g

…(i) …(ii)

…(i)

Let x fraction of volume is immersed in mercury in the second case. Then,

w = xVρ Hg g + (1 − x ) Vρ w g Equating Eqs. (i) and (ii), we have ρ Hg = xρ Hg + (1 − x )ρ w 4 ∴ 12.6x = 2.4 or x = 0.19

…(ii)

Ans.

Chapter 16 1 2

a = πr2 = π (2 × 10−3 )2

7. (a) pA − ρv 2 + ρgh = pD

= 1.26 × 10−5 m 2

pA = pD = p0

But

Substituting these values in Eq. (i), we have

1 2 ρv = ρgh 2 v = 2gh

∴ ∴

(1.26 × 10−5 ) 2 × 9.8 × y = π (1.11 × 10−5 ) x 2 or y = 0.4 x 4 This is the desired x-y relation. 10. Initially, kx = mg

Here, h = (4 + 1) = 5 m v = 2 × 9.8 × 5 = 9.9 m/s

∴

(b) Applying Bernoulli’s equation at A and B, 1 pA + 0 + 0 = pB + ρv 2 + ρg (1.5) 2 1 2 or p0 = pB + ρv + 1.5 ρg 2 1 ∴ pB = 1.01 × 105 − × 900 × (9.9)2 2 − 1.5 × 900 × 9.8 4 Ans. = 4.36 × 10 Pa (c) Applying Bernoulli’s equation at A and C, 1 p0 = pC + ρv 2 − ρg (1.0) 2 1 ∴ pC = p0 + ρg − ρv 2 2 1 = 1.01 × 105 + 900 × 9.8 – × 900 × (9.9)2 2 4 Ans. = 6.6 × 10 Pa 1 2 mg 8. ρv = ρgh + 2 A Here, h = 1.0 – 0.5 = 0.5 m, A = Area of piston = 0.5 m 2 v = 2gh +

∴

2mg ρA

= 2 × 9.8 × 0.5 +

2 × 20 × 9.8 103 × 0.5

= 3.25 m/s ∴ Speed with which it hits the surface is v′ = v 2 + 2gh′ = (3.25)2 + (2 × 9.8 × 0.5) = 4.51 m /s

9. a 2gy = πx 2  − 

Here, −

Fluid Mechanics — 645

dy  dt 

dy 4 × 10−2 = = 1.11 × 10–5 m/s dt 3600

Ans. …(i)

…(i)

In the second case, kx

mg

F

mg + 2kx

F = mg + 2kx F − mg or kx = 2 From Eqs. (i) and (ii), we have F − mg mg = 2 or F = 3 mg

…(ii)

Let V be the total volume then, ∴

Vρ w g = 3mg 3mg (3)(8) 3 m V = = ρw g 103

= 0.024 m 3 mass Volume of wood = density 8 = = 0.0095 840 ∴Volume of cavity = 0.024 – 0.0095 = 0.0145 ∴ Percentage volume of cavity 0.0145 = × 100 0.024 = 60.41% 11. v = 2g (10 − h)

…(i)

Component of its velocity parallel to the plane is v cos 30°. Let the stream strikes the plane after time t. Then 0 = v cos 30° − g sin 30° t v cot 30° t= ∴ g

646 — Mechanics - II x = vt =

Further

v 2 cot 30° g

(b) The motion of the ball is periodic but not simple harmonic because the acceleration of the ball is  d − d g in air and  L  g inside the liquid which is  d 

= 3y or ∴

v 2 cot 30° 1   = 3  h − gt 2  g 2   3v 2 g v 2 cot 2 30° = 3 h −  g 2 g2   v2 3 v2 =h− g 2 g

or ∴

not proportional to the displacement, which is necessary and sufficient condition for SHM. (c) When dL = d , retardation or acceleration inside

5 v2 = h or 2 g

∴

5(10 − h) = h

h = 8.33 m

Ans.

12. In elastic collision with the surface, direction of velocity is reversed but its magnitude remains the same. Therefore, time of fall = time of rise. t or time of fall = 1 2 Hence, velocity of the ball just before it collides with liquid is t …(i) v=g 1 2 Retardation inside the liquid upthrust − weight VdLg − Vdg a= = mass Vd  dL − d  …(ii) = g  d  Time taken to come to rest under this retardation will be v gt gt1 t= = 1= a 2a  dL − d  2 g  d  =

14. (a) Fr = 6πηrv = 6π (0.8)(10−3 )(10−2 ) = 1.5 × 10−4 N

Ans.

(b) Hydrostatic force = Upthrust 4 4  =  πr3 ρg = π (10−3 )3 × 1260 × 9.8 3  3 = 5.2 × 10−5 N

Ans.

(c) At terminal velocity w = Upthrust + viscous force or (50 × 10−3 × 9.8) = (5.2 × 10−5 ) + 6π (0.8)(10−3 ) vT Solving we get vT = 32.5 m/s

Ans.

15. The loop will take circular shape after pricking.

dt1 2(dL − d )

Radius of which is given by the relation.

Same will be the time to come back on the liquid surface. Therefore, (a) t2 = time the ball takes to came back to the position from where it was released dt1 = t1 + 2t = t1 + dL − d  d  = t1 1 +  or d  L − d

the liquid becomes zero (upthrust = weight). Therefore, the ball will continue to move with gt constant velocity v = 1 inside the liquid. 2 2T 13. Pressure inside the bubble = p0 + ρgh + r ∴ Amount of pressure inside the bubble greater 2T than the atmospheric pressure = ρgh + r Substituting the values we get, 2 × 0.075 ∆p = 103 × 9.8 × 0.1 + 10−3 = 980 + 150 = 1130 N/m 2

td t2 = 1 L dL − d

dθ

P

Q

dθ T

T dθ dθ

l = 2πR or

R=

l 6.28 = = 1 cm 2π 2 × 3.14

= 10−2 m

Fluid Mechanics — 647

Chapter 16 2T sin (dθ ) force in inward direction is balanced by surface tension force in outward direction. ∴ 2T sin (dθ ) = (Surface tension) × (length of arc) For small angles, sin dθ ≈ dθ ∴ 2T dθ = S (2Rd θ ) (S = Surface tension) ∴ T = SR

18.

F 1 2 = ρv A 2 1 …(i) or F = ρAv 2 2 Here, v is the velocity of liquid, with which it comes out of the hole.

= (0.030)(10−2 ) = 3.0 × 10

−4

N

A

Ans.

v F

16. (a) Time taken to empty the tank (has been derived in theory) is t=

2A a 2g

x

V = Ax V t= sv and w = F ⋅x From the above four equations, 1  V  w =  Aρv 2   2   A

H

Further

A = 400 a Substituting the values we have, 2 × 400 t= 1 2 × 9.8 Given,

= 180 s = 3 min

Ans.

=

(b) Rate of flow of water Q = a 2gH = constant Total volume of water V = AH ∴ Time take to empty the tank with constant rate V AH t= = Q a 2gH 400 × 1 = 2 × 9.8 × 1 = 90 s = 1.5 min

Ans.

17. (a) ∆p = h (ρ w − ρ 0 ) g = (10)(1000 − 500) 9.8 Now, ∴

= 49000 N/m 2 1 ∆p = ρ w v 2 2 2∆p 2 × 49000 v= = ρw 1000 = 9.8 m/s

…(iii) …(iv)

1 V2 1 ρV 3 ρ 2 2 ⋅V = 2 st 2 s2t 2

Ans.

19. (a) p0 A = kx p0 A ( p0 )(πr2 ) = k k (1.01 × 105 )(π )(0.025)2 = 3600 = 0.055 m = 5.5 cm

∴

x=

Ans.

(b) Work done by atmospheric pressure 1 1 W = kx 2 = (3600)(0.055)2 2 2 = 5.445 J −3

−3

Ans.

20. Given, A1 = 4 × 10 m , A2 = 8 × 10 m , 2

2

h1 = 2 m, h2 = 5 m , v1 = 1 m/s and ρ = 103 kg/m 3 A2 v2

Ans. A1

The flow will stop when, (b) (10 + 5)ρ 0g = 5ρ 0g + hρ w g 10ρ 0 = hρ w 10 × 500 =5m 1000 i.e. flow will stop when the water-oil interface is at a height of 5.0 m. Ans.

∴

…(ii)

v1

2 h2

h1

1

h=

From continuity equation, we have A1v1 = A2v2

648 — Mechanics - II or

A  v2 =  1  v1 or  A2 

 4 × 10−3  v2 =   (1 m/s)  8 × 10−3 

1 v2 = m/s 2 Applying Bernoulli’s equation at sections 1 and 2 1 1 p1 + ρv12 + ρgh1 = p2 + ρv22 + ρgh2 2 2 or

p1 − p2 = ρg (h2 − h1 ) +

1 2 ρ (v2 − v12 ) 2

…(i)

(i) Work done per unit volume by the pressure as the fluid flows from P to Q. w1 = p1 − p2 1 [From Eq. (i)] = ρg (h2 − h1 ) + ρ (v22 − v12 ) 2 1  1  = (103 ) (9.8)(5 − 2) + (10)3  − 1  J/m 3 4  2  = [ 29400 − 375 ] J/m 3 = 29025 J/m 3 (ii) Work done per unit volume by the gravity as the fluid flows from P to Q. W 2 = ρg (h2 − h1 ) = {(103 )(9.8)(5 − 2)} J/m 3 or W 2 = 29400 J/m 3

21. Volume of the portion of the plate immersed in

22. Given that r1 =

3.0 = 15 . mm = 15 . × 10−3 m, 2

6.0 = 3.0 mm = 3.0 × 10−3m, 2 T = 7.3 × 10−2 Nm −1, θ = 0° ρ = 1.0 × 103 kg/m 3, g = 9.8 m/s2 r2 =

When angle of contact is zero degree, the radius of the meniscus equals radius of bore. 2T Excess pressure in the first bore, p1 = r1 =

2 × 7.3 × 10−2 = 97.3 Pa 15 . × 10−3

Excess pressure in the second bore, p2 = =

2 × 7.3 × 10−2 = 48.7 Pa 3 × 10−3

Hence, pressure difference in the two limbs of the tube ∆p = p1 − p2 = hρg p − p2 or h= 1 ρg 97.3 − 48.7 = 10 . × 103 × 9.8 ≈ 5 × 10−3 m

water is 1 (1.54 ) × 0.2 = 1.54 cm 3 2 Therefore, if the density of water is taken as 1, then upthrust = weight of the water displaced = 1.54 × 1 × 980 = 1509.2 dynes Now, the total length of the plate in contact with the water surface is 2(10 + 0.2) = 20.4 cm ∴Downward pull upon the plate due to surface tension = 20.4 × 73 = 1489.2 dynes ∴Resultant upthrust = 1509.2 − 1489.2 20 = 20.0 dynes = 980 10 ×

= 0.0204 gm-wt ∴Apparent weight of the plate in water = weight of the plate in air − resultant upthrust Ans. = 8.2 − 0.0204 = 8.1796gm

2T r2

= 5.0 mm

23. p1V1 = p2V2 4T   ∴  p0 +   r 

4 3  4T   4  3   πR   2 × πr  =  p0 +     R  3 3

From here we can find expression of T .

24. p1V1 = p2V2 4T   4 3  4T σ 2   4   − ∴  p0 +   πR 3   πr  =  p0 +     r 3 R 2ε 0   3 σR ε0 Vε 0 σ= R

V =

Here, or

25. p1V1 = p2V2 4T   4 3  4T σ 2   4   −   π  (2R )3  p0 +   πR  =  p0 +     R 3 2R 2ε 0   3   Here,

σ=

q q = A 4 π (2R )2

JEE Main and Advanced Previous Years’ Questions (2018-13) JEE Main 1. Seven identical circular planar discs, each of mass M and radius R are welded symmetrically as shown in the figure. The moment of inertia of the arrangement about the axis normal to the plane and passing through the point P is (2018)

Ka 3mg

(c)

mg 3Ka

(d)

mg Ka

(b) 4.70 × 103 N/m 2 (d) 4.70 × 102 N/m 2

5. The moment of inertia of a uniform

2. From a uniform circular disc of radius R and 2R 3 mass 9 M, a small disc of R radius is removed as R 3 shown in the figure. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through centre of disc is (2018) 40 MR 2 (c) 10MR 2 9

(b)

(a) 2.35 × 103 N/m 2 (c) 2.35 × 102 N/m 2

55 (b) MR 2 2 181 (d) MR 2 2

(b)

Ka mg

3.32 × 10 −27 kg. If 1023 hydrogen molecules strike per second, a fixed wall of area 2 cm 2 at an angle of 45° to the normal and rebound elastically with a speed of 103 m/s, then the pressure on the wall is nearly (2018)

O

(a) 4MR 2

(a)

4. The mass of a hydrogen molecule is

P

19 (a) MR 2 2 73 (c) MR 2 2

When a mass m is placed on the surface of the piston to compress the liquid, the fractional decrement in the radius of the  dr  sphere,   is  r  (2018)

(d)

37 MR 2 9

3. A solid sphere of radius r made of a soft material of bulk modulus K is surrounded by a liquid in a cylindrical container. A massless piston of area a floats on the surface of the liquid, covering entire cross-section of cylindrical container.

cylinder of length l and radius R about its perpendicular bisector is I. What is the ratio l/ R such that the moment of inertia is minimum? (2017) (a)

3 2

(b) 1

(c)

3 2

(d)

3 2

6. A slender uniform rod of mass M and length l is pivoted at one end so that it can rotate in a vertical plane (see the figure). There is negligible friction at the pivot. The free end is held vertically above the pivot and then released. The angular acceleration of the rod when it makes an angle θ with the vertical, is (2017) 2g sin θ 3l 2g (c) cosθ 3l (a)

3g cos θ 2l 3g (d) sinθ 2l

(b)

2

Mechanics Vol. 2 7. The variation of acceleration due to gravity g with distance d from centre of the Earth is best represented by (R = Earth’s radius) (2017) g

g

d

d

R

O g

O g

R

O

R

(d)

(c)

d

d O

R

8. A particle is executing simple harmonic motion with a time period T . At time t = 0, it is at its position of equilibrium. The kinetic energy-time graph of the particle will look, like (2017) KE

(a)

KE

O

T

(b)

t

KE

(c)

O

O

T/2

T

t

KE

T/4 T/2

t

T

(d)

O

T/2 T

t

T

9. A magnetic needle of magnetic moment

6.7 × 10−2 Am 2 and moment of inertia 7.5 × 10−6 kg m 2 is performing simple harmonic oscillations in a magnetic field of 0.01 T. Time taken for 10 complete oscillations is (2017)

(a) 8.89 s

(b) 6.98 s

(c) 8.76 s

(d) 6.65s

10. The following observations were taken for determining surface tension T of water by capillary method. Diameter of capillary, d = 1.25 × 10−2 m rise of water, h = 1.45 × 10−2m. Using g = 9.80 m/s 2 and rhg the simplified relation T = × 103 N / m, 2 the possible error in surface tension is closest to (2017) (a) 1.5%

(b) 2.4%

1 9 1 (c) 81

(a)

12. A particle of mass m

(b)

(a)

same, the stress in the leg will change by a factor of (2017)

(c) 10%

(d) 0.15%

11. A man grows into a giant such that his linear dimensions increase by a factor of 9. Assuming that his density remains

(b) 81 (d) 9 y

D

a v

C

v a a v is moving along the v A side of a square of B a side a, with a R uniform speed v in 45º x O the xy-plane as shown in the figure. Which of the following statements is false for the angular momentum L about the origin? (2016)

−mv $ R k when the particle is moving 2 from A to B R (b) L = mv  − a  k$ when the particle is  2  moving from C to D R (c) L = mv  + a  k$ when the particle is  2  moving from B to C mv $ (d) L = R k when the particle is moving from 2 D to A (a) L =

13. A roller is made

B

D

by joining together two O corners at their vertices O. It is kept on two rails C A AB and CD which are placed asymmetrically (see the figure), with its axis perpendicular to CD and its centre O at the centre of line joining AB and CD (see the figure). It is given a light push, so that it starts rolling with its centre O moving parallel to CD in the direction shown. As it moves, the roller will tend to (2016) (a) turn left (b) turn right (c) go straight (d) turn left and right alternately

3

Previous Years’ Questions (2018-13) 14. A satellite is revolving in a circular orbit at a height h from the Earth’s surface (radius of earth R , h < < R ). The minimum increase in its orbital velocity required, so that the satellite could escape from the Earth’s gravitational field, is close to (Neglect the effect of atmosphere) (2016) (b) gR (d) gR ( 2 − 1)

(a) 2gR (c) gR / 2

15. A student measures the time period of 100 oscillations of a simple pendulum four times. The data set is 90 s, 91s, 92s and 95s. If the minimum division in the measuring clock is 1s, then the reported mean time should be (2016) (a) (92 ± 2 s) (c) (92 ± 1.8s)

(b) (92 ± 5 s) (d) (92 ± 3 s)

16. A particle performs simple harmonic motion with amplitude A. Its speed is trebled at the instant that it is at a 2 distance A from equilibrium position. 3 The new amplitude of the motion is (2016) (a)

A 3

41

(b) 3A

(c) A 3

(d)

uniform cone from its vertex is z 0. If the radius of its base is R and its height is h, then z 0 is equal to (2015) 3h 4

(b)

h2 4R

(c)

5h 8

(d)

3h 2 8R

18. A particle of mass m moving in the x-direction with speed 2v is hit by another particle of mass 2 m moving in the y-direction with speed v. If the collision is perfectly inelastic, the percentage loss in the energy during the collision is close to (2015)

(a) 50 %

(b) 56 %

(c) 62 %

(d) 44 %

19. From a solid sphere of mass M and radius R, a cube of maximum possible volume is cut. Moment of inertia of cube about an axis passing through its centre and perpendicular to one of its faces is (2015) (a)

MR 2 4MR 2 (b) 32 2 π 9 3π

(c)

mass M and radius R, a spherical portion of radius  R   is removed as shown  2 in the figure. Taking gravitational potential V = 0 at r = ∞, the potential at the centre of the cavity thus formed is (G = gravitational constant) (2015) − GM R − 2 GM (c) 3R

MR 2 4MR 2 (d) 16 2 π 3 3π

− GM 2R − 2 GM (d) R

(a)

(b)

21. For a simple pendulum, a graph is plotted between its Kinetic Energy (KE) and Potential Energy (PE) against its displacement d. Which one of the following represents these correctly? (graphs are schematic and not drawn to scale) (2015) E

E PE

(a)

7 A 3

17. Distance of the centre of mass of a solid

(a)

20. From a solid sphere of

KE E

PE

d E

KE

(c)

KE

(b)

d

(d)

d

PE KE

PE

22. A pendulum made of a uniform wire of cross-sectional area A has time period T. When an additional mass M is added to its bob, the time period changes to TM . If the Young’s modulus of the material of the wire is Y, then 1/Y is equal to (2015) (g = gravitational acceleration) 2  T   A (a) 1 −     TM   Mg   T 2  Mg (b)   M  − 1    T  A 2   A T (c) 1 −  M    T   Mg   T 2  A (d)   M  − 1   T   Mg

4

Mechanics Vol. 2

23. A bob of mass m attached to an inextensible string of length l is suspended from a vertical support. The bob rotates in a horizontal circle with an angular speed ω rad/s about the vertical support. About the point of suspension (2014) (a) angular momentum is conserved (b) angular momentum changes in magnitude but not in direction (c) angular momentum changes in direction but not in magnitude (d) angular momentum changes both in direction and magnitude

24. A mass m supported by a massless string wound R m around a uniform hollow cylinder of mass m and radius R. If the string does m not slip on the cylinder, with what acceleration will the mass fall on release? (2014) (a) 2g / 3 (c) 5g / 6

25. Four particles, each of mass M and equidistant from each other, move along a circle of radius R under the action of their mutual gravitational attraction, the speed of each particle is (2014) GM R

(b)

(c)

GM (1 + 2 2 ) R

(d)

2 2

GM R

1 GM (1 + 2 2 ) 2 R

26. A particle moves with simple harmonic motion in a straight line. In first τ sec, after starting from rest it travels a distance a and in next τ sec, it travels 2a, in same direction, then (2014) (a) (b) (c) (d)

(2014)

1+ 1− 1+ (c) 1−

(a)

amplitude of motion is 3a time period of oscillations is 8 π amplitude of motion is 4a time period of oscillations is 6π

27. There is a circular tube in a vertical plane. Two liquids which do not mix and of densities d1 and d2 are filled in the

sinα sinα tanα tanα

α

d2

d1

1 + cosα 1 − cosα 1 + sinα (d) 1 − cosα (b)

28. On heating water, bubbles beings formed at the bottom of the vessel detach and rise. Take the bubbles to be R spheres of radius R and making a circular contact of 2r radius r with the bottom of the vessel. If r ∆t y ≠ 0 for 0 < t < ∆t Let us represent this function by f ( t ). Thus, y ( x = 0, t ) = f ( t ) Here, x = 0 is the extreme left end of the string. The disturbance travels on the string towards right with a constant speed v. Thus, the displacement x produced at the left end at time t, reaches the point x at time t + . v x But the displacement of the particle at point x at time t was originated at x = 0 at time t − . v x x   y ( x, t ) = y  x = 0, t −  = f  t −    v v

∴

 x y ( x, t ) = f  t −   v

Thus,

…(iii)

Eq. (iii) represents a wave travelling in the positive x -direction with a constant speed v. The time t and x the position x must appear in the wave equation in the combination t − . If the wave travels in the v negative x-direction, its general equation may be written as  x y ( x, t ) = f  t +   v Note

In the wave equation,  x y = (x , t ) = f t −   v 1 Coefficient of t is 1 and coefficient of x is . v coefficient of t 1 wave speed = = =v ∴ coefficient of x ( 1 / v )

6 — Waves and Thermodynamics V

Example 17.1 In a wave motion y = a sin ( kx – ωt), y can represent: (a) electric field (c) displacement Solution

(JEE 1999)

(b) magnetic field (d) pressure

( a , b , c, d )

In case of sound wave, y can represent pressure and displacement, while in the case of electromagnetic wave, it represents electric and magnetic fields. V

Example 17.2 Show that the equation, y = a sin (ωt – kx ) satisfies the wave ∂2 y ∂2 y equation 2 = v2 . Find speed of wave and the direction in which it is ∂t ∂x 2 travelling. Solution

∂2 y

= – ω 2 a sin (ωt – kx ) and

∂t 2 We can write these two equations as

∂2 y

∂2 y ∂x 2

= – k 2 a sin (ωt – kx )

ω 2 ∂2 y ⋅ ∂t 2 k 2 ∂x 2 ∂2 y ∂2 y = v2 2 2 ∂t ∂x

Comparing this with,

=

ω Ans. k The negative sign between ωt and kx implies that wave is travelling along positive x-direction.

We get wave speed v =

INTRODUCTORY EXERCISE

17.1

1. Prove that the equation y = a sin ωt does not satisfy the wave equation and hence it does not represent a wave. 2

2. A wave pulse is described by y ( x , t ) = ae – ( bx – ct ) , where a, b and c are positive constants. What is the speed of this wave?

3. You have learnt that a travelling wave in one dimension is represented by a function y = f ( x , t ) where x and t must appear in the combination ax ± bt or x − vt or x + vt , i.e. y = f ( x ± vt ).Is the converse true? Examine if the following functions for y can possibly represent a travelling wave (a) ( x − vt )2

(b) log [( x + vt )/ x 0 ]

(c) 1/( x + vt )

4. The equation of a wave travelling on a string stretched along the X -axis is given by y = Ae (a) (b) (c) (d)

t  x − +  a T

2

Write the dimensions of A, a and T. Find the wave speed. In which direction is the wave travelling? Where is the maximum of the pulse located at t = T and at t = 2T ?

Chapter 17

Wave Motion — 7

17.4 Sine Wave If oscillations of yare simple harmonic in nature, then y ( x, t ) function is sine or cosine function. Such type of wave is called sine wave or sinusoidal wave. For better understanding, we can visualize a sine wave travelling on a string. The general expression of a sine wave is y = A sin (ωt ± kx ± φ )

y = A cos (ωt ± kx ± φ )

and

In the above equations, (i) A is the amplitude of oscillations of y. (ii) ω is called the angular frequency. Here, 2π T

ω = 2π f =

and

f =

1 T

where, f is normal frequency of oscillations and T is the time period of oscillations. SI unit of ω is radian/sec while that of f is Hz or s −1 . (iii) k is called the angular wave number, where k=

2π λ

Here, λ is the wavelength of wave.

Wavelength ( λ ) In a transverse wave motion, the particles of the medium oscillate about their mean or equilibrium position at right angles to the direction of propagation of wave motion itself. Crest

λ

Oscillation Wave motion λ

Trough

Mean position

Fig. 17.3

This form of wave motion travels in the form of crests and troughs, for example, waves travelling along a stretched string. The distance between two successive crests or troughs is known as wavelength ( λ ) of the wave. In a longitudinal wave motion, particles of the medium oscillate about their mean or equilibrium position along the direction of propagation of the wave motion itself. This type of wave motion travels in the form of compressions and rarefactions. Mean position

Wave motion

Oscillations λ

λ

Fig. 17.4

The distance between two successive compressions or rarefactions constitute one wavelength.

8 — Waves and Thermodynamics Note

2π or angular wave number is defined as the number of waves in 2π length. Similarly, λ 1 there is one another word called wave number. This is equal to and numerically this is equal to number of λ waves in unit length. Numerically, k =

(iv) Wave speed v, = ∴

coefficient of t coefficient of x ω v= k ω = 2π f 2π k= λ

But and

Substituting in the above equation, we also get v= fλ (v)

∂y is SHM velocity of the particle executing oscillations (lying between +ωA and −ωA). ∂t ∂2 y Similarly, 2 is SHM acceleration (lying between +ω 2 A and −ω 2 A). ∂t Let us take an example: suppose a sine wave travelling on a string is

Then, and

y = A sin(ωt − kx + φ )

…(i)

∂y = ωA cos (ωt − kx + φ ) ∂t

…(ii)

∂2 y ∂τ 2

= − ω 2 A sin (ωt − kx + φ )

…(iii)

From Eq. (i), we can determine y-displacement of SHM of any particle at position x and at time t. Value of y lies between + A and −A. ∂y ∂2 y From Eq. (ii), we can determine or SHM velocity and from Eq. (iii), we can find 2 or SHM ∂t ∂t acceleration. Angle (ωt − kx + φ ) is same in all three equations which is also called phase angle of three ∂y ∂2 y functions. With the help of this phase angle, we can determine values of y, and 2 at any ∂t ∂t position x at time t. If we put x = 0 and t = 0, then this angle is only φ. This is called initial phase angle of the particle at co-ordinate x = 0. With the help of this angle, we can find the initial values ∂y ∂2 y of y, and 2 of this string particle (which is executing SHM at position x = 0). ∂t ∂t

Chapter 17

Wave Motion — 9

Extra Points to Remember ˜

˜

Alternate expressions of a sine wave travelling along positive x -direction are 2π x t ( x − vt ) = A sin 2 π  −  y = A sin k ( x − vt ) = A sin (kx − ωt ) = A sin  λ T λ x t Similarly, the expression, y = A sin k ( x + vt ) = A sin (kx + ωt ) = A sin 2 π  +  etc.  λ T represent a sine wave travelling along negative x-direction. Difference between two equations y = A sin(kx − ωt ) and y = A sin(ωt − kx) It hardly matters whether we write the first or the second equation. Both the equations represent a wave ω travelling in positive x-direction with speed v = ⋅ The difference between them is that they are out of k phase, i.e. phase difference between them is π. It means, if a particle in position x = 0 at time t = 0 is in its mean position and moving upwards (represented by first wave) then the same particle will be in its mean position but moving downwards (represented by the second wave). Similarly, the waves y = A sin (kx – ωt ) and y = – A sin (kx – ωt ) are also out of phase.

V

Example 17.3

A wave travelling along a string is described by y ( x , t) = 0.005 sin ( 80.0 x − 3.0 t). in which the numerical constants are in SI units (0.005 m, 80.0 rad m −1 and 3.0 rad s −1 ). Calculate (a) the amplitude. (b) the wavelength (c) the period and frequency of the wave. Also, calculate the displacement y of the wave at a (NCERT Solved Example) distance x = 30.0 cm and time t = 20 s? Solution

On comparing the given equation with

y ( x, t ) = A sin ( kx − ωt ), we find (a) the amplitude A = 0.005 m = 5 mm. (b) the angular wave number k and angular frequency ω are k = 80.0 m −1 ∴

λ = 2π / k =

and ω = 3.0 s −1

2π = 0.0785 m 80.0

= 7.85 cm 2π = 2.09 s 3.0 and frequency, f = 1/ T = 0.48 Hz

(c) T = 2π / ω =

Ans. Ans. Ans.

The displacement y at x = 30.0 cm and time t = 20 s is given by y = ( 0.005 m ) sin ( 80.0 × 0.3 − 3.0 × 20) = ( 0.005 m ) sin ( −36) = ( 0.005 m ) sin ( −36 + 12π ) = ( 0.005 m ) sin (1.714 ) = ( 0.005 m ) sin ( 98.27° ) = 4.94 mm

Ans.

10 — Waves and Thermodynamics V

Example 17.4 The equation of a wave is π π y ( x , t) = 0.05 sin  ( 10 x – 40 t) –  m 2 4   Find : (a) the wavelength, the frequency and the wave velocity (b) the particle velocity and acceleration at x = 0.5 m and t = 0.05 s. Solution (a) The equation may be rewritten as π  y ( x, t ) = 0.05 sin  5πx – 20πt –  m  4 Comparing this with equation of plane progressive harmonic wave, y ( x, t ) = A sin ( kx – ωt + φ ) we have, 2π Wave number, k= = 5π rad / m λ ∴ λ = 0.4 m

Ans.

The angular frequency is ω = 2π f = 20π rad /s f = 10 Hz

∴

Ans.

The wave velocity is, ω = 4 m/s in + x-direction k (b) The particle velocity and particle acceleration at the given values of x and t are ∂y π  5π = – ( 20π ) ( 0.05) cos  –π–   2 ∂t 4 v= f λ =

= 2.22 m/s ∂ y 2

∂t

2

Ans.

Ans.

π  5π = – ( 20π ) 2 ( 0.05) sin  –π–   2 4 = 140 m/s 2

INTRODUCTORY EXERCISE

Ans.

17.2

−1

1. Consider the wave y = (5 mm ) sin [1cm x − (60 s −1) t ] . Find (a) the amplitude, (b) the angular wave number, (c) the wavelength, (d) the frequency, (e) the time period and (f) the wave velocity. x t  2. A wave is described by the equation y = (1.0 mm )sin π  − .  2.0 cm 0.01 s  (a) (b) (c) (d)

Find time period and wavelength. Find the speed of particle at x = 1.0 cm and time t = 0.01 s. What are the speeds of the particles at x = 3.0 cm, 5.0 cm and 7.0 cm at t = 0.01 s? What are the speeds of the particles at x = 1.0 cm at t = 0.011,0.012 and 0.013 s?

Chapter 17

Wave Motion — 11

17.5 Two Graphs in Sine Wave As we have learned in the above article that y ( x, t ) equation of a sine wave is either sine or cosine equation. Now, corresponding to this equation we can have two graphs and two simple equations.

First Graph In y ( x, t ) equation, if value of t is fixed (or substituted), then the equation left is y ( x ) equation.So, we can plot y - x graph corresponding to this equation. And obviously the graph (or the equation) will be a sine or cosine graph. For example Suppose a sine wave is travelling along positive x -direction on a string. At a given time (say at 9 AM), the y - x graph may be as shown in the figure. y(mm)

+10 +4 2 a

c

b 6

d

e

x(m)

–5 –10

Fig. 17.5

The important points in the above graph are (i) amplitude of oscillation is 10 mm. (ii) at 9 AM, y -displacement of the string particle at x = 2 m is −5 mm and of the particle at x = 6 m is + 4 mm. Or, we can say that this graph represents SHM y -displacements of different string particles (at different x -coordinates) at 9 AM. Hence, this is a photograph (or snapshot) of the string at 9 AM. ∂y  dy  (iii) Slope of this graph at any point is  not  , as y has two variables x and t. ∂x  dx  (iv) Two string particles at different locations are in different phases. The phase difference between them is given by  2π  ∆φ =   ( ∆x )  λ Here, ∆x is the path difference between them. For example

λ . So, phase difference between them is π. 2 Particle ‘a’ and ‘c’ have a path difference of λ. So, phase difference is 2π. λ π Similarly, particles ‘c’ and ‘d’ have a path difference of . Therefore, the phase difference is . 4 2 (v) From the above graph, we cannot determine the direction of wave velocity. For that, wave equation will be required. Particles ‘a’ and ‘b’ have a path difference of

12 — Waves and Thermodynamics Second Graph In y ( x, t ) equation, if the value of x is fixed (or substituted) then the equation left is y ( t ) equation. So, we can plot y - t graph corresponding to this equation. Again, the graph (or the equation) will be a sine or cosine graph. For example Suppose a sine wave is travelling along positive x -direction on a string. At a given position (say x = 4 m), the y - t graph may be as shown in the figure. y (mm) +6 +4 t4

8 4

t1

t2

t3

t (s)

–5 –6

Fig. 17.6

The important points in this graph are (i) amplitude of oscillations is 6 mm. (ii) y -displacement of the particle at x = 4 m is 4 mm at 4 s. Similarly, y -displacement of the particle at x = 4 m is −5 mm at 8 s or we can say that this graph is basically SHM y - t graph of the string particle at x = 4 m, which shows different SHM y -displacements of this particle at different times or it is videography of SHM oscillations of this particle. ∂y (iii) Slope of this particle at any time is and this is SHM velocity ( = ± ω A 2 − y 2 ) of the particle ∂t at x = 4 m at that time. (iv) The same particle at two different times will have different phase angles. The phase difference in a time interval of ∆t is given by  2π  ∆φ =   ∆t T  For example, T ⋅ So, phase difference is π. 2 T π t 2 and t 3 have a time interval of ⋅ So, the phase difference is . 4 2 Similarly, t1 and t 4 have a time interval of T. Therefore, phase difference is 2π. t1 and t 2 have a time interval of

Chapter 17

Wave Motion — 13

Wave velocity ( v ), particle velocity ( v p ) and particle acceleration (a p ) in a Sinusoidal wave Wave velocity (v) is the velocity by which oscillations of y (or energy) transfer from one point to another point. Later, we will see that this velocity depends on characteristics of medium. Particle velocity ( v P ) and particle acceleration (a P ) are different. In a sinusoidal wave, particles of the medium oscillate simple harmonically about their mean position. Therefore, all the formulae we have read in SHM apply to the particles here also. For example, maximum particle velocity is ± Aω at mean position and it is zero at extreme positions etc. Similarly, maximum particle acceleration is ± ω 2 A at extreme positions and zero at mean position. However, the wave velocity is different from the particle velocity. This depends on certain characteristics of the medium. Unlike the particle velocity which oscillates simple harmonically (between + Aω and – Aω) the wave velocity is constant for given characteristics of the medium. Suppose, a sine wave travelling along positive x- axis is …(i) y ( x, t ) = A sin ( kx – ωt ) Let us differentiate this function partially with respect to t and x. ∂y (x, t ) …(ii) = – Aω cos ( kx – ωt ) ∂t ∂y (x, t ) …(iii) = Ak cos ( kx – ωt ) ∂x Now, these can be written as ∂y ( x, t )  ω  ∂y ( x, t ) =–   k  ∂x ∂t ∂y ( x, t ) = particle velocity v P ∂t ω = wave velocity v k ∂y ( x, t ) = slope of the wave ∂x

Here,

and

v P = – v (slope )

Thus,

…(iv)

i.e. particle velocity at a given position and time is equal to negative of the product of wave velocity with slope of the wave at that point at that instant. Acceleration of the particle is the second partial derivative of y ( x, t ) with respect to t, ∴

aP =

∂ 2 y ( x, t ) ∂t 2

= – ω 2 A sin ( kx – ωt ) = – ω 2 y( x, t )

i.e. acceleration of the particle equals – ω 2 times its displacement, which is the same result we obtained in SHM. Thus, a P = – ω 2 (displacement )

…(v)

14 — Waves and Thermodynamics We can also show that, ∂ 2 y ( x, t ) ∂t 2 ∂ 2 y ( x, t )

or

∂t 2

 ω 2  ∂ 2 y ( x, t ) = 2⋅ k  ∂x 2 = v2

∂ 2 y ( x, t )

…(vi)

∂x 2

which is also the wave equation. y v

vP

1

2 vP,aP

aP

x

Fig. 17.7

Fig. 17.7 shows velocity ( v P ) and acceleration ( a P ) given by Eqs. (iv) and (v) for two points 1 and 2 on a string. The sinusoidal wave is travelling along positive x-direction. At 1 Slope of the curve is positive. Hence, from Eq. (iv) particle velocity ( v P ) is negative or downwards. Similarly, displacement of the particle is positive, so from Eq. (v) acceleration will also be negative or downwards. At 2 Slope is negative while displacement is positive. Hence, v P will be positive (upwards) and a P is negative (downwards). Note

Direction of v P will change if the wave travels along negative x-direction. V

Example 17.5 Under what condition, maximum particle velocity is four times the wave velocity corresponding to the equation, y = A sin( ωt − kx ) Solution

Maximum particle velocity is ωA and wave velocity is

Now, given that maximum particle velocity = 4 (wave velocity) or ∴ So, this is the required condition.

ω ωA = 4    k A=

4 k

ω . k

Chapter 17 V

Example 17.6 A transverse sinusoidal wave moves along a string in the positive x-direction at a speed of 10 cm/s. The wavelength of the wave is 0.5 m and its amplitude is 10 cm. At a particular time t, the snapshot of the wave is shown in figure. The velocity of point P when its displacement is 5 cm is

Wave Motion — 15

y

P x

(JEE 2008)

3π ^ 3π ^ Fig. 17.8 (b) − j m/s j m/s 50 50 3π ^ 3π ^ (c) (d) − i m/s i m/s 50 50 Solution Particle velocity v P = − v (slope of y-x graph) Here, v = + ve, as the wave is travelling in positive x-direction.Slope at P is negative. (a)

^

∴ Velocity of particle is in positive y (or j ) direction. ∴ Correct option is (a). Note We can also find the magnitude of particle velocity by the relation. vP = ω A2 − y 2 v  0.1  2 π Here, A = 10 cm = 0.1 m, y = 5 cm = 0.05 m and ω = 2 πf = 2 π   = 2 π   = rad s−1  λ  0.5  5 V

Example 17.7 Equation of a transverse wave travelling in a rope is given by y = 5 sin ( 4.0 t − 0.02 x ) where y and x are expressed in cm and time in seconds. Calculate (a) the amplitude, frequency, velocity and wavelength of the wave. (b) the maximum transverse speed and acceleration of a particle in the rope.

Solution (a) Comparing this with the standard equation of wave motion, 2π   y = A sin (ωt − kx ) = A sin  2πft − x  λ  where A, f and λ are amplitude, frequency and wavelength respectively. Thus, amplitude A = 5 cm 4 = 0.637 Hz ⇒ 2πf = 4 ⇒ Frequency, f = 2π 2π 2π Again = 0.02 or Wavelength, λ = = (100 π ) cm λ 0.02 4 2π Velocity of the wave, v = fλ = = 200 cm/s 2π 0.02 (b) Transverse velocity of the particle, ∂y vP = = 5 × 4 cos ( 4.0 t − 0.02 x ) = 20 cos ( 4.0 t − 0.02 x ) ∂t

Ans.

16 — Waves and Thermodynamics Maximum velocity of the particle = 20 cm/s ∂2 y Particle acceleration, a P = 2 = − 20 × 4 sin ( 4.0 t − 0.02 x ) ∂t Maximum particle acceleration = 80 cm/s 2 V

Example 17.8 In the above example, find phase difference ∆φ. (a) of same particle at two different times with a time interval of 1 s (b) of two different particles located at a distance of 10 cm at same time 2π  2π  Solution (a) ∆φ =   ∆t, but = ω = 4.0 rad /s T T ∴

∆φ = ω ∆t = ( 4.0) (1.0) = 4 rad

 2π  (b) ∆φ =   ∆x, but λ ∴

Ans.

2π = k = 0.02 cm −1 λ ∆φ = ( 0.02) (10) = 0.2 rad

INTRODUCTORY EXERCISE

Ans.

17.3

1. The equation of a wave travelling on a string is y = (0.10 mm )sin [(31.4 m −1) x + (314 s −1) t ] (a) In which direction does the wave travel? (b) Find the wave speed, the wavelength and the frequency of the wave. (c) What is the maximum displacement and the maximum speed of a portion of the string?

2. The equation for a wave travelling in x -direction on a string is y = (3.0 cm )sin [(3.14 cm −1) x − (314 s −1) t ]. (a) Find the maximum velocity of a particle of the string. (b) Find the acceleration of a particle at x = 6.0 cm at time t = 0.11 s

3. The equation of a travelling wave is t   x y ( x , t ) = 0.02 sin  + m  0.05 0.01 Find (a) the wave velocity and (b) the particle velocity at x = 0.2 m and t = 0.3 s. Given cos θ = − 0.85, where θ = 34 rad

4. A wave of frequency 500 Hz has a wave velocity of 350 m /s. (a) Find the distance between two points which are 60° out of phase. (b) Find the phase difference between two displacements at a certain point at time 10−3 s apart.

Chapter 17

Wave Motion — 17

17.6 Wave Speed Wave speed (v) depends on the medium or characteristics of the medium. Waves of all frequencies or all wavelengths travel in a given medium with same speed. For examples, all electromagnetic waves travel in vacuum with same speed ( ≈ 3 × 10 8 m /s ). In water, this speed will be different. Similarly, all longitudinal waves (including the sound wave) travel in air with the same speed ( ≈ 330 m /s ). Now, frequency of a wave depends on source and wavelength is self adjusted in a value, v λ= f In the above expression, v is same (obviously in a given medium) for all sources (or frequencies). So, with increase in value of f , wavelength λ automatically decreases. For example, usually frequency of female voice is more than frequency of male voice. So, wavelength of female voice will be less. V

Example 17.9 Speed of sound in air is 330 m/ s. Find maximum and minimum wavelength of audible sound in air. Solution Frequency of audible sound varies from 20 Hz to 20000 Hz. Speed of all frequencies in air will be same ( = 330 m/s ). v 330 ∴ λ min = = f max 20000 = 0.0165 m

Ans.

Similarly, λ max =

v 330 = f min 20

= 16.5 m

INTRODUCTORY EXERCISE

Ans.

17.4

1. Speed of light in vacuum is 3 × 108m /s. Range of wavelength of visible light is 4000 Å - 7000 Å. Find the range of frequency of visible light.

2. Speed of sound in air is 330 m/s. Frequency of Anoop's voice is 1000 Hz and of Shubham's voice is 2000 Hz. Find the wavelength corresponding to their voice.

Speed of Different Waves Normally, two wave speeds are required at this stage. (i) Transverse wave speed on a string. (ii) Longitudinal wave speed in all three states: solid, liquid and gas.

18 — Waves and Thermodynamics Transverse Wave Speed on a String Speed of transverse wave on a string is given by v=

T µ

Here, µ = mass per unit length of the string m mA = l lA  m =  A V 

=

(A = area of cross-section of the string) (V = volume of string)

= ρA Hence, the above expression can also be written as v=

(ρ = density of string)

T ρA

Proof Consider a pulse travelling along a string with a speed v to the right. If the amplitude of the pulse is small compared to the length of the string, the tension T will be approximately constant along the string. In the reference frame moving with speed v to the right, the pulse is stationary and the string moves with a speed v to the left. ∆l

v ar =

∆l

v

v2 R

θ

θ ΣFr

T

T

R R

θ θ

O

O (a) (b) Fig. 17.9 (a) To obtain the speed v of a wave on a stretched string, it is convenient to describe the motion of a small segment of the string in a moving frame of reference. (b) In the moving frame of reference, the small segment of length ∆l moves to the left with speed v. The net force on the segment is in the radial direction because the horizontal components of the tension forces are cancelled.

Figure shows a small segment of the string of length ∆ l. This segment forms a part of a circular arc of radius R. Instantaneously, the segment is moving with speed v in a circular path, so it has a centripetal v2 acceleration . The forces acting on the segment are the tension T at each end. The horizontal R components of these forces are equal and opposite and thus cancel. The vertical components of these forces point radially inward toward the centre of the circular arc. These radial forces provide the centripetal acceleration. Let the angle subtended by the segment at centre be 2θ. The net radial force acting on the segment is ΣFr = 2T sin θ = 2T θ

Chapter 17

Wave Motion — 19

Where we have used the approximation sin θ ≈ θ for small θ. If µ is the mass per unit length of the string, the mass of the segment of length ∆ l is m = µ∆l = 2 µRθ  v2  mv 2 or 2T θ = (2 µRθ )   From Newton’s second law, ΣFr = ma = R R ∴

v=

(as ∆l = 2Rθ)

T µ

Longitudinal Wave Speed in Three States Speed of longitudinal wave through a gas (or a liquid) is given by B ρ

v=

Here, B = Bulk modulus of the gas (or liquid) and ρ = density of the gas (or liquid) Now Newton, who first deduced this relation for v, assumed that during the passage of a sound wave through a gas (or air), the temperature of the gas remains constant, i.e. sound wave travels under isothermal conditions and hence took B to be the isothermal elasticity of the gas and which is equal to its pressure p. So, Newton’s formula for the velocity of a sound wave (or a longitudinal wave) in a gaseous medium becomes v=

p ρ

If, however, we calculate the velocity of sound in air at NTP with the help of this formula by substituting. p = 1.01 × 10 5 N/ m 2 and ρ = 1.29 × 10 −3 kg/ m 3 then v comes out to be nearly 280 m/s. Actually, the velocity of sound in air at NTP as measured by Newton himself, is found to be 332 m/s. Newton could not explain this large discrepancy between his theoretical and experimental results. Laplace after 140 years correctly argued that a sound wave passes through a gas (or air) very rapidly. So, adiabatic conditions are developed. So, he took B to be the adiabatic elasticity of the gas, which is equal to γ p where γ is the ratio of Cp (molar heat capacity at constant pressure) and CV (molar heat capacity at constant volume). Thus, Newton’s formula as corrected by Laplace becomes v= For air, γ =1.41, so that in air,

v=

γp ρ 1.41 p ρ

which gives 331.6 m/s as the velocity of sound (in air) at NTP which is in agreement with Newton’s experimental result.

20 — Waves and Thermodynamics Speed of longitudinal wave in a thin rod or wire is given by v=

Y ρ

Here, Y is the Young’s modulus of elasticity. Note In the chapter of sound wave (Chapter-19), we will discuss longitudinal wave speed in detail (with proof and examples). In the present chapter, we are taking examples of only transverse wave speed on a string. V

Example 17.10 One end of 12.0 m long rubber tube with a total mass of 0.9 kg is fastened to a fixed support. A cord attached to the other end passes over a pulley and supports an object with a mass of 5.0 kg. The tube is struck a transverse blow at one end. Find the time required for the pulse to reach the other end. ( g = 9.8 m/s 2 ) Tension in the rubber tube AB, T = mg or T = (5.0) (9.8) = 49 N 0.9 Mass per unit length of rubber tube, µ = = 0.075 kg / m 12 T 49 = = 25.56 m/s ∴ Speed of wave on the tube, v = µ 0.075 Solution

∴ The required time is, t = V

AB 12 = = 0.47 s v 25.56

B m

Ans.

A

Fig. 17.10

Example 17.11 A wire of uniform cross-section is stretched between two points 100 cm apart. The wire is fixed at one end and a weight is hung over a pulley at the other end. A weight of 9 kg produces a fundamental frequency of 750 Hz. (a) What is the velocity of the wave in wire? (b) If the weight is reduced to 4 kg, what is the velocity of wave?

Note Fundamental frequency is given by f =

Solution

(a) L = 100 cm,

v 2L

f1 = 750 Hz v1 = 2Lf1 = 2 × 100 × 750 = 150000 cms −1 = 1500 ms −1

v1 =

(b)

T1 µ

and

v2 =

Ans.

T2 µ v2 T2 = v1 T1

∴ ∴

v2 4 = 1500 9 v 2 = 1000 ms −1

Ans.

Chapter 17

INTRODUCTORY EXERCISE

Wave Motion — 21

17.5

1. Figure shows a string of linear mass density 1.0 g cm −1 on which a wave pulse is travelling. Find the time taken by the pulse in travelling through a distance of 50 cm on the string. Take g = 10 ms −2.

1 kg

Fig. 17.11

2. A steel wire of length 64 cm weighs 5 g. If it is stretched by a force of 8 N, what would be the speed of a transverse wave passing on it?

3. Two blocks each having a mass of 3.2 kg are connected by a wire CD and the system is suspended from the ceiling by another wire AB. The linear mass density of the wire AB is 10 gm −1 and that of CD is 8 g m −1. Find the speed of a transverse wave pulse produced in AB and in CD. A B C D

Fig. 17.12

4. In the arrangement shown in figure, the string has a mass of 4.5 g. How much time will it take for a transverse disturbance produced at the floor to reach the pulley? Take g = 10 ms −2.

2.5 cm 2.0 m 2 kg

Fig. 17.13

5. A copper wire2.4 mm in diameter is 3 m long and is used to suspend a 2 kg mass from a beam. If a transverse disturbance is sent along the wire by striking it lightly with a pencil, how fast will the disturbance travel? The density of copper is 8920 kg/m 3.

6. One end of a horizontal rope is attached to a prong of an electrically driven tuning fork that vibrates at 120 Hz. The other end passes over a pulley and supports a 1.50 kg mass. The linear mass density of the rope is 0.0550 kg/m. (a) What is the speed of a transverse wave on the rope? (b) What is the wavelength? (c) How would your answers to parts (a) and (b) change if the mass were increased to 3.00 kg?

22 — Waves and Thermodynamics

17.7 Energy in Wave Motion In a wave, many particles oscillate. In a sinusoidal wave, these oscillations are simple harmonic in nature. Each particle has some energy of oscillation. At the same time, energy transfer also takes place. Related to energy of oscillation and energy transfer, there are three terms, namely, energy density (u), power ( P ) and intensity ( I ).

Energy density (u ) Energy of oscillation per unit volume is called energy density ( u). Its SI unit is J/ m 3 . In case of SHM, 1 energy of oscillation (of a single particle) is E = m ω 2 A 2 . In a sinusoidal wave, each particle of the 2 string oscillates simple harmonically. Therefore, 1 mω 2 A 2 energy of oscillation E 2 or u = = Energy density = volume V V m But, = density or ρ V 1 ∴ u = ρω 2 A 2 2

Power (P) Energy transferred per unit time is called power. The expression of power is 1 P = ρω 2 A 2 Sv 2 Here, S is area of cross-section of the medium in which wave is v travelling and v is the wave speed. SI unit of power is J/s or Watt. M N Now, let us derive the above expression for a sinusoidal travelling wave on a string. Area of cross-section of string is S and the wave speed is v . Suppose at time t =0, wave is at point M. In 1 s, it will travel a t=0 t=1s distance v and it will reach at point N . Or, we can say that v new Fig. 17.14 length of string will start oscillating. Area of cross-section of string is S. Therefore, in 1 s, Sv new volume of string will start oscillating. Energy of oscillation per unit volume is called energy density (u). So, in 1 s, (uSv) new energy will be added or (uSv) energy will be transferred from the source. Energy transferred in one second is called power. ∴ P = uSv 1 1 (as u = ρω 2 A 2 ) = ρω 2 A 2 Sv 2 2 1 2 2 or P = ρω A Sv 2 This is the desired expression of power.

Chapter 17

Wave Motion — 23

Intensity ( I ) Energy transferred per unit cross-sectional area per unit time is called intensity. Thus, Energy transferred Power P I= = = ( time )(cross - sectional area ) cross - sectional area S 1 2 2 ρω A Sv 1 or =2 I = ρω 2 A 2 v 2 S 2 2 The SI unit of intensity is J/s-m or Watt / m .

Extra Points to Remember ˜

˜

˜

Although the above relations for power and intensity have been discussed for a transverse wave on a string, they hold good for other waves also. In SHM, potential energy is maximum (and kinetic energy is zero) at the extreme positions. For a string segment, the potential energy depends on the slope of the y string and is maximum when the slope is maximum, which is at the equilibrium position of the segment, the same position for which the kinetic energy is maximum. At A :

˜

V

Kinetic energy and potential energy both are zero.

A x

B

At B : Kinetic energy and potential energy both are maximum. Intensity due to a point source If a point source emits wave uniformly in Fig. 17.15 all directions, the energy at a distance r from the source is distributed 2 uniformly on a spherical surface of radius r and area S = 4 πr . If P is the power emitted by the source, the P power per unit area at a distance r from the source is . The average power per unit area that is 4 πr 2 incident perpendicular to the direction of propagation is called the intensity. Therefore, P 1 or I ∝ 2 I= 4 πr 2 r

Example 17.12 A stretched string is forced to transmit transverse waves by means of an oscillator coupled to one end. The string has a diameter of 4 mm. The amplitude of the oscillation is 10 −4 m and the frequency is 10 Hz. Tension in the string is 100 N and mass density of wire is 4.2 × 103 kg / m3 . Find (a) (b) (c) (d)

the equation of the waves along the string the energy per unit volume of the wave the average energy flow per unit time across any section of the string and power required to drive the oscillator.

Solution

(a) Speed of transverse wave on the string is v=

T ρS

(as µ = ρS )

24 — Waves and Thermodynamics Substituting the values, we have v=

100 = 43.53 m/s  π ( 4.2 × 103 )   ( 4.0 × 10–3 ) 2  4

ω = 2πf = 20π = 62.83 rad/s ω k = = 1.44 m –1 v ∴ Equation of the waves along the string, y ( x, t ) = A sin ( kx – ωt ) = (10–4 m ) sin [(1.44 m –1 ) x – ( 62.83 rad/s ) t ]

Ans.

1 (b) Energy per unit volume of the string, u = energy density = ρω 2 A 2 2 Substituting the values, we have  1 u =   ( 4.2 × 103 ) ( 62.83) 2 (10–4 ) 2  2 = 8.29 × 10–2 J/ m 3

Ans.

1  (c) Average energy flow per unit time, P = power =  ρω 2 A 2  ( Sv ) = ( u ) ( Sv ) 2   π Substituting the values, we have P = ( 8.29 × 10–2 )   ( 4.0 × 10–3 ) 2 ( 43.53)  4 = 4.53 × 10–5 J/s (d) Therefore, power required to drive the oscillator is 4.53 × 10

INTRODUCTORY EXERCISE

Ans. –5

W.

Ans.

17.6

1. Spherical waves are emitted from a 1.0 W source in an isotropic non-absorbing medium. What is the wave intensity 1.0 m from the source?

2. A line source emits a cylindrical expanding wave. Assuming the medium absorbs no energy, find how the amplitude and intensity of the wave depend on the distance from the source?

3. A certain 120 Hz wave on a string has an amplitude of 0.160 mm. How much energy exists in an 80 g length of the string?

4. A taut string for which µ = 5.00 × 10−2 kg/m is under a tension of 80.0 N. How much power must be supplied to the string to generate sinusoidal waves at a frequency of 60.0 Hz and an amplitude of 6.00 cm?

5. A 200 Hz wave with amplitude 1 mm travels on a long string of linear mass density 6 g/m kept under a tension of 60 N. (a) Find the average power transmitted across a given point on the string. (b) Find the total energy associated with the wave in a 2.0 m long portion of the string. 6. A transverse wave of amplitude 0.50 mm and frequency 100 Hz is produced on a wire stretched to a tension of 100 N . If the wave speed is 100 m /s. What average power is the source transmitting to the wire?

Chapter 17

Wave Motion — 25

List of Formulae 1. Wave Equation Only those functions of x and t or y ( x , t ) represent a wave function which satisfies the following three conditions  ∂ 2y   ∂ 2y   2  =k  2  ∂t   ∂x 

Condition 1

k =v 2

Here,

(v = wave speed)

Condition 2

y ( x , t ) should be a single valued function for all values of x and t.

Condition 3

The wave function and its first derivative must be continuous.

2. If y ( x , t ) is of type f (ax ± bt ), then wave speed v =

coefficient of t coefficient of x

Further, wave travels in positive x-direction if ax and bt are of opposite sign and it travels along negative x -direction if ax and bt are of same sign.

3. Sine wave General equation of this wave is y = A sin(ωt ± kx ± φ ) y = A cos(ωt ± kx ± φ )

or In these equations, (i) A is amplitude of oscillation, (ii) ω is angular frequency,

2π ⇒ ω 1 ω f = = T 2π

T = and

ω = 2 πf

(iii) k is angular wave number, k =

2π λ

(λ → wavelength)

ω =fλ k (v) φ is initial phase angle at x = 0 and (vi) (ωt ± kx ± φ ) is phase angle at time t at coordinate x. (iv) Wave speed, v =

4. Particle velocity ( vP ) and wave velocity ( v ) in sine wave (i) y = f ( x , t ) Then,

vP =

∂y ∂x

(ii) In sine wave, particles are executing SHM. Therefore, all equations of SHM can be applied for particles also. (iii) Relation between vP and v vP = −v ⋅

∂y ∂x

Here, ∂y / ∂x is the slope of y - x graph when t is kept constant.

26 — Waves and Thermodynamics 5. Phase Difference (∆φ) ∆φ = ω (t1 − t 2 ) 2π ∆φ = ⋅ ∆t T

Case I or

= phase difference of one particle at a time interval of ∆t. ∆φ = k ( x1 ~ x 2 ) 2π = ⋅ ∆x λ

Case II

= phase difference at one time between two particles at a path difference of ∆x.

6. Wave Speed (i) Speed of transverse wave on a stretched wire T T = µ ρS

v = (ii) Speed of longitudinal wave v =

E ρ

(a) In solids, E = Y = Young’s modulus of elasticity v =

∴

Y ρ

(b) In liquids, E = B = Bulk modulus of elasticity v =

∴

B ρ

(c) In gases, E = adiabatic bulk modulus = γ p ∴

v =

γp = ρ

γ RT M

7. Energy Density (u), Power (P) and Intensity (I) in sine Wave (i) Energy density, u =

1 ρω 2A 2 = energy of oscillation per unit volume 2

1 ρω 2A 2Sv = energy transferred per unit time 2 1 (iii) Intensity, I = ρω 2A 2v = energy transferred per unit time per unit area 2 (ii) Power,

P=

Solved Examples TYPED PROBLEMS Type 1. Based on symmetry of a wave pulse

Concept To check symmetry of a wave pulse or wave velocity, always concentrate on maximum or minimum value of y. For example, y

y t → Fixed

x0 – ∆x

x0 + ∆x

x0 (i)

x

x

x0 (ii)

Wave pulse shown in Fig. (i) is symmetric but the wave pulse shown in Fig. (ii) is asymmetric. Now, the question is how will you check whether the pulse is symmetric or not. The answer is In both cases maximum value of y at the given time (say t 0 ) is at x = x 0 . Now, if y ( t = t 0 , x = x 0 + ∆x) = y (t = t 0 , x = x 0 − ∆x), then the pulse is symmetric otherwise not. So, at a given time (or any other time of your choice) first of all you have to find that x coordinate where you are getting the maximum value of y. y

y

v

x1 t1

x2 t2

x

Further, suppose that maximum value of y is at x1 at time t1 and at x 2 at time t 2 (> t1 ). Then, from the figure we can see that peak of the wave pulse is travelling towards positive x - direction. In time ( t 2 − t1 ), it has travelled a distance ( x 2 − x1 ) along positive x - direction. Hence, the wave velocity is ( x − x1 ) v=+ 2 ( t 2 − t1 )

28 — Waves and Thermodynamics V

Example 1

0.8

y( x , t) =

represents a moving pulse where x and y are [( 4x + 5t)2 + 5] in metre and t in second. Then, choose the correct alternative(s): (JEE 1999) (a) pulse is moving in positive x-direction (b) in 2 s it will travel a distance of 2.5 m (c) its maximum displacement is 0.16 m (d) it is a symmetric pulse Solution (b), (c) and (d) are correct options. y 0.16 m

–x

x

0 x t=0

The shape of pulse at x = 0 and t = 0 would be as shown in figure 0.8 y (0, 0) = = 0.16 m 5 From the equation it is clear that ymax = 0.16 m Pulse will be symmetric (symmetry is checked about ymax ) if At t = 0 ; y (x) = y (– x) From the given equation 0.8  y (x) = 16x2 + 5   at t = 0 0.8  and y (– x) = 16x2 + 5  y (x) = y (– x)

or Therefore, pulse is symmetric. Speed of pulse y

y 0.16 m

–x

x = –1.25 m t=1s

0.16 m

x=0 t=0

x

At t = 1 s and x = – 1.25 m, value of y is again 0.16 m, i.e. pulse has travelled a distance of 1.25 m in 1 second in negative x-direction or we can say that the speed of pulse is 1.25 m/s and it is travelling in negative x-direction. Therefore, it will travel a distance of 2.5 m in 2 seconds.

Chapter 17

Wave Motion — 29

Type 2. To find wave velocity from two y ( x ) equations given at two different times.

Concept In y ( x, t) equation, if value of t is substituted then the equation left is y ( x) equation. In this type of problem, y ( x) equation will be given at two different times and we have to find the wave velocity.

How to Solve? l

At the given time find the x - coordinates where you are getting the maximum (or minimum) value of y. From these two x - coordinates and two times we can calculate the wave velocity by the method discussed in Type 1.

V

Example 2 At time t =0, y( x ) equation of a wave pulse is y=

10 2 + ( x − 2) 2

and at t = 2 s , y ( x) equation of the same wave pulse is 10 y= 2 + ( x + 4) 2 Here, y is in mm and x in metres. Find the wave velocity. Solution From the given y (x) equations at two different times we can see that value of y is  10  maximum  = or 5 mm at x = 2 m at time t = 0 and at x = − 4 m at time t = 2 s.  2  So, peak of the wave pulse has travelled a distance of 6 m (from x = 2 m to x = − 4 m) in 2 s along negative x -direction. Hence, the wave velocity is 6 Ans. v = − = − 3 m /s 2

Type 3. To make complete y ( x, t ) function if y ( x ) function at some given time and wave velocity are given

Concept coefficient of t coefficient of x ∴Coefficient of t = ( v)( coefficient of x) (ii) Sign of coefficient of t will be opposite to the sign of coefficient of x if wave travels along positive x- direction and they are of same sign if the wave travels along negative x -direction. (i) Wave speed, v =

30 — Waves and Thermodynamics V

Example 3 A wave is travelling along positive x - direction with velocity 2 m/s. Further, y( x ) equation of the wave pulse at t = 0 is 10 y= 2 + ( 2x + 4) 2 (a) From the given information make complete y( x , t ) equation. (b) Find y ( x ) equation at t = 1 s Solution (a) Here, coefficient to x is 2. Wave speed is 2 m/s. Therefore, coefficient of t = v (coefficient of x ) = 2 × 2 = 4 units. Further, coefficient of x is positive and the wave is travelling along positive x -direction. Hence, coefficient of t must be negative. Now, suppose the y(x, t ) function is 10 …(i) y= 2 + (2x − 4t + α )2 Here, α is a constant. At time t = 0, Eq. (i) becomes y=

10 2 + (2x + α )2

y=

10 2 + (2x + 4)2

y=

10 2 + (2x − 4t + 4)2

and the given function is

Therefore, the value of α is 4. Substituting in Eq. (i), we have

(b) At t = 1 s 10 2 + (2x − 4 × 1 + 4)2 10 y= 2 + 4 x2

y= or

Ans.

Type 4. Based on transverse wave speed on a string

Concept (i) We know that transverse wave speed is given by T T or v= µ ρS (ii) If tension is uniform, then v is also uniform and we can calculate the time taken by the wave pulse in travelling from one point to another point by the direct relation distance t= speed If tension is non-uniform, then v will be non-uniform and in that case time can be obtained by integration.

Chapter 17 V

Example 4 ceiling.

Wave Motion — 31

A uniform rope of mass 0.1 kg and length 2.45 m hangs from a

(a) Find the speed of transverse wave in the rope at a point 0.5 m distant from the lower end. (b) Calculate the time taken by a transverse wave to travel the full length of the rope. Solution (a) As the string has mass and it is suspended vertically, tension in it will be different at different points. For a point at a distance x from the free end, tension will be due to the weight of the string below it. So, if m is the mass of string of length l, the mass of length x  m of the string will be   x.  l

l x

∴ ∴ or

 m T =   xg = µxg  l T = xg µ T v= = xg µ

 m  = µ   l

…(i)

At x = 0.5 m,

v = 0.5 × 9.8 = 2.21 m/s (b) From Eq. (i), we can see that velocity of the wave is different at different points. So, if at point x the wave travels a distance dx in time dt, then dx dx dt = = v gx t l dx ∴ ∫0dt = ∫0 gx or

t =2

l 2.45 =2 g 9.8

= 1.0 s

Ans.

Type 5. To write the equation corresponding to given y - x graph (or a snapshot) at a given time

Concept From the given y - x graph, we can easily determine A, ω and k. Secondly, we have to check whether the wave is travelling along positive x - direction or negative x -direction. Because this factor will decide whether, ωt and kx should be of same sign or opposite sign.

32 — Waves and Thermodynamics V

Example 5 Figure shows a snapshot of a sinusoidal travelling wave taken at t = 0.3 s. The wavelength is 7.5 cm and the amplitude is 2 cm. If the crest P was at x = 0 at t = 0, write the equation of travelling wave. y P 2 cm

t = 0.3 s x

1.2 cm

Solution Given,

A = 2 cm, λ = 7.5 cm

2π = 0.84 cm−1 λ The wave has travelled a distance of 1.2 cm in 0.3 s. Hence, speed of the wave, 1.2 v= = 4 cm /s 0.3 ∴ Angular frequency ω = (v) (k) = 3.36 rad /s Since the wave is travelling along positive x-direction and crest (maximum displacement) is at x = 0 at t = 0, we can write the wave equation as k=

∴

y (x, t ) = A cos (kx – ωt ) or y (x, t ) = A cos (ωt – kx) as cos (– θ ) = cos θ Therefore, the desired equation is y (x, t ) = ( 2 cm) cos [(0.84 cm−1 ) x – (3.36 rad /s ) t ] cm V

Ans.

Example 6 For the wave shown in figure, write the equation of this wave if its position is shown at t = 0. Speed of wave is v = 300 m/ s. y (m)

0.06 0.2 v

Solution The amplitude A = 0.06 m 5 λ = 0.2 m 2 ∴

λ = 0.08 m v 300 f = = = 3750 Hz λ 0.08 2π k= = 78.5 m−1 λ

x (m)

Chapter 17

Wave Motion — 33

ω = 2πf = 23562 rad /s ∂y = positive ∂x

and At t = 0, x = 0, and the given curve is a sine curve.

Hence, equation of wave travelling in positive x-direction should have the form, y (x, t ) = A sin (kx – ωt ) Substituting the values, we have y (x, t ) = (0.06 m) sin [(78.5 m−1 ) x – (23562 s –1 ) t ] m

Ans.

Miscellaneous Examples V

Example 7 A block of mass M = 2 kg is suspended from a string AB of mass 6 kg as shown in figure. A transverse wave pulse of wavelength λ 0 is produced at point B. Find its wavelength while reaching at point A. A

B M

Concept While moving from B to A, tension will increase. So, wave speed will also increase (as v = T / µ ). Frequency will remain unchanged because it depends on source. Therefore, v wavelength will also increase (as λ = or λ ∝ v). f Solution λ =

T /µ v = f f

or

λ∝ T

∴

λB TB = λA TA

∴

 T  λB =  B  λ A  TA 

(as µ and f are constants)

 (2 + 6) g  =  2g   =2λ

Ans.

34 — Waves and Thermodynamics V

Example 8 A wave moves with speed 300 m/ s on a wire which is under a tension of 500 N . Find how much tension must be changed to increase the speed to 312 m/ s ? Solution Speed of a transverse wave on a wire is, v=

T µ

…(i)

Differentiating with respect to tension, we have dv 1 = dT 2 µT Dividing Eq. (ii) by Eq. (i), we have dv 1 dT = v 2 T Substituting the proper values, we have

or

dT =

…(ii)

dT = (2T )

dv v

(2) (500) (312 – 300) 300

= 40 N

Ans.

i.e. tension should be increased by 40 N. V

Example 9 For a wave described by y = A sin (ωt – kx ), consider the following π π 3π points (a) x = 0 (b) x = (c) x = and (d) x = . 4k 2k 4k For a particle at each of these points at t = 0, describe whether the particle is moving or not and in what direction and describe whether the particle is speeding up, slowing down or instantaneously not accelerating? Solution

y = A sin (ωt – kx)

∂y = ωA cos (ωt – kx) and particle acceleration ∂t ∂ 2y aP (x, t ) = 2 = – ω 2A sin (ωt – kx) ∂t (a) t = 0, x = 0 : vP = + ωA and aP = 0 i.e. particle is moving upwards but its acceleration is zero. Particle velocity vP (x, t ) =

Note Direction of velocity can also be obtained in a different manner as under, At t = 0, y = A sin (– kx ) = – A sin kx y +A x At t = 0 –A

i.e. y-x graph is as shown in figure. At x = 0, slope is negative. Therefore, particle velocity is positive (v P = – v × slope), as the wave is travelling along positive x-direction.

Chapter 17 (b) t = 0, x =

π 4k

x=

Wave Motion — 35

π 4k kx =

∴

π 4

ωA  π vP = ωA cos  –  = +  4 2 ω 2A  π aP = – ω 2A sin  –  = +  4 2

and

Velocity of particle is positive, i.e. the particle is moving upwards (along positive y-direction). Further vP and aP are in the same direction (both are positive). Hence, the particle is speeding up. π π π (c) t = 0, x = x= ∴ kx = 2k 2k 2 vP = ωA cos (– π /2) = 0 aP = – ω 2A sin (– π /2) = ω 2A i.e. particle is stationary or at its extreme position ( y = – A ). So, it is speeding up at this instant. (d) t = 0, x =

3π 4k

x=

3π 4k

∴

kx =

3π 4

ωA  3π  vP = ωA cos  –  =–  4 2

∴

ω 2A  3π  aP = – ω 2A sin  –  =+  4 2 Velocity of particle is negative, i.e. the particle is moving downwards. Further vP and aP are in opposite directions, i.e. the particle is slowing down. V

Example 10 A thin string is held at one end and oscillates vertically so that, y ( x = 0, t) = 8 sin 4t ( cm) Neglect the gravitational force. The string’s linear mass density is 0.2 kg/m and its tension is 1 N. The string passes through a bath filled with 1 kg water. Due to friction heat is transferred to the bath. The heat transfer efficiency is 50%. Calculate how much time passes before the temperature of the bath rises one degree kelvin? Solution Comparing the given equation with equation of a travelling wave,

Speed of travelling wave,

y = A sin (kx ± ωt ) at x = 0 A = 8 cm = 8 × 10–2 m ω = 4 rad /s T 1 v= = = 2.236 m/s µ 0.2

Further, ρS = µ = 0.2 kg/m The average power over a period is P=

1 (ρS ) ω 2A 2v 2

we find,

36 — Waves and Thermodynamics Substituting the values, we have P=

1 (0.2) (4)2 (8 × 10–2)2 (2.236) 2

= 2.29 × 10−2 J/s The power transferred to the bath is, P ′ = 0.5 P = 1.145 × 10−2 J /s Now let, it takes t second to raise the temperature of 1 kg water by 1 degree kelvin. Then P ′ t = ms∆t Here, s = specific heat of water = 4.2 × 103 J/kg-° K ∴

t=

ms∆t (1) (4.2 × 103 ) (1) = P′ 1.145 × 10−2

= 3.6 × 105 s ≈ 4.2 day V

Ans.

Example 11 Consider a wave propagating in the negative x-direction whose frequency is 100 Hz. At t = 5 s, the displacement associated with the wave is given by y = 0.5 cos ( 0.1 x) where x and y are measured in centimetres and t in seconds. Obtain the displacement (as a function of x) at t = 10 s. What is the wavelength and velocity associated with the wave? Solution A wave travelling in negative x-direction can be represented as y (x, t ) = A cos (kx + ωt + φ ) At t = 5 s, y (x, t = 5) = A cos (kx + 5ω + φ ) Comparing this with the given equation, We have, and Now,

∴ From Eq. (i),

A = 0.5 cm, k = 0.1 cm−1 5ω + φ = 0 2π 2π λ= = = (20π ) cm k 0.1 ω = 2πf = (200π ) rad /s ω 200π v= = = (2000 π ) cm/s k 0.1

…(i) Ans.

Ans.

φ = – 5ω

At t = 10 s, y (x, t = 10) = 0.5 cos (0.1 x + 10 ω – 5 ω ) = 0.5 cos (0.1 x + 5 ω ) Substituting ω = 200 π , y (x, t = 10) = 0.5 cos (0.1 x + 1000π ) = 0.5 cos (0.1 x)

Ans.

Chapter 17 V

Wave Motion — 37

Example 12 A simple harmonic wave of amplitude 8 units travels along positive x-axis. At any given instant of time, for a particle at a distance of 10 cm from the origin, the displacement is + 6 units, and for a particle at a distance of 25 cm from the origin, the displacement is + 4 units. Calculate the wavelength. Solution

y = A sin

2π (vt – x) λ y x t = sin 2π  –  T λ A

or In the first case,

y1 = sin 2π A Here,

x1  t  –  T λ

y1 = + 6, A = 8, x1 = 10 cm

∴

6  t 10 = sin 2π  –  T 8 λ

…(i)

4  t 25 = sin 2π  –  T 8 λ

…(ii)

Similarly in the second case,

From Eq. (i),

or

 t 10  6 2π  –  = sin –1   = 0.85 rad T  8 λ t 10 – = 0.14 T λ

…(iii)

Similarly from Eq. (ii),

or

 4 π  t 25 2π  –  = sin –1   = rad  8 6 T λ t 25 – = 0.08 T λ

…(iv)

Subtracting Eq. (iv) from Eq. (iii), we get 15 = 0.06 λ ∴ V

λ = 250 cm

Ans.

Example 13 A wave pulse on a horizontal string is represented by the function y( x, t) =

5.0 1.0 + ( x – 2t ) 2

(CGS units)

Plot this function at t = 0, 2.5 and 5.0 s. Solution At the given times the function representing the wave pulse is y (x, 0) = y (x, 2.5 s ) =

5.0 1.0 + x2 5.0 1.0 + (x – 5.0)2

38 — Waves and Thermodynamics 5.0 1.0 + (x − 10.0)2

y (x, 5.0 s ) = y (cm)

10.0

y (x,0)

5.0

y (x,2.5s) y (x,5.50s)

x (cm) –5.0

0

5.0

10.0

15.0

The maximum of y(x, 0) is 5.0 cm; at t = 0, it is located at x = 0. At t = 2.5 and 5.0 s, the maximum of the pulse has moved to x = 5.0 and 10.0 cm, respectively. So, in each 2.5 s time interval, the pulse moves 5.0 cm in the positive x-direction. Its velocity is therefore +2.0 cm/s.

Example 14 A uniform circular hoop of string is rotating clockwise in the absence of gravity. The tangential speed is v0 . Find the speed of the wave travelling on this string. Solution v0 θ

T cos θ

AB θ

T

θ

B

°–

T cos θ θ A

T

90

V

T

θ θ

90°

T θ

C (a)

(b)

Let T be the tension in the string. Consider a small circular element AB of the string of length, ∆l = R (2θ )

(R = radius of hoop)

The components of tension T cos θ are equal and opposite and thus cancel out. The components towards centre C (i.e. T sin θ) provides the necessary centripetal force to element AB. mv02 …(i) ∴ 2T sin θ = R mass   Here, m = µ∆l = 2 µRθ µ =   length  As θ is small, Substituting in Eq. (i), we get

sin θ ≈ θ 2T θ =

2 µRθ v02 R

Chapter 17 or

T = v02 µ

or

T = v0 µ

Wave Motion — 39

…(ii)

Speed of wave travelling on this string, v=

T = v0 µ

[from Eq. (ii)]

i.e. the velocity of the transverse wave along the hoop of string is the same as the velocity of rotation of the hoop, viz. v0. Ans. V

Example 15 A sinusoidal wave travelling in the positive direction on a stretched string has amplitude 2.0 cm, wavelength 1.0 m and wave velocity 5.0 m/s. At ∂y < 0. Find the wave function y ( x , t). x = 0 and t = 0, it is given that y = 0 and ∂t Solution We start with a general equation for a rightward moving wave, y (x, t ) = A sin (kx – ωt + φ ) The amplitude given is A = 2.0 cm = 0.02 m The wavelength is given as λ = 1.0 m ∴ Angular wave number, k=

2π = (2π ) m–1 λ

Angular frequency, ∴ We are given that for x = 0, t = 0, and

ω = vk = (10 π ) rad /s y (x, t ) = (0.02) sin [2π (x – 5.0 t ) + φ ] y=0 ∂y A2 ). Later, we will see that the time t 0 is 2 ( f1 – f 2 ) Thus, the resultant amplitude oscillates between A1 + A2 and A1 – A2 with a time period 1 or with a frequency f = f 1 – f 2 known as beat frequency. Thus, T = 2t 0 = f1 – f 2 Beat frequency, f = f 1 – f 2

Calculation of Beat Frequency Suppose two waves of frequencies f 1 and f 2 ( < f 1 ) are meeting at some point in space. The corresponding periods are T1 and T2 ( > T1 ). If the two waves are in phase at t = 0, they will again be in phase when the first wave has gone through exactly one more cycle than the second. This will happen at a time t = T , the period of the beat. Let n be the number of cycles of the first wave in time T, then the number of cycles of the second wave in the same time is ( n – 1). Hence, …(i) T = nT1 …(ii) and T = ( n – 1) T2 Eliminating n from these two equations, we have TT 1 1 T= 1 2 = = 1 1 T2 – T1 f1 – f 2 – T1 T2 The reciprocal of the beat period is the beat frequency 1 f = = f1 – f 2 . T

Alternate Method Let the oscillations at some point in space (say x = 0) due to two waves be y1 = A1 sin 2πf 1 t and y2 = A2 sin 2πf 2 t If they are in phase at some time t, then 2πf 1 t = 2πf 2 t or f 1 t = f 2 t They will be again in phase at time ( t + T ) if, 2π f 1 ( t + T ) = 2π f 2 ( t + T ) + 2π or f 1 (t + T ) = f 2 (t + T ) + 1 1 Solving Eqs. (i) and (ii), we get T = f1 – f 2

(ω = 2πf ) …(i)

…(ii)

Note If a tuning fork is loaded with wax, its frequency decreases. On the other hand, when tuning fork is filed, its frequency increases.

Chapter 19

Sound Waves — 121

V

Example 19.14 Two tuning forks A and B produce 6 beats per second. Frequency of A is 300 Hz. When B is slightly loaded with wax, beat frequency decreases. Find original frequency of B. Solution Since, A and B produce 6 beats per second. Therefore, original frequency of B may be 306 Hz or 294 Hz. When B is loaded with wax, its frequency will decrease (suppose it decreases by 1Hz). So, if it is 306 Hz, it will become 305 Hz. If it is 294 Hz, it will become 293 Hz. Frequency of A is unchanged ( = 300 Hz ). If f B becomes 305 Hz, then beat frequency will become 5 Hz. If f B becomes 293 Hz, then it will become 7 Hz. But in the question it is given that beat frequency has decreased. So, the correct answer is 306 Hz.

V

Example 19.15 The string of a violin emits a note of 400 Hz at its correct tension. The string is bit taut and produces 5 beats per second with a tuning fork of frequency 400 Hz. Find frequency of the note emitted by this taut string. The frequency of vibration of a string increases with increase in the tension. Thus, the note emitted by the string will be a little more than 400 Hz. As it produces 5 beats per second with the 440 Hz tuning fork, the frequency will be 405 Hz. Solution

V

Example 19.16 Two tuning forks P and Q when set vibrating, give 4 beats per second. If a prong of the fork P is filed, the beats are reduced to 2 per second. Determine the original frequency of P, if that of Q is 250 Hz. There are four beats between P and Q, therefore the possible frequencies of P are 246 or 254 (that is 250 ± 4) Hz. When the prong of P is filed, its frequency becomes greater than the original frequency. If we assume that the original frequency of P is 254, then on filing its frequency will be greater than 254. The beats between P and Q will be more than 4. But it is given that the beats are reduced to 2, therefore, 254 is not possible. Therefore, the required frequency must be 246 Hz. (This is true, because on filing the frequency may increase to 248, giving 2 beats with Q of frequency 250 Hz) Ans.

Solution

INTRODUCTORY EXERCISE

19.6

1. A tuning fork produces 4 beats per second with another tuning fork of frequency 256 Hz. The first one is now loaded with a little wax and the beat frequency is found to increase to 6 per second. What was the original frequency of the first tuning fork?

2. A tuning fork of unknown frequency makes three beats per second with a standard fork of frequency 384 Hz. The beat frequency decreases when a small piece of wax is put on a prong of the first fork. What is the frequency of this fork?

122 — Waves and Thermodynamics

19.9 The Doppler’s Effect If a wave source and a receiver are moving relative to each other, the frequency observed by the receiver ( f ′) is different from the actual source frequency ( f ). This phenomenon is called the Doppler’s effect, named after the Austrian physicist Christian Johann Doppler (1803–1853), who discovered it in light waves. Perhaps you might have noticed how the sound of a vehicle’s horn changes as the vehicle moves past you. The frequency (pitch) of the sound you hear as the vehicle approaches you is higher than the frequency you hear as it moves away from you. The Doppler’s effect applies to waves in general. Let us apply it to sound waves. We consider the special case in which the source and observer move along the line joining them. We will use the following symbols, v = speed of sound, v s = speed of source and v o = speed of observer and v m = velocity of medium in which sound travels. For example, if Doppler’s effect is observed in air, it is wind velocity. If Doppler’s effect is observed inside a river, then it is river velocity. The general formula of the changed frequency is  v ± vm ± vo  f ′=   f  v ± vm ± vs 

Sign Convention We are talking about that sound which is travelling from source to observer (S to O). If the medium is also travelling in the same direction, then it means medium is supporting the sound. So, take positive sign both in numerator and denominator. If the medium travels in opposite direction, then take negative sign. If nothing is given in the question, then take it zero. v o andv s The concept is approaching nature always increases the frequency. So, take positive sign with v o (because it is in numerator) and negative sign with v s (as it is in denominator). On the other hand, receding nature decreases the frequency. So, take negative sign with v o and positive sign with v s . Now, let us make some different cases. Case 1

vm

Wind O

vo

S

vs

Sound

Fig. 19.10

In the given figure, sound is travelling from right to left but wind is blowing from left to right, so we will have to take negative sign with wind velocity. Observer is approaching towards source, so take positive sign with v o . Source is receding from the observer, so take positive sign with v s . The correct formula is given below  v − v wind + v o  f ′=   f  v − v wind + v s 

Sound Waves — 123

Chapter 19 Case 2 Wind O

S Sound

Fig. 19.11

Sound and wind both are travelling in the same direction (from S to O), so take positive sign with wind velocity. Source and observer both are approaching towards each other. So, take positive sign with v o and negative sign with v s . Therefore, the correct formula is  v + v wind + v o  f ′=  f  v + v wind − v s  Now, let us derive two special cases when medium velocity is zero.

Source at Rest, Observer Moves Suppose that the observer O moves towards the source S at speed v o . The speed of the sound waves v relative to O is v r = v + v o , but wavelength has its normal value λ = . Thus, the frequency heard by f O is

S

v

O

S

v

vo

vo

λ

Fig. 19.12

f′=

vr  v + vo  =  f λ  v 

If O were moving away from S, the frequency heard by O would be  v – vo  f′=  f  v  Combining these two expressions, we find  v ± vo  f′=  f  v 

…(i)

124 — Waves and Thermodynamics Source Moves, Observer at Rest O S Suppose that the source S moves towards O as shown in figure. vs If S were at rest, the distance between two consecutive wave pulses emitted by Fig. 19.13 v sound would be λ = = vT. However, in one time period S moves a distance f v sT before it emits the next pulse. As a result the wavelength is modified. Directly ahead of S the effective wavelength (for both S and O) is  v – vs  λ′ = vT – v sT = ( v − v s ) T =    f 

S

O vs

v

λ′ The wavelength in front of the source is less than the normal whereas in the rear it is larger than normal.

Fig. 19.14

The speed of sound waves relative to O is simply v. Thus, the frequency observed by O is v  v  f′= =  f λ ′  v – vs   v + vs  If S were moving away from O, the effective wavelength would be λ′ =   and the apparent  f  frequency would be  v  f′=  f  v + vs  Combining these two results, we have  v  f′=  f  v ± vs 

…(ii)

All four possibilities can be combined into one equation  v ± vo  f′=  f  v m vs 

…(iii)

where the upper signs (+ numerator, – denominator) correspond to the source and observer along the line joining the two in the direction toward the other, and the lower signs in the direction away from the other.

Chapter 19

Sound Waves — 125

Alternate Method The above formulae can be derived alternately as discussed below. l S

vo vs

S′

O′

O

vsT

vo t1

Fig. 19.15

Assume that the source and observer are moving along the same line and that the observer O is to the right of the source S. Suppose that at time t = 0, when the source and the observer are separated by a distance SO = l, the source emits a wave pulse (say p1 ) that reaches the observer at a later time t1 . In that time the observer has moved a distance v o t1 and the total distance travelled by p1 in the time t1 has been l + v o t1 . If v is the speed of sound, this distance is also vt1 . Then, vt1 = l + v 0 t1 or

t1 =

l v – v0

…(iv)

At time t = T , the source is at S ′ and the wave pulse (say p2 ) emitted at this time will reach the observer at a time say t 2 , measured from the same time t = 0, as before. The total distance travelled by p2 till it is received by the observer (measured from S ′ ) is ( l – v sT ) + v o t 2 . The actual travel time for p2 is (t 2 – T ) and the distance travelled is v ( t 2 – T ). Therefore, v ( t 2 – T ) = ( l – v sT ) + v 0 t 2 or

t2 =

l + (v – v s ) T v – vo

…(v)

The time interval reckoned by the observer between the two pulses emitted by the source at S and at S ′ is  v – vs  T ′ = t 2 – t1 =  T  v – vo  This is really the changed time period as observed by the observer. Hence, the new frequency is f′=

or

1  v – vo  1 =  T ′  v – vs  T

 v – vo  f′=  f  v – vs 

 v ± vo  This is a result which we can expect by using equation f ′ =   f in the case when source is  v m vs  moving towards observer and observer is moving away from the source.

126 — Waves and Thermodynamics Extra Points to Remember ˜

We have derived equation number (iii) by assuming that vo and v s are along the line joining source and observer. If the motion is along some other direction, the components of velocities along the line joining source and observer are considered. O

θ

v0 S

θ vs

Fig. 19.16

For example in the figure shown, ˜

 v + vo cos θ  f′ =   f  v + v s cos θ 

Change in frequency depends on the fact that whether the source is moving towards the observer or the observer is moving towards the source. But when the speed of source and observer are much lesser than that of sound, the change in frequency becomes independent of the fact whether the source is moving or the observer. This can be shown as under. Suppose a source is moving towards a stationary observer, with speedu and the speed of sound is v,then  v   1 f′ =   f= u  v – u  1–  v  Using the binomial expansion, we have 1 – u     v ∴

–1

u    f =  1 –  v   

u v

if u T2 ∴ v1 > v2 or f1 > f2 and f1 − f2 = 6 Hz Now, if T1 is increased, f1 will increase or f1 − f2 will increase. Therefore, (d) option is wrong. If T1 is decreased, f1 will decrease and it may be possible that now f2 − f1 become 6 Hz.Therefore, (c) option is correct. Similarly, when T2 is increased, f2 will increase and again f2 − f1 may become equal to 6 Hz. So, (b) is also correct. But (a) is wrong. V

Example 4 A sonometer wire under tension of 64 N vibrating in its fundamental mode is in resonance with a vibrating tuning fork. The vibrating portion of the sonometer wire has a length of 10 cm and mass of 1 g. The vibrating tuning fork is now moved away from the vibrating wire with a constant speed and an observer standing near the sonometer hears one beat per second. Calculate the speed with which the tuning fork is moved, if the speed of sound in air is 300 m/s. (JEE 1983) Solution Fundamental frequency of sonometer wire, f = =

v 1 T = 2l 2l µ 1 64 × 0.1 2 × 0.1 10−3

= 400 Hz fb = f − f ′ = 1 Hz f ′ = 399 Hz  v  f′= f    v + vs 

Given beat frequency, ∴ Using

 300  399 = 400    300 + vs 

or

vs = 0.75 m/s

Solving we get,

Ans.

Type 2. Reflection of a sound from a wall and beats

Concept

S

O

Wall

A source of sound is approaching towards a wall. An observer hears two sounds. One is direct from the source and the other reflected from wall. If both frequencies are same, then the observer will hear no beats and if there is a frequency difference then beats are heard to

134 — Waves and Thermodynamics the observer. This all depends on the position of observer. For finding the frequency of reflected sound we have the following two approaches. Approach 1 Take wall as a plane mirror. Image of S on this plane mirror will behave like another source S′ for the reflected sound.

O

S

S′

Wall

For example, in the shown figure both direct sound source S and reflected sound source S′ are approaching towards the observer. Approach 2 First consider wall as an observer for the sound received to it from the source S. Now, consider this wall as a source for reflected sound. V

Example 5 A siren emitting a sound of frequency 1000 Hz moves away from you toward a cliff at a speed of 10 m/s. (a) What is the frequency of the sound you hear coming directly from the siren? (b) What is the frequency of the sound you hear reflected off the cliff? (c) What beat frequency would you hear? Take the speed of sound in air as 330 m/s. Solution The situation is as shown in figure.

S

S'

O

Cliff

(a) Frequency of sound reaching directly to us (by S)  v  f1 =    v + vs   330  f =  (1000)  330 + 10 = 970.6 Hz (b) Frequency of sound which is reflected from the cliff (from S′ )  v   330  f2 =   f =  (1000)  v – vs   330 – 10

Ans.

= 1031.3 Hz

Ans.

= 60.7 Hz

Ans.

(c) Beat frequency = f2 – f1

Note Numerically the beat frequency comes out to be 60.7 Hz . But, beats between two tones can be detected by ear upto a frequency of about 7 per second. At higher frequencies beats cannot be distinguished in the sound produced. Hence, the correct answer of part (c) should be zero.

Sound Waves — 135

Chapter 19 V

Example 6 A source of sound of frequency f is approaching towards a wall with speed vs . Speed of sound is v. Three observers O1 , O2 and O3 are at different locations as shown. Find the beat frequency as observed by three different observers.

vs

S O1

O2

O3 Wall

Solution For reflected sound we can take wall as a plane mirror. O2

S O1

vs

vs

S′

O3 Wall

Observer O1 S is receding from O1 and S′ is approaching towards him. So, f s ′ > fs fb = fs′ − fs  v   v  =f  −f  v + vs   v − vs   2vv  =  2 s 2 f  v − vs 

Ans.

Observer O2 O2 has no relative motion with S. Therefore, there will be no change in the frequency of direct sound from S. Further, O2 and S′ are approaching towards each other. Therefore, the frequency of reflected sound will increase. fb = fs′ − fs  v + vo2  =f −f  v − vs  But vo2 = vs ∴

 v + vs  fb = f  −f  v − vs   2vs  =  f  v − vs 

Ans.

Observer O3 S and S′ both are approaching towards O3 . So, both frequencies will increase. But after the increase they are equal. So, beat frequency will be zero.  v  fs = fs′ = f    v − vs  ∴

fb = fs − fs ′ = 0

136 — Waves and Thermodynamics V

Example 7 A sound wave of frequency f travels horizontally to the right. It is reflected from a large vertical plane surface moving to left with a speed v. The speed of sound in medium is c (JEE 1995) (a) The number of waves striking the surface per second is f c ( c − v) f ( c + v) ( c + v) (c) The frequency of the reflected wave is f ( c − v)

( c + v) c

(b) The wavelength of reflected wave is

(d) The number of beats heard by a stationary listener to the left of the reflecting surface is vf c−v Solution Moving plane is like a moving observer. Therefore, number of waves encountered by moving plane,  v + v0   c + v f1 = f   =f    v   c  Frequency of reflected wave,

 v  f2 = f1    v − vs 

v

 c + v =f    c − v

C

O

Sound

c c  c − v λ2 = =   f2 f  c + v

Wavelength of reflected wave,

 c + v 2 fv Beat frequency, fb = f2 − f = f   −f = c− v  c − v Therefore, the correct options are (a), (b) and (c).

Type 3. Based on the experiment of finding speed of sound using resonance tube

Concept (i) If a vibrating tuning fork (of known frequency) is held over the open end of a resonance tube, then resonance is obtained at some position as the level of water is lowered. e

e λ — 4

l1

3λ — 4

l2

(i)

(ii)

If e is the end correction of the tube and l1 is the length from the water level to the top of the tube, then

Chapter 19

Sound Waves — 137

λ …(i) 4 Now, the water level is further lowered until a resonance is again obtained. If l 2 is new length of the air column, then 3λ …(ii) l2 + e = 4 Subtracting Eq. (i) from Eq. (ii), we get λ l 2 − l1 = 2 v v  or λ = 2 ( l 2 − l1 ) = λ =   f f l1 + e =

∴

v = 2 f ( l 2 − l1 )

(ii) The above result is independent of the end correction e. λ (iii) In the above two figures v, λ, f and loop size  =  are same in both figures. Length of  2 closed organ pipe is different. In the second figure, length is more. So, the fundamental v frequency  f =  will be less.  4l  Suppose frequency of given tuning fork is 300 Hz. Then, fundamental frequency of first pipe is also 300Hz. But fundamental frequency of the second pipe should be 100 Hz. So, that its next higher frequency (or first overtone frequency) which is three times the fundamental ( = 3 × 100 = 300 Hz) should also be in resonance with the given tuning fork. V

Example 8 A tuning fork of 512 Hz is used to produce resonance in a resonance tube experiment. The level of water at first resonance is 30.7 cm and at second (JEE 2005) resonance is 63.2 cm. The error in calculating velocity of sound is (a) 204.1 cm/s (b) 110 cm/s (c) 58 cm/s (d) 280 cm/s Solution Actual speed of sound in air is 330 m/s λ = (l2 − l1 ) = (63.2 − 30.7) cm = 32.5 cm = 0.325 m 2 λ = 0.65 m ∴ Speed of sound observed, v0 = f λ = 512 × 0.65 = 332.8 m/s ∴ Error in calculating velocity of sound = 2.8 m/s = 280 cm/s ∴ The correct option is (d). or

V

Example 9 In the experiment for the determination of the speed of sound in air using the resonance column method, the length of the air column that resonates in the fundamental mode, with a tuning fork is 0.1 m. When this length is changed to 0.35 m, the same tuning fork resonates with the first overtone. Calculate the end (JEE 2003) correction. (a) 0.012 m

(b) 0.025 m

(c) 0.05 m

(d) 0.024 m

138 — Waves and Thermodynamics Solution Let e be the end correction. Given that, fundamental tone for a length 0.1m = first overtone for the length 0.35 m. v 3v = 4 (0.1 + e) 4 (0.35 + e) Solving this equation, we get e = 0.025 m = 2.5 cm. ∴ The correct option is (b). Alternate method l2 + e = 3λ /4 and l1 + e = λ /4 Solving these two equations, we get l − 3l1 0.35 − 3 × 0.1 e= 2 = = 0.025 m 2 2 V

Example 10 A student is performing the experiment of resonance column. The diameter of the column tube is 4 cm. The frequency of the tuning fork is 512 Hz. The air temperature is 38° C in which the speed of sound is 336 m/s. The zero of the meter scale coincides with the top end of the resonance column tube. When the first resonance occurs, the reading of the water level in the column is (JEE 2012) (a) 14.0 cm (c) 16.4 cm Solution With end correction,

(b) 15.2 cm (d) 17.6 cm  v  f =n , 4 (l + e) 

(where, n = 1, 3, …)

  v =n  4 ( l + 0.6 r )  Because, e = 0.6 r, where r is radius of pipe. For first resonance, n = 1 ∴

f =

v or 4 (l + 0.6 r )

l=

 336 × 100  v − 0.6 r =   − 0.6 × 2 cm 4f 4 × 512    

= 15.2 cm ∴ The correct option is (b).

Type 4. Based on a situation which is similar to Doppler’s effect

Concept Let us take an example. 5 m/s Child-1

Child-2

A belt shown in figure is moving with velocity 5 m/s. Child-1 is placing some coins over the belt at a regular interval of 1sec. Child -2 is receiving the coins on the other hand. Now, let us compare this situation with Doppler’s effect.

Chapter 19

Sound Waves — 139

Child-1 is source, child-2 is observer, velocity of belt is net velocity of sound ( = v ± v m ) by which sound travels from source to observer. Actually, in this case v = velocity of coin (similar to velocity of sound) =0

v m = velocity of medium (or velocity of belt) = 5 m/s

T = time interval of placing two successive coins = 1 sec

1 = number of coins placed per second T λ = distance between two coins = v m T = 5 m. Now, let us consider the different cases.

∴

f =

Case 1 When source (child-1) and observer (child-2) both are at rest only belt (or medium) is in motion. In this case, distance between two coins moving towards child-2 with 5 m/ s is 5m. If child-2 receives first coin at time t1 = 0, then he will receive the second coin at 5m t2 = = 1sec. 5 m/ s ∴ ∴

T ′ = time interval of receival of two successive coins = t 2 − t1 = 1 sec 1 f′ = = 1 Hz T′

Result By the motion of only medium, there is no change in frequency. Case 2 If child-1 (or source) starts moving towards child-2 with 2 m/s. Then, in T = 1 sec, belt will move a distance of 5 m towards child-2 and child-1 will also travel a distance of 2m towards child-2 at the time of placing the second coin. Hence, λ = new distance between two successive coins = ( 5 − 2) m = 3 m Now, if child-2 receives the first coin at time t1 = 0, then he will receive the second coin (at a 3 distance of 3m from first coin) moving towards him with 5 m/s at time t 2 = sec. 5 3 1 5 ∴ T ′ = t 2 − t1 = sec or f ′ = = Hz or f ′ > f T′ 3 5

Result By the motion of source only wavelength changes therefore frequency changes. Case 3 Child-1 (or source) is stationary and child-2 starts moving towards child-1 with 2 m/s. This time, λ = distance between two successive coins is unchanged or 5 m. But coins and child-2 are moving towards each other with 5 m/s and 2 m/s, respectively. Therefore, relative velocity between them is 7m/s. So, if child-2 receives the first coin at 5 time t1 = 0 , then he will receive the second coin at time t 2 = sec 7 5 1 7 T ′ = t 2 − t1 = sec or f ′ = = Hz or f ′ > f ∴ 7 T′ 5

140 — Waves and Thermodynamics Result By the motion of observer relative velocity between sound and observer changes. Therefore, frequency changes. 5 7 Hz and in case-3, f ′ = Hz . These two are not same, although speeds are same in both 3 5 cases but change in frequency is different.

Note In case-2, f ′ =

Type 5. When v s or v o are not along the line joining S and O but they are very far from each other.

Concept In this case, we take components of vs and vo along the line SO. For example in the figure shown below, vs θ vs cosθ

O

  v f′ = f    v − vs cos θ 

If source is very far from observer, then θ→ 0° or cosθ→ 1  v  ∴ f′ = f    v − vs  V

Example 11 A source of frequency f is moving towards the observer along the line SO with a constant velocity vs as shown in figure. Plot f ′ versus t graph. Where f ′ is the changed frequency observed by the observer. S

vs

O

Solution During approach  v  f′ = f   = f1  v − vs 

(say)

f′ > f but f′ is constant with time. During recede  v  f′ = f   = f2 (say)  v + vs  f′ < f but f′ is constant with time. The correct f′ versus t graph is as shown. In the figure, to is the time when source is crossing the observer.

f′ f1 f f2 to

t

Chapter 19 V

Sound Waves — 141

Example 12 Repeat example-11 if source does not move along the line SO. vs

P θ

S

90° O

Solution The changed frequency f′ is now,   v f′ = f    v − vs cos θ  This time f′ is not constant but it is a function of θ, which is variable. But, at a far distance, θ → 0° or cos θ → 1 or  v  f′ = f  ∴   v − vs  and this is the maximum value of f′ (say fmax ) At point P, θ = 90° ∴ During receding (to the right of P)

f′ = f   v f′ = f    v + vs cos θ 

Again, at large distance from P, θ → 0° or cos θ → 1 ∴

 v  f′ = f    v + vs 

This time, this is the minimum value of f′ (say fmin ). Hence, f′ varies from fmax to fmin , with f ′ = f at the time of crossing P. The correct f′ versus t graph is as shown below. f′ fmax f fmin tp

V

t

Example 13 A whistle emitting a sound of frequency 440 Hz is tied to a string of 1.5 m length and rotated with an angular velocity of 20 rad/s in the horizontal plane. Calculate the range of frequencies heard by an observer stationed at a large distance from the whistle. (Speed of sound = 330 m/s). (JEE 1996) Solution

vs = Speed of source (whistle) = Rω = (1.5) (20) m/s, vs = 30 m/s

Maximum frequency will be heard by the observer in position P and minimum in position Q.

142 — Waves and Thermodynamics Now,  v  fmax = f    v − vs  where, v = speed of sound in air = 330 m/s P vs O

Q

 330  = (440)   Hz  330 − 30 fmax = 484 Hz  v   330  fmin = f   = (440)    330 + 30  v + vs 

or and

fmin = 403.33 Hz Therefore, range of frequencies heard by observer is from 484 Hz to 403.33 Hz.

Miscellaneous Examples V

Example 14 The water level in a vertical glass tube 1.0 m long can be adjusted to any position in the tube. A tuning fork vibrating at 660 Hz is held just over the open top end of the tube. At what positions of the water level will there be in resonance? Speed of sound is 330 m/s. Solution Resonance corresponds to a pressure antinode at closed end and pressure node at open end. Further, the distance between a pressure node and a pressure antinode is

λ , the 4

condition of resonance would be, λ  v Length of air column l = n = n   4f  4 Here, n = 1, 3, 5, …  330  l1 = (1)   = 0.125 m  4 × 660 l2 = 3l1 = 0.375 m l3 = 5l1 = 0.625 m l4 = 7l1 = 0.875 m l5 = 9l1 = 1.125 m Since, l5 > 1 m (the length of tube), the length of air columns can have the values from l1 to l4 only.

Chapter 19

Sound Waves — 143

Therefore, level of water at resonance will be (1.0 – 0.125) m = 0.875 m (1.0 – 0.375) m = 0.625 m (1.0 – 0.625) m = 0.375 m (1.0 – 0.875) m = 0.125 m

and

Ans.

0.875 m 0.625 m

0.375 m 0.125 m

In all the four cases shown in figure, the resonance frequency is 660 Hz but first one is the fundamental tone or first harmonic. Second is first overtone or third harmonic and so on. V

Example 15 A tube 1.0 m long is closed at one end. A stretched wire is placed near the open end. The wire is 0.3 m long and has a mass of 0.01 kg. It is held fixed at both ends and vibrates in its fundamental mode. It sets the air column in the tube into vibration at its fundamental frequency by resonance. Find (a) the frequency of oscillation of the air column and (b) the tension in the wire. Speed of sound in air = 330 m/ s. v 4l 330 = = 82.5 Hz 4 ×1

Solution (a) Fundamental frequency of closed pipe =

Ans.

(b) At resonance, given that : fundamental frequency of stretched wire (at both ends)

∴ ∴ or

= fundamental frequency of air column v = 82.5 Hz 2l T /µ = 82.5 2l T = µ (2 × 0.3 × 82.5)2  0.01 2 =  (2 × 0.3 × 82.5)  0.3  = 81.675 N

Ans.

144 — Waves and Thermodynamics V

Example 16 Two coherent narrow slits emitting sound of wavelength λ in the same phase are placed parallel to each other at a small separation of 2 λ. The sound is detected by moving a detector on the screen at a distance D ( >> λ ) from the slit S1 as shown in figure. Find the distance y such that the intensity at P is equal to intensity at O.

P

2λ S1

y S2

O

D Screen

Solution At point O on the screen the path difference between the sound waves reaching from S1 and S 2 is 2λ, i.e. constructive interference is obtained at O. At a very large distance from point O on the screen the path difference is zero. P ∆x is decreasing from 2λ to 0 M θ 90°

O ∆x = 2λ

S2

S1 2λ

D

Thus, we can conclude that as we move away from point O on the screen path difference decreases from 2λ to zero. At O constructive interference is obtained (where ∆x = 2λ ). So, next constructive interference will be obtained where ∆x = λ. Hence, S1P – S 2P = λ D + y – 2

or

2

y + (D – 2λ )2 = λ 2

∴

D 2 + y2 – λ =

y2 + (D – 2λ )2

On squaring both sides, we get D 2 + y2 + λ2 – 2λ

2 D 2 + y2 = 4D – 3λ

or as

D 2 + y2 = y2 + D 2 + 4λ2 – 4λD

D >> λ ,

4D – 3λ ≈ 4D

∴

2 D 2 + y2 = 4D

or

D 2 + y2 = 2D

Again squaring both sides, we get D 2 + y2 = 4D 2 or Ans. y= 3 D Alternate method Let ∆x = λ at angle θ as shown. Path difference between the waves is S1M = 2λ cos θ because S 2P ≈ MP ∴ or Now, or

2λ cos θ = λ

(∆x = λ )

θ = 60° PO = S1O tan θ = S1O tan 60° y= 3 D

Ans.

Chapter 19 V

Sound Waves — 145

Example 17 A fighter plane moving in a vertical loop with constant speed of radius R. The centre of the loop is at a height h directly overhead of an observer standing on the ground. The observer receives maximum frequency of the sound produced by the plane when it is nearest to him. Find the speed of the plane. Velocity of sound in air is v. Solution Let the speed of the plane (source) be vs. Maximum frequency will be observed by the observer when vs is along SO. The observer receives maximum frequency when the plane is nearest to him. That is as soon as the wave pulse reaches from S to O with speed v the plane reaches from S to S′ with speed vs. Hence,

C θ 90° S'

S h

vs

O

t=

SO SS ′ or = v vs = cos θ =

Here, V

R h

 SS ′  vs =   v  SO  Rθ 2

h – R2

v

Ans.

 R or θ = cos –1    h

Example 18 A source of sound of frequency 1000 Hz moves uniformly along a straight line with velocity 0.8 times velocity of sound. An observer is located at a distance l = 250 m from this line. Find (a) the frequency of the sound at the instant when the source is closest to the observer. (b) the distance of the source when he observes no change in the frequency. Solution (a) Suppose the pulse which is emitted when the source is at S reaches the observer O in the same time in which the source reaches from S to S′ , then vs = 0.8v S

S'

θ

v

O

cos θ =

SS ′ vst vs = = = 0.8 SO vt v

146 — Waves and Thermodynamics   v f′ =   f  v – vs cos θ 

Now,

  v =  (1000)  v – (0.8v) (0.8)    1 =  (1000)  1 – 0.64 Ans. = 2777.7 Hz (b) The observer will observe no change in the frequency when the source is at S as shown in figure. In the time when the wave pulse reaches from S to O, the source will reach from S to S′. Hence, S vs = 0.8v S'

v

250 m

O

t= ∴

SO SS ′ = v vs

v  SS ′ =  s  SO  v = (0.8) (250) = 200 m

Therefore, distance of observer from source at this instant is S ′ O = (SO )2 + (SS ′ )2 = (250)2 + (200)2 ≈ 320 m V

Ans.

Example 19 The air column in a pipe closed at one end is made to vibrate in its second overtone by a tuning fork of frequency 440 Hz. The speed of sound in air is 330 m/s. End corrections may be neglected. Let p0 denotes the mean pressure at any point in the pipe, and ∆p0 the maximum amplitude of pressure variation. (a) (b) (c) (d)

Find the length L of the air column. What is the amplitude of pressure variation at the middle of the column? What are the maximum and minimum pressures at the open end of the pipe? What are the maximum and minimum pressures at the closed end of the pipe?

Solution (a) Frequency of second overtone of the closed pipe  v = 5   = 440 Hz  4L  ∴

L=

5v m 4 × 440

(Given)

Chapter 19

Sound Waves — 147

Substituting v = speed of sound in air = 330 m/s 5 × 330 15 L= = m 4 × 440 16 λ=

(b)

Ans.

4L 4 (15 /16) 3 = = m 5 5 4

Open end is displacement antinode. Therefore, it would be a pressure node at x = 0; ∆p = 0

or

x = 0: ∆p = 0

L=

5λ 4

x = x : ∆p = ± ∆p 0 sinkx

Pressure amplitude at x = x can be written as ∆p = ± ∆p sin kx 2 π 2 π 8 π –1 k= = = m λ 3 /4 3

where,

L 15 /16 = m or (15/32) m will be 2 2  5π   8π   15 ∆p = ± ∆p0 sin     = ± ∆p0 sin    4  3   32

Therefore, pressure amplitude at x =

∆p = ±

∆p0 2

Ans.

(c) Open end is a pressure node, i.e. ∆p = 0 Hence, pmax = pmin = Mean pressure ( p0 ) (d) Closed end is a displacement node or pressure antinode.

V

Therefore,

pmax = p0 + ∆p0

and

pmin = p0 – ∆p0

Example 20 At a distance 20 m from a point source of sound the loudness level is 30 dB. Neglecting the damping, find (a) the loudness at 10 m from the source (b) the distance from the source at which sound is not heard. Solution (a) Intensity due to a point source varies with distance r from it as I∝

1 r2 2

or

I1  r2  =  I 2  r12

Now,

L1 = 10 log

I1 I0

148 — Waves and Thermodynamics L 2 = 10 log

and

I2 I0

 I I  L1 – L 2 = 10 log 1 – log 2  I0 I0  

∴

= 10 log

r  I1 = 10 log  2 I2  r1 

2

Substituting, L1 = 30 dB , r1 = 20 m and r2 = 10 m 2

 10 30 – L 2 = 10 log   = – 6.0  20

we have

L 2 = 36 dB

or r  (b) L1 – L 2 = 10 log  2  r1 

Ans.

2

Sound is not heard at a point where L 2 = 0 r  30 = 10 log  2  r1 

or

2

 r2   = 1000 or  r1 

∴ ∴ V

2

r2 = 31.62 r1

r2 = (31.62) (20) ≈ 632 m

Ans.

Example 21 A boat is travelling in a river with a speed 10 m/s along the stream flowing with a speed 2 m/ s. From this boat, a sound transmitter is lowered into the river through a rigid support. The wavelength of the sound emitted from the transmitter inside the water is 14.45 mm. Assume that attenuation of sound in water and air is negligible. (a) What will be the frequency detected by a receiver kept inside the river downstream? (b) The transmitter and the receiver are now pulled up into air. The air is blowing with a speed 5 m/ s in the direction opposite the river stream. Determine the frequency of the sound detected by the receiver. (Temperature of the air and water = 20° C; Density of river water = 103 kg/ m3 ; Bulk modulus of the water = 2.088 × 109 Pa; Gas constant, R = 8.31 J / mol -K ; Mean molecular mass of air = 28.8 × 10–3 kg/ mol; C p/ CV for air = 1.4 ) (JEE 2001) Solution Velocity of sound in water is Source

Observer(At rest)

vs = 10 m/s

vw =

B 2.088 × 109 = ρ 103

= 1445 m/s

Chapter 19

Sound Waves — 149

Frequency of sound in water will be f0 =

1445 vw = Hz λ w 14.45 × 10–3

f0 = 105 Hz (a) Frequency of sound detected by receiver (observer) at rest would be  vw + vr  f1 = f0    vw + vr – vs   1445 + 2  = (105 )   Hz  1445 + 2 – 10 f1 = 1.0069 × 105 Hz

Ans.

(b) Velocity of sound in air is va =

γRT M

Wind speed vm = 5 m/s

Observer (At rest)

Source vs = 10 m/s

=

(1.4) (8.31) (20 + 273) 28.8 × 10–3

= 344 m/s Frequency does not depend on the medium. Therefore, frequency in air is also f0 = 105 Hz. ∴ Frequency of sound detected by receiver (observer) in air would be  va – vw  344 – 5  5  f 2 = f0  Hz  = 10  344 – 5 – 10   va – vw – vs   f2 = 1.0304 × 105 Hz V

Ans.

Example 22 Three sound sources A, B and C have frequencies 400, 401 and 402 Hz, respectively. Calculate the number of beats noted per second. Solution Let us make the following table.  1  Beat time period   second  f1 – f2 

Group

Beat frequency (f1 – f2 ) Hz

A and B

1

B and C

1

1

A and C

2

0.5

1

Beat time period for A and B is 1 s. It implies that if A and B are in phase at time t = 0, they are again in phase after 1 s. Same is the case with B and C. But beat time period for A and C is 0.5 s. Therefore, beat time period for all together A, B and C will be 1 s. Because if, at t = 0, A, B and C all are in phase then after 1 s. (A and B) and (B and C) will again be in phase for the first

150 — Waves and Thermodynamics time while (A and C) will be in phase for the second time. Or we can say that all A , B and C are again in phase after 1 s. Beat time period, Tb = 1 s ∴ 1 Ans. or Beat frequency, fb = = 1 Hz Tb Alternate method

Suppose at time t, the equations of waves are

y1 = A1 sin 2πfAt y2 = A2 sin 2πfB t and y3 = A3 sin 2πfC t If they are in phase at some given instant of time t, then

(ω = 2πf )

…(i) 2π fAt = 2π fB t = 2π fC t Let Tb be the beat time period, i.e. after time Tb they all are again in phase. As fC > fB > fA , so …(ii) 2πfC (t + Tb ) = 2πfA (t + Tb ) + 2mπ and …(iii) 2πfB (t + Tb ) = 2πfA (t + Tb ) + 2nπ Here, m and n (< m) are positive integers. From Eqs. (i) and (ii), …(iv) ( fC – fA ) Tb = m Similarly, from Eqs. (i) and (iii) …(v) ( fB – fA ) Tb = n Dividing Eq. (iv) by Eq. (v), m fC – fA 402 – 400 2 = = = n fB – fA 401 – 400 1 Thus, letting m = 2 and n = 1 Tb =

m fC – fA

2 = 1 Hz 2 1 Beat frequency, fb = = 1 Hz Tb

[from Eq. (iv)]

=

∴

Ans.

Exercises LEVEL 1 Assertion and Reason Directions : Choose the correct option. (a) If both Assertion and Reason are true and the Reason is correct explanation of the Assertion. (b) If both Assertion and Reason are true but Reason is not the correct explanation of Assertion. (c) If Assertion is true, but the Reason is false. (d) If Assertion is false but the Reason is true.

1. Assertion : A closed pipe and an open organ pipe are of same length. Then, neither of their frequencies can be same. Reason : In the above case fundamental frequency of closed organ pipe will be two times the fundamental frequency of open organ pipe.

2. Assertion : A sound source is approaching towards a stationary observer along the line joining them. Then, apparent frequency to the observer will go on increasing. Reason : If there is no relative motion between source and observer, apparent frequency is equal to the actual frequency.

3. Assertion : In longitudinal wave pressure is maximum at a point where displacement is zero. Reason :

There is a phase difference of

π between y( x , t ) and ∆P ( x , t ) equation in case of 2

longitudinal wave.

4. Assertion : A train is approaching towards a hill. The driver of the train will hear beats. Reason : Apparent frequency of reflected sound observed by driver will be more than the frequency of direct sound observed by him.

5. Assertion : Sound level increases linearly with intensity of sound. Reason :

If intensity of sound is doubled, sound level increases approximately 3 dB.

6. Assertion : Speed of sound in gases is independent of pressure of gas. Reason :

With increase in temperature of gas speed of sound will increase.

7. Assertion : Beat frequency between two tuning forks A and B is 4 Hz. Frequency of A is greater than the frequency of B. When A is loaded with wax, beat frequency may increase or decrease. Reason : When a tuning fork is loaded with wax, its frequency decreases.

8. Assertion : Two successive frequencies of an organ pipe are 450 Hz and 750 Hz. Then, this pipe is a closed pipe. Reason : Fundamental frequency of this pipe is 150 Hz.

152 — Waves and Thermodynamics 9. Assertion : Fundamental frequency of a narrow pipe is more. Reason : more.

According to Laplace end correction if radius of pipe is less, frequency should be

10. Assertion : In the experiment of finding speed of sound by resonance tube method, as the level of water is lowered, wavelength increases. Reason : By lowering the water level number of loops increases.

Objective Questions 1. Velocity of sound in vacuum is (b) greater than 330 m/s (d) None of these

(a) equal to 330 m/s (c) less than 330 m/s

2. Longitudinal waves are possible in (a) solids (c) gases

(b) liquids (d) All of these

3. If the fundamental frequency of a pipe closed at one end is 512 Hz. The frequency of a pipe of the same dimension but open at both ends will be (a) 1024 Hz (c) 256 Hz

(b) 512 Hz (d) 128 Hz

4. The temperature at which the velocity of sound in oxygen will be the same as that of nitrogen at 15°C is (a) 112°C (c) 56°C

(b) 72°C (d) 17°C

5. A closed organ pipe is excited to vibrate in the third overtone. It is observed that there are (a) three nodes and three antinodes (c) four nodes and three antinodes

(b) three nodes and four antinodes (d) four nodes and four antinodes

6. When temperature is increased, the frequency of organ pipe (b) decreases (d) Nothing can be said

(a) increases (c) remains same

7. When a sound wave travels from water to air, it (a) bends towards normal (c) may bend in any direction

(b) bends away from normal (d) data insufficient

8. A closed organ pipe and an open organ pipe are tuned to the same fundamental frequency. The ratio of their lengths is (a) 1 : 2 (c) 1 : 4

(b) 2 : 1 (d) 4 : 1

9. A sonometer wire under a tension of 10 kg weight is in unison with a tuning fork of frequency 320 Hz. To make the wire vibrate in unison with a tuning fork of frequency 256 Hz, the tension should be altered by (a) 3.6 kg decreased (c) 6.4 kg decreased

(b) 3.6 kg increased (d) 6.4 kg increased

10. A tuning fork of frequency 256 Hz is moving towards a wall with a velocity of 5 m/ s.If the speed of sound is 330 m/ s, then the number of beats heard per second by a stationary observer lying between tuning fork and the wall is (a) 2

(b) 4

(c) zero

(d) 8

Chapter 19

Sound Waves — 153

11. Two sound waves of wavelength 1 m and 1.01 m in a gas produce 10 beats in 3 s. The velocity of sound in the gas is (a) 330 m/s

(b) 337 m/s

(c) 360 m/s

(d) 300 m/s

12. When a source is going away from a stationary observer with the velocity equal to that of sound in air, then the frequency heard by observer is n times the original frequency. The value of n is (a) 0.5 (c) 1.0

(b) 0.25 (d) No sound is heard

13. When interference is produced by two progressive waves of equal frequencies, then the maximum intensity of the resulting sound are N times the intensity of each of the component waves. The value of N is (a) 1 (c) 4

(b) 2 (d) 8

14. A tuning fork of frequency 500 Hz is sounded on a resonance tube. The first and second resonances are obtained at 17 cm and 52 cm. The velocity of sound is (a) 170 m/s (c) 520 m/s

(b) 350 m/s (d) 850 m/s

15. A vehicle, with a horn of frequency n is moving with a velocity of 30 m/s in a direction perpendicular to the straight line joining the observer and the vehicle. The observer perceives the sound to have a frequency ( n + n1 ). If the sound velocity in air is 300 m/s, then (a) n1 = 10 n (c) n1 = 0.1 n

(b) n1 = 0 (d) n1 = − 0.1 n

16. How many frequencies below 1 kHz of natural oscillations of air column will be produced if a pipe of length 1 m is closed at one end? [velocity of sound in air is 340 m/s] (b) 6 (d) 8

(a) 3 (c) 4

17. A sound source emits frequency of 180 Hz when moving towards a rigid wall with speed 5 m/s and an observer is moving away from wall with speed 5 m/s. Both source and observer moves on a straight line which is perpendicular to the wall. The number of beats per second heard by the observer will be [speed of sound = 355 m/s] (a) 5 beats/s (c) 6 beats/s

(b) 10 beats/s (d) 8 beats/s

18. Two sound waves of wavelengths λ 1 and λ 2 ( λ 2 > λ 1 ) produce n beats/s, the speed of sound is (a)

 1 1 (b) n  −   λ 1 λ 2

nλ 1λ 2 λ 2 − λ1

(c) n(λ 2 − λ 1 )

(d) n(λ 2 + λ 1 )

19. A, B and C are three tuning forks. Frequency of A is 350 Hz. Beats produced by A and B are 5/s and by B and C are 4/s. When a wax is put on A beat frequency between A and B is 2 Hz and between A and C is 6 Hz. Then, frequency of B and C respectively, are (a) 355 Hz, 349 Hz (c) 355 Hz, 341 Hz

(b) 345 Hz, 341 Hz (d) 345 Hz, 349 Hz

20. The first resonance length of a resonance tube is 40 cm and the second resonance length is 122 cm. The third resonance length of the tube will be (a) 200 cm

(b) 202 cm

(c) 203 cm

(d) 204 cm

154 — Waves and Thermodynamics 21. Two identical wires are stretched by the same tension of 100 N and each emits a note of frequency 200 Hz. If the tension in one wire is increased by 1 N, then the beat frequency is 1 Hz 2 (d) None of these

(a) 2 Hz

(b)

(c) 1 Hz

22. A tuning fork of frequency 340 Hz is sounded above an organ pipe of length 120 cm. Water is now slowly poured in it. The minimum height of water column required for resonance is (speed of sound in air = 340 m/s) (a) 25 cm (c) 75 cm

(b) 95 cm (d) 45 cm

23. In a closed end pipe of length 105 cm, standing waves are set up corresponding to the third overtone. What distance from the closed end, amongst the following is a pressure node? (a) 20 cm (c) 85 cm

(b) 60 cm (d) 45 cm

24. Oxygen is 16 times heavier than hydrogen. At NTP equal volume of hydrogen and oxygen are mixed. The ratio of speed of sound in the mixture to that in hydrogen is (a)

8

(b)

1 8

(c)

2 17

(d)

32 17

25. A train is moving towards a stationary observer. Which of the following curve best represents the frequency received by observer f as a function of time? f

f

(a)

(b) t

t f

f

(c)

(d) t

t

26. A closed organ pipe and an open organ pipe of same length produce 4 beats when they are set into vibrations simultaneously. If the length of each of them were twice their initial lengths, the number of beats produced will be (a) 2 (c) 1

(b) 4 (d) 8

27. One train is approaching an observer at rest and another train is receding from him with the same velocity 4 m/s. Both trains blow whistles of same frequency of 243 Hz. The beat frequency in Hz as heard by the observer is (speed of sound in air = 320 m/ s) (a) 10 (c) 4

(b) 6 (d) 1

Chapter 19

Sound Waves — 155

28. Speed of sound in air is 320 m/s. A pipe closed at one end has a length of 1 m and there is another pipe open at both ends having a length of 1.6 m. Neglecting end corrections, both the air columns in the pipes can resonate for sound of frequency (a) 80 Hz

(b) 240 Hz

(c) 320 Hz

(d) 400 Hz

29. Four sources of sound each of sound level 10 dB are sounded together in phase, the resultant intensity level will be (log10 2 = 0.3)

(a) 40 dB

(b) 26 dB

(c) 22 dB

(d) 13 dB

π ( x − 600 t ) ( p is in N/ m 2, x is in metre and t is 2 in second) is sent down a closed organ pipe. If the pipe vibrates in its second overtone, the length of the pipe is

30. A longitudinal sound wave given by p = 2.5 sin

(a) 6 m

(b) 8 m

(c) 5 m

(d) 10 m

31. Sound waves of frequency 600 Hz fall normally on perfectly reflecting wall. The distance from the wall at which the air particles have the maximum amplitude of vibration is (speed of sound in air = 330 m/ s) (a) 13.75 cm

(b) 40.25 cm

(c) 70.5 cm

(d) 60.75 cm

32. The wavelength of two sound waves are 49 cm and 50 cm, respectively. If the room temperature is 30°C, then the number of beats produced by them is approximately (velocity of sound in air at 30°C = 332 m/ s) (a) 6

(b) 10

(c) 13

(d) 18

33. Two persons A and B, each carrying a source of frequency 300 Hz, are standing a few metre apart. A starts moving towards B with velocity 30 m/s. If the speed of sound is 300 m / s, which of the following is true? (a) (b) (c) (d)

Number of beats heard by A is higher than that heard by B The number of beats heard by B are 30 Hz Both (a) and (b) are correct Both (a) and (b) are wrong

34. A fixed source of sound emitting a certain frequency appears as fa when the observer is approaching the source with speed v0 and fr when the observer recedes from the source with the same speed. The frequency of the source is (a)

f r + fa 2

(b)

f r − fa 2

(c)

fa fr

(d)

2 fr fa f r + fa

Subjective Questions 1. Determine the speed of sound waves in water, and find the wavelength of a wave having a frequency of 242 Hz. Take Bwater = 2 × 109 Pa.

2. If the source and receiver are at rest relative to each other but the wave medium is moving relative to them, will the receiver detect any wavelength or frequency shift.

3. Using the fact that hydrogen gas consists of diatomic molecules with M = 2 kg / K - mol. Find the speed of sound in hydrogen at 27° C.

4. About how many times more intense will the normal ear perceive a sound of 10−6 W/ m 2 than one of 10−9 W/ m 2?

156 — Waves and Thermodynamics 5. A 300 Hz source, an observer and a wind are moving as shown in the figure with respect to the ground. What frequency is heard by the observer? Take speed of sound in air = 340 m/ s. Observer

Source 5 m/s

20 m/s

10 m/s Wind

6. A person standing between two parallel hills fires a gun. He hears the first echo after

3 s, and a 2

5 s. If speed of sound is 332 m/ s, calculate the distance between the hills. 5 When will he hear the third echo? 7. Helium is a monoatomic gas that has a density of 0.179 kg / m3 at a pressure of 76 cm of mercury and a temperature of 0° C. Find the speed of compressional waves (sound) in helium at this temperature and pressure. second echo after

8. (a) In a liquid with density 1300 kg / m3 , longitudinal waves with frequency 400 Hz are found to have wavelength 8.00 m. Calculate the bulk modulus of the liquid. (b) A metal bar with a length of 1.50 m has density 6400 kg / m3 . Longitudinal sound waves take 3.90 × 10−4 s to travel from one end of the bar to the other. What is Young’s modulus for this metal? 9. What must be the stress ( F / A) in a stretched wire of a material whose Young’s modulus is Y for the speed of longitudinal waves equal to 30 times the speed of transverse waves?

10. A gas is a mixture of two parts by volume of hydrogen and one part by volume of nitrogen at STP. If the velocity of sound in hydrogen at 0° C is 1300 m/ s. Find the velocity of sound in the gaseous mixture at 27° C.

11. The explosion of a fire cracker in the air at a height of 40 m produces a 100 dB sound level at ground below. What is the instantaneous total radiated power? Assuming that it radiates as a point source.

12. (a) What is the intensity of a 60 dB sound? (b) If the sound level is 60 dB close to a speaker that has an area of 120 cm 2. What is the acoustic power output of the speaker?

13. (a) By what factor must the sound intensity be increased to increase the sound intensity level by 13.0 dB? (b) Explain why you do not need to know the original sound intensity?

14. The speed of a certain compressional wave in air at standard temperature and pressure is 330 m/ s. A point source of frequency 300 Hz radiates energy uniformly in all directions at the rate of 5 Watt. (a) What is the intensity of the wave at a distance of 20 m from the source? (b) What is the amplitude of the wave there? [Density of air at STP = 1.29 kg/m3 ]

15. What is the amplitude of motion for the air in the path of a 60 dB, 800 Hz sound wave? Assume that ρair = 1.29 kg / m3 and v = 330 m/ s.

16. A rock band gives rise to an average sound level of 102 dB at a distance of 20 m from the centre of the band. As an approximation, assume that the band radiates sound equally into a sphere. What is the sound power output of the band?

Chapter 19

Sound Waves — 157

17. If it were possible to generate a sinusoidal 300 Hz sound wave in air that has a displacement amplitude of 0.200 mm. What would be the sound level of the wave? (Assume v = 330 m/ s and ρair = 1.29 kg / m3 )

18. (a) A longitudinal wave propagating in a water-filled pipe has intensity 3.00 × 10−6 W/ m 2 and

frequency 3400 Hz. Find the amplitude A and wavelength λ of the wave. Water has density 1000 kg / m3 and bulk modulus 2.18 × 109 Pa. (b) If the pipe is filled with air at pressure 1.00 × 105 Pa and density 1.20 kg / m3 , what will be the amplitude A and wavelength λ of a longitudinal wave with the same intensity and frequency as in part (a)? (c) In which fluid is the amplitude larger, water or air? What is the ratio of the two amplitudes? Why is this ratio so different from one? Consider air as diatomic.

19. For a person with normal hearing, the faintest sound that can be heard at a frequency of 400 Hz has a pressure amplitude of about 6.0 × 10−5 Pa. Calculate the corresponding intensity and sound intensity level at 20° C. (Assume v = 330 m/ s and ρair = 1.29 kg / m3 ).

20. Find the fundamental frequency and the frequency of the first two overtones of a pipe 45.0 cm long. (a) If the pipe is open at both ends. (b) If the pipe is closed at one end. Use v = 344 m/ s.

21. A uniform tube of length 60 cm stands vertically with its lower end dipping into water. First two air column lengths above water are 15 cm and 45 cm, when the tube responds to a vibrating fork of frequency 500 Hz. Find the lowest frequency to which the tube will respond when it is open at both ends.

22. Write the equation for the fundamental standing sound waves in a tube that is open at both ends. If the tube is 80 cm long and speed of the wave is 330 m/ s. Represent the amplitude of the wave at an antinode by A.

23. A long glass tube is held vertically, dipping into water, while a tuning fork of frequency 512 Hz is repeatedly struck and held over the open end. Strong resonance is obtained, when the length of the tube above the surface of water is 50 cm and again 84 cm, but not at any intermediate point. Find the speed of sound in air and next length of the air column for resonance.

24. A wire of length 40 cm which has a mass of 4 g oscillates in its second harmonic and sets the air column in the tube to vibrations in its fundamental mode as shown in figure. Assuming the speed of sound in air as 340 m/ s, find the tension in the wire.

25. In a resonance tube experiment to determine the speed of sound in air, a pipe of diameter 5 cm is used. The column in pipe resonates with a tuning fork of frequency 480 Hz when the minimum length of the air column is 16 cm. Find the speed of sound in air column at room temperature.

158 — Waves and Thermodynamics 26. On a day when the speed of sound is 345 m/ s,the fundamental frequency of a closed organ pipe is 220 Hz. (a) How long is this closed pipe? (b) The second overtone of this pipe has the same wavelength as the third harmonic of an open pipe. How long is the open pipe?

27. A closed organ pipe is sounded near a guitar, causing one of the strings to vibrate with large amplitude. We vary the tension of the string until we find the maximum amplitude. The string is 80% as long as the closed pipe. If both the pipe and the string vibrate at their fundamental frequency, calculate the ratio of the wave speed on the string to the speed of sound in air.

28. A police siren emits a sinusoidal wave with frequency fS = 300 Hz. The speed of sound is 340 m/ s. (a) Find the wavelength of the waves if the siren is at rest in the air. (b) If the siren is moving at 30 m/ s, find the wavelength of the waves ahead of and behind the source.

29. Two identical violin strings, when in tune and stretched with the same tension, have a fundamental frequency of 440.0 Hz. One of the string is retuned by adjusting its tension. When this is done, 1.5 beats per second are heard when both strings are plucked simultaneously. (a) What are the possible fundamental frequencies of the retuned string? (b) By what fractional amount was the string tension changed if it was (i) increased (ii) decreased?

30. A swimming duck paddles the water with its feet once every 1.6 s, producing surface waves with this period. The duck is moving at constant speed in a pond where the speed of surface waves is 0.32 m/ s, and the crests of the waves ahead of the duck are spaced 0.12 m apart. (a) What is the duck’s speed? (b) How far apart are the crests behind the duck?

31. A railroad train is travelling at 30.0 m/ s in still air. The frequency of the note emitted by the train whistle is 262 Hz. What frequency is heard by a passenger on a train moving in the opposite direction to the first at 18.0 m/ s and (a) approaching the first? (b) receding from the first? Speed of sound in air = 340 m/s.

32. A boy is walking away from a wall at a speed of 1.0 m/ s in a direction at right angles to the wall. As he walks, he blows a whistle steadily. An observer towards whom the boy is walking hears 4.0 beats per second. If the speed of sound is 340 m/ s, what is the frequency of the whistle?

33. A tuning fork P of unknown frequency gives 7 beats in 2 seconds with another tuning fork Q. When Q runs towards a wall with a speed of 5 m/ s it gives 5 beats per second with its echo. On loading P with wax, it gives 5 beats per second with Q. What is the frequency of P? Assume speed of sound = 332 m/ s.

34. A stationary observer receives sonic oscillations from two tuning forks one of which approaches and the other recedes with the same velocity. As this takes place, the observer hears the beats of frequency f = 2.0 Hz. Find the velocity of each tuning fork if their oscillation frequency is f0 = 680 Hz and the velocity of sound in air is v = 340 m/ s.

35. Sound waves from a tuning fork A reach a point P by two separate paths ABP and ACP. When ACP is greater than ABP by 11.5 cm, there is silence at P. When the difference is 23 cm the sound becomes loudest at P and when 34.5 cm there is silence again and so on. Calculate the minimum frequency of the fork if the velocity of sound is taken to be 331.2 m/ s.

36. Two loudspeakers S1 and S 2 each emit sounds of frequency 220 Hz uniformly in all directions.

S1 has an acoustic output of 1.2 × 10−3 W and S 2 has 1.8 × 10−3 W . S1 and S 2 vibrate in phase. Consider a point P such that S1P = 0.75 m and S 2P = 3 m. How are the phases arriving at P related? What is the intensity at P when both S1 and S 2 are on? Speed of sound in air is 330 m/ s.

37. A source of sound emitting waves at 360 Hz is placed in front of a vertical wall, at a distance 2 m from it. A detector is also placed in front of the wall at the same distance from it. Find the

Chapter 19

Sound Waves — 159

minimum distance between the source and the detector for which the detector detects a maximum of sound. Take speed of sound in air = 360 m/ s. Assume that there is no phase change in reflected wave.

38. The atomic mass of iodine is 127 g/mol. A standing wave in iodine vapour at 400 K has nodes that are 6.77 cm apart when the frequency is 1000 Hz. At this temperature, is iodine vapour monoatomic or diatomic.

39. A tuning fork whose natural frequency is 440 Hz is placed just above the open end of a tube that contains water. The water is slowly drained from the tube while the tuning fork remains in place and is kept vibrating. The sound is found to be enhanced when the air column is 60 cm long and when it is 100 cm long. Find the speed of sound in air.

40. A piano wire A vibrates at a fundamental frequency of 600 Hz. A second identical wire B produces 6 beats per second with it when the tension in A is slightly increased. Find the ratio of the tension in A to the tension in B.

41. A tuning fork of frequency 256 Hz produces 4 beats per second with a wire of length 25 cm vibrating in its fundamental mode. The beat frequency decreases when the length is slightly shortened. What could be the minimum length by which the wire be shortened so that it produces no beats with the tuning fork?

42. Show that when the speed of the source and the observer are small compared to the speed of sound in the medium, the change in frequency becomes independent of the fact whether the source is moving or the observer. 43. A sound source moves with a speed of 80 m/s relative to still air toward a stationary listener. The frequency of sound is 200 Hz and speed of sound in air is 340 m/s. (a) Find the wavelength of the sound between the source and the listener. (b) Find the frequency heard by the listener.

44. A railroad train is travelling at 30 m/s in still air. The frequency of the note emitted by the locomotive whistle is 500 Hz. What is the wavelength of the sound waves : (a) in front of the locomotive?

(b) behind the locomotive?

What is the frequency of the sound heard by a stationary listener (c) in front of the locomotive?

(d) behind the locomotive?

Speed of sound in air is 344 m/s.

45. For a certain organ pipe, three successive resonance frequencies are observed at 425, 595 and 765 Hz respectively. Taking the speed of sound in air to be 340 m/s, (a) explain whether the pipe is closed at one end or open at both ends. (b) determine the fundamental frequency and length of the pipe.

46. Two tuning forks A and B sounded together give 8 beats per second. With an air resonance tube closed at one end, the two forks give resonances when the two air columns are 32 cm and 33 cm respectively. Calculate the frequencies of forks.

LEVEL 2 Single Correct Option 1. A plane wave of sound travelling in air is incident upon a plane water surface. The angle of incidence is 60°. If velocity of sound in air and water are 330 m/ s and 1400 m/ s, then the wave undergoes (a) (b) (c) (d)

refraction only reflection only Both reflection and refraction neither reflection nor refraction

2. An organ pipe of ( 3.9 π ) m long, open at both ends is driven to third harmonic standing wave. If

the amplitude of pressure oscillation is 1% of mean atmospheric pressure [ p0 = 105 N/ m 2 ]. The maximum displacement of particle from mean position will be [Given, velocity of sound = 200 m/s and density of air = 1.3 kg/m3 ]

(a) 2.5 cm (c) 1 cm

(b) 5 cm (d) 2 cm

3. A plane sound wave passes from medium 1 into medium 2. The speed of sound in medium 1 is 200 m/s and in medium 2 is 100 m/s. The ratio of amplitude of the transmitted wave to that of incident wave is 3 4 5 (c) 6

(a)

4 5 2 (d) 3 (b)

4. Two sources of sound are moving in opposite directions with velocities v1 and v2 ( v1 > v2 ). Both are moving away from a stationary observer. The frequency of both the sources is 1700 Hz. What is the value of ( v1 − v2 ) so that the beat frequency observed by the observer is 10 Hz? vsound = 340 m/s and assume that v1 and v2 both are very much less than vsound . (a) 1 m/s (c) 3 m/s

(b) 2 m/s (d) 4 m/s

5. A sounding body emitting a frequency of 150 Hz is dropped from a height. During its fall under gravity it crosses a balloon moving upwards with a constant velocity of 2 m/s one second after it started to fall. The difference in the frequency observed by the man in balloon just before and just after crossing the body will be (velocity of sound = 300 m/s, g = 10 m/s2) (a) 12 (c) 8

(b) 6 (d) 4

6. A closed organ pipe has length L. The air in it is vibrating in third overtone with maximum amplitude a. The amplitude at distance (a) 0 (b) a a (c) 2 (d) Data insufficient

L from closed end of the pipe is 7

Chapter 19

Sound Waves — 161

7. S1 and S 2 are two coherent sources of sound having no initial phase difference. The velocity of sound is 330 m/s. No maxima will be formed on the line passing through S 2 and perpendicular to the line joining S1 and S 2. If the frequency of both the sources is

3m

S1

S2

(a) 330 Hz

(b) 120 Hz

(c) 100 Hz

(d) 220 Hz

8. A source is moving with constant speed vs = 20 m/s towards a stationary observer due east of the source. Wind is blowing at the speed of 20 m/s at 60° north of east. The source has frequency 500 Hz. Speed of sound = 300 m/s. The frequency registered by the observer is approximately (b) 552 Hz (d) 517 Hz

(a) 541 Hz (c) 534 Hz

9. A car travelling towards a hill at 10 m/s sounds its horn which has a frequency 500 Hz. This is heard in a second car travelling behind the first car in the same direction with speed 20 m/s. The sound can also be heard in the second car by reflection of sound from the hill. The beat frequency heard by the driver of the second car will be (speed of sound in air = 340 m/s) (a) 31 Hz (c) 21 Hz

(b) 24 Hz (d) 34 Hz

10. Two sounding bodies are producing progressive waves given by y1 = 2 sin ( 400πt ) and y2 = sin ( 404 πt ) where t is in second, which superpose near the ears of a person. The person will hear

(a) (b) (c) (d)

2 beats/s with intensity ratio 9/4 between maxima and minima 2 beats/s with intensity ratio 9 between maxima and minima 4 beats/s with intensity ratio 16 between maxima and minima 4 beats/s with intensity ratio 16/9 between maxima and minima

11. The air in a closed tube 34 cm long is vibrating with two nodes and two antinodes and its temperature is 51°C. What is the wavelength of the waves produced in air outside the tube, when the temperature of air is 16°C? (a) 42.8 cm

(b) 68 cm

(c) 17 cm

(d) 102 cm

12. A police car moving at 22 m/s, chase a motorcyclist. The police man sounds his horn at 176 Hz, while both of them move towards a stationary siren of frequency 165 Hz. Calculate the speed of the motorcyclist, if he does not observe any beats. (velocity of sound in air = 330 m/s) (JEE 2003)

Police car 22 m/s 176 Hz

(a) 33 m/s

(b) 22 m/s

Motorcycle V

(c) zero

Stationary siren (165 Hz)

(d) 11 m/s

13. A closed organ pipe resonates in its fundamental mode at a frequency of 200 Hz with O2 in the pipe at a certain temperature. If the pipe now contains 2 moles of O2 and 3 moles of ozone, then what will be the fundamental frequency of same pipe at same temperature? (a) 268.23 Hz

(b) 175.4 Hz

(c) 149.45 Hz

(d) None of these

162 — Waves and Thermodynamics 14. A detector is released from rest over a source of sound of frequency f0 = 103 Hz. The frequency observed by the detector at time t is plotted in the graph. The speed of sound in air is ( g = 10 m/ s2 ) f (Hz)

1100 1000 t (s)

3.0

(a) 330 m/s

(b) 350 m/s

(c) 300 m/s

15. Sound

waves are travelling along positive x-direction. Displacement of particle at any time t is as shown in figure. Select the wrong statement.

(a) (b) (c) (d)

Particle located at E has its velocity in negative x-direction Particle located at D has zero velocity Both (a) and (b) are correct Both (a) and (b) are wrong

(d) 310 m/s y

A

C B

D

E

x

More than One Correct Options 1. An air column in a pipe, which is closed at one end, is in resonance with a vibrating tuning fork of frequency 264 Hz. If v = 330 m/ s, the length of the column in cm is (are) (b) 62.50 (d) 125

(a) 31.25 (c) 93.75

2. Which of the following is/are correct? (Velocity of sound in air)2

(Velocity of sound in air) (T = constant)

(a)

(b)

(Parabola) (Pressure)

O

(Velocity of transverse wave in a string)

(c)

Temperature (in °C)

O

(Fundamental frequency of an organ pipe)

(d) (parabola) O

Tension

O

3. Choose the correct options for longitudinal wave (a) (b) (c) (d)

maximum pressure variation is BAk maximum density variation is ρAk pressure equation and density equation are in phase pressure equation and displacement equation are out of phase

Length of organ pipe

Chapter 19

Sound Waves — 163

4. Second overtone frequency of a closed pipe and fourth harmonic frequency of an open pipe are same. Then, choose the correct options. (a) Fundamental frequency of closed pipe is more than the fundamental frequency of open pipe (b) First overtone frequency of closed pipe is more than the first overtone frequency of open pipe (c) Fifteenth harmonic frequency of closed pipe is equal to twelfth harmonic frequency of open pipe (d) Tenth harmonic frequency of closed pipe is equal to eighth harmonic frequency of open pipe

5. For fundamental frequency f of a closed pipe, choose the correct options. (a) (b) (c) (d)

If radius of pipe is increased, f will decrease If temperature is increased, f will increase If molecular mass of the gas filled in the pipe is increased, f will decrease. If pressure of gas (filled in the pipe) is increased without change in temperature, f will remain unchanged

6. A source is approaching towards an observer with constant speed along the line joining them. After crossing the observer, source recedes from observer with same speed. Let f is apparent frequency heard by observer. Then, (a) (b) (c) (d)

f f f f

will keep on increasing during approaching will keep on decreasing during receding will remain constant during approaching will remain constant during receding

Comprehension Based Questions Passage A man of mass 50 kg is running on a plank of mass 150 kg with speed of 8 m/s relative to plank as shown in the figure (both were initially at rest and the velocity of man with respect to ground any how remains constant). Plank is placed on smooth horizontal surface. The man, while running, whistles with frequency f0. A detector ( D ) placed on plank detects frequency. The man jumps off with same velocity (w.r.t. to ground) from point D and slides on the smooth horizontal surface [Assume coefficient of friction between man and horizontal is zero]. The speed of sound in still medium is 330 m/s. Answer the following questions on the basis of above situations.

D µ=0

1. The frequency of sound detected by detector D, before man jumps off the plank is 332 f0 324 328 (c) f0 336

(a)

330 f0 322 330 (d) f0 338 (b)

2. The frequency of sound detected by detector D, after man jumps off the plank is 332 f0 324 328 (c) f0 336

(a)

330 f0 322 330 (d) f0 338 (b)

164 — Waves and Thermodynamics 3. Choose the correct plot between the frequency detected by detector versus position of the man relative to detector. Frequency detected

Frequency detected

f0

(a)

(b)

Position of man relative to detector

f0

Position of man relative to detector

Frequency detected

(c)

Frequency detected

f0

f0

(d)

Position of man relative to detector

Position of man relative to detector

Match the Columns 1. Fundamental frequency of an open organ pipe is f. Match the following two columns for a closed pipe of double the length. Column I

Column II

(a) Fundamental frequency

(p) 1.25 f

(b) Second overtone frequency

(q) f

(c) Third harmonic frequency

(r) 0.75 f

(d) First overtone frequency

(s) None of these

1 th the speed of 4 sound. There are three observers O1 , O2 and O3 as shown. Match the following two columns.

2. A train T horns a sound of frequency f. It is moving towards a wall with speed O2 O1

O3 T

Column I

Column II

(b) Beat frequency observed to O2

2 f 3 8 f (q) 15

(c) Beat frequency observed to O3

(r) None of these

(d) If train moves in opposite direction with same speed, then beat frequency observed to O3

(s) Zero

(a) Beat frequency observed to O1

(p)

Chapter 19

Sound Waves — 165

3. A tuning fork is placed near a vibrating stretched wire. A boy standing near the two hears a beat frequency f. It is known that frequency of tuning fork is greater than frequency of stretched wire. Match the following two columns. Column I (a) If tuning fork is loaded with wax, (b) If prongs of tuning fork are filed, (c) If tension in stretched wire is increased, (d) If tension in stretched wire is decreased,

Column II (p) beat frequency must increase. (q) beat frequency must decrease. (r) beat frequency may increase. (s) beat frequency may decrease.

4. I represents intensity of sound wave, A the amplitude and r the distance from the source. Then, match the following two columns. Column I (a) Intensity due to a point source.

Column II (p) proportional to r −1/ 2

(b) Amplitude due to a point source. (q) proportional to r −1 (c) Intensity due to a line source.

(r) proportional to r −2

(d) Amplitude due to a line source.

(s) proportional to r −4

5. The equation of longitudinal stationary wave in second overtone mode in a closed organ pipe is y = ( 4 mm) sin π x cos πt Here, x is in metre and t in second. Then, match the following two columns. Column I

Column II

(a) Length of pipe

(p) 1 m

(b) Wavelength

(q) 1.5 m

(c) Distance of displacement node from the closed end (r) 2.0 m (d) Distance of pressure node from the closed end

(s) None of these

Subjective Questions 1. A window whose area is 2 m 2 opens on a street where the street noise results at the window an intensity level of 60 dB. How much acoustic power enters the window through sound waves? Now, if a sound absorber is fitted at the window, how much energy from the street will it collect in a day?

2. A point A is located at a distance r = 1.5 m from a point source of sound of frequency 600 Hz. The power of the source is 0.8 W. Speed of sound in air is 340 m/s and density of air is 1.29 kg/ m3 . Find at the point A, (a) the pressure oscillation amplitude ( ∆p)m (b) the displacement oscillation amplitude A.

3. A flute which we treat as a pipe open at both ends is 60 cm long. (a) What is the fundamental frequency when all the holes are covered? (b) How far from the mouthpiece should a hole be uncovered for the fundamental frequency to be 330 Hz? Take speed of sound in air as 340 m/s.

166 — Waves and Thermodynamics 4. A source S and a detector D of high frequency waves are a distance d apart on the ground. The direct wave from S is found to be in phase at D with the wave from S that is reflected from a horizontal layer at an altitude H. The incident and reflected rays make the same angle with the reflecting layer. When the layer rises a distance h, no signal is detected at D. Neglect absorption in the atmosphere and find the relation between d , h , H and the wavelength λ of the waves. h

H S

d

D

5. Two sound speakers are driven in phase by an audio amplifier at frequency 600 Hz. The speed

of sound is 340 m/s. The speakers are on the y-axis, one at y = + 1.0 m and the other at y = – 1.0 m. A listener begins at y = 0 and walks along a line parallel to the y-axis at a very large distance x away.

(a) At what angle θ (between the line from the origin to the listener at the x-axis) will she first hear a minimum sound intensity? (b) At what angle will she first hear a maximum (after θ = 0°) sound intensity? (c) How many maxima can she possibly hear if she keeps walking in the same direction?

6. Two speakers separated by some distance emit sound of the same frequency. At some point P the intensity due to each speaker separately is I 0. The path difference from P to one of the 1 speakers is λ greater than that from P to the other speaker. What is the intensity at P if 2 (a) the speakers are coherent and in phase; (b) the speakers are incoherent; and (c) the speakers are coherent but have a phase difference of 180°?

7. Two loudspeakers radiate in phase at 170 Hz. An observer sits at 8 m from one speaker and 11m from the other. The intensity level from either speaker acting alone is 60 dB. The speed of sound is 340 m/ s. (a) Find the observed intensity when both speakers are on together. (b) Find the observed intensity level when both speakers are on together but one has its leads

reversed so that the speakers are 180° out of phase. (c) Find the observed intensity level when both speakers are on and in phase but the frequency is 85 Hz.

8. Two identical speakers emit sound waves of frequency 680 Hz uniformly in all directions with a total audio output of 1 mW each. The speed of sound in air is 340 m/s. A point P is a distance 2.00 m from one speaker and 3.00 m from the other. (a) (b) (c) (d)

Find the intensities I1 and I 2 from each speaker at point P separately. If the speakers are driven coherently and in phase, what is the intensity at point P? If they are driven coherently but out of phase by 180°, what is the intensity at point P? If the speakers are incoherent, what is the intensity at point P?

9. A train of length l is moving with a constant speed v along a circular track of radius R. The engine of the train emits a sound of frequency f. Find the frequency heard by a guard at the rear end of the train.

Chapter 19

Sound Waves — 167

10. A 3 m long organ pipe open at both ends is driven to third harmonic standing wave. If the

amplitude of pressure oscillations is 1 per cent of mean atmospheric pressure ( p0 = 105 Nm 2 ). Find the amplitude of particle displacement and density oscillations. Speed of sound v = 332 m / s and density of air ρ = 1.03 kg / m3 .

11. A siren creates a sound level of 60 dB at a location 500 m from the speaker. The siren is powered by a battery that delivers a total energy of 1.0 kJ. Assuming that the efficiency of siren is 30%, determine the total time the siren can sound.

12. A cylinder of length 1 m is divided by a thin perfectly flexible diaphragm in the middle. It is closed by similar flexible diaphragms at the ends. The two chambers into which it is divided contain hydrogen and oxygen. The two diaphragms are set in vibrations of same frequency. What is the minimum frequency of these diaphragms for which the middle diaphragm will be motionless? Velocity of sound in hydrogen is 1100 m/ s and that in oxygen is 300 m/ s.

13. A conveyor belt moves to the right with speed v = 300 m/ min. A very fast pieman puts pies on the belt at a rate of 20 per minute and they are received at the other end by a pieeater. (a) If the pieman is stationary find the spacing x between the pies and the frequency with which they are received by the stationary pieeater. (b) The pieman now walks with speed 30 m/min towards the receiver while continuing to put pies on the belt at 20 per minute. Find the spacing of the pies and the frequency with which they are received by the stationary pieeater.

14. A point sound source is situated in a medium of bulk modulus 1.6 × 105 N/ m 2. An observer standing at a distance 10 m from the source writes down the equation for the wave as y = A sin (15πx − 6000 πt ). Here y and x are in metres and t is in second. The maximum pressure amplitude received to the observer’s ear is ( 24 π ) Pa, then find. (a) the density of the medium, (b) the displacement amplitude A of the waves received by the observer and (c) the power of the sound source.

15. Two sources of sound S1 and S 2 vibrate at the same frequency and are in phase. The intensity of sound detected at a point P (as shown in figure) is I 0. P

θ S1

θ S2

(a) If θ = 45° what will be the intensity of sound detected at this point if one of the sources is switched off ? (b) What will be intensity of sound detected at P if θ = 60° and both the sources are now switched on?

16. Two narrow cylindrical pipes A and B have the same length. Pipe A is open at both ends and is filled with a monoatomic gas of molar mass M A . Pipe B is open at one end and closed at the other end and is filled with a diatomic gas of molar mass M B . Both gases are at the same temperature. (JEE 2002)

168 — Waves and Thermodynamics (a) If the frequency of the second harmonic of the fundamental mode in pipe A is equal to the frequency of the third harmonic of the fundamental mode in pipe B, determine the value of MA . MB (b) Now, the open end of pipe B is also closed (so that the pipe is closed at both ends.) Find the ratio of the fundamental frequency in pipe A to that in pipe B.

17. A boat is travelling in a river with a speed 10 m/ s along the stream flowing with a speed 2 m/ s. From this boat a sound transmitter is lowered into the river through a rigid support. The wavelength of the sound emitted from the transmitter inside the water is 14.45 mm. Assume that attenuation of sound in water and air is negligible. (a) What will be the frequency detected by a receiver kept inside the river downstream? (b) The transmitter and the receiver are now pulled up into the air. The air is blowing with a speed 5 m/s in the direction opposite to the river stream. Determine the frequency of the sound detected by the receiver. (Temperature of the air and water = 20° C; Density of river water = 103 kg/m3 ; bulk modulus the water = 2.088 × 109 Pa; Gas constant R = 8.31 J/mol-K; Mean molecular mass of air = 28.8 × 10−3 kg per mol and C p /CV for air = 1.4 )

18. A string 25 cm long and having a mass of 2.5 g is under tension. A pipe closed at one end is 40 cm long. When the string is set vibrating in its first overtone and the air in the pipe in its fundamental frequency, 8 beats per second are heard. It is observed that decreasing the tension in the string decrease the beat frequency. If the speed of sound in air is 320 m/ s, find the tension in the string.

19. A source emits sound waves of frequency 1000 Hz.The source moves to the right with a speed of 32 m/ s relative to ground. On the right a reflecting surface moves towards left with a speed of 64 m/ s relative to the ground. The speed of sound in air is 332 m/ s. Find (a) (b) (c) (d)

the wavelength of sound in ahead of the source, the number of waves arriving per second which meets the reflecting surface, the speed of reflected waves and the wavelength of reflected waves.

Answers Introductory Exercise 19.1 1. 1.4 × 10 5 N/ m2 4. 1.04 × 10

–5

2. 7.25 cm, 72.5 m

3. (a) Zero (b) 3.63 × 10 –6 m

3. 3.6 × 10 9 Pa

4. 315 m/s

m

Introductory Exercise 19.2 1. 819°C

2. 19.6 m/s

Introductory Exercise 19.3 1. (a) 4.67 Pa

(b) 2.64 × 10 −2 W / m2

4. Faintest (a) 4.49 × 10

−13

(c) 104 dB

2

W / m , –3.48 dB

Loudest (a) 0.881 W/m2 , +119 dB

2. 7.9

(b) 1.43 × 10

−11

3. 20 dB m

(b) 2.01 × 10 –5 m

Introductory Exercise 19.4 1. 1375 Hz

2. (a) 11.7 cm

(b) 180°

Introductory Exercise 19.5 1. (a) 2. (c) 3. (a) 4. (a) 0.392 m (b) 0.470 m 5. (a) Fundamental 0.8 m, first overtone 0.267 m, 0.8 m, second overtone 0.16 m, 0.48 m, 0.8 m (b) Fundamental 0, first overtone 0, 0.533 m. Second overtone 0, 0.32 m, 0.64 m 6. (a) closed (b) 5, 7 (c) 1.075 m

Introductory Exercise 19.6 1. 252 Hz

2. 387 Hz

Introductory Exercise 19.7 1. (d)

2. (d)

3. (b)

4. (a)

Exercises LEVEL 1 Assertion and Reason 1. (c)

2. (d)

3. (d)

4. (a)

5. (d)

6. (d)

7. (b)

8. (b)

9. (a)

10. (d)

Objective Questions 1.(d)

2.(d)

3.(a)

4.(c)

5.(d)

6.(a)

7.(a)

8.(a)

9.(a)

10.(c)

11.(b)

12.(a)

13.(c)

14.(b)

15.(b)

16.(b)

17.(a)

18.(a)

19.(b)

20.(d)

21.(c)

22.(d)

23.(d)

24.(c)

25.(b)

26.(a)

27.(b)

28.(d)

29.(c)

30.(c)

31.(a)

32.(c)

33.(d)

34.(a)

Subjective Questions 1. 1414 m/s, 5.84 m

2. No

6. 664 m, 4 s

8. (a) 1.33 × 1010 Pa

7. 972 m/s

3. 1321 m/s

4. Two times

(b) 9.47 × 1010 Pa

5. 274 Hz 9. Y /900

170 — Waves and Thermodynamics 10. 591 m/s

12. (a) 10 −6 W/m2

11. 201 W

14. (a) 9.95 × 10

−4

(b) 1.15 × 10

2

W/m

−6

15. 13.6 nm

m

(b) 5.66 × 10 −9 m, 0.1 m

18. (a) 9.44 × 10 −11 m, 0.43 m 19. 4.2 × 10 −12 W/m2 , 6.23 dB

(c) air,

13. (a) 20

16. 80 W

17. 134.4 dB

Aair = 60 Awater

20. (a) 382.2Hz, 764.4 Hz, 1146.6 Hz (b) 191.1 Hz, 573.3 Hz, 955.5 Hz

22. y = A cos (3.93x ) sin (1297t )

21. 250 Hz

(b) 1.2 × 10 −8 W

25. 336 m/s 26. (a) 0.392 m (b) 0.470 m 29. (a) 441.5 Hz, 438.5 Hz (b) (i) + 0.68%

23. 348.16 m/s, 118 cm

27. 0.40 (ii) –0.68%

24. 11.56 N

28. (a) 1.13 m (b) 1.03 m, 1.23 m 30. (a) 0.245 m/s (b) 0.904 m

31. (a) 302 Hz (b) 228 Hz 32. 680 Hz 33. 160 Hz 34. 0.5 m/s 35. 1440 Hz 36. ∆φ1 = π , ∆φ 2 = 4 π , Resultant intensity = 8.2 ×10 −5 Wm−2 38. Diatomic 39. 352 m/s 43. (a) 1.3 m (b) 262 Hz 45. (a) closed (b) 85 Hz, 1 m

40. 1.02 41. 0.4 cm 44. (a) 0.628 m (b) 0.748 m 46. f1 = 264 Hz and f2 = 256 Hz

37. 7.5 m

(c) 548 Hz (d) 460 Hz

LEVEL 2 Single Correct Option 1.(b)

2.(a)

3.(d)

4.(b)

5.(a)

11.(a)

12.(b)

13.(b)

14.(c)

15.(c)

6.(b)

7.(c)

8.(c)

9.(a)

10.(b)

More than One Correct Options 1. (a,c)

2. (c,d)

3. (a,b,c,d)

4. (b,c,d)

5. (a,b,c,d)

6. (c,d)

Comprehension Based Questions 1. (a)

2. (c)

3. (a)

Match the Columns 1. (a) → s

(b) → p

(c) → r

(d) → r

2. (a) → q

(b) → p

(c) → s

(d) → s

3. (a) → r,s

(b) → p

(c) → r, s

(d) → p

4. (a) → r

(b) → q

(c) → q

(d) → p

5. (a) → s

(b) → r

(c) → p,r

(d) → q

Subjective Questions 2. (a) 4.98 N/m2 (b) 3.0 × 10 −6 m

1. 2 µW , 0.173 J 4. λ = 2 4(H + h) + d 2

2

− 2 4H

2

+ d

3. (a) 283.33 Hz (b) 51.5 cm

2

5. (a) 8.14° (b) 16.5° (c) Three maxima beyond the maximum corresponding to θ = 0° 6. (a) 0 (b) 2I0

7. (a) 0 (b) 66 dB

(c) 4I0

8. (a) I1 = 19.9 µW / m2 , I2 = 8.84 µW / m2 10. 0.28 cm, 9.0 × 10

9. f

−1

13. (a) 15 m, 20 min 15. (a)

I0 4

(b) I0

18. 27.04 N

−3

3

kg/m −1

(b) 13.5 m, 22.22 min

16. (a)

400 189

19. (a) 0.3 m

(b)

3 4

(c) 63 dB

(b) 55.3 µW / m2

(c) 2.2 µW / m2

11. 95.5 s

(d) 28.7 µW / m2

12. 1650 Hz 3

14. (a) 1 kg/m

(b) 10 µm (c) 288 π3 W

17. (a) 1.0069 × 10 5 Hz (b) 1.0304 × 10 5 Hz

(b) 1320 (c) 332 m/s

(d) 0.2 m

Thermometry, Thermal Expansion and Kinetic Theory of Gases

Chapter Contents 20.1 Thermometers and The Celsius Temperature Scale 20.2 The Constant Volume Gas Thermometer and The Absolute Temperature Scale 20.3 Heat and Temperature 20.4 Thermal Expansion 20.5 Behaviour of Gases 20.6 Degree of Freedom 20.7 Internal Energy of an Ideal Gas 20.8 Law of Equipartition of Energy 20.9 Molar Heat Capacity 20.10 Kinetic Theory of Gases

172 — Waves and Thermodynamics

20.1 Thermometers and The Celsius Temperature Scale Thermometers are devices that are used to measure temperatures. All thermometers are based on the principle that some physical properties of a system change as the system’s temperature changes. Some physical properties that change with temperature are 1. the volume of a liquid 2. the length of a solid 3. the pressure of a gas at constant volume 4. the volume of a gas at constant pressure and 5. the electric resistance of a conductor. A common thermometer in everyday use consists of a mass of liquid, usually mercury or alcohol that expands in a glass capillary tube when heated. In this case, the physical property is the change in volume of the liquid. Any temperature change is proportional to the change in length of the liquid column. The thermometer can be calibrated accordingly. On the celsius temperature scale, a thermometer is usually calibrated between 0°C (called the ice point of water) and 100°C (called the steam point of water). Once the liquid levels in the thermometer have been established at these two points, the distance between the two points is divided into 100 equal segments to create the celsius scale. Thus, each segment denotes a change in temperature of one celsius degree (1°C). A practical problem in this type of thermometer is that readings may vary for two different liquids. When one thermometer reads a temperature, for example 40°C the other may indicate a slightly different value. This discrepancies between thermometers are especially large at temperatures far from the calibration points. To surmount this problem we need a universal thermometer whose readings are independent of the substance used in it. The gas thermometer used in the next article meets this requirement.

20.2 The Constant Volume Gas Thermometer and The Absolute

Temperature Scale The physical property used by the constant volume gas thermometer is the change in pressure of a gas at constant volume. p

0°C

100°C

t (°C)

Fig. 20.1

The pressure versus temperature graph for a typical gas taken with a constant volume is shown in figure. The two dots represent the two reference temperatures namely, the ice and steam points of water. The line connecting them serves as a calibration curve for unknown temperatures. Experiments show that the thermometer readings are nearly independent of the type of gas used as long as the gas pressure is low and the temperature is well above the point at which the gas liquefies.

Chapter 20

Thermometry, Thermal Expansion and Kinetic Theory of Gases — 173

If you extend the curves shown in figure toward negative temperatures, you find, in every case, that the pressure is zero when the temperature is –273.15° C. This significant temperature is used as the basis for the absolute temperature scale, which sets –273.15° C as its zero point. This temperature is often referred to as absolute zero. The size of a degree on the absolute temperature scale is identical to the size of a degree on the celsius scale. Thus, the conversion between these temperatures is TC = T – 273.15

…(i)

In 1954, by the International committee on weights and measures, the triple point of water was chosen as the reference temperature for this new scale. The triple point of water is the single combination of temperature and pressure at which liquid water, gaseous water and ice (solid water) coexist in equilibrium. This triple point occurs at a temperature of approximately 0.01° C and a pressure of 4.58 mm of mercury. On the new scale, which uses the unit kelvin, the temperature of water at the triple point was set at 273.16 kelvin, abbreviated as 273.16 K. (No degree sign is used with the unit kelvin). p

Gas 1 Gas 2 Gas 3

–273.15°C

t (°C)

0°C

Fig. 20.2

This new absolute temperature scale (also called the kelvin scale) employs the SI unit of absolute 1 of the difference between absolute zero and temperature, the kelvin which is defined to be “ 273.16 the temperature of the triple point of water”.

The Celsius, Fahrenheit and Kelvin Temperature Scales Eq. (i) shows the relation between the temperatures in celsius scales and kelvin scale. Because the size of a degree is the same on the two scales, a temperature difference of 10°C is equal to a temperature difference of 10 K. The two scales differ only in the choice of the zero point. The ice point temperature on the kelvin scale, 273.15 K, corresponds to 0.00°C and the kelvin steam point 373.15 K, is equivalent to 100.00°C. A common temperature scale in everyday use in US is the Fahrenheit scale. The ice point in this scale is 32°F and the steam point is 212°F. The distance between these two points are divided in 180 equal parts. The relation between celsius scale and Fahrenheit scale is as derived below. 0°C

100°C

32°F

212°F

Fig. 20.3

(100 equal parts)

(180 equal parts)

174 — Waves and Thermodynamics 100 parts of celsius scale =180 parts of Fahrenheit scale 9 parts of Fahrenheit scale ∴ 1 part of celsius scale = 5 K

C

373.15

100K

100°

100°C

273.15

0°C

F

212°

180°F

32°C

Relation among Kelvin, Celsius and Fahrenheit temperature scales

Fig. 20.4

9 TF = 32 + TC 5 5 ∆TC = ∆T = ∆TF 9

Hence, Further,

…(ii) …(iii)

Extra Points to Remember ˜

Different Thermometers Thermometric property It is the property that can be used to measure the temperature. It is represented by any physical quantity such as length, volume, pressure and resistance etc., which varies linearly with a certain range of temperature. Let X denotes the thermometric physical quantity and X 0 , X100 and Xt be its values at 0°C, 100°C and t °C respectively. Then,  X − X0  t = t  × 100° C  X100 − X 0  (i) Constant volume gas thermometer The pressure of a gas at constant volume is the thermometric property. Therefore,  p − p0  t = t  × 100° C  p100 − p0  (ii) Platinum resistance thermometer The resistance of a platinum wire is the thermometric property. Hence,  R − R0  t = t  × 100° C  R100 − R 0  (iii) Mercury thermometer In this thermometer, the length of a mercury column from some fixed point is taken as thermometric property. Thus,  l − l0  t = t  × 100° C  l100 − l0 

˜

Two other thermometers, commonly used are thermocouple thermometer and total radiation pyrometer.

Chapter 20 ˜

˜

Thermometry, Thermal Expansion and Kinetic Theory of Gases — 175

Total radiation pyrometer is used to measure very high temperatures. When a body is at a high temperature, it glows brightly and the radiation emitted per second from unit area of the surface of the body is proportional to the fourth power of the absolute temperature of the body. If this radiation is measured by some device, the temperature of the body is calculated. This is the principle of a total radiation pyrometer. The main advantage of this thermometer is that the experimental body is not kept in contact with it. Hence, there is no definite higher limit of its temperature range. It can measure temperature from 800°C to 3000°C–4000°C. However, it cannot be used to measure temperatures below 800°C because at low temperatures the emission of radiation is so poor that it cannot be measured directly. Ranges of different thermometers

Table 20.1 Thermometer

˜

Lower limit

Upper limit

Mercury thermometer

–30°C

300°C

Gas thermometer

–268°C

1500°C

Platinum resistance thermometer

–200°C

1200°C

Thermocouple thermometer

–200°C

1600°C

Radiation thermometer

800°C

No limit

Reaumur’s scale Other than Celsius, Fahrenheit and Kelvin temperature scales Reaumur’s scale was designed by Reaumur in 1730. The lower fixed point is 0°R representing melting point of ice. The upper fixed point is 80°R, which represents boiling point of water. The distance between the two fixed points is divided into 80 equal parts. Each part represents 1°R. If TC , TF and TR are temperature values of a body on Celsius scale, Fahrenheit scale and Reaumur scale respectively, then TC − 0 TF − 32 TR − 0 = = 100 180 80

˜

V

A substance is found to exist in three states solid, liquid and gas. For each substance, there is a set of temperature and pressure at which all the three states may coexist. This is called triple point of that substance. For water, the values of pressure and temperature corresponding to triple point are 4.58 mm of Hg and 273.16°K.

Example 20.1 Express a temperature of 60°F in degrees Celsius and in Kelvin. Solution Substituting TF = 60° F in Eq. (ii), 5 5 TC = (TF – 32) = ( 60° – 32° ) 9 9 = 15.55° C

Ans.

From Eq. (i), T = TC + 273.15 = 15.55° C + 273.15 = 288.7 K V

Ans.

Example 20.2 The temperature of an iron piece is heated from 30°C to 90°C. What is the change in its temperature on the Fahrenheit scale and on the kelvin scale?

176 — Waves and Thermodynamics ∆ TC = 90° C – 30° C = 60° C Using Eq. (iii),

Solution

9 9 ∆TC = ( 60° C) 5 5 = 108° F

∆TF =

and

∆T = ∆TC = 60 K

INTRODUCTORY EXERCISE

Ans. Ans.

20.1

1. What is the value of (a) 0°F in Celsius scale? (b) 0 K on Fahrenheit scale?

2. At what temperature is the Fahrenheit scale reading equal to (a) twice

(b) half of Celsius?

3. A faulty thermometer reads 5°C in melting ice and 99°C in steam. Find the correct temperature in °F when this faulty thermometer reads 52°C.

4. At what temperature the Fahrenheit and Kelvin scales of temperature give the same reading? 5. At what temperature the Fahrenheit and Celsius scales of temperature give the same reading?

20.3 Heat and Temperature The word heat is always used during transfer of thermal energy from hot body to cold body (due to temperature difference between them). Following are given some statements. Some of them are right and some are wrong. From those statements, you will be able to find the difference between heat and temperature.

Wrong Statements (i) This body has large quantity of heat. (ii) Temperature transfer is taking place from body A to body B.

Correct Statements (i) Temperature of body A is more than temperature of body B. (ii) Heat transfer is taking place from body A to body B because A is at higher temperature than body B. Note One calorie (1 cal ) is defined as the amount of heat required to raise the temperature of one gram of water from 14.5° C to 15.5° C. Experiments have shown that 1 cal = 4.186 J Similarly, 1 kcal = 1000 cal = 4186 J The calorie is not a fundamental SI unit.

Chapter 20

Thermometry, Thermal Expansion and Kinetic Theory of Gases — 177

20.4 Thermal Expansion Most substances expand when they are heated. Thermal expansion is a consequence of the change in average separation between the constituent atoms of an object. Atoms of an object can be imagined to be connected to one another by stiff springs as shown in Fig. 20.5. At ordinary temperatures, the atoms in a solid oscillate about their equilibrium positions with an amplitude of approximately 10 –11 m. The average spacing between the atoms is about 10 –10 m. As the temperature of solid increases, the atoms oscillate with greater amplitudes, as a result the average separation between them increases, consequently the object expands.

Fig. 20.5

More precisely, thermal expansion arises from the asymmetrical nature of the potential energy curve. At the atomic level, thermal expansion may be understood by considering how the potential energy of the atoms varies with distance. The equilibrium position of an atom will be at the minimum of the potential energy well if the well is symmetric. At a given temperature, each atom vibrates about its equilibrium position and its average position remains at the minimum point. If the shape of the well is not symmetrical, as shown in figure, the average position of an atom will not be at the minimum point. When the temperature is raised the amplitude of the vibrations increases and the average position is located at a greater interatomic separation. This increased separation is manifested as expansion of the material.

Linear Expansion Suppose that the temperature of a thin rod of length l is changed from T to T + ∆T . It is found experimentally that, if ∆T is not too large, the corresponding change in length ∆l of the rod is directly proportional to ∆T and l. Thus, ∆l ∝ ∆T and

∆l ∝ l

Introducing a proportionality constant α (which is different for different materials) we may write ∆l as ∆l = lα∆T

…(i)

Here, the constant α is called the coefficient of linear expansion of the material of the rod and its units are K –1 or [( ° C) –1 ]. Remember that ∆T = ∆TC . Actually, α does depend slightly on the temperature, but its variation is usually small enough to be negligible, even over a temperature range of 100°C. We will always assume that α is a constant.

178 — Waves and Thermodynamics Volume Expansion Because the linear dimensions of an object change with temperature, it follows that surface area and volume change as well. Just as with linear expansion, experiments show that if the temperature change ∆T is not too great (less than 100°C or so), the increase in volume ∆V is proportional to both, the temperature change ∆T and the initial volume V . Thus, ∆V ∝ ∆T and ∆V ∝ V Introducing a proportionality constant γ, we may write ∆V as … (ii) ∆V = V × γ × ∆T Here, γ is called the coefficient of volume expansion. The units of γ are K –1 or ( ° C) –1 .

Relation Between γ and α For an isotropic solid (which has the same value of α in all directions) γ = 3α. To see that γ = 3α for a solid, consider a cube of length l and volume V = l 3 . When the temperature of the cube is increased by dT, the side length increases by dl and the volume increases by an amount dV given by  dV  2 dV =   ⋅ dl = 3l ⋅ dl  dl  dl = lαdT dV = 3l 3αdT = (3α )VdT

Now, ∴ This is consistent with Eq. (ii),

dV = γVdT , only if …(iii) γ = 3α Average values of α and γ for some materials are listed in Table 20.2. You can check the relation γ = 3α, for the materials given in the table. Table 20.2 Material

α [K

−1

or (°C) −1]

γ [ K −1 or (°C) −1]

Steel

1.2 × 10−5

3.6 × 10−5

Copper

1.7 × 10−5

5.1 × 10−5

Brass

2.0 × 10−5

6.0 × 10−5

Aluminium

2.4 × 10−5

7.2 × 10−5

The Anomalous Expansion of Water Most liquids also expand when their temperatures increase. Their expansion can also be described by Eq. (ii). The volume expansion coefficients for liquids are about 100 times larger than those for solids. Some substances contract when heated over a certain temperature range. The most common example is water.

Chapter 20

Thermometry, Thermal Expansion and Kinetic Theory of Gases — 179

Figure shows how the volume of 1 g of water varies with temperature at atmospheric pressure. The volume decreases as the temperature is raised from 0°C to about 4°C, at 4°C the volume is a minimum and the density is a maximum (1000 kg/ m 3 ). Above 4° C, water expands with increasing temperature like most substances. 3

V(cm ) 1.0003 1.0002

1.000 2

T (°C)

4 6 Fig. 20.6

This anomalous behaviour of water causes ice to form first at the surface of a lake in cold weather. As winter approaches, the water temperature decreases initially at the surface. The water there sinks because of its increased density. Consequently, the surface reaches 0°C first and the lake becomes covered with ice. Now ice is bad conductor of heat so water at the bottom remains at 4°C, the highest density of water. Aquatic life is able to survive the cold winter as the bottom of the lake remains unfrozen at a temperature of about 4°C.

Extra Points to Remember ˜

If a solid object has a hole in it, what happens to the size of the hole, when the temperature of the object increases? A common misconception is that if the object expands, the hole will shrink because material expands into the hole. But the truth is that if the object expands, the hole will expand too, because every linear dimension of an object changes in the same way when the temperature changes. a + ∆a a

Ti + ∆T

Ti b + ∆b

b

Fig. 20.7 ˜

Expansion of a bimetallic strip As Table 20.2 indicates, each substance has its own characteristic average coefficient of expansion. Steel

Brass Room temperature

Higher temperature

Fig. 20.8

180 — Waves and Thermodynamics For example, when the temperatures of a brass rod and a steel rod of equal length are raised by the same amount from some common initial value, the brass rod expands more than the steel rod because brass has a greater average coefficient of expansion than steel. Such type of bimetallic strip is found in practical devices such as thermostats to break or make electrical contact.

Bimetallic strip

On ˜

25°C

Off Fig. 20.9

30°C

Variation of density with temperature Most substances expand when they are heated, i.e. volume of a 1 given mass of a substance increases on heating, so the density should decrease  as ρ ∝  . Let us see  V how the density (ρ) varies with increase in temperature. m ρ= V 1 or (for a given mass) ρ∝ V ρ′ V V V 1 ∴ = = = = ρ V ′ V + ∆V V + γV∆T 1 + γ∆T ∴

ρ′ =

ρ 1 + γ∆T

This expression can also be written as ρ′ = ρ (1 + γ∆T )–1 If γ is very small, ∴ ˜

(1 + γ∆T )–1 ≈ 1 – γ∆T ρ′ ≈ ρ (1 – γ∆T )

Effect of temperature on apparent weight when immersed in a liquid When a solid body is completely immersed in a liquid, its apparent weight gets decreased due to an upthrust acting on it by the liquid. The apparent weight is given by wapp = w – F Here, F = upthrust = VS ρL g

where, VS = volume of solid and ρL = density of liquid

Now, as the temperature is increased VS increases while ρL decreases. So, F may increase or decrease (or may remain constant also) depending upon the condition that which factor dominates on the other. We can write F ∝ VS ρL

or

 (V + ∆VS )  F ′ VS′ ρL′ 1 = ⋅ = S ⋅  F VS ρL VS  1 + γ L ∆T   V + γS VS ∆T    1 = S    VS    1 + γ L ∆T 

Chapter 20

Thermometry, Thermal Expansion and Kinetic Theory of Gases — 181  1 + γ S ∆T  F′ = F    1 + γ L ∆T 

or

γS > γ L , F ′ > F

Now, if

′ < wapp and vice-versa. wapp

or

γS = γ L ,

And if or ˜

F′ = F

′ = wapp wapp

Effect of temperature on immersed fraction of a solid in floating condition When a solid, whose density is less than the density of liquid is floating in it, then

Fig. 20.10

Weight of the solid = Upthrust on solid from liquid ∴

…(i)

Vρ sg = Vi ρl g ρ s = density of solid

Here,

ρl = density of liquid Vi = immersed volume of solid and V = total volume of solid From Eq. (i), vi ρs = =f ρl v

…(ii)

where, f = immersed fraction of solid. With increase in temperature, ρ s and ρl both will decrease. Therefore, this fraction may increase, decrease or remain constant. At some higher temperature, ρ′ …(iii) f′ = s ρl ′ From Eqs. (ii) and (iii), we have f ′  ρ s ′   ρl  =    f  ρ s   ρl ′    1 =  (1 + γ l ∆θ)  1 + γ s∆θ  or

f ′  1 + γ l ∆θ  =  f  1 + γ s∆θ 

Now, if γ l > γ s, f ′> f or immersed fraction will increase. If γ l = γ s, f ′ = f or immersed fraction will remain unchanged and if, γ l < γ s, then f ′< f or immersed fraction will decrease.

182 — Waves and Thermodynamics ˜

Effect of temperature on the time period of a pendulum The time period of a simple pendulum is given by l T = 2π g T∝ l

or

As the temperature is increased, length of the pendulum and hence, time period gets increased or a pendulum clock becomes slow and it loses the time. l + ∆l T′ l′ = = T l l Here, we put ∆l = lα ∆θ in place of lα ∆T so as to avoid the confusion with change in time period. Thus, l + lα∆θ T′ = = (1 + α∆θ)1/ 2 T l 1 or ( if α ∆θ α Fe. 5. An iron ball has a diameter of 6 cm and is 0.010 mm too large to pass through a hole in a brass plate when the ball and plate are at a temperature of 30°C. At what temperature, the same for ball and plate, will the ball just pass through the hole? Take the values of α from Table 20.2.

6. (a) An aluminium measuring rod which is correct at 5°C, measures a certain distance as 88.42 cm at 35 ° C. Determine the error in measuring the distance due to the expansion of the rod. (b) If this aluminium rod measures a length of steel as 88.42 cm at 35°C, what is the correct length of the steel at 35° C?

7. A steel tape is calibrated at 20° C . On a cold day when the temperature is −15° C, what will be the percentage error in the tape?

20.5 Behaviour of Gases Gases are the most diffused form of matter. In a gas, the molecules are highly energetic and they can be widely separated from each other. The behaviour of gases and their properties derive from these facts. Sometimes the term vapour is used to describe a gas. Strictly speaking, a gas is a substance at a temperature above its boiling point. A vapour is the gaseous phase of a substance that under ordinary conditions, exists as a liquid or solid. One of the more obvious characteristics of gases is their ability to expand and fill any volume they are placed in. This contrasts with the behaviour exhibited by solids (fixed shape and volume) and liquids (fixed volume but indeterminate shape).

Ideal and Real Gas Laws Gases, unlike solids and liquids have indefinite shape and indefinite volume. As a result, they are subjected to pressure changes, volume changes and temperature changes. Real gas behaviour is simpler. By understanding ideal gas behaviour, real gas behaviour becomes more tangible.

186 — Waves and Thermodynamics Ideal Gases How do we describe an ideal gas? An ideal gas has the following properties : 1. An ideal gas is considered to be a "point mass". A point mass is a particle so small, its volume is very nearly zero. This means an ideal gas particle has virtually no volume. 2. Collisions between ideal gases are "elastic". This means that no attractive or repulsive forces are involved during collisions. Also, the kinetic energy of the gas molecules remains constant since these intermolecular forces are lacking. Volume and temperature are by now familiar concepts. Pressure, however, may need some explanation. Pressure is defined as a force per unit area. When gas molecules collide with the sides of a exert, they exert a force over that area of the container. This gives rise to the pressure inside the container.

Gas Laws Assuming permanent (or real) gases to be ideal, through experiments, it was established that gases irrespective of their nature obey the following laws :

Boyle’s Law According to this law, for a given mass of a gas, the volume of a gas at constant temperature (called isothermal process) is inversely proportional to its pressure, i.e. 1 (T = constant) V∝ p or pV = constant or piVi = p f V f Thus, p -V graph in an isothermal process is a rectangular hyperbola. Or pV versus p or V graph is a straight line parallel to p or V axis. p

pV T = constant

T = constant

V

p or V

Fig. 20.12

Charles’ Law According to this law, for a given mass of a gas the volume of a gas at constant pressure (called isobaric process) is directly proportional to its absolute temperature, i.e. V

V T

p = constant

T (in K)

Fig. 20.13

p = constant

p or T

Chapter 20

Thermometry, Thermal Expansion and Kinetic Theory of Gases — 187 V ∝T V = constant T

or

( p = constant) Vi V f = Ti T f

or

Thus, V-T graph in an isobaric process is a straight line passing through origin. OrV / T versus V or T graph is a straight line parallel to V or T axis.

Gay Lussac’s Law or Pressure Law According to this law, for a given mass of a gas the pressure of a gas at constant volume (called isochoric process) is directly proportional to its absolute temperature, i.e. pf pi p p ∝ T ( V = constant) or = constant or = T Ti T f p

p T V = constant T (in K)

p or T

Fig. 20.14

Thus, p - T graph in an isochoric process is a straight line passing through origin or p/ T versus p or T graph is a straight line parallel to p or T axis.

Ideal Gas Equation All the above four laws can be written in one single equation known as ideal gas equation. According to this equation. m pV = nRT = RT M In this equation, n = number of moles of the gas m = M m = total mass of the gas M = molecular mass of the gas and R = universal gas constant = 8.31 J/ mol-K = 2.0 cal/ mol-K The above three laws can be derived from this single equation. For example, for a given mass of a gas (m = constant) (Boyle’s law) pV = constant at constant temperature p (Pressure law) = constant at constant volume T V (Charles' law) = constant at constant pressure T

188 — Waves and Thermodynamics Avogadro's Law The next empirical gas law, we will look at, is called Avogadro's law. This law deals with the relationship between the volume and moles of a gas at constant pressure and temperature. According to this law, at same temperature and pressure equal volumes of all gases contain equal number of molecules. Let's derive this law from the ideal gas law. Give the moles and volume subscripts, since their conditions will change. pV1 = n1 RT and pV2 = n2 RT Collect terms. Divide each equation by pressure, p and divide each equation by their respective mole term n1 n2 p = = V1 V2 RT Thus, the number of molecules per unit volume is same for all gases at a fixed temperature and pressure. The number in 22.4 litres of any gas is 6.02 ×10 23 . This is known as Avogadro number and is denoted by NA . The mass of 22.4 litres of any gas is equal to its molecular weight in grams at S.T.P. (standard temperature 273 K and pressure 1 atm). This amount of substance is called a mole. k=

Note

R is called Boltzmann constant. Its value in SI unit is 1.38 × 10 −23 J / K. NA

Extra Points to Remember ˜

In our previous discussion, we have discussed Charles' law and pressure law in absolute temperature scale. In centigrade scale, these laws are as under : Charles' law When a given mass of a gas is heated at constant Vt pressure then for each 1°C rise in temperature the volume of the gas increases by a fraction α of its volume at 0°C. Thus, if the V0 volume of a given mass of a gas at 0°C is V0 , then on heating at constant pressure to t °C its volume will increase by V0αt . Therefore, if its volume at t °C be Vt , then t (°C) Vt = V0 + V0αt

or

Vt = V0 (1 + αt )

–273

Here, α is called the ‘volume coefficient’ of the gas. For all gases, 1 the experimental value of α is nearly per ° C. 273 t  ∴ Vt = V0  1 +   273  Thus, Vt versus t graph is a straight line with slope

V0 and positive intercept V0 . Further Vt = 0 at 273

t = – 273 °C. Pressure law According to this law, when a given mass of a gas is heated at constant volume then for each 1°C rise in temperature, the pressure of the gas increases by a fraction β of its pressure at 0°C. Thus, if the pressure of a given mass of a gas at 0°C be p0 , then on heating at constant volume to t °C, its pressure will increase by p0 βt . Therefore, if its pressure at t °C be pt , then pt = p0 + p0 βt or

O Fig. 20.15

pt = p0 (1 + βt )

pt

p0

–273

O Fig. 20.16

t (°C)

Chapter 20

Thermometry, Thermal Expansion and Kinetic Theory of Gases — 189

Here, β is called the ‘pressure coefficient’ of the gas. For all gases the experimental value of β is also t  1 per °C ⇒ pt = p0  1 +   273  273 pt versus t graph is as shown in Fig 20.16. ˜

The above forms of Charles' law and pressure law can be simply expressed in terms of absolute temperature. Let at constant pressure, the volume of a given mass of a gas at 0°C, t 1 °C and t 2 °C be V0 , V1 and V2 respectively. Then, t 273 + t 1  V1 = V0  1 + 1  = V0      273 273  t 273 + t 2  V2 = V0  1 + 2  = V0     273  273  V1 273 + t 1 T1 = = V2 273 + t 2 T2

∴

where, T1 and T2 are the absolute temperatures corresponding to t 1 °C and t 2 °C . Hence, V1 V2 V or = constant or V ∝ T = T1 T2 T This is the form of Charles’ law which we have already studied in article 20.5. In the similar manner, we can prove the pressure law. ˜

Under isobaric conditions (p = constant), V-T graph is a straight line passing through origin (where, T is in n  nR   nR  kelvin). The slope of this line is   T or slope of the line is directly proportional to .  as V =   p  p p slope =

˜

nR p

or slope ∝

n p

Similarly, under isochoric conditions (V = constant), p-T graph is a straight line passing through origin nR n whose slope is or slope is directly proportional to . V V Density of a gas The ideal gas equation is m pV = nRT = RT M m pM ( ρ = density) =ρ= ∴ V RT pM ∴ ρ= RT ρ

ρ

ρ

m = constant T = constant m = constant

p = constant m = constant T (K)

p

V

Fig. 20.17

From this equation, we can see that ρ- p graph is straight line passing through origin at constant 1 temperature ( ρ ∝ p) and ρ-T graph is a rectangular hyperbola at constant pressure  ρ ∝  . Similarly, for a  T 1 given mass of a gas ρ-V graph is a rectangular hyperbola  ρ ∝  .  V

190 — Waves and Thermodynamics V

Example 20.8 p-V diagrams of same mass of a gas are drawn at two different temperatures T1 and T2 . Explain whether T1 > T2 or T2 > T1 . p

T2 T1

V

Fig. 20.18

Solution

The ideal gas equation is

pV nR T ∝ pV if number of moles of the gas are kept constant. Here, mass of the gas is constant, which implies that number of moles are constant, i.e. T ∝ pV . In the given diagram, product of p and V for T2 is more than T1 at all points (keeping either p or V same for both graphs). Hence, pV = nRT

or T =

T2 > T1 V

Ans.

Example 20.9 The p-V diagram of two different masses m1 and m2 are drawn (as shown) at constant temperature T. State whether m1 > m2 or m2 > m1 . p T

T

m2 m1 V

Fig. 20.19

Solution

∴

m RT M M m = ( pV )   or  RT 

pV = nRT =

m ∝ pV if T = constant

From the graph, we can see that p 2V2 > p1V1 (for same p or V). Therefore, m2 > m1 V

Example 20.10 The p-T graph for the given mass of an ideal gas is shown in figure. What inference can be drawn regarding the change in volume (whether it is constant, increasing or decreasing)? How to Proceed Definitely, it is not constant. Because when volume of the gas is constant p-T graph is a straight line passing through origin. The given line does not pass through origin, hence volume is not constant.

Ans. p B A T (K)

Fig. 20.20

Chapter 20

Thermometry, Thermal Expansion and Kinetic Theory of Gases — 191 T  V = ( nR )    p

Now, to see the volume of the gas we will have to see whether Solution

T is increasing or decreasing. p

From the given graph, we can write the p-T equation as

p = aT + b Here, a and b are positive constants. Further, p b =a+ T T Now, TB > T A b b  p  p or   <   ∴ < T  B T  A TB T A or

T   p   >   p B  T  A

or

VB > V A

( y = mx + c )

Ans.

Thus, as we move from A to B, volume of the gas is increasing. V

Example 20.11 A gas at 27° C in a cylinder has a volume of 4 litre and pressure 100 Nm −2 . (i) Gas is first compressed at constant temperature so that the pressure is 150 Nm −2 . Calculate the change in volume. (ii) It is then heated at constant volume so that temperature becomes 127° C. Calculate the new pressure. Solution (i) Using Boyle's law for constant temperature, ∴

p1V1 = p 2V2 pV 100 × 4 V2 = 1 1 = = 2.667 L p2 150

∴

Change in volume = V2 − V1 = 2.667 − 4 = −1.333 L (ii) Using Gay Lussac's law for constant volume, p 2 T2 T (127 + 273) × 150 or p 2 = 2 × p1 = = ( 27 + 273) p1 T1 T1 = 200 Nm −2 V

Example 20.12 A balloon partially filled with helium has a volume of 30 m3 , at the earth's surface, where pressure is 76 cm of Hg and temperature is 27° C. What will be the increase in volume of gas if balloon rises to a height, where pressure is 7.6 cm of Hg and temperature is −54° C?

192 — Waves and Thermodynamics Solution

As

p1V1 pV = 2 2 T1 T2 V2 =

∴

=

p1V1T2 T1 p 2 76 × 30 × ( −54 + 273 ) ( 27 + 273 ) × 7.6

= 219 m 3 ∴ Increase in volume of gas = V2 − V1 = 219 − 30 = 189 m 3

INTRODUCTORY EXERCISE

20.3

1. From the graph for an ideal gas, state whether m1 or m 2 is greater. T V m2

V m1 p

Fig. 20.21

2. A vessel is filled with an ideal gas at a pressure of 20 atm and is at a temperature of 27°C. One-half of the mass is removed from the vessel and the temperature of the remaining gas is increased to 87°C. At this temperature, find the pressure of the gas.

3. A vessel contains a mixture of 7 g of nitrogen and 11 g of carbon dioxide at temperature T = 290 K. If pressure of the mixture is 1 atm ( = 1.01 × 105 N / m 2 ), calculate its density (R = 8.31J/mol -K ).

4. An electric bulb of volume 250 cm 3 was sealed off during manufacture at a pressure of 10–3 mm of mercury at 27°C. Compute the number of air molecules contained in the bulb. Given that R = 8.31J/mol -K and NA = 6.02 × 1023 per mol.

5. State whether p1 > p 2 or p 2 > p1 for given mass of a gas? V p2 p1

T

Fig. 20.22

6. For a given mass of a gas what is the shape of p versus

1 graph at constant temperature? V

7. For a given mass of a gas, what is the shape of pV versus T graph in isothermal process?

Chapter 20

Thermometry, Thermal Expansion and Kinetic Theory of Gases — 193

20.6 Degree of Freedom ( f ) The minimum number of ways in which motion of a body (or a system) can be described completely is called its degree of freedom. For example In Fig. (a), block has one degree of freedom, because it is confined to move in a straight line and has only one translational degree of freedom. f=1

f=2 f=2

Fig. 20.23

In Fig. (b), the particle has two degrees of freedom because it is confined to move in a plane and so it has two translational degrees of freedom. In Fig. (c), the sphere has two degrees of freedom one rotational and another translational. Similarly, a particle free to move in space will have three translational degrees of freedom.

Degree of Freedom of Gas Molecules A gas molecule can have, the following types of energies (i) translational kinetic energy (ii) rotational kinetic energy (iii) vibrational energy (potential + kinetic)

Vibrational Energy The forces between different atoms of a gas molecule (interatomic force) may be visualized by imagining every atom as being connected to its neighbours by springs. Each atom can vibrate along the line joining the atoms. Energy associated with this is called vibrational energy. Degree of Freedom of Monoatomic Gas A monoatomic gas molecule (like He) consists of a single atom. It can have translational motion in any direction in space. Thus, it has 3 translational degrees of freedom. f =3

( all translational)

It can also rotate but due to its small moment of inertia, rotational kinetic energy is neglected.

Degree of Freedom of a Diatomic and Linear Polyatomic Gas The molecules of a diatomic and linear polyatomic gas (like O 2 , CO 2 and H 2 ) cannot only move bodily but also rotate about anyone of the three coordinate axes as shown in figure. However, its moment of inertia about the axis joining the two atoms (x-axis) is negligible. Hence, it can have only two rotational degrees of freedom. Thus, a diatomic molecule has 5 degrees of freedom : 3 translational and 2 rotational. At sufficiently high temperatures, it has vibrational energy as well providing it two more degrees of freedom (one vibrational kinetic energy and another vibrational

194 — Waves and Thermodynamics potential energy). Thus, at high temperatures a diatomic molecule has 7 degrees of freedom, 3 translational, 2 rotational and 2 vibrational. Thus, z y

x

Fig. 20.24

f = 5 (3 translational + 2 rotational) at room temperatures and f = 7 (3 translational + 2 rotational + 2 vibrational) at high temperatures.

Degree of Freedom of Non-linear Polyatomic Gas A non-linear polyatomic molecule (such as NH 3 ) can rotate about any of three coordinate axes. Hence, it has 6 degrees of freedom 3 translational and 3 rotational. At room temperatures, a polyatomic gas molecule has insignificant vibrational energy. But at high enough temperatures it is also significant. So, it has 8 degrees of freedom 3 rotational, 3 translational and 2 vibrational. Thus, z y

x

Fig. 20.25

f = 6 (3 translational + 3 rotational) at room temperatures and f = 8 (3 translational + 3 rotational + 2 vibrational) at high temperatures.

Degree of Freedom of a Solid An atom in a solid has no degrees of freedom for translational and rotational motion. At high temperatures, due to vibration along 3 axes it has 3 × 2 = 6 degrees of freedom. f = 6 (all vibrational) at high temperatures Note

(i) Degrees of freedom of a diatomic and polyatomic gas depends on temperature and since there is no clear cut demarcation line above which vibrational energy become significant. Moreover, this temperature varies from gas to gas. On the other hand, for a monoatomic gas there is no such confusion. Degree of freedom here is 3 at all temperatures. Unless and until stated in the question you can take f = 3 for monoatomic gas, f = 5 for a diatomic gas and f = 6 for a non-linear polyatomic gas. (ii) When a diatomic or polyatomic gas dissociates into atoms it behaves as a monoatomic gas. Whose degrees of freedom are changed accordingly.

Chapter 20

Thermometry, Thermal Expansion and Kinetic Theory of Gases — 195

20.7 Internal Energy of an Ideal Gas Disordered Suppose a gas is contained in a closed vessel as shown in motion figure. If the container as a whole is moving with some speed, then this motion is called the ordered motion of the gas. Ordered motion Source of this motion is some external force. The zig-zag motion of gas molecules within the vessel is known as the disordered motion. This motion is directly related to the Fig. 20.26 temperature of the gas. As the temperature is increased, the disordered motion of the gas molecules gets fast. The internal energy (U) of the gas is concerned only with its disordered motion. It is in no way concerned with its ordered motion. When the temperature of the gas is increased, its disordered motion and hence its internal energy is increased.

Intermolecular forces in an ideal gas is zero. Thus, PE due to intermolecular forces of an ideal gas is zero. A monoatomic gas is having a single atom. Hence, its vibrational energy is zero. For dia and polyatomic gases vibrational energy is significant only at high temperatures. So, they also have only translational and rotational KE. We may thus conclude that at room temperature the internal energy of an ideal gas (whether it is mono, dia or poly) consists of only translational and rotational KE. Thus, U (of an ideal gas ) = KT + K R at room temperatures. Internal Energy (U )

Kinetic Energy

Potential Energy

Due to intermolecular forces

Due to interatomic forces (vibrational)

Translational Rotational KE KE

Vibrational KE

Fig. 20.27

Later in the next article, we will see that KT (translational KE) and K R (rotational KE) depends on T only. They are directly proportional to the absolute temperature of the gas. Thus, internal energy of an ideal gas depends only on its absolute temperature (T) and is directly proportional to T. U ∝T

20.8 Law of Equipartition of Energy An ideal gas is just like an ideal father. An ideal father distributes whole of its properties and assets equally among his children. Same is the case with an ideal gas. It distributes its internal energy equally in all degrees of freedom. In each degree of freedom energy of one mole of an ideal gas is

196 — Waves and Thermodynamics 1 RT, where T is the absolute temperature of the gas. Thus, if f be the number of degrees of freedom, 2 f the internal energy of 1 mole of the gas will be RT or internal energy of n moles of the gas will be 2 n fRT . Thus, 2 n …(i) U = fRT 2 For a monoatomic gas, f = 3. 3 (for 1 mole of a monoatomic gas) U = RT 2 For a dia and linear polyatomic gas at low temperatures, f = 5, so, 5 (for 1 mole) U = RT 2 and for non-linear polyatomic gas at low temperatures, f = 6, so 6 (for 1 mole) U = RT = 3RT 2

Therefore,

Note

(i) For one molecule, R (the gas constant) is replaced by k (the Boltzmann constant). For example, internal 1 energy of one molecule in one degree of freedom will be kT. 2 (ii) Ignoring the vibrational effects we can summarise the above results in tabular form as below.

Table 20.3 Degree of freedom

Nature of gas Total

Internal energy of 1 mole

Translational Rotational

Monoatomic

3

3

0

Dia or linear polyatomic

5

3

2

Non-linear polyatomic

6

3

3

Total

Translational Rotational

3 RT 2 5 RT 2 3 RT

3 RT 2 3 RT 2 3 RT 2

0 RT 3 RT 2

Internal energy of 1 molecule Total

3 kT 2 5 kT 2 3 kT

Translational Rotational

3 kT 2 3 kT 2 3 kT 2

0 kT 3 kT 2

(iii) From the above table, we can see that translational kinetic energy of all types of gases is same. The difference is in rotational kinetic energy. V

Example 20.13 temperature T. Solution

Find total internal energy of 3 moles of hydrogen gas at

Using the relation,

U=

nf RT 2

we have, n = 3, f = 5 for diatomic H2 gas. 3× 5 U= RT = 7.5 RT ∴ 2

Ans.

Chapter 20 V

Thermometry, Thermal Expansion and Kinetic Theory of Gases — 197

Example 20.14 Ten moles of O2 gas are kept at temperature T. At some higher temperature 2T, forty percent of molecular oxygen breaks into atomic oxygen. Find change in internal energy of the gas. Solution

Initial energy

Using the relation,U =

nf RT 2

we have, n = 10, f = 5 for diatomic O2 gas 10 × 5 Ui = ( RT ) = 25 RT 2 Final energy Forty percent means 4 moles O2 breaks into O. So, it will become 8 moles of monoatomic gas O. Remaining 6 moles are of diatomic gas O2 . But now the new temperature is 2T. 8× 3 6× 5 ∴ Uf = ( R ) ( 2T ) + ( R )( 2T ) = 54 RT 2 2 So, change in internal energy, V

∆U = U f − U i = 29 RT

Ans.

Example 20.15 At a given temperature internal energy of a monoatomic, diatomic and non-linear polyatomic gas is U 0 each. Find their translational and rotational kinetic energies separately. Monoatomic gas Translational degree of freedom of monoatomic gas is 3 and rotational degree of freedom is zero. Therefore, whole internal energy is in the form of translational kinetic energy. ∴ K T = U 0 and K R = 0 Diatomic gas Translational degree of freedom of diatomic gas is 3 and rotational degree of freedom is 2. Hence, ratio of translational and rotational kinetic energy will be 3 : 2 3 2 K T = U 0 and K R = U 0 ∴ 5 5 Solution

Non-linear polyatomic gas In non-linear polyatomic gas translational and rotational both degrees of freedom are 3 each. So, translational and rotational kinetic energy are equal. Hence, U KT = KR = 0 2

INTRODUCTORY EXERCISE

20.4

1. A gas mixture consists of 2 moles of oxygen and 4 moles of argon at temperature T. Neglecting (JEE 1999) all vibrational modes, the total internal energy of the system is (a) 4 RT (b) 15 RT (c) 9 RT (d) 11RT 2. The average translational kinetic energy of O2 (molar mass 32) molecules at a particular temperature is 0.048 eV. The translational kinetic energy of N2 (molar mass 28) molecules in eV at the same temperature is (JEE 1997) (a) 0.0015 (b) 0.003 (c) 0.048 (d) 0.768 3. At a given temperature, rotational kinetic energy of diatomic gas is K0. Find its translational and total kinetic energy.

198 — Waves and Thermodynamics

20.9 Molar Heat Capacity “Molar heat capacity C is the heat required to raise the temperature of 1 mole of a gas by 1°C (or 1 K).” Thus, ∆Q or ∆Q = nC∆T C= n∆T For a gas, the value of C depends on the process through which its temperature is raised. For example, in an isothermal process ∆T`= 0 or C iso = ∞. In an adiabatic process, (we will discuss it later) ∆Q = 0.Hence, C adi = 0. Thus, molar heat capacity of a gas varies from 0 to ∞ depending on the process. In general, experiments are made either at constant volume or at constant pressure. In case of solids and liquids, due to small thermal expansion, the difference in measured values of molar heat capacities is very small and is usually neglected. However, in case of gases molar heat capacity at constant volume CV is quite different from that at constant pressure C p. Later in the next chapter, we will derive the following relations, (for an ideal gas) f dU R CV = = R= dT 2 γ –1 C p = CV + R Cp 2 γ= =1 + CV f Here, U is the internal energy of one mole of the gas. The most general expression for C in the process pV x = constant is R R (we will derive it later) C= + γ –1 1– x For example : For isobaric process p = constant or x = 0 and R C =Cp = + R = CV + R γ –1 For an isothermal process, pV = constant or x =1 ∴ C = ∞ and For an adiabatic process pV γ = constant or x = γ ∴ C =0 Values of f , U , CV , C p and γ for different gases are given in Table 20.4. Table 20.4 Nature of gas

f

Monoatomic

3

Dia and linear polyatomic

5

Non-linear polyatomic

6

U=

f RT 2

CV = dU / dT =

f R 2

Cp = CV + R

3 RT 2 5 RT 2

3 R 2 5 R 2

5 R 2 7 R 2

3 RT

3R

4R

γ=

Cp CV

= 1+

5 = 1.67 3 7 = 1.4 5 4 = 1.33 3

2 f

Chapter 20

Thermometry, Thermal Expansion and Kinetic Theory of Gases — 199

Extra Points to Remember ˜

Mixture of non-reactive gases (a) n = n1 + n2 (b) p = p1 + p2 (c) U = U1 + U 2 (d) ∆ U = ∆ U1 + ∆U 2 n1C V1 + n2C V2 n C + n2C p 2 (f) C p = 1 p1 (e) C V = = CV + R n1 + n2 n1 + n2 C n + n2 n1 n2 n M + n2 M 2 n (h) M = 1 1 (g) γ = p or ⇒ 1 = + γ −1 γ −1 γ1 − 1 γ 2 − 1 CV n1 + n2

V

Example 20.16 Two moles of helium (He) are mixed with four moles of hydrogen ( H 2 ). Find (a) CV of the mixture (b) C p of the mixture and (c) γ of the mixture. Solution Helium is a monoatomic gas. 3 5 5 CV = R , C P = R and γ = ∴ 2 2 3 Hydrogen is a diatomic gas. 5 7 7 ∴ CV = R and CV = R and γ = 2 2 5 (a) C V of the mixture CV = =

n1CV1 + n 2CV2 n1 + n 2

=

3  5  ( 2) R + 4  R 2  2  2+ 4

13 R 6

Ans.

(b) C p of the mixture C p = CV + R = =

13 R+R 6

19 R 6

Ans.

Alternate method Cp =

= =

n1C p1 + n 2C p2 n1 + n 2 5  7  ( 2) R + ( 4 )  R 2  2  2+ 4 19 R 6

Ans.

200 — Waves and Thermodynamics (c) γ of the mixture γ= =

Cp CV

=

(19/ 6)R (13/ 6)R

19 13

Ans.

Alternate method n1 + n 2 n1 n2 = + γ −1 γ1 − 1 γ 2 − 1 2+ 4 2 4 = + 7 γ −1 5 −1 −1 3 5

∴

Solving this equation, we get γ= V

19 13

Example 20.17 Temperature of two moles of a monoatomic gas is increased by 300 K in the process p ∝ V . Find (a) molar heat capacity of the gas in the given process (b) heat required in the given process. Solution (a) Molar heat capacity C is given by R …(i) C = CV + 1− x in the process pV x = constant. Now, CV =

3 R for monoatomic gas. 2

Further, the given process can be written as PV −1 = constant ∴

x = −1

Substituting the values in Eq. (i), we have C=

3 R R+ = 2R 2 1+ 1

Ans.

(b) Q or ∆Q = nC∆T Substituting the values, we have ∆Q = ( 2)( 2R )( 300) = 1200R

Ans.

Chapter 20

Thermometry, Thermal Expansion and Kinetic Theory of Gases — 201

20.10 Kinetic Theory of Gases We have studied the mechanics of single particles. When we approach the mechanics associated with the many particles in systems such as gases, liquids and solids, we are faced with analyzing the dynamics of a huge number of particles. The dynamics of such many particle systems is called statistical mechanics. The game involved in studying a system with a large number of particles is similar to what happens after every physics test. Of course we are interested in our individual marks, but we also want to know the class average. The kinetic theory that we study in this article is a special aspect of the statistical mechanics of large number of particles. We begin with the simplest model for a monatomic ideal gas, a dilute gas whose particles are single atoms rather than molecules. Macroscopic variables of a gas are pressure, volume and temperature and microscopic properties are speed of gas molecules, momentum of molecules, etc. Kinetic theory of gases relates the microscopic properties to macroscopic properties. Furthermore, the kinetic theory provides us with a physical basis for our understanding of the concept of pressure and temperature.

The Ideal Gas Approximation We make the following assumptions while describing an ideal gas : 1. The number of particles in the gas is very large. 2. The volume V containing the gas is much larger than the total volume actually occupied by the gas particles themselves. 3. The dynamics of the particles is governed by Newton’s laws of motion. 4. The particles are equally likely to be moving in any direction. 5. The gas particles interact with each other and with the walls of the container only via elastic collisions. 6. The particles of the gas are identical and indistinguishable.

The Pressure of an Ideal Gas y

v m

d

vx z

d

d

x

A cubical box with sides of length d containing an ideal gas. The molecule shown moves with velocity v.

Fig. 20.28

202 — Waves and Thermodynamics Consider an ideal gas consisting of N molecules in a container of volume V. The container is a cube with edges of length d. Consider the collision of one molecule moving with a velocity v toward the right hand face of the cube. The molecule has velocity components v x , v y and v z .Previously, we used m to represent the mass of a sample, but in this article we shall use m to represent the mass of one molecule. As the molecule collides with the wall elastically its x-component of velocity is reversed, while its y and z- components of velocity remain unaltered. Because the x-component of the momentum of the molecule is mv x before the collision and –mv x after the collision, the change in momentum of the molecule is ∆px = – mv x – ( mv x ) = – 2mv x Applying impulse = change in momentum to the molecule F∆t = ∆px = – 2mv x v

vy

–vx

vy

v

vx A molecule makes an elastic collision with the wall of the container. Its x-component of momentum is reversed, while its y-component remains unchanged. In this construction, we assume that the molecule moves in the x-y plane.

Fig. 20.29

where, F is the magnitude of the average force exerted by the wall on the molecule in time ∆t. For the molecules to collide second time with the same wall, it must travel a distance 2d in the x-direction. 2d Therefore, the time interval between two collisions with the same wall is ∆t = . Over a time vx interval that is long compared with ∆t, the average force exerted on the molecules for each collision is –2mv x –2mv x – mv x2 F= = = ∆t 2d / v x d mv x2 . d Each molecule of the gas exerts a force on the wall. We find the total force exerted by all the molecules on the wall by adding the forces exerted by the individual molecules.

According to Newton’s third law, the average force exerted by the molecule on the wall is,

Chapter 20

Thermometry, Thermal Expansion and Kinetic Theory of Gases — 203

∴

m 2 ( v + v x22 +… + v x2N ) d x1 2 2 2 mN  v x1 + v x2 + K + v xN = d  N

F wall =

   

This can also be written as F wall = where,

v x2

=

Nm 2 v d x v x21 + v x22 +… + v x2N

N Since, the velocity has three components v x , v y and v z , we can have v 2 = v x2 + v y2 + v z2

(as v 2 = v x2 + v y2 + v z2 )

Because the motion is completely random, the average values v x2 , v y2 and v z2 are equal to each other. So, v 2 = 3 v x2 Therefore,

F wall

1 v x2 = v 2 3 2 N  mv  = 3  d  or

∴ Pressure on the wall p=

F wall F wall 1  N  = 2 =  3 mv 2   3 d A d

1N 2  N  1  =   mv 2 =    mv 2   3V  3  V  2 ∴

p=

1 mN 2 2  N  v =   3 V 3 V 

1 2  mv  2 

…(i)

This result indicates that the pressure is proportional to the number of molecules per unit volume 1 ( N /V ) and to the average translational kinetic energy of the molecules mv 2 . This result relates the 2 large scale quantity (macroscopic) of pressure to an atomic quantity (microscopic)—the average value of the square of the molecular speed. The above equation verifies some features of pressure with which you are probably familiar. One way to increase the pressure inside a container is to increase the number of molecules per unit volume in the container.

The Meaning of the Absolute Temperature Rewriting Eq. (i) in the more familiar form pV =

2 N 3

1 2  mv  2 

204 — Waves and Thermodynamics Let us now compare it with the ideal gas equation pV = nRT nRT =

2 1  N  mv 2   3 2 N NA

Here,

n=

∴

2  N  1  T =  A   mv 2      3 R 2

or

T=

2 1 2  mv   3k  2

( N A = Avogadro number)

…(ii)

where, k is Boltzmann’s constant which has the value R k= = 1.38 × 10 –23 J/ K NA By rearranging Eq. (ii) we can relate the translational molecular kinetic energy to the temperature 1 3 mv 2 = kT 2 2 3 That is, the average translational kinetic energy per molecule is kT. Further, 2 1 v x2 = v 2 , it follows that 3 1 1 mv 2 = kT 2 x 2 In the similar manner, it follows that 1 1 mvy2 = kT 2 2 1 1 and mv z2 = kT 2 2 1 Thus, in each translational degree of freedom one gas molecule has an energy kT.One mole of a gas 2 1 1 has N A number of molecules. Thus, one mole of the gas has an energy ( kN A ) T = RT in each 2 2 degree of freedom. Which is nothing but the law of equipartition of energy. The total translational 3 kinetic energy of one mole of an ideal gas is therefore, RT. 2 3 (KE)Trans = RT (of one mole) 2

Chapter 20

Thermometry, Thermal Expansion and Kinetic Theory of Gases — 205

Root Mean Square Speed The square root of v 2 is called the root mean square (rms) speed of the molecules. From Eq. (ii), we obtain, for the rms speed v rms = v 2 = k=

Using

R , NA

3kT m mNA = M

RT p = M ρ

and we can write,

v rms =

3p 3kT 3RT = = m M ρ

Mean Speed or Average Speed The particles of a gas have a range of speeds. The average speed is found by taking the average of the speeds of all the particles at a given instant. Remember that the speed is a positive scalar since it is the magnitude of the velocity. v + v 2 +… + v N v av = 1 N From Maxwellian speed distribution law, we can show that v av =

8p 8kT 8RT = = πm πM πρ

Most Probable Speed This is defined as the speed which is possessed by maximum fraction of total number of molecules of the gas. For example, if speeds of 10 molecules of a gas are, 1, 2, 2, 3, 3, 3, 4, 5, 6, 6 km/s, then the most probable speed is 3 km/s, as maximum fraction of total molecules possess this speed. Again from Maxwellian speed distribution law : v mp = Note

2p 2kT 2RT = = m M ρ

1. In the above expressions of v rms, v av and v mp , M is the molar mass in kilogram per mole. For example, molar mass of hydrogen is 2 × 10 –3 kg/mol. 2. v rms > v av > v mp (RAM) 8 3. v rms : v av : v mp = 3 : : 2 π 8 and since, ≈ 2.5, we have v rms : v av : v mp = 3 : 2.5 : 2 π

206 — Waves and Thermodynamics Extra Points to Remember ˜

Pressure exerted by an ideal gas is numerically equal to two-third of the mean kinetic energy of translation per unit volume of (E ) the gas. Thus, p=

2 1 mN 2 E= vrms 3 3 V

Here, m = mass of one gas molecule N = total number of molecules ˜

Mean Free Path Every gas consists of a very large number of molecules. These molecules are in a state of continuous rapid and random motion. They undergo perfectly elastic collisions against one another. Therefore, path of a single gas molecule consists of a series of short zig-zag paths of different lengths. The mean free path of a gas molecule is the average distance between two successive collisions. It is represented by λ. λ=

kT 2 πσ 2ρ

Here, σ = diameter of the molecule k = Boltzmann’s constant ˜

Avogadro’s Hypothesis At constant temperature and pressure equal volumes of different gases contain equal number of molecules. In 1 g-mole of any gas there are 6.02 × 1023 molecules of that gas. This is called Avogadro’s number. Thus, N = 6.02 × 1023 per g-mole Therefore, the number of molecules in mass m of the substance Number of molecules = nN =

˜

m ×N M

Dalton’ Law of Partial Pressure According to this law if the gases filled in a vessel do not react chemically, then the combined pressure of all the gases is due to the partial pressure of the molecules of the individual gases. If p1, p2 , … represent the partial pressures of the different gases, then the total pressure is, p = p1 + p2 …

V

Example 20.18 Find the rms speed of hydrogen molecules at room temperature ( = 300 K ). Solution

Mass of 1 mole of hydrogen gas = 2g = 2 × 10−3 kg

⇒

v rms = =

3RT M 3 × 8.31 × 300 2 × 10−3

= 1.93 × 103 m/s

Ans.

Thermometry, Thermal Expansion and Kinetic Theory of Gases — 207

Chapter 20 V

Example 20.19 A tank used for filling helium balloons has a volume of 0.3 m3 and contains 2.0 mol of helium gas at 20.0°C. Assuming that the helium behaves like an ideal gas. (a) What is the total translational kinetic energy of the molecules of the gas? (b) What is the average kinetic energy per molecule? 3 Solution (a) Using (KE) Trans = nRT 2 with n = 2.0 mol and T = 273 + 20 = 293 K, we find that 3 ( KE) Trans = ( 2.0) ( 8.31) ( 293) 2 = 7.3 × 103 J (b) The average kinetic energy per molecule is

3 kT. 2

1 1 3 2 mv 2 = mv rms = kT 2 2 2 3 = (1.38 × 10–23 ) ( 293) 2

or

= 6.07 × 10–21 J V

Ans.

Ans.

Example 20.20 Consider an 1100 particles gas system with speeds distribution as follows : 1000 particles each with speed 100 m/s 2000 particles each with speed 200 m/s 4000 particles each with speed 300 m/s 3000 particles each with speed 400 m/s and 1000 particles each with speed 500 m/s Find the average speed, and rms speed. Solution

The average speed is (1000) (100) + ( 2000) ( 200) + ( 4000) ( 300) + ( 3000) ( 400) + (1000) ( 500) v av = 1100 = 309 m/s

Ans.

The rms speed is v rms =

(1000) (100) 2 + ( 2000) ( 200) 2 + ( 4000) ( 300) 2 + ( 3000) ( 400) 2 + (1000) ( 500) 2 1100

= 328 m/ s Note Here,

v rms 3 as values and gas molecules are arbitrarily taken. ≠ v av 8/ π

Ans.

208 — Waves and Thermodynamics V

Example 20.21 Calculate the change in internal energy of 3.0 mol of helium gas when its temperature is increased by 2.0 K. Solution Helium is a monatomic gas. Internal energy of n moles of the gas is 3 U = nRT 2 3 ∴ ∆U = nR ( ∆T ) 2 Substituting the values,  3 ∆U =   ( 3) ( 8.31) ( 2.0)  2 = 74.8 J

V

Ans.

Example 20.22 In a crude model of a rotating diatomic molecule of chlorine (Cl2 ), the two Cl atoms are 2.0 × 10 −10 m apart and rotate about their centre of mass with angular speed ω = 2.0 × 1012 rad/s. What is the rotational kinetic energy of one molecule of Cl2 , which has a molar mass of 70.0 g/mol? ω

m

m

Cl

Cl r

r

Fig. 20.30

Solution

Moment of inertia, I = 2 ( mr 2 ) = 2mr 2

Here,

m=

70 × 10–3 2 × 6.02 × 1023

= 5.81 × 10–26 kg and

r=

2.0 × 10–10 2

= 1.0 × 10–10 m ∴

I = 2 ( 5.81 × 10–26 ) (1.0 × 10–10 ) 2 = 1.16 × 10–45 kg -m 2

∴

1 2 Iω 2 1 = × (1.16 × 10–45 ) × ( 2.0 × 1012 ) 2 2

KR =

= 2.32 × 10–21 J

Ans.

Chapter 20 V

Thermometry, Thermal Expansion and Kinetic Theory of Gases — 209

Example 20.23 Prove that the pressure of an ideal gas is numerically equal to two third of the mean translational kinetic energy per unit volume of the gas. Solution

Translational KE per unit volume 1 E = ( mass per unit volume ) ( v 2 ) 2 =

or

p=

 3 p 3 1 (ρ )   = p 2  ρ 2 2 E 3

INTRODUCTORY EXERCISE

Hence Proved.

20.5

1. Calculate the root mean square speed of hydrogen molecules at 373.15 K. 2. Five gas molecules chosen at random are found to have speed of 500, 600, 700, 800 and 900 m/s. Find the rms speed. Is it the same as the average speed?

3. The average speed of all the molecules in a gas at a given instant is not zero, whereas the average velocity of all the molecules is zero. Explain why?

4. A sample of helium gas is at a temperature of 300 K and a pressure of 0.5 atm. What is the average kinetic energy of a molecule of a gas?

5. A sample of helium and neon gases has a temperature of 300 K and pressure of 1.0 atm. The molar mass of helium is 4.0 g/mol and that of neon is 20.2 g/mol. (a) Find the rms speed of the helium atoms and of the neon atoms. (b) What is the average kinetic energy per atom of each gas?

6. At what temperature will the particles in a sample of helium gas have an rms speed of 1.0 km/s? 7. For any distribution of speeds vrms ≥ v av. Is this statement true or false?

210 — Waves and Thermodynamics

Final Touch Points 1. Departures from ideal gas behaviour for a real gas An ideal gas is a simple theoretical model of a gas. No real gas is truly ideal. Figure shows departures from ideal gas behaviour for a real gas at three different temperatures. Notice that all curves approach the ideal gas behaviour for low pressures and high temperatures. Ideal gas

pV RT

T1 T2

T1 >T2 >T3 T3

0

200

400 600 p

800

At low pressures or high temperatures the molecules are far apart and molecular interactions are negligible. Without interactions, the gas behaves like an ideal one .

2. van der Waal’s Equation Experiments have proved that real gases deviate largely from ideal behaviour. The reason of this deviation is two wrong assumptions in the kinetic theory of gases. (i) The size of the molecules is much smaller in comparison to the volume of the gas, hence, it may be neglected. (ii) Molecules do not exert intermolecular force on each other. van der Waal made corrections for these assumptions and gave a new equation. This equation is known as van der Waal’s equation for real gases. (a) Correction for the finite size of molecules: Molecules occupy some volume. Therefore, the volume in which they perform thermal motion is less than the observed volume of the gas. It is represented by (V − b ). Here, b is a constant which depends on the effective size and number of molecules of the gas. Therefore, we should use (V − b ) in place of V in gas equation. (b) Correction for intermolecular attraction: Due to the intermolecular force between gas molecules the molecules which are very near to the wall experiences a net inward force. Due to this inward force there is a decrease in momentum of the particles of a gas. Thus, the pressure exerted by real gas molecules is less than the pressure exerted by the molecules of a  an ideal gas. So, we use  p + 2  in place of p in gas equation. Here, again a is a constant.  V  van der Waal’s equation of state for real gases thus becomes, a   p + 2  (V − b ) = RT  V 

3. Critical Temperature, Pressure and Volume : Gases can’t be liquified above a temperature called critical temperature (TC ) however large the pressure may be. The pressure required to liquify the gas at critical temperature is called critical pressure ( pC ) and the volume of the gas at critical temperature and pressure is called critical volume (VC ). Value of critical constants in terms of van der Waal’s constants a and b are as under : VC = 3 b, Further,

pC =

a 27b 2

and

TC =

8a 27Rb

RTC 8 = is called critical coefficient and is same for all gases. pCVC 3

Chapter 20

Thermometry, Thermal Expansion and Kinetic Theory of Gases — 211

4. Detailed Discussion on Molar Heat Capacity 3 R. No variation is observed in this. So, value of CV ,Cp ,Cp – CV 2 and γ comes out to be same for different monatomic gases (Table 20.5).

For monatomic gases value of CV is

Table 20.5 Cp

CV

Cp − CV

γ = Cp /CV

He

20.8

12.5

8.33

1.67

Ar

20.8

12.5

8.33

1.67

H2

28.8

20.4

8.33

1.41 1.40

Gas Monoatomic Gases

Diatomic Gases N2

29.1

20.8

8.33

O2

29.4

21.1

8.33

1.40

CO 2

29.3

21.0

8.33

1.40

Cl 2

34.7

25.7

8.96

1.35

CO 2

37.0

28.5

8.50

1.30

SO 2

40.4

31.4

9.00

1.29

H2O

35.4

27.0

8.37

1.30

CH2

35.5

27.1

8.41

1.31

Polyatomic Gases

*All values except that for water were obtained at 300 K. SI units are used for Cp and CV . For dia and polyatomic gases these values are not equal for different gases. These values vary from gas to gas. Even for one gas values are different at different temperatures. Molar specific heat at constant volume (CV)

4R

Vibration

3R

Rotation

2R

7R (7 degrees 2 of freedom active) 5R (5 degrees 2 of freedom active) 3R (3 degrees 2 of freedom active)

R Translation 0 20 50

500 100 2000 10000 200 1000 5000 Temperature (K)

Figure illustrates the variation in the molar specific heat (at constant volume) for H2 over a wide range 3R which is in temperatures. (Note that T is drawn on a logarithmic scale). Below about 100 K, CV is 2

212 — Waves and Thermodynamics 5R which 2 includes the two rotational degrees of freedom. It seems, therefore, that at low temperatures, rotation 7R is not allowed. At high temperatures, CV starts to rise toward the value . Thus, the vibrational 2 degrees of freedom contribute only at these high temperatures. In Table 20.4 the large values of CV for some polyatomic molecules show the contributions of vibrational energy. In addition, a molecule with three or more atoms that are not in a straight line has three, not two, rotational degrees of freedom. characteristic of three translational degrees of freedom. At room temperature (300 K) it is

From this discussion, we expect heat capacities to be temperature-dependent, generally increasing with increasing temperature.

5. Solids In crystalline solids (monoatomic), the atoms are arranged in a three dimensional array, called a lattice. Each atom in a lattice can vibrate along three mutually perpendicular directions, each of which has two degrees of freedom. One corresponding to vibrations KE and the other vibrational PE. Thus, each atom has a total of six degrees of freedom. The volume of a solid does not change significantly with temperature, and so there is little difference between CV and Cp for a solid. The molar heat capacity is expected to be

C=

f 6 R = R 2 2

C = 3R

or

(ideal monatomic solid)

Its numerical value is C ≈ 25 J / mol -K ≈ 6 cal/ mol -K. This result was first found experimentally by Dulong and Petit. Figure shows that the Dulong and Petit’s law is obeyed quite well at high ( > 250 K ) temperatures. At low temperatures, the heat capacities decreases. C (J/mol-K) 3R

100 200 300 400

T(K)

Solved Examples TYPED PROBLEMS Type 1. Conversion of graph

Concept (i) In heat, we normally find the following graphs: S.No

Equation type

Nature of graph

1.

y∝x 1 y∝ x

Straight line passing through origin

y = constant

Straight line parallel to x -axis

4.

x = constant

Straight line parallel to y -axis

5.

x = constant and y = constant

Dot

2. 3.

Rectangular hyperbola

(ii) Density of any substance is given by m V For given mass of any substance, density only depends on V . It varies with V as 1 ρ∝ V If V increases, ρ decreases and vice-versa. (iii) For an ideal gas, density is also given as pM ρ= RT For a given gas, p ρ∝ T (iv) For an ideal gas, pV = nRT nf and U = RT 2 For given moles of the gas, U ∝ T and T ∝ pV If product of p and V is increasing, it means temperature is increasing. So, U is increasing. For example, in isothermal process, T is constant. Therefore, U is constant. Hence, U versus T graph is a dot. ρ=

214 — Waves and Thermodynamics V

Example 1

p d

c

b a T

Corresponding to p-T graph as shown in figure, draw (a) p-V graph (c) ρ-T graph and

(b) V-T graph (d) U-T graph

Solution ab process From the given graph, we can see that p∝T ⇒

V = constant

or

ρ = constant

(isochoric) 1  as ρ ∝   V

p and T both are increasing. Therefore, U is also increasing (as U ∝ T ) Now, (i) p - V graph is straight line parallel to p-axis as V is constant. (ii) V - T graph is a straight line parallel to T -axis as V is constant. (iii) ρ -T graph is a straight line parallel to T -axis as ρ is constant. (iv)U - T graph is a straight line passing through origin as U ∝ T . bc process From the given graph, we can see that T = constant ⇒

U = constant

⇒

pV = constant 1 p∝ V

or

(isothermal)

p is increasing. Therefore, V will decrease. Hence, ρ will increase. Now, 1  (i) p - V graph is a rectangular hyperbola as P ∝  .  V (ii) V - T graph is a straight line parallel to V -axis, as T = constant (iii) ρ -T graph is a straight line parallel to ρ -axis, as T = constant (iv)U - T graph is a dot, as U and T both are constants. cd process From the given graph, we can see that p = constant ⇒

V ∝T

Temperature is decreasing. So, volume will also decrease. But density will increase.

(isobaric)

Chapter 20

Thermometry, Thermal Expansion and Kinetic Theory of Gases — 215

Now, (i) p - V graph is a straight line parallel to V -axis because p is constant. (ii) V - T graph is a straight line passing through origin, as V ∝ T . pM 1 as p, M and R all are constants. Hence, ρ -T graph is rectangular (iii) ρ = ⇒ ρ∝ RT T hyperbola. (iv)U - T graph is a straight line passing through origin as U ∝ T . da process This process is just inverse of bc process. So, this process will complete the cycle following the steps discussed in process bc. The four graphs are as shown below. p d

V

c

a

b

b c d

a V ρ

T U

d

b, c c a, d a

b

T

T

Type 2. When ordered motion of a gas converts into disordered motion

Concept In article 20.7, we have seen that if a container filled with a gas is in motion, then the gas molecules have some ordered motion also. When the container is suddenly stopped, this ordered motion of gas molecules converts into disordered motion. Therefore, internal energy (and hence the temperature) of the gas will increase. In such situation, we can apply the equation, 1 mv 2 = ∆U 2 nf where, U = RT 2 nf ∆U = R∆ T ∴ 2 m where, n= M

216 — Waves and Thermodynamics V

Example 2 An insulated box containing a monoatomic gas of molar mass M moving with a speed v0 is suddenly stopped. Find the increment in gas temperature as a result of stopping the box. (JEE 2003) Solution Decrease in kinetic energy = increase in internal energy of the gas 1 nf  m mv02 = R∆ T =    M 2 2

∴

∆T =

∴

3   R ∆T 2 

Mv02 3R

Type 3. Based on the concept of pressure exerted by gas molecules striking elastically with walls of a container.

Concept (i) Gas molecules are assumed to be moving with v rms . (ii) They strike with the wall elastically. So, in each collision change in linear momentum is 2mv rms . 2L (iii) Time interval between two successive collisions is if we have a cubical box of side v rms L and, ∆p F (iv) F = , Pressure = ∆t A V

Example 3 A cubical box of side 1 m contains helium gas (atomic weight 4) at a pressure of 100 N / m2 . During an observation time of 1 second, an atom travelling with the root-mean-square speed parallel to one of the edges of the cube, was found to make 500 hits with a particular wall, without any collision with 25 other atoms. Take R = J / mol- K and k = 1.38 × 10 –23 J / K . 3 (a) Evaluate the temperature of the gas. (b) Evaluate the average kinetic energy per atom. (c) Evaluate the total mass of helium gas in the box. Solution Volume of the box = 1 m3 , pressure of the gas = 100 N/m2. Let T be the temperature of the gas. (a) Time between two consecutive collisions with one wall = This time should be equal to

2l , where l is the side of the cube. vrms 2l 1 = vrms 500

or ∴

1 s. 500

vrms = 1000 m/s 3RT = 1000 M

l

(as l = 1 m)

Chapter 20

Thermometry, Thermal Expansion and Kinetic Theory of Gases — 217 T=

or

(1000)2M (106 ) (4 × 10–3 ) = 3R  25 3  3

= 160 K 3 (b) Average kinetic energy per atom = kT 2 3 = (1.38 × 10–23 ) (160) J 2 = 3.312 × 10−21 J

Ans.

Ans.

m (c) From pV = nRT = RT M We get mass of helium gas in the box, m=

pVM RT

Substituting the values, we get m=

(100) (1) (4 × 10–3 )  25   (160) 3

= 3.0 × 10−4 kg V

Example 4 in a vessel.

Ans.

1 g mole of oxygen at 27°C and 1 atmospheric pressure is enclosed

(a) Assuming the molecules to be moving with vrms , find the number of collisions per second which the molecules make with one square metre area of the vessel wall. (b) The vessel is next thermally insulated and moved with a constant speed v0. It is then suddenly stopped. The process results in a rise of temperature of the gas by 1°C. Calculate the speed v0. [k = 1.38 × 10–23 J / K and NA = 6.02 × 1023 / mol ] (JEE 2002) Solution (a) Mass of one oxygen molecule, m= =

vrms

M NA 32 g 6.02 × 1023

= 5.316 × 10–23 g = 5.316 × 10–26 kg 3kT = m =

3 × 1.38 × 10–23 × 300 5.316 × 10–26

= 483.35 m/s Change in momentum per collision, ∆p = mvrms – (–mvrms ) = 2mvrms = (2) (5.316 × 10–26 ) (483.35) = 5.14 × 10–23 kg -m/s

218 — Waves and Thermodynamics Now, suppose n particles strike per second F = n∆p = (n ) (5.14 × 10–23 ) N p=

Now, as

dp   Fext =   dt 

F , for unit area F = p A

∴ (n ) (5.14 × 10–23 ) = 1.01 × 105 or Ans. n = 1.965 × 1027 per second (b) When the vessel is stopped, the ordered motion of the vessel converts into disordered motion and temperature of the gas is increased. 1 …(i) mv02 = ∆U ∴ 2 5 (for 1 g mole of O2) U = RT 2 5 ∴ ∆ U = R∆ T 2 Here, m is not the mass of one gas molecule but it is the mass of the whole gas. m = mass of 1 mol = 32 × 10–3 kg Substituting these values in Eq. (i), we get 5 R∆ T v0 = m =

5 × 8.31 × 1 32 × 10–3

= 36 m/s

Ans.

Miscellaneous Examples V

5R occupies a volume V i at a 2 pressure pi . The gas undergoes a process in which the pressure is proportional to the volume. At the end of the process, it is found that the rms speed of the gas molecules has doubled from its initial value. Determine the amount of energy transferred to the gas by heat.

Example 5 An ideal diatomic gas with CV =

Solution

Given that,

p∝V

or

pV –1 = constant

As we know, molar heat capacity in the process pV x = constant is R R R C= + = CV + γ–1 1–x 1–x In the given problem, 5R and x = – 1 2 5R R C= + = 3R 2 2

CV = ∴

…(i)

Chapter 20

Thermometry, Thermal Expansion and Kinetic Theory of Gases — 219

At the end of the process rms speed is doubled, i.e. temperature has become four times (vrms ∝ T ). Now, ∆Q = nC∆T = nC (Tf – Ti ) = nC (4Ti – Ti ) = 3TinC = (3Ti ) (n ) (3R) = 9 (nRTi ) or ∆Q = 9PV i i V

Ans.

Example 6 Given, Avogadro's number N = 6.02 × 1023 and Boltzmann’s constant k = 1.38 × 10 −23 J/K. (a) Calculate the average kinetic energy of translation of the molecules of an ideal gas at 0°C and at 100°C. (b) Also calculate the corresponding energies per mole of the gas. Solution

(a) According to the kinetic theory, the average kinetic energy of translation per  3 molecule of an ideal gas at kelvin temperature T is   kT, where k is Boltzmann's constant.  2 3 At 0° C (T = 273 K), the kinetic energy of translation = kT 2 3 −23 = × (1.38 × 10 ) × 273 = 5.65 × 10−21 J/mole 2 At 100° C (T = 373 K), the energy is 3 × (1.38 × 10−23 ) × 373 = 7.72 × 10−21 J/mole 2 (b) 1 mole of gas contains N (= 6.02 × 1023 ) molecules. Therefore, at 0° C, the kinetic energy of translation of 1 mole of the gas is = (5.65 × 10−21 ) (6.02 × 1023 ) ≈ 3401 J/mol and at 100° C kinetic energy of translation of 1 mole of gas is = (7.72 × 10−21 ) (6.02 × 1023 ) ≈ 4647 J/mol V

Example 7 An air bubble starts rising from the bottom of a lake. Its diameter is 3.6 mm at the bottom and 4 mm at the surface. The depth of the lake is 250 cm and the temperature at the surface is 40°C. What is the temperature at the bottom of the lake? Given atmospheric pressure = 76 cm of Hg and g = 980 cm/s 2 . Solution

At the bottom of the lake, volume of the bubble 4 4 V1 = πr13 = π (0.18)3 cm3 3 3

Pressure on the bubble p1 = Atmospheric pressure + Pressure due to a column of 250 cm of water = 76 × 13.6 × 980 + 250 × 1 × 980 = (76 × 13.6 + 250) 980 dyne/cm 2

220 — Waves and Thermodynamics At the surface of the lake, volume of the bubble 4 4 V 2 = πr23 = π (0.2)3 cm3 3 3 Pressure on the bubble, p2 = atmospheric pressure = (76 × 13.6 × 980) dyne/cm 2 T2 = 273 + 40° C = 313° K p1V1 p2V 2 = T1 T2

Now

 4  4 (76 × 13.6 + 250) 980 ×   π (0.18)3 (76 × 13.6) × 980   π (0.2)3  3  3 = 313 T1

or

T1 = 283.37 K T1 = 283.37 − 273 = 10.37° C

or ∴

V

Example 8 p-V diagram of n moles of an ideal gas is as shown in figure. Find the maximum temperature between A and B. p 2p0

A

B

p0 V0

How to Proceed

V 2V0

For given number of moles of a gas,

( pV = nRT ) T ∝ pV Although ( pV )A = ( pV )B = 2 p0V 0 or TA = TB , yet it is not an isothermal process. Because in isothermal process p-V graph is a rectangular hyperbola while it is a straight line. So, to see the behaviour of temperature, first we will find either T-V equation or T-p equation and from that equation we can judge how the temperature varies. From the graph, first we will write p-V equation, then we will convert it either in T-V equation or in T-p equation with the help of equation, pV = nRT .

Solution

or or or

From the graph the p-V equation can be written as p  p = –  0  V + 3 p0  V0 

( y = – mx + c)

p  pV = –  0  V 2 + 3 p0V  V0  p  nRT = 3 p0V –  0  V 2  V0  T=

 p0  2 1  3 p0V –   V  nR   V0  

(as pV = nRT )

Chapter 20

Thermometry, Thermal Expansion and Kinetic Theory of Gases — 221

This is the required T-V equation. This is quadratic in V. Hence, T-V graph is a parabola. Now, to find maximum or minimum value of T, we can substitute. dT =0 dV 2p  3 or 3 p0 –  0  V = 0 or V = V 0 2  V0  d 2T 3 is negative at V = V 0. 2 2 dV 3 Hence, T is maximum at V = V 0 and this maximum value is 2 2 1   3 V 0   p0   3 V 0   Tmax =  –     (3 p0 )   2   V 0   2   nR  9p V or Tmax = 0 0 4nR Thus, T-V graph is as shown in figure.

Further

Ans.

T

Tmax

TA = TB

A

V0

TA = TB =

2 p0V 0 nR

B

3V0 2

2V0

and Tmax = = 2.25

V

9 p0V 0 4nR

p0V 0 nR

Note Most of the problems of Tmax, pmax and Vmax are solved by differentiation. Sometimes graph will be given and sometimes direct equation will be given. For pmax you will require either p-V or p-T equation. V

Example 9 Plot p-V, V-T and ρ-T graph corresponding to the p-T graph for an ideal gas shown in figure. p B

C

A

D T

222 — Waves and Thermodynamics Process AB is an isothermal process with T = constant and pB > pA .

Solution p

B

ρ

D

V

B

A

C

C

C A

B

D

A T

V

D

T

1 i.e. p-V graph is a rectangular hyperbola with pB > pA V and VB < V A. V-T graph : T = constant. Therefore, V-T graph is a straight line parallel to V -axis with VB < V A. p-V graph :

p∝

ρ-T graph : ρ =

pM RT ρ∝ p

or

As T is constant. Therefore, ρ-T graph is a straight line parallel to ρ-axis with ρB > ρ A as pB > pA. Process BC is an isobaric process with P = constant TC > TB .

and

p-V graph : As p is constant. Therefore, p-V graph is a straight line parallel to V-axis with VC > VB (because V ∝ T in an isobaric process) V-T graph : with TC > TB

In isobaric process V ∝ T , i.e. V-T graph is a straight line passing through origin,

and ρ-T graph : ρ ∝ and

VC > VB . 1 (when P = constant), i.e. ρ-T graph is a hyperbola with TC > TB T ρC < ρB

There is no need of discussing C-D and D-A processes. As they are opposite to AB and BC respectively. The corresponding three graphs are shown above.

Exercises LEVEL 1 Assertion and Reason Directions :

Choose the correct option.

(a) If both Assertion and Reason are true and the Reason is correct explanation of the Assertion.

(b) If both Assertion and Reason are true but Reason is not the correct explanation of Assertion. (c) If Assertion is true, but the Reason is false. (d) If Assertion is false but the Reason is true. 1. Assertion : Straight line on p-T graph for an ideal gas represents isochoric process. Reason :

If p ∝ T , V = constant.

2. Assertion : Vibrational kinetic energy is insignificant at low temperatures. Reason :

Interatomic forces are responsible for vibrational kinetic energy. 2 3. Assertion : In the formula p = E, the term E represents translational kinetic energy per 3 unit volume of gas. Reason : In case of monoatomic gas, translational kinetic energy and total kinetic energy are equal.

4. Assertion : If a gas container is placed in a moving train, the temperature of gas will increase. Reason :

Kinetic energy of gas molecules will increase.

5. Assertion : According to the law of equipartition of energy, internal energy of an ideal gas at a given temperature, is equally distributed in translational and rotational kinetic energies. Reason : Rotational kinetic energy of a monoatomic gas is zero

6. Assertion : Real gases behave as ideal gases most closely at low pressure and high temperature. Reason : Intermolecular force between ideal gas molecules is assumed to be zero.

7. Assertion : A glass of water is filled at 4°C. Water will overflow, if temperature is increased or decreased. (Ignore expansion of glass). Reason : Density of water is minimum at 4°C.

8. Assertion : If pressure of an ideal gas is doubled and volume is halved, then its internal energy will remain unchanged. Reason : Internal energy of an ideal gas is a function of only temperature. 1 9. Assertion : In equation p = α v 2rms , the term α represents density of gas. 3 3RT Reason : v rms = . M

224 — Waves and Thermodynamics 10. Assertion : In isobaric process, V-T graph is a straight line passing through origin. Slope of this line is directly proportional to mass of the gas. V is taken on y-axis.  nR  Reason : V =  T  p ∴

slope ∝ n

or

slope ∝ m

Objective Questions 1. The average velocity of molecules of a gas of molecular weight M at temperature T is (a)

3RT M

(b)

8RT πM

(c)

2RT M

(d) zero

2. Four particles have velocities 1, 0, 2 and 3 m/s. The root mean square velocity of the particles (definition wise) is (a) 3.5 m/s

(b)

(c) 1.5 m/s

(d)

3.5 m/s 14 m/s 3

3. The temperature of an ideal gas is increased from 27°C to 927°C. The rms speed of its molecules becomes (a) twice (c) four times

(b) half (d) one-fourth

4. In case of hydrogen and oxygen at NTP, which of the following is the same for both? (a) Average linear momentum per molecule (c) KE per unit volume

(b) Average KE per molecule (d) KE per unit mass

5. The average kinetic energy of the molecules of an ideal gas at 10°C has the value E. The temperature at which the kinetic energy of the same gas becomes 2E is (a) 5°C (c) 40°C

(b) 10°C (d) None of these

6. A polyatomic gas with n degrees of freedom has a mean energy per molecule given by n RT 2 n (c) kT 2 (a)

1 RT 2 1 (d) kT 2 (b)

7. In a process, the pressure of a gas remains constant. If the temperature is doubled, then the change in the volume will be (a) 100% (c) 50%

(b) 200% (d) 25%

8. A steel rod of length 1 m is heated from 25° to 75°C keeping its length constant. The longitudinal strain developed in the rod is (Given, coefficient of linear expansion of steel = 12 × 10−6/ ° C ) (a) 6 × 10−4 (c) − 6 × 10−4

(b) − 6 × 10−5 (d) zero

Chapter 20

Thermometry, Thermal Expansion and Kinetic Theory of Gases — 225

9. The coefficient of linear expansion of steel and brass are 11 × 10−6/°C and 19 × 10−6/°C, respectively. If their difference in lengths at all temperatures has to kept constant at 30 cm, their lengths at 0°C should be (a) 71.25 cm and 41.25 cm (c) 92 cm and 62 cm

(b) 82 cm and 52 cm (d) 62.25 cm and 32.25 cm

10. The expansion of an ideal gas of mass m at a constant pressure p is given by the straight line B. Then, the expansion of the same ideal gas of mass 2m at a pressure 2p is given by the straight line Volume A B C

Temperature

(a) C (c) B

(b) A (d) data insufficient

Subjective Questions 1. Change each of the given temperatures to the Celsius and Kelvin scales: 68° F, 5° F and 176° F. 2. Change each of the given temperatures to the Fahrenheit and Reaumur scales: 30° C, 5° C and −20° C.

3. At what temperature do the Celsius and Fahrenheit readings have the same numerical value? 4. You work in a materials testing lab and your boss tells you to increase the temperature of a

sample by 40.0° C. The only thermometer you can find at your workbench reads in °F. If the initial temperature of the sample is 68.2 °F. What is its temperature in °F, when the desired temperature increase has been achieved?

5. The steam point and the ice point of a mercury thermometer are marked as 80° and 20°. What will be the temperature in centigrade mercury scale when this thermometer reads 32°?

6. A platinum resistance thermometer reads 0° C when its resistance is 80 Ω and 100° C when its resistance is 90 Ω. Find the temperature at which the resistance is 86 Ω.

7. The steam point and the ice point of a mercury thermometer are marked as 80° and 10°. At what temperature on centigrade scale the reading of this thermometer will be 59°?

8. Find the temperature at which oxygen molecules would have the same rms speed as of hydrogen molecules at 300 K.

9. Find the mass (in kilogram) of an ammonia molecule NH3 . 10. Three moles of an ideal gas having γ = 1.67 are mixed with 2 moles of another ideal gas having γ = 1.4. Find the equivalent value of γ for the mixture.

11. How many degrees of freedom have the gas molecules, if under standard conditions the gas density is ρ = 1.3 kg/m3 and velocity of sound propagation on it is v = 330 m/s?

12. 4 g hydrogen is mixed with 11.2 litre of He at STP in a container of volume 20 litre. If the final temperature is 300 K, find the pressure.

226 — Waves and Thermodynamics 13. One mole of an ideal monoatomic gas is taken at a temperature of 300 K. Its volume is doubled keeping its pressure constant. Find the change in internal energy.

14. Two perfect monoatomic gases at absolute temperatures T1 and T2 are mixed. There is no loss of energy. Find the temperature of the mixture if the number of moles in the gases are n1 and n 2.

15. If the water molecules in 1.0 g of water were distributed uniformly over the surface of earth, how many such molecules would there be in 1.0 cm 2 of earth’s surface?

16. If the kinetic energy of the molecules in 5 litres of helium at 2 atm is E. What is the kinetic energy of molecules in 15 litres of oxygen at 3 atm in terms of E ?

17. At what temperature is the “effective” speed of gaseous hydrogen molecules (molecular weight = 2) equal to that of oxygen molecules (molecular weight = 32 ) at 47° C ?

18. At what temperature is vrms of H 2 molecules equal to the escape speed from earth’s surface. What is the corresponding temperature for escape of hydrogen from moon’s surface? Given gm = 1.6 m/ s2, Re = 6367 km and Rm = 1750 km.

19. The pressure of the gas in a constant volume gas thermometer is 80 cm of mercury in melting ice. When the bulb is placed in a liquid, the pressure becomes 160 cm of mercury. Find the temperature of the liquid.

20. The resistances of a platinum resistance thermometer at the ice point, the steam point and the boiling point of sulphur are 2.50, 3.50 and 6.50 Ω respectively. Find the boiling point of sulphur on the platinum scale. The ice point and the steam point measure 0° and 100°, respectively.

21. In a constant volume gas thermometer, the pressure of the working gas is measured by the difference in the levels of mercury in the two arms of a U-tube connected to the gas at one end . When the bulb is placed at the room temperature 27.0 °C, the mercury column in the arm open to atmosphere stands 5.00 cm above the level of mercury in the other arm. When the bulb is placed in a hot liquid, the difference of mercury levels becomes 45.0 cm. Calculate the temperature of the liquid.(Atmospheric pressure = 75.0 cm of mercury.)

22. A steel wire of 2.0 mm 2cross-section is held straight (but under no tension) by attaching it

firmly to two points a distance 1.50 m apart at 30 ° C. If the temperature now decreases to −10° C and if the two points remain fixed, what will be the tension in the wire? For steel, Y = 20,0000 MPa.

23. A metallic bob weighs 50 g in air. If it is immersed in a liquid at a temperature of 25° C, it weighs 45 g. When the temperature of the liquid is raised to 100° C, it weighs 45.1 g. Calculate the coefficient of cubical expansion of the liquid. Given that coefficient of cubical expansion of the metal is 12 × 10 − 6 ° C −1.

24. An ideal gas exerts a pressure of 1.52 MPa when its temperature is 298.15 K and its volume is

10−2 m3 . (a) How many moles of gas are there? (b) What is the mass density if the gas is molecular hydrogen? (c) What is the mass density if the gas is oxygen?

25. A compressor pumps 70 L of air into a 6 L tank with the temperature remaining unchanged. If all the air is originally at 1 atm. What is the final absolute pressure of the air in the tank?

26. A partially inflated balloon contains 500 m3 of helium at 27° C and 1 atm pressure. What is the volume of the helium at an altitude of 18000 ft, where the pressure is 0.5 atm and the temperature is −3° C?

27. A cylinder whose inside diameter is 4.00 cm contains air compressed by a piston of mass m = 13.0 kg which can slide freely in the cylinder. The entire arrangement is immersed in a

Chapter 20

Thermometry, Thermal Expansion and Kinetic Theory of Gases — 227

water bath whose temperature can be controlled. The system is initially in equilibrium at temperature ti = 20° C. The initial height of the piston above the bottom of the cylinder is hi = 4.00 cm. The temperature of the water bath is gradually increased to a final temperature t f = 100° C. Calculate the final height h f of the piston. m

h

28. The closed cylinder shown in figure has a freely moving piston separating chambers 1 and 2. Chamber 1 contains 25 mg of N 2 gas and chamber 2 contains 40 mg of helium gas. When equilibrium is established what will be the ratio L1 / L2? What is the ratio of the number of moles of N 2 to the number of moles of He? (Molecular weights of N 2 and He are 28 and 4). Piston

1

L1

2

L2

29. Two gases occupy two containers A and B. The gas in A of volume 0.11 m3 exerts a pressure of 1.38 MPa. The gas in B of volume 0.16 m3 exerts a pressure of 0.69 MPa. Two containers are united by a tube of negligible volume and the gases are allowed to intermingle. What is the final pressure in the container if the temperature remains constant ?

30. A glass bulb of volume 400 cm3 is connected to another of volume 20 cm3 by means of a tube of negligible volume. The bulbs contain dry air and are both at a common temperature and pressure of 20°C and 1.00 atm. The larger bulb is immersed in steam at 100°C and the smaller in melting ice at 0°C. Find the final common pressure.

31. The condition called standard temperature and pressure (STP) for a gas is defined as temperature of 0° C = 273.15 K and a pressure of 1 atm = 1.013 × 105 Pa. If you want to keep a mole of an ideal gas in your room at STP, how big a container do you need?

32. A large cylindrical tank contains 0.750 m3 of nitrogen gas at 27° C and 1.50 × 105 Pa (absolute pressure). The tank has a tightfitting piston that allows the volume to be changed. What will be the pressure if the volume is decreased to 0.480 m3 and the temperature is increased to 157°C.

33. A vessel of volume 5 litres contains 1.4 g of N 2 and 0.4 g of He at 1500 K. If 30% of the nitrogen molecules are dissociated into atoms then find the gas pressure.

34. Temperature of diatomic gas is 300 K. If moment of inertia of its molecules is 8.28 × 10−38g-cm 2. Calculate their root mean square angular velocity.

35. Find the number of degrees of freedom of molecules in a gas. Whose molar heat capacity (a) at constant pressure C p = 29 J mol−1K −1 (b) C = 29 J mol−1K −1 in the process pT = constant.

228 — Waves and Thermodynamics 2 th of the energy of molecules is associated with the rotation of molecules and 5 the rest of it is associated with the motion of the centre of mass.

36. In a certain gas

(a) What is the average translational energy of one such molecule when the temperature is 27°C? (b) How much energy must be supplied to one mole of this gas at constant volume to raise the temperature by 1°C?

37. A mixture contains 1 mole of helium (C p = 2.5 R , CV = 1.5 R ) and 1 mole of hydrogen (C p = 3.5R , CV = 2.5 R ). Calculate the values of C p , CV and γ for the mixture.

38. An ideal gas (C p / CV = γ ) is taken through a process in which the pressure and the volume vary as P = aV b. Find the value of b for which the specific heat capacity in the process is zero.

39. An ideal gas is taken through a process in which the pressure and the volume are changed

according to the equation p = kV . Show that the molar heat capacity of the gas for the process R is given by C = CV + . 2

40. The pressure of a gas in a 100 mL container is 200 kPa and the average translational kinetic

energy of each gas particle is 6 × 10–21 J. Find the number of gas particles in the container. How many moles are there in the container?

41. One gram mole NO2 at 57° C and 2 atm pressure is kept in a vessel. Assuming the molecules to be moving with rms velocity. Find the number of collisions per second which the molecules make with one square metre area of the vessel wall.

42. A 2.00 mL volume container contains 50 mg of gas at a pressure of 100 kPa. The mass of each gas particle is 8.0 × 10−26 kg. Find the average translational kinetic energy of each particle.

43. Call the rms speed of the molecules in an ideal gas v0 at temperature T0 and pressure p0. Find

the speed if (a) the temperature is raised from T0 = 293 K to 573 K (b) the pressure is doubled and T = T0 (c) the molecular weight of each of the gas molecules is tripled.

44. (a) What is the average translational kinetic energy of a molecule of an ideal gas at temperature of 27° C ? (b) What is the total random translational kinetic energy of the molecules in one mole of this gas? (c) What is the rms speed of oxygen molecules at this temperature?

45. At 0°C and 1.0 atm ( = 1.01 × 105 N/ m 2 ) pressure the densities of air, oxygen and nitrogen are 1.284 kg/ m3 , 1.429 kg/ m3 and 1.251 kg/ m3 respectively. Calculate the percentage of nitrogen in the air from these data, assuming only these two gases to be present.

46. An air bubble of 20 cm3 volume is at the bottom of a lake 40 m deep where the temperature is 4°C. The bubble rises to the surface which is at a temperature of 20°C. Take the temperature to be the same as that of the surrounding water and find its volume just before it reaches the surface.

47. For a certain gas the heat capacity at constant pressure is greater than that at constant volume by 29.1 J/K. (a) How many moles of the gas are there? (b) If the gas is monatomic, what are heat capacities at constant volume and pressure? (c) If the gas molecules are diatomic which rotate but do not vibrate, what are heat capacities at constant volume and at constant pressure.

Chapter 20

Thermometry, Thermal Expansion and Kinetic Theory of Gases — 229

48. The heat capacity at constant volume of a sample of a monatomic gas is 35 J/K. Find (a) the number of moles (b) the internal energy at 0°C (c) the molar heat capacity at constant pressure.

LEVEL 2 Single Correct Option 1. Two thermally insulated vessels 1 and 2 are filled with air at temperatures (T1 , T2 ), volumes (V1 , V 2 ) and pressures ( p1 , p2 ) respectively. If the valve joining the two vessels is opened, the temperature inside the vessel at equilibrium will be (P = common pressure) (a) T1 + T2 (c)

(b) (T1 + T2 )/2

T1T2p(V1 + V 2) p1V1T2 + p2V 2T1

(d)

T1T2( p1V1 + p2V 2) p1V1T2 + p2V 2T1

2. Two marks on a glass rod 10 cm apart are found to increase their distance by 0.08 mm when the rod is heated from 0°C to 100°C. A flask made of the same glass as that of rod measures a volume of 100 cc at 0°C. The volume it measures at 100°C in cc is (a) 100.24 (c) 100.36

(b) 100.12 (d) 100.48

3. The given curve represents the variation of temperature as a function of volume for one mole of an ideal gas. Which of the following curves best represents the variation of pressure as a function of volume? T 45°

V p

p

(a)

V

p

V

(b)

(c)

p

V

(d)

V

4. A gas is found to be obeyed the law p V = constant. The initial temperature and volume are 2

T0 and V 0. If the gas expands to a volume 3 V 0, then the final temperature becomes (a)

3 T0 T0 (c) 3

(b)

2 T0 T0 (d) 2

5. Air fills a room in winter at 7°C and in summer at 37°C. If the pressure is the same in winter and summer, the ratio of the weight of the air filled in winter and that in summer is (a) 2.2 (c) 1.1

(b) 1.75 (d) 3.3

230 — Waves and Thermodynamics 6. Three closed vessels A, B and C are at the same temperature T and contain gases which obey Maxwell distribution law of velocities. Vessel A contains O2 , B only N2 and C mixture of equal quantities of O2 and N2. If the average speed of the O2 molecules in vessel A is v1 that of N2 molecules in vessel B is v2, then the average speed of the O2 molecules in vessel C is

(v1 + v2) 2 (c) v1v2

(b) v1

(a)

(d) None of these

7. In a very good vacuum system in the laboratory, the vacuum attained was 10−13 atm. If the temperature of the system was 300 K, the number of molecules present in a volume of 1 cm3 is (a) 2.4 × 106 (c) 2.4 × 109

(b) 24 (d) zero

8. If nitrogen gas molecule goes straight up with its rms speed at 0°C from the surface of the earth and there are no collisions with other molecules, then it will rise to an approximate height of (b) 12 km (d) 8 m

(a) 8 km (c) 12 m

9. The given p-U graph shows the variation of internal energy of an ideal gas with increase in pressure. Which of the following pressure-volume graph is equivalent to this graph? p

U p

(a)

p

p

V

(b)

V

(c)

p

V

(d)

V

10. 28 g of N 2 gas is contained in a flask at a pressure of 10 atm and at a temperature of 57°C. It is found that due to leakage in the flask, the pressure is reduced to half and the temperature to 27°C. The quantity of N2 gas that leaked out is (a) 11/20 g (c) 5/63 g

(b) 20/11 g (d) 63/5 g

11. A mixture of 4 g of hydrogen and 8 g of helium at NTP has a density about (a) 0.22 kg/m3 (c) 1.12 kg/m3

(b) 0.62 kg/m3 (d) 0.13 kg/m3

12. The pressure ( p) and the density (ρ) of given mass of a gas expressed by Boyle’s law, p = Kρ holds true (a) (b) (c) (d)

for any gas under any condition for same gas under any condition only if the temperature is kept constant None of the above

Chapter 20

Thermometry, Thermal Expansion and Kinetic Theory of Gases — 231

More than One Correct Options p2 = constant. (ρ = density of ρ the gas). The gas is initially at temperature T, pressure p and density ρ. The gas expands such that density changes to ρ/2.

1. During an experiment, an ideal gas is found to obey a condition

(a) (b) (c) (d)

The pressure of the gas changes to 2 p The temperature of the gas changes to 2 T The graph of the above process on p-T diagram is parabola The graph of the above process on p-T diagram is hyperbola

2. During an experiment, an ideal gas is found to obey a condition Vp2 = constant. The gas is initially at a temperature T, pressure p and volume V. The gas expands to volume 4V. p 2 (b) The temperature of gas changes to 4T (c) The graph of the above process on p-T diagram is parabola (d) The graph of the above process on p-T diagram is hyperbola (a) The pressure of gas changes to

3. Find the correct options. (a) Ice point in Fahrenheit scale is 32°F (c) Steam point in Fahrenheit scale is 212°F

(b) Ice point in Fahrenheit scale is 98.8°F (d) Steam point in Fahrenheit scale is 252°F

4. In the P-V diagram shown in figure, choose the correct options for the

P b

process a -b : (a) (b) (c) (d)

density of gas has reduced to half temperature of gas has increased to two times internal energy of gas has increased to four times T-V graph is a parabola passing through origin

a

V0

5. Choose the wrong options

2V0

V

(a) Translational kinetic energy of all ideal gases at same temperature is same 1 (b) In one degree of freedom all ideal gases has internal energy = RT 2 (c) Translational degree of freedom of all ideal gases is three 3 (d) Translational kinetic energy of one mole of all ideal gases is RT 2

6. Along the line-1, mass of gas is m1 and pressure is p1. Along the line-2 mass of same gas is m2 and pressure is p2. Choose the correct options. V

1 2

T

(a) m1 may be less than m2 (c) p1 may be less than p2

(b) m2 may be less than m1 (d) p2 may be less than p1

232 — Waves and Thermodynamics 7. Choose the correct options. m RT , m is mass of gas per unit volume M m (b) In pV = RT , m is mass of one molecule of gas M 1 mN 2 (c) In p = vrms , m is total mass of gas. 3 V 3kT (d) In vrms = , m is mass of one molecule of gas m

(a) In p =

Match the Columns 1. Match the following two columns for 2 moles of an ideal diatomic gas at room temperature T . Column I (a) (b) (c) (d)

Column II

Translational kinetic energy Rotational kinetic energy Potential energy Total internal energy

(p) (q) (r) (s)

2 RT 4 RT 3 RT None of these

2. In the graph shown, U is the internal energy of gas and ρ the density. Corresponding to given graph, match the following two columns. U

ρ

Column I

Column II

Pressure Volume Temperature Ratio T /V

(a) (b) (c) (d)

(p) (q) (r) (s)

is constant is increasing is decreasing data insufficient

3. At a given temperature T , x1RT x2RT = rms speed of gas molecules, v2 = = average speed of gas molecules M M x RT x4RT = most probable speed of gas molecules, v4 = = speed of sound v3 = 3 M M v1 =

Column I (a) (b) (c) (d)

x1 x2 x3 x4

Column II (p) (q) (r) (s)

1.5 2.0 3.0 data insufficient

Chapter 20

Thermometry, Thermal Expansion and Kinetic Theory of Gases — 233

4. With increase in temperature, match the following two columns. Column I

Column II

(a) Density of water (b) Fraction of a solid floating in a liquid (c) Apparent weight of a solid immersed in water (d) Time period of pendulum

(p) will increase (q) will decrease (r) will remain unchanged (s) may increase or decrease

5. Corresponding to isobaric process, match the following two columns. Column I (a)

Column II

p-T graph (p)

(b)

U -ρ graph (q)

(c)

T -V graph (r)

(d)

T -ρ graph (s)

Note First physical quantity is along y-axis.

Subjective Questions 1. Show that the volume thermal expansion coefficient for an ideal gas at constant pressure is

1 . T

2. The volume of a diatomic gas ( γ = 7/ 5) is increased two times in a polytropic process with molar heat capacity C = R. How many times will the rate of collision of molecules against the wall of the vessel be reduced as a result of this process? 3. A perfectly conducting vessel of volume V = 0.4 m3 contains an ideal gas at constant temperature T = 273 K. A portion of the gas is let out and the pressure of the gas falls by ∆p = 0.24 atm. ( Density of the gas at STP is ρ = 1.2 kg/ m3 ) . Find the mass of the gas which escapes from the vessel. 4. A thin-walled cylinder of mass m, height h and cross-sectional area A is filled with a gas and floats on the surface of water. As a result of leakage from the lower part of the cylinder, the depth of its submergence has increased by ∆h . Find the initial pressure p1 of the gas in the cylinder if the atmospheric pressure is p0 and the temperature remains constant.

234 — Waves and Thermodynamics 5. Find the minimum attainable pressure of an ideal gas in the process T = T0 + αV 2, where T0 and α are positive constants and V is the volume of one mole of gas.

6. A solid body floats in a liquid at a temperature t = 50° C being completely submerged in it. What percentage of the volume of the body is submerged in the liquid after it is cooled to t0 = 0° C , if the coefficient of cubic expansion for the solid is γ s = 0.3 × 10−5 ° C−1 and of the liquid γ l = 8 × 10−5 °C−1.

7. Two vessels connected by a pipe with a sliding plug contain mercury. In one vessel, the height of mercury column is 39.2 cm and its temperature is 0° C, while in the other, the height of mercury column is 40 cm and its temperature is 100° C. Find the coefficient of cubical expansion for mercury. The volume of the connecting pipe should be neglected.

8. Two steel rods and an aluminium rod of equal length l0 and equal cross-section are joined rigidly at their ends as shown in the figure below. All the rods are in a state of zero tension at 0° C. Steel Aluminium Steel

Find the length of the system when the temperature is raised to θ. Coefficient of linear expansion of aluminium and steel are α a and α s , respectively. Young’s modulus of aluminium is Y a and of steel is Y s .

9. A metal rod A of 25 cm length expands by 0.050 cm when its temperature is raised from 0° C to 100° C. Another rod B of a different metal of length 40cm expands by 0.040 cm for the same rise in temperature. A third rod C of 50 cm length is made up of pieces of rods A and B placed end to end expands by 0.03 cm on heating from 0° C to 50° C. Find the lengths of each portion of the composite rod.

Answers Introductory Exercise 20.1 1. (a) −17.8° C (b) – 459.67°F 4. 574.25

2. (a) 160°C 5. –40°C

3. 122°F

(b) – 24.6°C

Introductory Exercise 20.2 1. Gains, 15.55 s

2. It will first increase and then decrease

4. Cool the system 7. – 0.042%

5. 50.8°C

6. (a) 0.064 cm

3. 1.5 kg/ m3

4. 8 × 1015

3. (γ 2 − γ 1) ∆T

(b) 88.48 cm

Introductory Exercise 20.3 1. m1 > m2

2. 12 atm

6. Straight line passing through origin

5. p1 > p2

7. A dot

Introductory Exercise 20.4 1. (d)

2. (c)

3.

3 5 K0 , K0 2 2

Introductory Exercise 20.5 2. vrms = 714 m/s, vav = 700 m/s

1. 2.15 km/s

3. Speed is a scalar quantity while velocity is a vector quantity. 4. 6.21 × 10 –21 J 5. (a) 1368 m /s, 609 m/s (b) 6.21 × 10 –21 J

6. 160 K

7. True

Exercises LEVEL 1 Assertion and Reason 1. (d)

2. (b)

3. (b)

4. (d)

5. (d)

6. (b)

7. (c)

8. (b)

4. (b)

5. (d)

6. (c)

7. (a)

8. (c)

9. (b)

10. (a)

Objective Questions 1. (b)

2. (b)

3. (a)

9. (a)

10. (c)

Subjective Questions 1. 20°C, –15°C, 80°C, 293K, 258K, 353K 3. – 40°F = − 40°C 4. 140.2°F

2. 86°F, 41°F, – 4°F, 546°R, 501°R, 456°R 5. 20°C 6. 60° C 7. 70° C

9. 2.82 × 10 −26 kg

8. 4800 K

13. 450 R

16. 7.5 E

18. TE = 10059 K, TM = 449 K

21. 177.07°C

2

17. –253°C 22. 192 N

23. 3.1 × 10

14. T =

10. 1.53 nT 1 1 + n2T 2

12. 3.12 × 10 N/m 5

−4

11. 5 15. 6.5 × 103

n1 + n2 19. 546.30 K

20. 400°

per °C

24. (a) 6.135 mol (b) 1.24 kg/m3 (c) 19.6 kg/m3 26. 900 m3 27. 5.09 cm 28. 0.089, 0.089

25. 11.7 atm absolute pressure 29. 0.97 MPa

236 — Waves and Thermodynamics 30. 1.13 atm 35. (a) 5

31. 22.4 L

(b) 3

32. 3.36 × 10 5Pa 33. 4.1 × 10 5N/ m2

34. 1013 rad/s

36. (a) 6.21 × 10 −21 J

37. 3R , 2R , 1.5

40. 5 × 10 21, 8.3 × 10 −3 mol

38. −γ 43. (a) 1.40 v0

(b) v0

(b) 20.8 J

42. 4.8 × 10 −22 J

41. 3.1 × 10 27 44. (a) 6.21 × 10 −21 J

(c) 0.58 v0 or v0 / 3

(b) 3740 J

(c) 484 m/s

3

45. 76.5% by mass

46. 105 cm

47. (a) 3.5 mol (b) 43.65 J/K, 72.75 J/K (c) 72.75 J/K, 101.85 J/K 48. (a) 2.81 mol (b) 9.56 kJ (c) 20.8 J/mol-K

LEVEL 2 Single Correct Option 1.(c)

2.(a)

11.(d)

12.(c)

3.(a)

4.(a)

5.(c)

6.(b)

7.(a)

8.(b)

9.(b)

10.(d)

More than One Correct Options 1. (b,d)

2. (a,d)

3. (a,c)

4. (a,c,d)

5. (a,b)

6. (a,b,c,d)

7. (a,d)

Match the Columns 1. (a) → r

(b) → p

(c) → s

(d) → s

2. (a) → q

(b) → r

(c) → q

(d) → q

3. (a) → r

(b) → s

(c) → q

(d) → s

4. (a) → s

(b) → s

(c) → s

(d) → p

5. (a) → q

(b) → r

(c) → p

(d) → r

Subjective Questions 2. (2)4 /3 times

3. 115.2 g

mg   ∆h  4. p1 =  p0 +  1 −   A  h

5. 2R αT 0

6. 99.99%

7. 2.0 × 10 −4 per ° C

 8. l0 1 + 

9. 10 cm, 40 cm

 α aYa + 2α sYs     θ  Ya + 2Ys  

Laws of Thermodynamics Chapter Contents 21.1 The First Law of Thermodynamics 21.2 Further Explanation of Three Terms Used in First Law 21.3 Different Thermodynamic Processes 21.4 Heat Engine and its Efficiency 21.5 Refrigerator 21.6 Zeroth Law of Thermodynamics 21.7 Second Law of Thermodynamics

238 — Waves and Thermodynamics

21.1 The First Law of Thermodynamics The first law of thermodynamics is basically law of conservation of energy. This law can be applied for any type of system like solid, liquid and gas. But, in most of the cases the system will be an ideal gas. A process in which there are changes in the state of a thermodynamic system (like p, V, T, U and ρ etc.) is called a thermodynamic process. We now come to the first law Suppose Q heat is given to a system, then part of it is used by the system in doing work W against the surroundings (like atmosphere) and part is used by the system in increasing its internal energy ∆U . Thus, …(i)

Q = W + ∆U

Let us take a real life situation similar to first law. Consider a person X .Suppose his monthly income is Rs. 50,000 (Q ). He spends Rs. 30,000 (W ) as his monthly expenditure. Then, obviously the remaining Rs. 20,000 goes to his savings ( ∆U ). In some month it is also possible that he spends more than his income. In that case he will withdraw it from his bank and his savings will get reduced ( ∆U < 0). In the similar manner, other combinations can be made.

Sign Convention (i) Q If heat is given to the system, then Q is positive and if heat is taken from the system, then it is negative. (ii) W Work done used in Eq. (i) is the work done by the system (not work done on the system). This work done is positive if volume of the system increases. Sign of work done in different situations is given in tabular form as below Table 21.1 S.No Volume of the system

Work done by the system used in Eq. (i)

Work done on the system

1.

is increasing

positive

negative

2.

is decreasing

negative

positive

3.

is constant

zero

zero

(iii) ∆U Internal energy of a system is due to disordered motion of its constituent particles. It mainly depends on state (solid, liquid and gas) and temperature. For example, two different states at same temperature will have different internal energy. Ice at 0° C and water at 0° C have different energies. Similarly, same state at different temperatures will have different energies. For example, internal energy of an ideal gas is given by nf U = RT 2 or U ∝T If temperature increases, internal energy also increases and ∆U = U f − U i will be positive.

Chapter 21

Laws of Thermodynamics — 239

Extra Points to Remember If a thermodynamic system changes from an initial equilibrium state A to final equilibrium state B through three different paths 1, 2 and 3, then Q and W will be different along these three paths. But Q − W or ∆U will be same along all three paths. This is because, Q and W are path functions. But U is a state function. Thus, A

˜

Q1 − W1 = Q2 − W2 = Q3 − W3 = ∆U

1

B

2 3

Fig. 21.1

In a closed path, ∆U = 0 V

Example 21.1 When a system goes from state A to state B, it is supplied with 400 J of heat and it does 100 J of work. (a) For this transition, what is the system’s change in internal energy? (b) If the system moves from B to A, what is the change in internal energy? ′ = 400 J of work is (c) If in moving from A to B along a different path in which W AB done on the system, how much heat does it absorb? Solution (a) From the first law, ∆UAB = QAB – WAB = ( 400 – 100) J = 300 J (b) Consider a closed path that passes through the state A and B. Internal energy is a state function so ∆U is zero for a closed path. Thus, ∆U = ∆UAB + ∆UBA = 0 or ∆UBA = – ∆UAB = – 300 J (c) The change in internal energy is the same for any path, so ′ = QAB ′ ′ – WAB ∆UAB = ∆ UAB ′ – (– 400 J ) 300 J = Q AB ′ = – 100 J and the heat exchanged is Q AB The negative sign indicates that the system loses heat in this transition.

INTRODUCTORY EXERCISE

21.1

1. The quantities in the following table represent four different paths for the same initial and final states. Find a, b, c, d, e, f and g.

Table 21.2 Q(J)

W(J)

∆U(J)

−80

−120

d

90

c

e

a

40

f

b

−40

g

2. In a certain chemical process, a lab technician supplies 254 J of heat to a system. At the same time, 73 J of work are done on the system by its surroundings. What is the increase in the internal energy of the system?

240 — Waves and Thermodynamics

21.2 Further Explanation of Three Terms Used in First Law First law of thermodynamics basically revolves round the three terms Q, ∆U and W. If you substitute these three terms correctly with proper signs in the equation Q = ∆U + W , then you are able to solve most of the problems of first law. Let us take each term one by one. Here, we are taking the system an ideal gas. (i) Heat Transfer (Q or ∆Q)

There are two methods of finding Q or ∆Q. Q = nC∆T

Method 1.

∆Q = nC∆T

or

where, C is the molar heat capacity of the gas and n is the number of moles of the gas. Always take, ∆T = T f – Ti where, Tf is the final temperature and Ti the initial temperature of the gas. Further, we have discussed in chapter 20, that molar heat capacity of an ideal gas in the process pV x = constant is R R R C= + = CV + γ –1 1– x 1– x C = CV =

R in isochoric process and γ –1

C = C p = CV + R in isobaric process Mostly C p and CV are used. Note For finding C, nature of gas and process should be known. Nature of gas will give us CV and process the value of x.

Method 2. We can also find Q by finding ∆U and W by the equation Q = W + ∆U (ii) Change in internal energy ( ∆U ) There are two methods of finding ∆U .

Method 1. For change in internal energy of the gas ∆U = nCV ∆T Students are often confused that the result ∆U = nCV ∆T can be applied only in case of an isochoric process (as CV is here used). However, it is not so. It can be applied in any process, whether it is isobaric, isothermal, adiabatic or else. Note In the above expression, value of CV depends on the nature of gas.

Method 2. We can also find ∆U from the basic equation of first law of thermodynamics, ∆U = Q − W (iii) Work done ( W ) This is the most important of the three.

Chapter 21

Laws of Thermodynamics — 241

Work Done During Volume Changes A gas in a cylinder with a movable piston is a simple example of a thermodynamic system.

F = pA dx

Fig. 21.2

Figure shows a gas confined to a cylinder that has a movable piston at one end. If the gas expands against the piston, it exerts a force and does work on the piston. If the piston compresses the gas as it is moved inward, work is done on the gas. The work associated with such volume changes can be determined as follows. Let the gas pressure on the piston face be p. Then, the force on the piston due to the gas is pA, where A is the area of the face. When the piston is pushed outward an infinitesimal distance dx, the work done by the gas is dW = Fdx = pA dx which, since the change in volume of the gas is dV = Adx, becomes dW = pdV For a finite change in volume fromVi toV f , this equation is then integrated between Vi to Vf to find the net work W = ∫ dW = ∫

Vf Vi

pdV

Now, there are five methods of finding work done by a gas.

Method 1. This is used when p - V equation is known to us. Suppose p as a function of V is known to us. p = f (V ) then work done can be found by W =∫

Vf Vi

f (V ) dV

Method 2. The work done by a gas is also equal to the area under p - V graph. Following different cases are possible : Case 1 When volume is constant p

p B

A or

A

B V

V

Fig. 21.3

242 — Waves and Thermodynamics V = constant WAB = 0

∴ Case 2

When volume is increasing p

p A

B or

A

B

V

V

Fig. 21.4

V is increasing WAB > 0 WAB = Shaded area

∴ Case 3

When volume is decreasing p

p A

B or

A

B

V

V

Fig. 21.5

V is decreasing WAB < 0 WAB = – Shaded area

∴ Case 4

Cyclic process p

p

V

V (a)

(b)

Fig. 21.6

Wclockwise cycle = + Shaded area Wanticlockwise cycle = – Shaded area

[in figure (a)] [in figure (b)]

Laws of Thermodynamics — 243

Chapter 21 Case 5

Incomplete cycle p

p C

B

A

B C

D

A V

V

Fig. 21.7

WABC = + Shaded area WABCD = – Shaded area.

Method 3. Sometimes work done by the gas is also obtained by finding the forces against which work is done by the gas.

k x p0

p0 m, A Gas

Fig. 21.8

For example, in the figure shown, work is done by the gas against the following forces (when the piston is displaced upwards) (i) Against gravity force mg. Since, mg is a constant force. ∴ W1 = Force × displacement = mgx (ii) Against the force p0 A. Further, this is also a constant force. ∴ W2 = Force × displacement = p0 Ax But, Ax = ∆V ∴ W2 = p0 ∆V (iii) Against spring force kx. This is a variable force. Hence, x 1 W3 = ∫ ( kx ) dx = kx 2 0 2 Note While calculating W3 , we have assumed that initially spring is in its natural length.

Method 4. In some cases, one process and limits of temperature (or temperature change) is given. In those cases, with the help of given process and the standard ideal gas equation pV = nRT , first we convert pdV into f (T ) dT and then integrate this expression with the limits of temperature. Thus, W = ∫ pdV = ∫

Tf

Ti

f (T ) dT

244 — Waves and Thermodynamics Method 5. The last method of finding work done is from the fundamental equation of first law. or W = Q − ∆U Now, let us take example of each one of them. V

Example 21.2 Method 1 of Q Temperature of two moles of a monoatomic gas is increased by 300 K in the process p ∝ V . (a) Find molar heat capacity of the gas in the given process. (b) Find heat given to the gas in that. Solution

(a) p ∝ V

⇒

pV −1 = constant

If we compare with pV x = constant, then x = −1 Now, C = CV +

R 1− x

3 R for a monoatomic gas 2 3 R C= R+ = 2R 2 1 − ( −1)

CV = ∴

Ans.

(b) Q = nC∆T Substituting the values, we get Q = ( 2) ( 2R ) ( 300) = 1200R V

Ans.

Example 21.3 Method 2 of Q In a given process work done on a gas is 40 J and increase in its internal energy is 10 J . Find heat given or taken to/from the gas in this process. Solution Given, ∆U = + 10 J Work done on the gas is 40 J. Therefore, work done by the gas used in the equation, Q = W + ∆U will be −40 J. Now, putting the values in the equation,

Q = W + ∆U We have, Q = − 40 + 10 = − 30 J

Ans.

Here, negative sign indicates that heat is taken out from the gas. V

Example 21.4 Method 1 of ∆U Temperature of two moles of a monoatomic gas is increased by 600 K in a given process. Find change in internal energy of the gas.

Laws of Thermodynamics — 245

Chapter 21 Solution

Using the equation, ∆U = nCV ∆T 3 CV = R 2

for change in internal energy for monoatomic gas

3  ∆U = ( 2)  R ( 600) 2 

∴

= 1800 R V

Ans.

Example 21.5 Method 2 of ∆U Work done by a gas in a given process is −20 J . Heat given to the gas is 60 J . Find change in internal energy of the gas. ∆U = Q − W Substituting the values we have,

Solution

∆U = 60 − ( −20) = 80 J

Ans.

∆U is positive. Hence, internal energy of the gas is increasing. V

Example 21.6 Method 1 of W By integration, make expressions of work done by gas in (a) Isobaric process ( p = constant) (b) Isothermal process ( pV = constant) (c) Adiabatic process ( pV γ = constant) Solution

(a) Isobaric process W =∫

Vf

Vi

pdV = p ∫

Vf

Vi

(as p = constant )

dV

V

= p [V ]V f = p (V f − Vi ) i

= p∆V

Ans.

Note Any process freely taking place in atmosphere is considered isobaric. For example, melting of ice, boiling of water etc. Here, the constant pressure is p0 . Therefore, W = p0 ∆V = p0 (Vf − Vi )

(b) Isothermal process W =∫

Vf Vi

= nRT

pdV = ∫ Vf

∫V

i

Vf Vi

 nRT    dV  V 

nRT   as p =   V 

dV V

 Vf  = nRT ln   = nRT ln  Vi 

(as T = constant)  pi   pf

  

 Vf p  as p i Vi = p f Vf So, = i  Vi pf  

246 — Waves and Thermodynamics (c) Adiabatic process pV γ = constant = k (say ) = p i Viγ = p f V fγ Further,

p=

k V

γ

= kV − γ V

W =∫

Vf

Vi

= = V

pdV = ∫

Vf

Vi

kV f– γ

+1

 kV – γ + 1  f kV – γ dV =    – γ + 1  Vi

– kVi– γ

–γ + 1 p f V f – p iVi 1– γ

=

+1

=

p f V fγ V f– γ

nRT f – nRTi 1– γ

+1

+1

1– γ =

nR∆T 1– γ p

Example 21.7 Method 2 of W In the given p-V diagram, find (a) pressures at c and d (b) work done in different processes separately (c) work done in complete cycle abcd. Solution (a) Line bc is passing through origin. Hence, p ∝V From b to c, volume is doubled. Hence, pressure is also doubled. ∴

– p iViγ Vi– γ

pc = 2 pb = 4 p0

2p0 p0

c

b

d a

V0

2V0

V

Fig. 21.9

Ans.

Similarly, line ad is also passing through origin. ∴

pd = 2 pa = 2 p0

Ans.

(b) ab and cd processes are isochoric (V = constant ). Hence, Wab = Wcd = 0 Wbc = area under the line bc = area of trapezium 1 = ( p b + p c ) ( 2V0 − V0 ) 2 1 = ( 2 p 0 + 4 p 0 ) (V0 ) 2 = 3 p 0V0 Since, volume is increasing. Therefore, Wbc is positive. Wda = − area under the line da = − area of trapezium 1 = − ( p a + p d ) ( 2V0 − V0 ) 2

Ans.

Laws of Thermodynamics — 247

Chapter 21 =−

1 ( p 0 + 2 p 0 ) (V0 ) 2

= − 1.5 p 0V0

Ans.

Volume is decreasing. Therefore, work done is negative. (d) Work done in complete cycle W net = Wab + Wbc + Wcd + Wda = 0 + 3 p 0V0 + 0 − 1.5 p 0V0 = 1.5 p 0V0

Ans.

Note Net work done is also equal to the area between the cycle. Since, cycle is clockwise. Hence, net work done is positive. V

Example 21.8 Method 3 of W Mass of a piston shown in Fig. 21.10 is m and area of cross-section is A. Initially spring is in its natural length. Find work done by the gas. In the given condition, work is done by the gas only against spring force kx. This force is a variable force. Hence, x 1 W = ∫ ( kx ) dx = kx 2 0 2

k Gas

Solution

V

vacuum x

A, m

Fig. 21.10

Example 21.9 Method 4 of W The temperature of n-moles of an ideal gas is increased from T 0 to 2T 0 through a α process p = . Find work done in this process. T pV = nRT α p= T

Solution

and

(ideal gas equation)

…(i) …(ii)

Dividing Eq. (i) by Eq. (ii), we get V= ∴

nRT 2 α

W =∫

Vf

Vi

2nRT dT α 2 T0  α   2nRT  p dV = ∫    dT T0  T   α 

= 2nRT0 V

or

dV =

Ans.

Example 21.10 Method 5 of W Heat taken from a gas in a process is 80 J and increase in internal energy of the gas is 20 J . Find work done by the gas in the given process. Heat is taken from the gas. Therefore, Q is negative. Or, Q = − 80 J Internal energy of the gas is increasing. Solution

248 — Waves and Thermodynamics Therefore, ∆U is positive. Or Using the first law equation, Q = W + ∆U

∆U = + 20 J or W = Q − ∆U = − 80 − 20 = − 100 J

Ans.

Here, negative sign indicates that volume of the gas is decreasing and work is done on the gas.

INTRODUCTORY EXERCISE

21.2

1. A gas in a cylinder is held at a constant pressure of 1.7 × 105 Pa and is cooled and compressed from 1.20 m 3 to 0.8 m 3. The internal energy of the gas decreases by 1.1 × 105 J. (a) Find the work done by the gas. (b) Find the magnitude of the heat flow into or out of the gas and state the direction of heat flow. (c) Does it matter whether or not the gas is ideal?

2. A thermodynamic system undergoes a cyclic process as shown in figure. p 1 2 V

Fig. 21.12

(a) over one complete cycle, does the system do positive or negative work. (b) over one complete cycle, does heat flow into or out of the system. (c) In each of the loops 1 and 2, does heat flow into or out of the system.

3. How many moles of helium at temperature 300 K and 1.00 atm pressure are needed to make the internal energy of the gas 100 J?

4. Temperature of four moles of a monoatomic gas is increased by 300 K in isochoric process. Find W , Q and ∆U.

5. Find work done by the gas in the process AB shown in the following figures. p p0

p A

p

B

B

B p0

A

A 2V0

4V0 (i)

V

V0 (ii)

V

V0 (iii)

2V0

V

Fig. 21.12

a T

6. Temperature of two moles of an ideal gas is increased by 300 K in a processV = , where a is positive constant. Find work done by the gas in the given process. p  7. Pressure and volume of a gas changes from ( p 0, V0 ) to  0 , 2V0 in a process pV 2 = constant.  4  Find work done by the gas in the given process.

Chapter 21

Laws of Thermodynamics — 249

21.3 Different Thermodynamic Processes Different thermodynamic processes and their important points are given below in tabular form. Table 21.3 S.No

1.

Name of the process

Isothermal

Important points in the process

Q = nC∆T = W + ∆U

∆U = nCV ∆T

W

T, pV, U = constant ∆T = ∆( pV ) = ∆U = 0 1 p1V1 = p2 V2 or p ∝ V

Q=W

0

V  p nRT ln  f  = nRT ln  i   Vi   pf 

C = CV ∴ Q = nC V ∆T

nC V ∆T

0

C = CP ∴ Q = nC P ∆T

nC V ∆T

p∆V → For any system Q − ∆U = n(C P − C V )∆T = nR∆T → for an ideal gas

0

nC V ∆T

W = −∆U ∴ W = − nC V ∆T  R  = −n   (Tf − Ti )  γ − 1 pV − pf Vf = i i γ −1

Qnet = Wnet

0

p = constant T p ∆V = ∆ρ = ∆   = 0 T p1 p2 = T1 T2 V, ρ,

2.

Isochoric

or

3.

Isobaric

p∝T

V p, = constant T V ∆p = ∆   = 0 T V1 V2 or V ∝ T = T1 T2

pV γ = constant TV γ−1 = constant T γ p1− γ = constant

4.

Adiabatic process

5.

( pi, Vi, Ti ) = ( pf , Vf , Tf ) Since Ti = Tf Cyclic process ⇒ Ui = Uf or ∆T = ∆U = 0

6.

7.

Wnet = area between cycle on p- V diagram

Polytropic process pV X = constant

R R C= + γ − 1 1− x R = CV + 1− x

nC∆T

nC V ∆T

nR∆T 1− x nR(Ti − Tf ) = 1− x ( pV − pf Vf ) = i i 1− x

Free expansion in vacuum

∆U = 0 ⇒ U, T and pV = constant 1 or p ∝ V

0

0

0

Q − ∆U =

250 — Waves and Thermodynamics Important Points in the Above Table In isobaric process In isobaric process, neither of the three terms (Q, ∆U and W ) is zero but they have a constant ratio which depends on nature of gas (like monoatomic or diatomic etc.) Q : ∆U : W = nC p ∆T : nCV ∆T : nR∆T = C p : CV : R 5 3 So, this ratio is C p : CV : R . For example, C p = R and CV = R for a monoatomic gas. Therefore, 2 2 5 3 for a monoatomic gas this ratio is R : R : R or 5 : 3 : 2. If Q is 50 J, then ∆U will be 30 J and W is 20 J. 2 2

In adiabatic process (i) In a thermodynamic process, there are three variables p, V and T. If relation between any two ( p- V , V - T or p- T ) are known, then other relations can be obtained using the ideal gas equation pV = nRT or (For given value of n ) pV ∝ T T or p∝ V T and V∝ p For example, p- V equation pV γ = constant can be converted into p- T or V - T equation. T p-T equation Replace V with p γ

∴ V -T equation ∴

T  p   = constant  p Replace p with

or

p1− γ T γ = constant

T V

T  γ   V = constant V 

or TV γ −1 = constant

(ii) An adiabatic process is defined as one with no heat transfer into or out of a system :Q = 0.We can prevent heat flow either by surrounding the system with thermally insulating material or by carrying out the process so quickly that there is not enough time for appreciable heat flow. From the first law, we find that for every adiabatic process, (as Q = 0) W = – ∆U Therefore, if the work done by a gas is positive (i.e. volume of the gas is increasing), then ∆U will be negative. Hence, U and therefore T will decrease. The cooling of air can be experienced practically during bursting of a tyre. The process is so fast that it can be assumed as adiabatic. As the gas expands. Therefore, it cools. On the other hand, the compression stroke in an internal combustion engine is an approximately adiabatic process. The temperature rises as the air fuel mixture in the cylinder is compressed. Note Contrary to adiabatic process which is very fast an isothermal process is very slow. Because the system needs sufficient time to interact with surroundings to keep its temperature constant.

Laws of Thermodynamics — 251

Chapter 21

In cyclic process In a cyclic process, initial and final points are same. p b

a

c V

Fig. 21.13

Therefore, ( pi , Vi , Ti ) = ( p f , V f , T f ) Internal energy is a state function which only depends on temperature (in case of an ideal gas). Ti = T f ⇒ Ui = U f or ∆U net = 0 If there are three processes in a cycle abc, then ∆U ab + ∆U bc + ∆U ca = 0 From first law of thermodynamics, Q = W + ∆U, if ∆U net = 0, then Q net = Wnet or Qab + Qbc + Qca = Wab + Wbc + Wca Further, Wnet = area under p- V diagram. For example, Wnet = + area of triangle ‘abc’ in the shown diagram. Cycle is clockwise. So, work done will be positive.

Free expansion in vacuum A gas in a closed adiabatic chamber is taken to vacuum and then chamber is opened. So, the gas expands. Since, there is no external force which opposes this expansion. So, work done by gas W =0 Further no heat is supplied or taken from the gas. Therefore, heat exchange is also zero. Or Q =0 From first law of thermodynamics, Q = W + ∆U change in internal energy is also zero. Or ∆U = 0 or U = constant ⇒ T or pV is also constant 1 or p∝ V This behaviour is similar to an isothermal process. The only difference is, in this process all three terms Q, ∆U and W are zero. But in isothermal process, only ∆U is zero.

252 — Waves and Thermodynamics Extra Points to Remember ˜

Slope of p - V diagram In a general polytropic process, pV x = constant or differentiating, we get p ( xV x −1 ) dV + V x (dp) = 0 or slope of p -V graph = − x

dp p =− x dV V

⇒

p V

In isobaric process p = constant ⇒ x = 0, therefore slope = 0 In isothermal process pV = constant ⇒ x = 1, therefore slope = −

p V

In adiabatic process pV γ = constant ⇒ x = γ , therefore slope = − γ

p V

Thus, slope of an adiabatic graph is γ times slope of isothermal graph at that point. Because γ > 1, the isothermal curve is not as steep as that for the adiabatic expansion. p

p

monoatomic diatomic polyatomic

1 2 3

Isothermal

Adiabatic and isothermal expansion of an ideal gas

γ = 1.33

3 1 Adiabatic

Adiabatic expansion of mono,dia and polyatomic gases.

γ = 1.4

2

γ = 1.67

V

V

Fig. 21.14

p - V diagram of different processes is shown in one graph as below. p 1 1 2 3 4

2 4

3

Isobaric Isothermal adiabatic Isochoric

V

Fig. 21.15 V

Example 21.11 p-V plots for two gases during adiabatic processes are shown in the figure. Plots 1 and 2 should correspond respectively to (JEE 2001)

p

(a) He and O 2 (b) O 2 and He (c) He and Ar (d) O 2 and N 2

1 2

Fig. 21.16

V

Solution

Chapter 21

Laws of Thermodynamics — 253

dp p =−γ dV V slope ∝ γ

(with negative sign)

In adiabatic process :

slope of p-V graph,

From the given graph, (slope) 2 > (slope)1 ∴ γ 2 > γ1 Therefore, 1 should correspond to O2 (γ = 1.4) and 2 should correspond to He (γ = 1.67 ). Hence, the correct option is (b). V

Example 21.12 Starting with the same initial conditions, an ideal gas expands from volume V1 to V2 in three different ways, the work done by the gas is W1 if the process is purely isothermal, W2 if purely isobaric and W3 if purely (JEE 2000) adiabatic, then (a) W 2 > W1 > W3 (b) W 2 > W3 > W1 (c) W1 > W 2 > W3 (d) W1 > W3 > W 2 Solution The corresponding p-V graphs (also called indicator diagram) in three different processes will be as shown below. p A

2

1 3

V1

V2

V

Fig. 21.17

Area under the graph gives the work done by the gas. ( Area ) 2 > ( Area )1 > ( Area ) 3 Therefore, the correct option is (a). V

⇒ W2 > W1 > W3

Example 21.13 When an ideal diatomic gas is heated at constant pressure, the fraction of the heat energy supplied, which increases the internal energy of the gas, is (JEE 1990) (a)

2 5

Solution

(b)

3 5

(c)

The desired fraction is ∆U nCV ∆T CV 1 or f= = = = ∆Q nC p ∆T C p γ

Therefore, the correct option is (d).

3 7 f=

(d)

5 7

5 7

7  as γ =   5

Ans.

254 — Waves and Thermodynamics V

Example 21.14 What is the heat input needed to raise the temperature of 2 moles of helium gas from 0°C to 100°C (a) at constant volume, (b) at constant pressure? (c) What is the work done by the gas in part (b)? Give your answer in terms of R. Solution Helium is monoatomic gas. Therefore, 3R 5R and C p = CV = 2 2 (a) At constant volume, Q = nCV ∆T  3R  = ( 2)   (100)  2 = 300R (b) At constant pressure, Q = nC p ∆T  5R  = ( 2)   (100)  2 = 500R (c) At constant pressure, W = Q – ∆U = nC p ∆T – nCV ∆T = nR∆T = ( 2) ( R ) (100) = 200R

V

Example 21.15 An ideal monoatomic gas at 300K expands adiabatically to twice its volume. What is the final temperature? Solution

For an ideal monoatomic gas, γ=

In an adiabatic process, ∴ or

TV γ

–1

= constant

T f V fγ

–1

= TiViγ

5 3

–1

V T f = Ti  i V f

   

γ –1

5

 1 3 = ( 300)    2

–1

= 189 K

Chapter 21 V

Laws of Thermodynamics — 255

Example 21.16 p - T graph of 2 moles of an ideal monoatomic gas 3 5    CV = R and C p = R  is as shown below. Find Q, W and ∆U for each of the  2 2  four processes separately and then show that, ∆U net = 0 and Q net = W net p b

2p0 p0

c

d

a

2T0

T0

T

Fig. 21.18

Solution

Table 21.4 Process

ab

Name of process

Isothermal (as T = constant)

∆U

Q

W p nRT ln  i   pf 

Q = W = −2 RT0 ln(2 )

0

 p  = 2 RT0 ln  0   2 p0  = −2 RT0 ln(2 )

bc

Isobaric (as p = constant )

∆U = nC V ∆T Q = nC P ∆T 5  3  = 2  R  (2T0 − T0 ) = 2  R  (2T0 − T0 ) 2  2  = 5 RT0

cd

Isothermal (as T = constant )

= 3 RT0

Q − ∆U or nR∆T = 2 RT0 p nRT ln  i   pf 

Q = W = 4RT0 ln(2 )

0

2 p  = 2 R (2T0 ) ln  0   p0  = 4RT0 ln(2 )

da

Isobaric (as p = constant )

∆U = nC V ∆T Q = nC p ∆T 5  3  = 2  R  (T0 − 2T0 ) = 2  R  (T0 − 2T0 ) 2  2  = −5 RT0

Net values

= −3RT0

2 RT0 ln (2 )

In the given table, we can see that and

Q − ∆U or nR∆T = − 2 RT0

∆U net = 0 Q net = W net

0

2 RT0 ln(2 )

256 — Waves and Thermodynamics INTRODUCTORY EXERCISE

21.3

1. One mole of an ideal monoatomic gas is initially at 300 K. Find the final temperature if 200 J of heat are added as follows : (a) at constant volume

(b) at constant pressure.

2. An ideal gas expands while the pressure is kept constant. During this process, does heat flow into the gas or out of the gas? Justify your answer.

3. Consider the cyclic process depicted in figure. If Q is negative for the process BC and if ∆U is

negative for the process CA, what are the signs of Q, W and ∆U that are associated with each process? p (kPa) B

8 6 4 2

A 6

C 8 10 Fig. 21.19

V(m3)

4. A well insulated box contains a partition dividing the box into two equal volumes as shown in figure. Initially, the left hand side contains an ideal Vacuum monoatomic gas and the other half is a vacuum. The partition is suddenly removed so that the gas expands throughout the entire box. (a) Does the temperature of the gas change? Fig. 21.20 (b) Does the internal energy of the system change? (c) Does the gas work? Cp ∆Q ∆Q and in an isobaric process. The ratio of molar heat capacities 5. Find the ratio of = γ. ∆U ∆W CV

6. A certain amount of an ideal gas passes from state A to B first by means of

p

process 1, then by means of process 2. In which of the process is the amount of heat absorbed by the gas greater?

1 A

B 2

V

Fig. 21.21 p

7. A sample of ideal gas is expanded to twice its original volume of 1.00 m in a quasi-static process for which p = αV , with α = 5.00 atm /m 6, as shown in Fig 21.22. How much work is done by the expanding gas? 3

2

f p = αV

2

i

8. As a result of the isobaric heating by ∆T = 72 K, one mole of a

certain ideal gas obtains an amount of heat Q = 1.6 kJ.Find the work performed by the gas, the increment of its internal energy and γ.

O

1.00 m3

2.00 m3

Fig. 21.22

V

Chapter 21

Laws of Thermodynamics — 257

21.4 Heat Engine and its Efficiency A heat engine is a device which converts heat energy into mechanical energy. In every heat engine, there are the following three components : (i) Working substance (which is normally a gas in a cylinder) (ii) Source (at temperature T1 ) (iii) Sink (at temperature T2 ), T2 < T1 and sink is normally atmosphere. Source T1

Working substance Q1

Q2

Sink T2

W

Fig. 21.23

The working substance absorbs some heat (Q1) from the source, converts a part of it into work (W ) and the rest (Q2 ) is rejected to the sink. From conservation of energy, Q1 = W + Q2 Now, the above work done is used by us for different purposes. It is just like a shopkeeper. He takes some money from you. (Suppose he takes Rs. 100/- from you). So, you are source. In lieu of this he provides services to you (suppose he provides services of worth Rs. 80/-). This is work done. The remaining Rs. 20/- is his profit which goes to his account and this is basically sink. Then, the efficiency of the shopkeeper is 80%. There can’t be a shopkeeper whose efficiency is 100%. Similarly, efficiency of a heat engine is defined as the ratio of net work done per cycle by the engine to the total amount of heat absorbed per cycle by the working substance from the source. It is denoted by η. Thus, Q W Q1 − Q2 η= = =1 − 2 Q1 Q1 Q1 Work is done by the working substance in a cyclic process. In the above expression of efficiency, W is net work done in the complete cycle which should be positive. So, cycle should be clockwise on p - V diagram. Q1 is the total heat given to the working substance or the total positive heat. Q2 is total heat rejected by the working substance or it is the magnitude of total negative heat. Thus, efficiency (η) of a cycle can also be defined as  Work done by the working substance   (an ideal gas in our case) during a cycle  η=  ×100  Heat supplied to the gas during the cycle    =

WTotal × 100 | Q+ve |

258 — Waves and Thermodynamics  Q  | Q+ve | – | Q–ve | × 100 = 1 – –ve  × 100 | Q+ve |  Q+ve   Q  W η = Total × 100 = 1 – –ve  × 100 | Q+ve |  Q+ve  =

Thus,

Depending on the number of processes in a cycle, it may be called a two-stroke engine or four-stroke engine. p

p

V

V Two stroke-engine

Four stroke-engine

Fig. 21.24

Note There cannot be a heat engine whose efficiency is 100%. It is always less than 100%. Thus, or or

η ≠ 100% W ≠ Q1 or Wnet ≠ Q+ve Q2 ≠ 0 or | Q−ve | ≠ 0

Efficiency of a Cycle By the similar method discussed above, we can also find efficiency of a cycle provided net work done in the whole cycle comes out to be positive or it is clockwise cycle on p - V diagram. But always remember that, in a cyclic process, ∆U net = 0 and Q net = Wnet If there are four processes in the cycle, then ∆U 1 + ∆U 2 + ∆U 3 + ∆U 4 = 0 and Q1 + Q2 + Q3 + Q4 = W1 + W2 + W3 + W4

Carnot Engine Carnot cycle consists of the following four processes : (i) Isothermal expansion (process AB) (ii) Adiabatic expansion (process BC) (iii) Isothermal compression (process CD) and (iv) Adiabatic compression (process DA) The p - V diagram of the cycle is shown in the figure.

p A

T1

B Q1

D

T2

C Q2 V

Fig. 21.25

Chapter 21

Laws of Thermodynamics — 259

Table 21.5 Process

Name of the process

Q

∆U

W

AB

Isothermal expansion T1 = constant

Q = W = positive = Q1

0

positive

BC

Adiabatic expansion

0

negative

positive

CD

Isothermal compression T2 = constant

Q = W = negative = Q2

0

negative

DA

Adiabatic compression

0

positive

negative

In process AB, heat Q1 is taken by the working substance at constant temperature T1 and in process CD heat Q2 is rejected from the working substance at constant temperature T2 . The net work done is area of graph ABCD. Note

(i) In the whole cycle only Q1 is the positive heat and Q2 the negative heat. Thus, Q+ ve = Q1 and | Q−ve | = Q2  Q  ∴ η =  1 − 2  × 100 Q1   Q T Specially for Carnot cycle, 2 also comes out to be 2 . Q1 T1  T2  η =  1 −  × 100 ∴  T1  (ii) Efficiency of Carnot engine is maximum (not 100%) for given temperatures T1 and T2 . But still Carnot engine is not a practical engine because many ideal situations have been assumed while designing this engine which can practically not be obtained.

21.5 Refrigerator Refrigerator is an apparatus which takes heat from a cold body, work is done on it and the work done together with the heat absorbed is rejected to the source. Source T1

Q1

Working substance Q2

Sink T2

W

Fig. 21.26

An ideal refrigerator can be regarded as Carnot ideal heat engine working in the reverse direction.

Coefficient of Performance Coefficient of performance (β) of a refrigerator is defined as the ratio of quantity of heat removed per cycle (Q2 ) to the work done on the working substance per cycle to remove this heat. Thus,

260 — Waves and Thermodynamics β=

Q2 Q2 = W Q1 − Q2

β=

T2 1− η = T1 − T2 η

We can also show that

Here, η is the efficiency of Carnot cycle. V

Example 21.17 An ideal gas is taken through a cyclic thermodynamic process through four steps. The amounts of heat involved in these steps are Q1 = 5960 J , Q2 = − 5585 J, Q3 = − 2980 J and Q4 = 3645 J respectively. The corresponding quantities of work involved are W1 = 2200 J, W2 = − 825 J, W3 = − 1100 J and (JEE 1994) W4 respectively. (a) Find the value of W 4 . (b) What is the efficiency of the cycle? Solution (a) In a cyclic process, ∆U = 0 Therefore, Q net = W net or Q1 + Q2 + Q3 + Q4 = W1 + W2 + W3 + W4 Hence, W4 = (Q1 + Q2 + Q3 + Q4 ) − (W1 + W2 + W3 ) = {( 5960 − 5585 − 2980 + 3645) − ( 2200 − 825 − 1100)} W4 = 765 J

or

Ans.

(b) Efficiency, η=

Total work done in the cycle × 100 Heat absorbed (positive heat) by the gas during the cycle

 W + W2 + W3 + W4  = 1  × 100 Q1 + Q4    ( 2200 − 825 − 1100 + 765) =  × 100 5960 + 3645   1040 = × 100 9605 η = 10.82%

Ans.

Note From energy conservation, Wnet = Q+ ve − Q−ve ∴

η=

Wnet Q+ ve



= 1 −



× 100 =

(in a cycle) (Q+ ve − Q− ve ) Q+ ve

× 100

Q− ve 

 × 100

Q+ ve 

In the above question, Q− ve = | Q2 | + | Q3 | = (5585 + 2980 ) J = 8565 J

Chapter 21

Laws of Thermodynamics — 261

Q+ ve = Q1 + Q4 = (5960 + 3645) J = 9605 J 8565  η = 1 −  × 100  9605

and ∴

η = 10.82% V

Example 21.18 The density versus pressure graph of one mole of an ideal monatomic gas undergoing a cyclic process is shown in figure. The molecular mass of the gas is M. ρ B

2ρ0

ρ0

A

C

p0

2p0

p

Fig. 21.27

(a) Find the work done in each process. (b) Find heat rejected by gas in one complete cycle. (c) Find the efficiency of the cycle. Solution

(a) Given, n = 1

∴ m= M

Process AB ρ ∝ p, i.e. It is an isothermal process (T = constant), because ρ =

pM . RT

p   1 W AB = RT A ln  A  = RT A ln    2  pB 

∴

=–

p0 M ln ( 2) ρ0

∆U AB = 0 Q AB = W AB = –

and

p0 M ln ( 2) ρ0

Process BC is an isobaric process ( p = constant) M M M M  p0 M WBC = p B (VC – VB ) = 2 p 0  – −  = 2 p0  = ρ0  ρC ρ B   ρ 0 2ρ 0  ∆U BC = CV ∆T

QBC

 3   2 p M 2 p0 M  3 p0 M =  R  0 – =  2   ρ0 R 2ρ 0 R  2ρ 0 5 p0 M = WBC + ∆U BC = 2ρ 0

262 — Waves and Thermodynamics Process CA As ρ = constant ∴ So, it is an isochoric process.

V = constant

WCA = 0 ∆UCA = CV ∆T 3  =  R (T A – TC ) 2   p0 M 2 p0 M  –   ρ0 R   ρ0 R 3p M =– 0 2ρ 0

3  =  R 2 

QCA = ∆UCA = –

3 p0 M 2ρ 0

(b) Heat rejected by gas = |Q AB | + |QCA | p M 3  = 0  + ln ( 2) ρ0  2 

Ans.

(c) Efficiency of the cycle (in fraction) Total work done WTotal η= = Heat supplied Q+ ve =

= V

p0 M [1 – ln ( 2)] ρ0 5  p0 M    2  ρ0  2 [1 – ln ( 2)] 5

Ans.

Example 21.19 Carnot engine takes one thousand kilo calories of heat from a reservoir at 827°C and exhausts it to a sink at 27°C. How, much work does it perform? What is the efficiency of the engine? Solution

Given,

Q1 = 106 cal T1 = ( 827 + 273) = 1100 K

and as,

T2 = ( 27 + 273) = 300 K Q2 T2 T  300  6 = ⇒ Q2 = 2 ⋅ Q1 =   (10 )   Q1 T1 T1 1100 = 2.72 × 105 cal W = Q1 − Q2 = 7.28 × 105 cal

Ans.

Chapter 21

Laws of Thermodynamics — 263

Efficiency of the cycle,  T  η = 1 − 2  × 100 or  T1 

300   η = 1 −  × 100  1100 = 72.72%

V

Ans.

Example 21.20 In a refrigerator, heat from inside at 277 K is transferred to a room at 300 K. How many joules of heat shall be delivered to the room for each joule of electrical energy consumed ideally? Solution

Coefficient of performance of a refrigerator, Q T2 β= 2 = W T1 − T2

∴

Q2 = W

T2 T1 − T2

But W = Energy consumed by the refrigerator = 1 J, T1 = 300 K, T2 = 277 K 277 277 ∴ Q2 = 1× = = 12 J 300 − 277 23 Heat rejected by the refrigerator, Q1 = W + Q2 = 1+ 12 = 13 J V

Ans.

Example 21.21 Calculate the least amount of work that must be done to freeze one gram of water at 0°C by means of a refrigerator. Temperature of surroundings is 27°C. How much heat is passed on the surroundings in this process? Latent heat of fusion L = 80 cal / g . Solution Q2 = mL = 1 × 80 = 80 cal

T2 = 0° C = 273 K and T1 = 27° C = 300 K Least amount of work will be needed for carnot’s type of cycle. T2 Q2 = W T1 − T2 ∴

W= =

Q2 (T1 − T2 ) T2 80 ( 300 − 273) 273

= 7.91 cal

Ans.

Q1 = Q2 + W = ( 80 + 7.91 ) = 87.91 cal

Ans.

264 — Waves and Thermodynamics

INTRODUCTORY EXERCISE

21.4

1. Carnot engine takes 1000 K cal of heat from a reservoir at 827°C and exhausts it to a sink at 27°C. How much heat is rejected to the sink? What is the efficiency of the engine?

2. One of the most efficient engines ever developed operated between 2100 K and 700 K. Its actual efficiency is 40%. What percentage of its maximum possible efficiency is this?

3. In a heat engine, the temperature of the source and sink are 500 K and 375 K. If the engine consumes 25 × 105 J per cycle, find (a) the efficiency of the engine, (b) work done per cycle, and (c) heat rejected to the sink per cycle.

4. A Carnot engine takes 3 × 106 cal of heat from a reservoir at 627 °C and gives it to a sink at 27 °C. Find the work done by the engine.

5. The efficiency of a Carnot cycle is 1/6. If on reducing the temperature of the sink by 65 °C, the efficiency becomes 1/3, find the source and sink temperatures between which the cycle is working.

6. Refrigerator A works between −10 °C and 27 °C, while refrigerator B works between −27 °C and 17 °C, both removing heat equal to 2000 J from the freezer. Which of the two is the better refrigerator?

7. A refrigerator has to transfer an average of 263 J of heat per second from temperature −10 °C to 25 °C. Calculate the average power consumed, assuming no energy losses in the process.

8. n moles of a monoatomic gas are taken around in a cyclic process consisting of four processes along ABCDA as shown. All the lines on the p-V diagram have slope of magnitude p 0 / V0 . The pressure at A and C is p 0 and the volumes at A and C areV0 / 2 and 3V0 / 2, respectively . Calculate the percentage efficiency of the cycle. p B p0

A

C D

V0 3V0 2 2 Fig. 21.28

V

21.6 Zeroth Law of Thermodynamics The zeroth law of thermodynamics states that if two system A and B are in thermal equilibrium with a third system C, then A and B are also in thermal equilibrium with each other. It is analogous to the transitive property in math (if A = C and B = C , then A = B ). Another way of stating the zeroth law is that every object has a certain temperature, and when two objects are in thermal equilibrium, their temperatures are equal. It is called the zeroth law because it came to light after the first and second laws of thermodynamics had already been established and named, but was considered more fundamental and thus was given a lower number zero.

Chapter 21

Laws of Thermodynamics — 265

21.7 Second Law of Thermodynamics The first law of thermodynamics is the principle of conservation of energy. Common experience shows that there are many conceivable processes that are perfectly allowed by the first law and yet are never observed. For example, nobody has ever seen a book lying on a table jumping to a height by itself. But such a thing would be possible if the principle of conservation of energy were the only restriction. Thus, the second law of thermodynamics is a general principle which places constraints upon the direction of and the attainable efficiency of or coefficient of performance. It also imposes restrictions on entropy. Second law of thermodynamics can be better understood by the following three statements

In Terms of Entropy The second law can be expressed in several ways, the simplest being that heat will naturally flow from a hotter to a colder body. As its heat is a property of thermodynamic systems called entropy represented by “S”  in loose terms, a measure of the amount of disorder within a system. This can be represented in many ways, for example in the arrangement of the molecules  water molecules in an ice cube are more ordered than the same molecules after they have been heated into a gas. The entropy of the ice cube is, therefore, lower than that of the gas. Similarly, the entropy of a plate is higher when it is in pieces on the floor compared with when it is in one piece in the sink. The second equation is a way to express the second law of thermodynamics in terms of entropy. The formula says that the entropy of an isolated natural system will always tend to stay the same or increase  in other words, the energy in the universe is gradually moving towards disorder.

Kelvin Planck’s Statement Kelvin-Planck’s statement is based on the fact that the efficiency of the heat engine cycle is never 100%. This means that in the heat engine cycle some heat is always rejected to the low temperature reservoir. The heat engine cycle always operates between two heat reservoirs and produces work.

Clausius Statement As we know from the previous statement, the natural tendency of the heat is to flow from high temperature reservoir to the low temperature reservoir. The Clausius statement says that, to transfer the heat from low temperature to high temperature reservoir some external work should be done on the cycle. This statement has been the basis for the working for all refrigerators, heat pumps and air-conditioners. This work cannot be zero or coefficient of performance of a refrigerator cannot be infinite.

266 — Waves and Thermodynamics

Final Touch Points 1. Adiabatic and Diathermic Wall Adiabatic wall An insulating wall (can be movable also) that does not allow flow of energy (heat) from one chamber to another is called an adiabatic wall. If two thermodynamic systems A and B are separated by an adiabatic wall then the thermodynamic state of A will be independent of the state of B and vice-versa if wall is fixed otherwise if wall is movable, only pressure will be same on both sides.

Insulated

A

B

Diathermic wall A conducting wall that allows energy flow (heat) from one chamber to another is called a diathermic wall. If two thermodynamic systems A and B are separated by a diathermic wall then thermal equilibrium is attained in due course of time (if wall is fixed). If wall is movable, then temperature and pressure on both sides will become same.

Note

In the above two cases, thermodynamic systems A and B are insulated from the external surroundings.

2. Quasi-Static Process Quasi means almost or near to.Quasi-static process means very nearly static process. Let us consider a system of gas contained in cylinder. The gas is held by a moving piston. A weight w is placed over the piston. Due to the weight , the gas in cylinder is compressed. After the gas reaches equilibrium, the properties of gas are denoted by p1, V1 and T1. The weight placed over the piston is balanced by upward force exerted by the gas. If the weight is suddenly removed, then there will be an unbalanced force between the system and the surroundings. The gas under pressure will expand and push the piston upwards. The properties at this state after reaching equilibrium are P2, V2 and T2. But the intermediate states passed through, by the system are non-equilibrium states which cannot be described by thermodynamic coordinates. In this case, we only have initial and final states and do not have a path connecting them. Piston

w Piston p1,V1,T1

Gas

Initial state of gas

Gas

p2,V2,T2

Final state of gas Sudden expansion of gas

1

2

3

4

Gas

Initial state of gas

5

p1,V1,T1

Gas

Final state of gas Quasi-static expansion of gas

p2,V2,T2

Chapter 21

Laws of Thermodynamics — 267

Suppose, the weight is made of large numbers of small weights. And one by one each of these small weights are removed and allowed the system to reach an equilibrium state. Then, we have intermediate equilibrium states and the path described by these states will not deviate much from the thermodynamic equilibrium state. Such a process, which is the locus of all the intermediate points passed by the system is known as quasi-static process. It means, this process is almost near to the thermodynamically equilibrium process. Infinite slowness is the characteristic feature of quasi-static process. 1 x x x x

x

x2

A quasi-static process

A quasi-static process is obviously a hypothetical concept. In practice, processes that are sufficiently slow and do not involve accelerated motion of the piston are reasonably approximation to an ideal quasi-static process. We shall from now onwards deal with quasi-static processes only, except when stated otherwise.

3. Reversible and Irreversible Process Reversible process The process in which the system and surroundings can be restored to the initial state from the final state without producing any changes in the thermodynamic properties of the universe is called a reversible process. In the figure below, let us suppose that the system has undergone a change from state A to state B. If the system can be restored from state B to state A, and there is no change in the universe, then the process is said to be a reversible process. The reversible process can be reversed completely and there is no trace left to show that the system had undergone thermodynamic change. Reversible process y A

B x

For the system to undergo reversible change, it should occur infinitely slowly or it should be quasi-static process. During reversible process, all the changes in state that occur in the system are in thermodynamic equilibrium with each other. Thus, there are two important conditions for the reversible process to occur. Firstly, the process should occur very slowly and secondly all of the initial and final states of the system should be in equilibrium with each other. For example, a quasi-static isothermal expansion of an ideal gas in a cylinder fitted with a frictionless movable piston is a reversible process. In actual practice, the reversible process never occurs, thus it is an ideal or hypothetical process. Irreversible process The process is said to be an irreversible process if it cannot return the system and the surroundings to their original conditions when the process is reversed. The irreversible process is not at equilibrium throughout the process. Several examples can be cited. For examples, (i) When we are driving the car uphill, it consumes a lot of fuel and this fuel is not returned when we are driving down the hill.

268 — Waves and Thermodynamics (ii) The base of a vessel on an oven is hotter than its other parts. When the vessel is removed, heat is transferred from the base to the other parts, bringing the vessel to a uniform temperature (which in due course cools to the temperature of the surroundings). The process cannot be reversed; a part of the vessel will not get cooler spontaneously and warm up the base. It will violate the second law of thermodynamics, if it did. (iii) The free expansion of a gas is irreversible. (iv) Cooking gas leaking from a gas cylinder in the kitchen diffuses to the entire room. The diffusion process will not spontaneously reverse and bring the gas back to the cylinder. Many factors contribute in making any process irreversible. The most common of these are: friction, viscosity and other dissipative effects. The irreversible process is also called the natural process because all the processes occurring in nature are irreversible processes. dU dU Let us derive the relation CV = 4. CV = , where U = internal energy of 1 mole of the gas. dT dT Consider 1 mole (n = 1) of an ideal monoatomic gas which undergoes an isochoric process (V = constant ). From the first law of thermodynamics. dQ = dW + dU Here,

…(i)

dW = 0 as V = constant dQ = CdT = CV dT

(In dQ = nCV dT , n = 1and C = CV )

Substituting in Eq. (i), we have

or

CV dT = dU dU CV = dT

Hence proved.

5. Cp – CV = R To prove this relation (also known as Mayor’s formula) let us consider 1 mole of an ideal gas which undergoes an isobaric (p = constant) process. From first law of thermodynamics,

dQ = dW + dU

Here,

dQ = CpdT

…(ii) (as n = 1and C = Cp )

dU = CV dT and

 RT  dW = pdV = pd    p = d (RT )

RT   as V =   P  (as P = constant)

= RdT Substituting these values in Eq. (ii) We have CpdT = RdT + CV dT R We have already derived, 6. CV = γ –1

or Cp – CV = R

Hence proved.

Cp – CV = R Dividing this equation by CV , we have Cp R –1= CV CV ∴

or

CV =

γ –1= R γ –1

R CV

  C as P = γ CV   Hence proved.

Chapter 21

Laws of Thermodynamics — 269

7. Polytropic process When p and V bear the relation pV x = constant, where x ≠ 1 or γ the process is called a polytropic one. In this process the molar heat capacity is, R R R C = CV + = + 1– x γ –1 1– x Let us now derive this relation. The molar heat capacity is defined as ∆Q for 1 mole C= ∆T ∆U + ∆W = ∆T ∆U ∆W = + ∆T ∆T ∆W C = CV + ∴ ∆T

 ∆U  = CV    ∆T  …(iii) V

Here,

Vf Vf kV – x + 1  f ∆W = ∫ pdV = ∫ kV – x dV =   Vi Vi  – x + 1 Vi

= =

∴

kVf – x

+1

– kVi – x –x + 1

pf Vf x Vf – x

+1

+1

– pi Vi xVi – x 1– x

=

pf Vf – pi Vi RTf – RTi = 1– x 1– x

=

R∆T 1– x

∆W R = ∆T 1 – x

Substituting in Eq. (iii), we get the result i.e. ∴

C = CV +

R R R = + 1– x γ –1 1– x

+1

Solved Examples TYPED PROBLEMS Type 1. Based on first law of thermodynamics applied to a general system

Concept First law of thermodynamics is simply law of conservation of energy which can be applied for any system. V

Example 1 Boiling water : Suppose 1.0 g of water vaporizes isobarically at atmospheric pressure ( 1.01 × 105 Pa ). Its volume in the liquid state is V i = V liquid = 1.0 cm3 and its volume in vapour state is V f = Vvapour = 1671 cm3 . Find the work done in the expansion and the change in internal energy of the system. Ignore any mixing of the steam and the surrounding air. Take latent heat of vaporization L v = 2.26 × 10 6 J / kg . Solution Because the expansion takes place at constant pressure, the work done is W =∫

Vf Vi

p0dV = p0

Vf

∫V

i

dV = p0 (V f – Vi )

= (1.01 × 105 ) (1671 × 10–6 – 1.0 × 10–6 ) = 169 J Q = mLv = (1.0 × 10–3 ) (2.26 × 106 ) = 2260 J Hence, from the first law, the change in internal energy ∆U = Q – W = 2260 – 169 = 2091 J

Note

V

Ans. Ans.

Ans.

The positive value of ∆U indicates that the internal energy of the system increases. We see that most  2091 J  = 93% of the energy transferred to the liquid goes into increasing the internal energy of the   2260 J  169 J system only = 7% leaves the system by work done by the steam on the surrounding atmosphere. 2260 J

Example 2 A metal of mass 1 kg at constant atmospheric pressure and at initial temperature 20°C is given a heat of 20000 J. Find the following (JEE 2005) (a) change in temperature, (b) work done and (c) change in internal energy.

(Given, specific heat = 400 J/kg-°C, coefficient of cubical expansion, γ = 9 × 10 −5 / °C, density ρ = 9000 kg/ m3 , atmospheric pressure = 105 N / m 2 )

Laws of Thermodynamics — 271

Chapter 21 Solution (a) From ∆Q = ms∆T ∆T =

∆Q 20000 = = 50°C ms 1 × 400

 1  −5 ∆V = Vγ ∆T =   (9 × 10 ) (50)  9000

(b)

∴ (c)

= 5 × 10−7 m3 W = p0 ⋅ ∆V = (105 ) (5 × 10−7 ) = 0.05 J ∆U = ∆Q − W = (20000 − 0.05) J = 19999.95 J

Type 2. To make p- V , V - T or p- T equation corresponding to a given process

Concept Suppose we wish to make p-V equation for a given process then with the help of equation of first law of thermodynamics and pV = nRT , first make an equation of type f ( p) dp + f (V ) dV = 0 Now, integrating this equation we will get the desired p -V equation. V

Example 3 Make p -V equation for an adiabatic process. Solution In adiabatic process, dQ = 0 and ∴ ∴

dW = – dU pdV = – CV dT pdV dT = – CV

(for n = 1) …(i)

Also, for 1 mole of an ideal gas, or or

d ( pV ) = d (RT ) pdV + Vdp = RdT pdV + Vdp dT = R

From Eqs. (i) and (ii), we get CV Vdp + (CV + R) pdV or CV Vdp + C p pdV Dividing this equation by PV, we are left with dp dV CV + Cp p V dp dV or +γ p V dp dV or ∫ p +γ∫ V or We can write this in the form

=0 =0 =0 =0 =0

ln ( p) + γ ln (V ) = constant pV γ = constant

…(ii)

272 — Waves and Thermodynamics Type 3. To find values of all three terms of first law of thermodynamics from a p- V diagram provided nature of gas is given.

Concept Use the equation nRT = pV V

Example 4 A cyclic process abcd is given for a monoatomic gas 3 5    CV = R and C p = R  as shown in figure. Find Q, W and ∆U in each of the  2 2  four processes separately. Also find the efficiency of cycle. p 2p0 p0

b

c

d

a V0

2V0

V

Solution Process ab V = constant

(∴ Isochoric process )

W ab = 0 ∴

Qab = ∆U ab = nCV ∆T 3  = n  R (Tb − Ta ) 2  3 (nRTb − nRTa ) 2 3 = ( pbVb − paV a ) 2 3 = (2 p0V 0 − p0V 0 ) 2 =

= 1.5 p0V 0 Process bc p = constant Qbc = nC p∆T 5  = n  R (Tc − Tb ) 2  5 (nRTc − nRTb ) 2 5 = ( pcV c − pbVb ) 2 5 = (4 p0V 0 − 2 p0V 0 ) 2 =

= 5 p0V 0

(∴ Isobaric process )

Chapter 21

Laws of Thermodynamics — 273

∆U bc = nCV ∆T 3  = n  R (Tc − Tb ) 2  3 (nRTc − nRTb ) 2 3 = ( pcV c − pbVb ) 2 3 = (4 p0V 0 − 2 p0V 0 ) 2 =

= 3 p0V 0 Wbc = Qbc − ∆U bc = 2 p0V 0 Process cd Again an isochoric process. ∴

W cd = 0 Qcd = ∆U cd = nCV ∆T 3  = n  R (Td − Tc ) 2  3 = (nRTd − nRTc ) 2 3 = ( pdV d − pcV c ) 2 3 = (2 p0V 0 − 4 p0V 0 ) 2 = − 3 p0V 0

Process da This is an isobaric process. ∴

Qda = nC p ∆T 5  = n  R (Ta − Td ) 2  5 (nRTa − nRTd ) 2 5 = ( paV a − pdV d ) 2 5 = ( p0V 0 − 2 p0V 0 ) 2 =

= − 2.5 p0V 0 ∆U da = nCV ∆T 3  = n  R (Ta − Td ) 2  3 (nRTa − nRTd ) 2 3 = ( paV a − pdV d ) 2 =

274 — Waves and Thermodynamics =

3 ( p0V 0 − 2 p0V 0 ) 2

= − 1.5 p0V 0 W da = Qda − ∆U da = − p0V 0 Efficiency of cycle In the complete cycle, W net = W ab + Wbc + W cd + W da = 0 + 2 p0V 0 + 0 − p0V 0 = p0V 0

Note This Wnet is also equal to area under the cycle. ΣQ+ ve = Qab + Qbc = 1.5p0V0 + 5 p0V0 = 6.5p0V0 Wnet η= × 100 ΣQ+ ve

∴

 pV  =  0 0  × 100  6.5 p0V0  = 15.38 %

Ans.

Miscellaneous Examples V

Example 5 For a Carnot cycle (or engine) discussed in article 21.4, prove that efficiency of cycle is given by  T  η = 1 − 2  T1   Solution Efficiency =

net work done by gas |W1| + |W 2| − |W3| − |W 4| = heat absorbed by gas |Q1|

...(i)

Process 1 On this isothermal expansion process, the constant temperature is T1 so work done by the gas V  ...(ii) W1 = nRT1 ln  b   Va  Remember that Vb > V a, so this quantity is positive, as expected. (In process 1, the gas does work by lifting something) In isothermal process, Q =W ...(iii) ∴ |Q1| = |W1| Process 2 On this adiabatic expansion process, the temperature and volume are related through TV γ−1 = constant

Chapter 21

Laws of Thermodynamics — 275

T1Vbγ −1 = T2V cγ −1

∴

T1  V c  =  T2  Vb 

or

γ −1

...(iv)

Work done by the gas in this adiabatic process is p V − pbVb nRTc − nRTb W2 = c c = 1−γ 1−γ T −T   T − T2 =  2 1  nR =  1  nR − 1 γ    γ −1 

...(v)

Once again, as expected, this quantity is positive. Process 3 In the isothermal compression process, the work done by the gas is V W3 = nRT2 ln d Vc

...(vi)

Because V d < V c, the work done by the gas is negative. The work done on the gas is V V |W3|= − nRT2 ln d = nRT2 ln c Vc Vd

...(vii)

Furthermore, just as in process 1, ...(viii) |Q3 | = |W3 | Process 4 In the adiabatic compression process, the calculations are exactly the same as they were in process 2 but of course with different variables. Therefore, T2  V a  =  T1  V d  W4 =

and

γ −1

...(ix)

nR (T2 − T1 ) γ −1

...(x)

As expected, this quantity is negative and |W 4| = − W 4 =

nR (T1 − T2) γ −1

...(xi)

We can now calculate the efficiency. Efficiency =

|W1| + |W 2| − |W3 | − |W 4| |Q1|

=1+

|W 2| − |W3 | − |W 4| |Q1|

(as |W1| = |Q1|)

But our calculations show that|W 2| = |W 4|. |W3| nRT2 ln (V c /V d ) T ln (V c /V d ) Efficiency = 1 − =1 − =1− 2 T1 ln (Vb /V a ) |Q1| nRT1 ln (Vb /V a ) We have seen that T1  V c  =  T2  Vb 

γ −1

V  =  d  Va 

γ −1

or

Vc Vd = Vb V a

Substituting in Eq. (xii), we get Efficiency = 1 −

T2 T1

or

V c Vb = Vd Va

...(xii)

276 — Waves and Thermodynamics V

Example 6 An ideal gas expands isothermally along AB and does 700 J of work.

p

A B

(a) How much heat does the gas exchange along AB? (b) The gas then expands adiabatically along BC and does 400 J of work. When the gas returns to A along CA, it exhausts 100 J of heat to its surroundings. How much work is done on the gas along this path? Solution (a) AB is an isothermal process. Hence, and (b) BC is an adiabatic process. Hence,

C V

∆U AB = 0 QAB = W AB = 700 J

Ans.

QBC = 0 WBC = 400 J ∴ ∆U BC = – WBC = – 400 J ABC is a cyclic process and internal energy is a state function. Therefore, (∆U )whole cycle = 0 = ∆U AB + ∆U BC + ∆U CA and from first law of thermodynamics, QAB + QBC + QCA = W AB + WBC + WCA Substituting the values, 700 + 0 – 100 = 700 + 400 + ∆WCA ∴ ∆WCA = – 500 J Negative sign implies that work is done on the gas. Table below shows different values in different processes.

Ans.

Table 21.6 Process

Q (J)

W(J)

∆U(J)

AB

700

700

0

BC

0

400

−400

CA

−100

−500

400

For complete cycle

600

600

0

Note Total work done is 600 J, which implies that area of the closed curve is also 600 J. V

Example 7 The p-V diagram of 0.2 mol of a diatomic ideal gas is shown in figure. Process BC is adiabatic. The value of γ for this gas is 1.4. (a) Find the pressure and volume at points A, B and C. (b) Calculate ∆Q , ∆W and ∆U for each of the three processes. (c) Find the thermal efficiency of the cycle. Take 1 atm = 1.0 × 105 N / m 2.

p 600 K B

1.0 atm

A 300 K

C

455 K V

Laws of Thermodynamics — 277

Chapter 21 Solution (a) pA = pC = 1 atm = 1.01 × 105 N/m2 Process AB is an isochoric process. ∴

p∝T

or

pB TB = pA TA

T   600 pB =  B  pA =   (1 atm) = 2 atm  300  TA 

∴

From ideal gas equation ∴

= 2.02 × 105 N/m2 nRT V = p nRTA V A = VB = pA (0.2) (8.31) (300) ≈ 5.0 × 10–3 m3 (1.01 × 105 )

=

and

=5 L nRTC (0.2) (8.31) (455) VC = = pC (1.01 × 105 ) = 7.5 × 10–3 m3 ≈ 7.5 L

Table 21.7 State

p

V

A

1 atm

5L

B

2 atm

5L

C

1 atm

7.5 L

(b) Process AB is an isochoric process. Hence, ∆W AB = 0 5  ∆QAB = ∆U AB = nCV ∆T = n  R (TB – TA ) 2   5 = (0.2)   (8.31) (600 – 300)  2 ≈ 1246 J Process BC is an adiabatic process. Hence, ∴

∆QBC = 0 ∆WBC = – ∆U BC ∆U BC = nCV ∆T = nCV (TC – TB ) 5  = (0.2)  R (455 – 600) 2   5 = (0.2)   (8.31) (– 145) J  2

∴

∆WBC

≈ – 602 J = – ∆U BC = 602 J

278 — Waves and Thermodynamics Process CA is an isobaric process. Hence, 7  ∆QCA = nC p∆T = n  R (TA – TC ) 2   7 = (0.2)   (8.31) (300 – 455)  2 ≈ – 902 J ∆U CA = nCV ∆T ∆QCA = γ =– ∴

Cp  as γ =  CV  

902 ≈ – 644 J 1.4

∆WCA = ∆QCA – ∆U CA = – 258 J

Table 21.8 Process

∆Q (in J)

∆W (in J)

∆U (in J)

AB

1246

0

1246

BC

0

602

− 602

CA

− 902

− 258

− 644

Total

344

344

0

(c) Efficiency of the cycle η=

WTotal 344 × 100 = × 100 |Q+ve| 1246

= 27.6% V

Example 8 Find the molar specific heat of the process p =

a for a monoatomic T

gas, a being constant. Solution

We know that

Specific heat, Since,

dQ = dU + dW dQ dU dW C= = + dT dT dT dU = CV dT dW C = CV + dT pdV = CV + dT

∴ For the given process,

pV = RT RT RT 2 V = = p a dV 2RT = dT a

…(i) …(ii)

Chapter 21

Laws of Thermodynamics — 279

 2RT  C = CV + p    a 

∴

= CV + 2R 3 7 = R + 2R = R 2 2 V

Example 9 At 27° C two moles of an ideal monatomic gas occupy a volume V . The gas expands adiabatically to a volume 2V . Calculate (a) final temperature of the gas (b) change in its internal energy and (c) the work done by the gas during the process. [R = 8.31 J/mol-K] Solution

(a) In case of adiabatic change TV γ − 1 = constant  5 T1V1γ − 1 = T2V 2γ − 1 with γ =    3

So that

300 × V 2 / 3 = T (2 V )2 / 3 300 T= = 189 K (2)2/ 3

i.e. or

3  ∆U = nCV ∆T = n  R ∆T 2 

(b) As

 3 ∆U = 2 ×   × 8.31 (189 − 300)  2

So,

= − 2767.23 J Negative sign means internal energy will decrease. (c) According to first law of thermodynamics Q = ∆U + ∆W And as for adiabatic change ∆Q = 0, ∆W = − ∆U = 2767.23 J V

Example 10 Two moles of a diatomic ideal gas is taken through pT = constant. Its temperature is increased from T to 2T . Find the work done by the system? Solution

pT = constant

∴ or Comparing with We have,

p ( pV ) = constant pV

1/ 2

= constant

pV x = constant 1 x= 2 nR∆T 2R(2T − T ) W = = 1 1−x 1− 2 = 4RT

(as T ∝ pV )

280 — Waves and Thermodynamics V

Example 11 An ideal monatomic gas at temperature 27°C and pressure 10 6 N / m2 occupies 10 L volume. 10,000 cal of heat is added to the system without changing the volume. Calculate the change in temperature of the gas. Given : R = 8.31 J/mol-K and J = 4.18 J/cal. Solution Here, ∴

For n moles of gas, we have pV = nRT p = 106 N/m 2, V = 10 L = 10−2 m3 and T = 27° C = 300 K pV 106 × 10−2 n= = = 4.0 RT 8.31 × 300

For monatomic gas, CV =

3 R 2

3 × 8.31 J/mol-K 2 3 8.31 = × ≈ 3 cal/mol-K 2 4.18 Let ∆T be the rise in temperature when n moles of the gas is given Q cal of heat at constant volume. Then, Q Q = nCV ∆T or ∆T = nCV CV =

Thus,

=

10000 cal 4.0 mole × 3 cal /mol - K

= 833 K V

Example 12 One mole of a monoatomic ideal gas is taken through the cycle shown in figure.

p

A → B Adiabatic expansion B → C Cooling at constant volume C → D Adiabatic compression. D → A Heating at constant volume

A B D

The pressure and temperature at A, B etc., are denoted by pA , TA ; pB , TB etc. respectively.  2  1 Given, TA = 1000K , pB =   pA and pC =   pA . Calculate  3  3

C V

(a) the work done by the gas in the process A → B (b) the heat lost by the gas in the process B → C 0.4  2 Given,   = 0.85 and R = 8.31 J/mol-K  3 Solution

(a) As for adiabatic change pV γ = constant γ

i.e. i.e.

 nRT  p  = constant  p  Tγ pγ − 1

(as pV = nRT ) γ

T  p  = constant so  B  =  B   TA   pA 

γ −1

, where γ =

5 3

Chapter 21  2 TB = TA    3

i.e.

W AB =

So,

1−

1 γ

 2 = 1000    3

Laws of Thermodynamics — 281 2/ 5

= 850 K

nR[TF − TI ] 1 × 8.31 [1000 − 850] = [1 − γ ]   5   3 − 1  

i.e. W AB = 1869.75 J (b) For B → C , V = constant so ∆W = 0 So, from first law of thermodynamics ∆Q = ∆U + ∆W = nCV ∆T + 0 3  ∆Q = 1 ×  R (TC − 850) 2 

or

3   as CV = R  2 

Now, along path BC , V = constant; p ∝ T pC TC i.e. = pB TB  1   pA  3 850 T × TB = B = = 425 K TC = 2 2  2 p   A  3

…(ii)

3 × 8.31 (425 − 850) = − 5297.625 J 2 [Negative heat means, heat is lost by the system] ∆Q = 1 ×

So,

V

1 . If the molar heat T capacity for this process is C = 33.24 J / mol - K , find the degree of freedom of the molecules of the gas.

Example 13 A gas undergoes a process such that p ∝

Solution

p∝

As

or We have for one mole of an ideal gas

1 T

pT = constant pV = RT

…(i) …(ii)

From Eqs. (i) and (ii), p2V = constant 1/ 2 pV 1/ 2 = K (say) = pV = pf V f1/ 2 i i

or

…(iii)

From first law of thermodynamics, or or Here,

∆Q = ∆U + ∆W C∆T = CV ∆T + ∆W ∆W C = CV + ∆T ∆W = ∫ pdV = K ∫

Vf Vi

V –1/ 2dV

1/ 2 1/ 2 = 2K [V f1/ 2 − Vi1/ 2] = 2 [ pf V f1/ 2V f1/ 2 − pV i i Vi ] = 2 [ pf V f − pV i i ] = 2 R [Tf − Ti ]

…(iv)

282 — Waves and Thermodynamics =

R∆ T 1 /2

⇒

∆W = 2R ∆T

Substituting in Eq. (iv), we have C = CV + 2R = Substituting the values,

 1   1  33.24 = R  + 2 = 8.31  + 2 γ – 1  γ –1 

Solving this we get

γ = 1.5

Now,

γ =1 +

or degree of freedom Alternate Solution

R + 2R γ–1

F =

2 F

2 2 = =4 γ – 1 1.5 – 1

In the process pV x = constant, molar heat capacity is given by R R C= + γ–1 1– x

1 2 R R R C= + = + 2R ∴ γ–1 1–1 γ–1 2 Now, we may proceed in the similar manner. The given process is pV 1/ 2 = constant or

V

x=

Example 14 A gaseous mixture enclosed in a vessel consists of one gram mole of 7 5 a gas A with γ =   and some amount of gas B with γ = at a temperature T . 3 5 The gases A and B do not react with each other and are assumed to be ideal. Find  19  the number of gram moles of the gas B if γ for the gaseous mixture is   .  13  Solution So, ∴

Cp As for an ideal gas, C p − CV = R and γ =    CV  R CV = (γ − 1) R 3 (CV )1 = = R; 5 2     −1  3 R 5 (CV )2 = = R 2  7   −1  5

and

(CV )mix =

13 R = R 6  19   −1  13

Now, from conservation of energy, i.e

∆U = ∆U 1 + ∆U 2

Chapter 21

Laws of Thermodynamics — 283

(n1 + n2) (CV )mix ∆T = [n1 (CV )1 + n2 (CV )2] ∆T (CV )mix =

i.e.

13 R= 6

We have

=

V

n1 (CV )1 + n2 (CV )2 n1 + n2 1×

3 5 R + n2 R 2 2 1 + n2

(3 + 5n2)R 2 (1 + n2)

or

13 + 13n2 = 9 + 15n2,

i.e.

n2 = 2

Ans.

Example 15 An ideal gas having initial pressure p, volume V and temperature T is allowed to expand adiabatically until its volume becomes 5.66 V , while its temperature falls to T /2. (a) How many degrees of freedom do the gas molecules have? (b) Obtain the work done by the gas during the expansion as a function of the initial pressure p and volume V . Given that ( 5.66 )0.4 = 2 Solution

(a) For adiabatic expansion, TV γ − 1 = constant TV γ − 1 = T ′ V ′ γ − 1 T = (5.66 V )γ − 1 2

i.e.

i.e.

(5.66)γ − 1 = 2 γ = 1.4

i.e. 2 Using γ = 1 + F We get degree of freedom,

F =5 (b) Work done during adiabatic process for one mole gas is 1 W = [ p′ V ′ − pV ] 1−γ From relation, We get ∴

Ans.

pV p′ V ′ = T T′ T ′ pV 1 1 p p′ = ⋅ = × p= T V ′ 2 5.66 11.32 V 1  p  W = × − pV  1 − 1.4 11.32 5.66  =

 1  1 1− pV  0.4  11.32 × 5.66 

= 2.461 pV

Ans.

Exercises LEVEL 1 Assertion and Reason Directions : (a) (b) (c) (d)

Choose the correct option.

If both Assertion and Reason are true and the Reason is correct explanation of the Assertion. If both Assertion and Reason are true but Reason is not the correct explanation of Assertion. If Assertion is true, but the Reason is false. If Assertion is false but the Reason is true.

1. Assertion : In adiabatic expansion, temperature of gas always decreases. Reason :

In adiabatic process exchange of heat is zero.

2. Assertion : In a thermodynamic process, initial volume of gas is equal to final volume of gas. Work done by gas in this process should be zero. Reason : Work done by gas in isochoric process is zero.

3. Assertion : First law of thermodynamics can be applied for ideal gases only. Reason :

First law is simply, law of conservation of energy.

4. Assertion : When ice melts, work is done by atmosphere on (ice + water) system. Reason :

On melting of ice volume of ( ice + water) system decreases.

5. Assertion : Between two thermodynamic states, the value of (Q − W ) is constant for any process. Reason : Q and W are path functions.

6. Assertion : Efficiency of a heat engine can’t be greater than efficiency of Carnot engine. Reason :

Efficiency of any engine is never 100%.

7. Assertion : In the process pT = constant, if temperature of gas is increased work done by the gas is positive. Reason : For the given process, V ∝ T . 8. Assertion : In free expansion of a gas inside an adiabatic chamber Q , W and ∆U all are zero. 1 Reason : In such an expansion p ∝ . V

9. Assertion : For an ideal gas in a cyclic process and in an isothermal process change in internal energy is zero. Reason : In both processes there is no change in temperature.

10. Assertion : Isothermal and adiabatic, two processes are shown on p-V diagram. Process-1 is adiabatic and process-2 is isothermal. Reason : At a given point, slope of adiabatic process = γ × slope of isothermal process.

p 1 2 V

Chapter 21

Laws of Thermodynamics — 285

Objective Questions 1. In a process, the pressure of an ideal gas is proportional to square of the volume of the gas. If the temperature of the gas increases in this process, then work done by this gas (a) is positive (c) is zero

(b) is negative (d) may be positive or negative

2. n moles of a gas are filled in a container at temperature T . If the gas is slowly and isothermally compressed to half its initial volume, the work done by the atmosphere on the gas is

nRT 2 (c) nRT ln 2

nRT 2 (d) − nRT ln 2

(b) −

(a)

3. A gas undergoes A to B through three different processes 1, 2 and 3 as shown in the figure. The heat supplied to the gas is Q1 , Q2 and Q3 respectively, then p B 1

3

2 A

(a) Q1 = Q2 = Q3 (c) Q1 > Q2 > Q3

V

(b) Q1 < Q2 < Q3 (d) Q1 = Q3 > Q2

4. For an adiabatic compression the quantity pV (b) decreases (d) depends on γ

(a) increases (c) remains constant

5. The cyclic process form a circle on a pV diagram as shown in figure. The work done by the gas is V V2

V1 p1

π ( p2 − p1 )2 4 π (c) ( p2 − p1 ) (V 2 − V1 ) 2

(a)

p2

p

π (V 2 − V1 )2 4 π (d) ( p2 − p1 ) (V1 − V 2) 4

(b)

6. An ideal gas has initial volume V and pressure p. In doubling its volume the minimum work done will be in the process (of the given processes) (a) isobaric process (c) adiabatic process

(b) isothermal process (d) same in all given processes

286 — Waves and Thermodynamics 7. Figure shows two processes a and b for a given sample of a gas. If

p

∆Q1 , ∆Q2 are the amounts of heat absorbed by the system in the two cases and ∆U 1 , ∆U 2 are changes in internal energies respectively, then

a

(a) ∆Q1 = ∆Q2; ∆U 1 = ∆U 2 (b) ∆Q1 > ∆Q2; ∆U 1 > ∆U 2

b

(c) ∆Q1 < ∆Q2; ∆U 1 < ∆U 2

O

(d) ∆Q1 > ∆Q2; ∆U 1 = ∆U 2

V

8. A Carnot engine works between 600 K and 300 K. The efficiency of the engine is (a) 50% (c) 20%

(b) 70% (d) 80%

9. Air in a cylinder is suddenly compressed by a piston which is then maintained at the same position. As the time passes pressure of the gas (a) increases (b) decreases (c) remains the same (d) may increase or decrease depending on the nature of the gas

10. A cycle pump becomes hot near the nozzle after a few quick strokes even if they are smooth because (a) the volume of air decreases (c) the compression is adiabatic

(b) the number of air molecules increases (d) collision between air particles increases

11. In an adiabatic change, the pressure p and temperature T of a diatomic gas are related by the relation p ∝ T α , where α equals

(b) 0.4 (d) 3.5

(a) 1.67 (c) 0.6

12. A diatomic gas obeys the law pV x = constant. For what value of x, it has negative molar specific heat ? (b) x < 1.4 (d) 0 < x < 1

(a) x > 1.4 (c) 1 < x < 1.4

13. The molar specific heat at constant volume of gas mixture is (a) 2 moles of O2 and 4 moles of H2 (c) 2 moles of argon and 4 moles of O2

13R .The gas mixture consists of 6

(b) 2 moles of O2 and 4 moles of argon (d) 2 moles of CO2 and 4 moles of argon

14. Heat energy absorbed by a system in going through a cyclic process as shown in the figure [V in litres and p in kPa] is 30 V 10 10

(a) 107 πJ

(b) 104 πJ

30

(c) 102πJ

p

(d) 103 πJ

Laws of Thermodynamics — 287

Chapter 21

15. If W ABC is the work done in process A → B → C and W DEF is work done in process D → E → F as shown in the figure, then 7p0 6p0 5p0 C

4p0

F

3p0 2p0 p0

A

D

B E

V0

2V0 3V0 4V0 5V0 6V0 7V0

(a) |WDEF | > |W ABC | (c) WDEF = W ABC

(b) |WDEF | < |W ABC | (d) WDEF = − W ABC

Subjective Questions 1. How many moles of helium at temperature 300 K and 1.00 atm pressure are needed to make the internal energy of the gas 100 J?

2. Show how internal energy U varies with T in isochoric, isobaric and adiabatic process? 3. A system is taken around the cycle shown in figure from state a to state b and then back to state a.The absolute value of the heat transfer during one cycle is 7200 J. (a) Does the system absorb or liberate heat when it goes around the cycle in the direction shown in the figure? (b) What is the work W done by the system in one cycle? (c) If the system goes around the cycle in a counter-clock wise direction, does it absorb or liberate heat in one cycle? What is the magnitude of the heat absorbed or liberated in one counter-clockwise cycle? p

b

a V

O

4. For the thermodynamic cycle shown in figure find (a) net output work of the gas during the cycle, (b) net heat flow into the gas per cycle. p (Pa) 4 × 105

2 × 105

0

B

A

C

D

1

2

3

4

5

V (cm3)

288 — Waves and Thermodynamics 5. A thermodynamic system undergoes a cyclic process as shown in figure. The cycle consists of two closed loops, loop I and loop II. (a) Over one complete cycle, does the system do positive or negative work? (b) In each of loops I and II, is the net work done by the system positive or negative ? (c) Over one complete cycle, does heat flow into or out of the system? (d) In each of loops I and II, does heat flow into or out of the system? p I

II V

O

6. A gas undergoes the cycle shown in figure. The cycle is repeated 100 times per minute. Determine the power generated. p (atm) 30

10

A

C

B 2

3

V (litre)

7. One mole of an ideal monoatomic gas is initially at 300 K. Find the final temperature if 200 J of heat is added (a) at constant volume (b) at constant pressure.

8. A closed vessel 10 L in volume contains a diatomic gas under a pressure of 105 N/ m 2. What amount of heat should be imparted to the gas to increase the pressure in the vessel five times?

9. One mole of an ideal monatomic gas is taken round the cyclic process ABCA as shown in figure. Calculate p 3p0

p0

B

A V0

(a) (b) (c) (d)

C 2V0

V

the work done by the gas. the heat rejected by the gas in the path CA and heat absorbed in the path AB. the net heat absorbed by the gas in the path BC. the maximum temperature attained by the gas during the cycle.

10. A diatomic ideal gas is heated at constant volume until its pressure becomes three times. It is again heated at constant pressure until its volume is doubled. Find the molar heat capacity for the whole process.

11. Two moles of a certain gas at a temperature T0 = 300 K were cooled isochorically so that the pressure of the gas got reduced 2 times. Then as a result of isobaric process, the gas is allowed to expand till its temperature got back to the initial value. Find the total amount of heat absorbed by gas in this process.

Chapter 21

Laws of Thermodynamics — 289

12. Five moles of an ideal monoatomic gas with an initial temperature of 127 °C expand and in the process absorb 1200 J of heat and do 2100 J of work . What is the final temperature of the gas?

13. Find the change in the internal energy of 2 kg of water as it is heated from 0°C to 4°C . The specific heat capacity of water is 4200 J/ kg-K and its densities at 0 °C and 4°C are 999.9 kg/ m3 and 1000 kg/ m3 , respectively. Atmospheric pressure = 105 Pa.

14. Calculate the increase in the internal energy of 10 g of water when it is heated from 0 °C to 100 °C and converted into steam at 100 kPa. The density of steam = 0.6 kg/ m3 . Specific heat capacity of water = 4200 J/ kg- °C and the latent heat of vaporisation of water = 2.5 × 106 J/ kg.

15. One gram of water (1 cm3 ) becomes 1671 cm3 of steam when boiled at a constant pressure of 1 atm (1.013 × 105 Pa ). The heat of vaporization at this pressure is Lv = 2.256 × 106 J / kg. Compute (a) the work done by the water when it vaporizes and (b) its increase in internal energy.

16. A gas in a cylinder is held at a constant pressure of 2.30 × 105 Pa and is cooled and compressed from 1.70 m3 to 1.20 m3 . The internal energy of the gas decreases by 1.40 × 105 J. (a) Find the work done by the gas. (b) Find the absolute value|Q| of the heat flow into or out of the gas and state the direction of the heat flow. (c) Does it matter whether or not the gas is ideal? Why or why not?

17. p-V diagram of an ideal gas for a process ABC is as shown in the figure. p 3p0

C

A

p0

B

O

V0

3V0

V

(a) Find total heat absorbed or released by the gas during the process ABC. (b) Change in internal energy of the gas during the process ABC. (c) Plot pressure versus density graph of the gas for the process ABC.

18. In the given graph, an ideal gas changes its state from A to C by two paths ABC and AC. p B

8 Pa 4 Pa

C

A

5 m3

15 m3

V

(a) Find the path along which work done is less. (b) The internal energy of gas at A is 10 J and the amount of heat supplied in path AC is 200 J. Calculate the internal energy of gas at C. (c) The internal energy of gas at state B is 20 J. Find the amount of heat supplied to the gas to go from A to B.

290 — Waves and Thermodynamics 19. When a gas expands along AB, it does 500 J of work and absorbs 250 J of heat. When the gas expands along AC, it does 700 J of work and absorbs 300 J of heat. p D

C

A

B V

(a) How much heat does the gas exchange along BC? (b) When the gas makes the transition from C to A along CDA, 800 J of work are done on it from C to D. How much heat does it exchange along CDA?

20. A 1.0 kg bar of copper is heated at atmospheric pressure (1.01 × 105 N/ m 2 ). If its temperature increases from 20°C to 50°C, calculate the change in its internal energy. α = 7.0 × 10−6 / ° C, ρ = 8.92 × 103 kg/ m3 and c = 387 J/ kg-° C

21. One mole of an ideal monoatomic gas occupies a volume of 1.0 × 10−2 m3 at a pressure of 2.0 × 105 N/ m 2. (a) What is the temperature of the gas? (b) The gas undergoes an adiabatic compression until its volume is decreased to 5.0 × 10−3 m3 . What is the new gas temperature? (c) How much work is done on the gas during the compression? (d) What is the change in the internal energy of the gas?

22. A bullet of mass 10 g travelling horizontally at 200 m/s strikes and embeds in a pendulum bob of mass 2.0 kg. (a) How much mechanical energy is dissipated in the collision? (b) Assuming that Cv for the bob plus bullet is 3 R, calculate the temperature increase of the system due to the collision. Take the molecular mass of the system to be 200 g/mol.

23. An ideal gas is carried through a thermodynamic cycle consisting of two isobaric and two isothermal processes as shown in figure. Show that the net work done in the entire cycle is p given by the equation. W net = p1 (V 2 – V1 ) ln 2 p1 p p2

p1

B

C

A V1

D V2

V

24. An ideal gas is enclosed in a cylinder with a movable piston on top. The piston has mass of 8000 g and an area of 5.00 cm 2 and is free to slide up and down, keeping the pressure of the gas constant. How much work is done as the temperature of 0.200 mol of the gas is raised from 200°C to 300°C?

LEVEL 2 Single Correct Option 1. The equation of a state of a gas is given by p (V − b) = nRT . If 1 mole of a gas is isothermally expanded from volume V and 2V, the work done during the process is 2V − b  (a) RT ln  V − b 

V − b (b) RT ln   V 

V − b   (c) RT ln  2 V − b

 V   (d) RT ln  V − b

2. A cyclic process for 1 mole of an ideal gas is shown in the V-T diagram. The work done in AB, BC and CA respectively is

V2

V  (a) 0, RT2 ln  2, R(T1 − T2) V1

V1

V  (b) R(T1 − T2), 0, RT1 ln  1 V 2

O

V  (c) 0, RT1 ln  1, R(T1 − T2) V 2

C

A

T1

B

T2

V  (d) 0, RT2 ln  2, R(T2 − T1 ) V1

3. Ten moles of a diatomic perfect gas are allowed to expand at constant pressure. The initial volume and temperature are V 0 and T0, respectively. If

7 RT0 heat is transferred to the gas, 2

then the final volume and temperature are (b) 0.9 V 0 , 0.9 T0 10 (d) 0.9 V 0 , T0 9

(a) 1.1 V 0 , 1.1 T0 10 (c) 1.1 V 0 , T0 11

4. An ideal monoatomic gas is carried around the cycle ABCDA as shown in the figure. The efficiency of the gas cycle is p 3p0

p0

B

C

D

A

2V0

V0

4 21 4 (c) 31 (a)

2 21 2 (d) 31 (b)

V

292 — Waves and Thermodynamics 5. In the process shown in figure, the internal energy of an ideal gas decreases by

3 p0V 0 in going 2

from point C to A. Heat transfer along the process CA is p B

2p0

p0

A V0

(a) (−3 p0V 0 ) (c) (−3 p0V 0 /2)

C 2V0

V

(b) (−5 p0V 0 /2) (d) zero

6. One mole of an ideal monoatomic gas at temperature T0 expands slowly according to the law p = constant. If the final temperature is 2T0, heat supplied to the gas is V

(a) 2RT0

(b)

3 RT0 2

(c) RT0

(d)

1 RT0 2

7. A mass of gas is first expanded isothermally and then compressed adiabatically to its original volume. What further simplest operation must be performed on the gas to restore it to its original state? (a) (b) (c) (d)

An isobaric cooling to bring its temperature to initial value An isochoric cooling to bring its pressure to its initial value An isothermal process to take its pressure to its initial value An isochoric heating to bring its temperature to initial value

8. A monatomic ideal gas, initially at temperature T1 , is enclosed in a cylinder fitted with a frictionless piston. The gas is allowed to expand adiabatically to a temperature T2 by releasing the piston suddenly. If L1 and L2 are the lengths of gas column before and after expansion T respectively, then 1 is given by T2 2

L 3 (a)  1   L 2

2

L (b) 1 L2

L 3 (d)  2  L1 

L (c) 2 L1

9. One mole of an ideal gas is taken through a cyclic process. The minimum temperature during the cycle is 300 K. Then, net exchange of heat for complete cycle is U 2U0 U0

A

D V0

(a) 600 R ln 2 (c) − 300 R ln 2

B

C

2V0

V

(b) 300 R ln 2 (d) 900 R ln 2

Chapter 21

Laws of Thermodynamics — 293

10. Two moles of an ideal gas are undergone a cyclic process 1-2-3-1. If net heat exchange in the process is 300 J, the work done by the gas in the process 2-3 is T 2

600 K 3 300 K

1 V

(a) –500 J

(b) – 5000 J

(c) – 3000 J

(d) None of these

11. Two cylinders fitted with pistons contain equal amount of an ideal diatomic gas at 300 K. The piston of A is free to move, while that of B is held fixed. The same amount of heat is given to the gas in each cylinder. If the rise in temperature of the gas in A is 30 K, then the rise in temperature of gas in B is (a) 30 K (c) 50 K

(b) 18 K (d) 42 K

12. A gas follows a process TV n − 1 = constant, where T = absolute temperature of the gas and V = volume of the gas. The bulk modulus of the gas in the process is given by (a) (n − 1) p (c) np

(b) p /(n − 1) (d) p /n

13. One mole of an ideal gas at temperature T1 expands slowly according to the law

p = constant. V

Its final temperature is T2. The work done by the gas is

(a) R (T2 − T1 ) R (c) (T2 − T1 ) 2

(b) 2R (T2 − T1 ) 2R (T2 − T1 ) 3

(d)

14. 600 J of heat is added to a monoatomic gas in a process in which the gas performs a work of 150 J. The molar heat capacity for the process is (a) 3R (c) 2R

(b) 4R (d) 6R

15. The internal energy of a gas is given byU = 2 pV . It expands from V 0 to 2V 0 against a constant pressure p0. The heat absorbed by the gas in the process is (a) 2 p0V 0 (c) 3 p0V 0

(b) 4 p0V 0 (d) p0V 0

16. The figure shows two paths for the change of state of a gas from A to B. The ratio of molar heat capacities in path 1 and path 2 is

p 2 A

B 1 V

(a) < 1 (c) 1

(b) > 1 (d) Data insufficient

294 — Waves and Thermodynamics 17. p-T diagram of one mole of an ideal monatomic gas is shown. Processes AB and CD are adiabatic. Work done in the complete cycle is p B

C A

O

T

D

3T

4T

T

5T

(b) −2 RT (d) −3.5 RT

(a) 2.5 RT (c) 1.5 RT

18. An ideal monoatomic gas undergoes a process in which its internal energyU and density ρ vary as Uρ = constant. The ratio of change in internal energy and the work done by the gas is

3 2 1 (c) 3

2 3 3 (d) 5

(a)

(b)

19. The given figure shows the variation of force applied by ideal gas on a piston which undergoes a process during which piston position changes from 0.1 to 0.4 m. If the internal energy of the system at the end of the process is 2.5 J higher, then the heat absorbed during the process is F (N) 100 50

0.1

0.2

0.3

0.4 x (m)

(b) 17.5 J (d) 22.5 J

(a) 15 J (c) 20 J

p = constant. Assuming that the V initial volume is same in both processes and the final volume which is two times the initial volume is also same in both processes, which of the following is true?

20. A gas can expand through two processes : (i) isobaric, (ii)

(a) (b) (c) (d)

Work done by gas in process (i) is greater than the work done by the gas in process (ii) Work done by gas in process (i) is smaller than the work done by the gas in process (ii) Final pressure is greater in process (i) Final temperature is greater in process (i)

21. An ideal gas of adiabatic exponent γ is expanded so that the amount of heat transferred to the gas is equal to the decrease of its internal energy. Then, the equation of the process in terms of the variables T and V is (a) TV (c)

( γ − 1) 2

( γ − 1) TV 4

=C

(b) TV

=C

(d)

( γ − 2) 2

( γ − 2) TV 4

=C =C

Laws of Thermodynamics — 295

Chapter 21

22. A thermodynamical process is shown in the figure with −4

pA = 3 × patm , V A = 2 × 10

−4

m , pB = 8 × patm , VC = 5 × 10 3

p B

3

m . pB

In the process AB and BC, 600 J and 200 J heat are added to the system. Find the change in internal energy of the system in the pA process CA. [ 1 patm = 105 N/ m 2 ]

A

(b) –560 J (d) +240 J

(a) 560 J (c) –240 J

C

VA

VC

V

23. A gas takes part in two processes in which it is heated from the same initial state 1 to the same final temperature. The processes are shown on the p-V diagram by the straight lines 1-3 and 1-2. 2 and 3 are the points on the same isothermal curve. Q1 and Q2 are the heat transfer along the two processes. Then, p 2 Isothermal 3

1

V

(a) Q1 = Q2 (c) Q1 > Q2

(b) Q1 < Q2 (d) Insufficient data

24. A closed system receives 200 kJ of heat at constant volume. It then rejects 100 kJ of heat while it has 50 kJ of work done on it at constant pressure. If an adiabatic process can be found which will restore the system to its initial state, the work done by the system during this process is (b) 50 kJ (d) 200 kJ

(a) 100 kJ (c) 150 kJ

25. 100 moles of an ideal monatomic gas undergoes the thermodynamic process as shown in the figure (105 Nm–2) A

2.4

B

1

D 1

C 2

(m3)

A → B : isothermal expansion

B → C : adiabatic expansion

C → D : isobaric compression

D → A : isochoric process

The heat transfer along the process AB is 9 × 104 J.The net work done by the gas during the cycle is [Take R = 8 JK −1 mol−1 ) (a) −0.5 × 104 J (c) −5 × 104 J

(b) + 0.5 × 104 J (d) + 5 × 104 J

296 — Waves and Thermodynamics 26. Two moles of an ideal monoatomic gas are expanded according to the equation pT = constant from its initial state ( p0 , V 0 ) to the final state due to which its pressure becomes half of the initial pressure. The change in internal energy is p

A

B T

3 p0V 0 4 9 p0V 0 (c) 2

3 p0V 0 2 5 p0V 0 (d) 2

(b)

(a)

27. The state of an ideal gas is changed through an isothermal process at temperature T0 as shown in figure. The work done by the gas in going from state B to C is double the work done by gas in p going from state A to B. If the pressure in the state B is 0 , then the pressure of the gas in state 2 C is p A

p0

B

p0/2

C

V0

p0 3 p (c) 0 6

a)

V

p0 4 p0 (d) 8 (b)

More than One Correct Options p , 2 volume 2V) along a straight line path in the p-V diagram. Select the correct statement(s) from the following.

1. An ideal gas is taken from the state A (pressure p, volume V ) to the state B (pressure

(a) (b) (c) (d)

The work done by the gas in the process A to B is negative In the T-V diagram, the path AB becomes a part of a parabola In the p-T diagram, the path AB becomes a part of a hyperbola In going from A to B, the temperature T of the gas first increases to a maximum value and then decreases

2. In the process pV 2 = constant, if temperature of gas is increased, then (a) (b) (c) (d)

change in internal energy of gas is positive work done by gas is positive heat is given to the gas heat is taken out from the gas

Laws of Thermodynamics — 297

Chapter 21

T

3. T-V diagram of two moles of a monoatomic gas is as shown in figure. For the process abcda choose the correct options given below (a) (b) (c) (d)

b

2T0

∆U = 0 work done by gas > 0 heat given to the gas is 4RT0 heat given to the gas is 2RT0

T0

a

c d

V0

2V0

V

4. Density (ρ) versus internal energy (U ) graph of a gas is as shown in figure. Choose the correct options. ρ a

b

c U

(a) Qbc = 0

(b) Wbc = 0

(c) W ca < 0

(d) Qab > 0

Here, W is work done by gas and Q is heat given to the gas.

5. Temperature of a monatomic gas is increased from T0 to 2T0 in three different processes : isochoric, isobaric and adiabatic. Heat given to the gas in these three processes are Q1 , Q2 and Q3 respectively. Then, choose the correct option. (a) Q1 > Q3 (c) Q2 > Q3

(b) Q2 > Q1 (d) Q3 = 0

6. A cyclic process 1-2-3-4-1 is depicted on V -T diagram. The p-T and p-V diagrams for this cyclic process are given below. Select the correct choices (more than one options is/are correct) V

4

3

1 2

T

p

p

2 2

(a)

3

1

3

(b) 1

4 T

V

V p 4

3

(d) None of these

(c) 2 1

V

4

298 — Waves and Thermodynamics Comprehension Based Questions Passage I (Q. No. 1 and 2) One mole of a monatomic ideal gas is taken along the cycle ABCA as shown in the diagram. B

2p A

p

C

V

2V

1. The net heat absorbed by the gas in the given cycle is pV 2 (d) 4 pV

(a) pV

(b)

(c) 2 pV

2. The ratio of specific heat in the process CA to the specific heat in the process BC is 5 3 (d) None of these

(a) 2

(b)

(c) 4

Passage II (Q. No. 3 to 5) One mole of a monatomic ideal gas is taken through the cycle ABCDA as shown in the figure. TA = 1000 K and 2 pA = 3 pB = 6 pC . p A

Adiabatic B

D Adiabatic

C V

0.4   25  2 JK −1 mol −1  Assume   = 0.85 and R =   3 3  

3. The temperature at B is (a) 350 K (c) 850 K

(b) 1175 K (d) 577 K

4. Work done by the gas in the process A → B is (a) 5312 J (c) 6251 J

(b) 1875 J (d) 8854 J

5. Heat lost by the gas in the process B → C is (a) 5312 J (c) 6251 J

(b) 1875 J (d) 8854 J

Chapter 21

Laws of Thermodynamics — 299

Match the Columns 1. Temperature of 2 moles of a monatomic gas is increased from T to 2T . Match the following two columns. Column I

Column II

(a) Work done by gas in isobaric process

(p) 3RT

(b) Change in internal energy in isobaric (q) 4RT process (c) Work done by gas in adiabatic process

(r) 2RT

(d) Change in internal energy in an adiabatic process

(s) None of these

2. For V-T diagrams of two processes a - b and c - d are shown in figure for same gas. Match the following two columns. V b a

d c T

Column I (a) Work done

Column II (p) is more in process ab

(b) Change in internal energy

(q) is more in process ca

(c) Heat exchange

(r) is same in both processes

(d) Molar heat capacity

(s) Can’t say anything

3. Temperature of a monatomic gas is increased by ∆T in process p2V = constant. Match the following two columns. Column I

Column II

(a) Work done by gas (p) 2nR∆T (b) Change in internal energy of gas (q) 5nR∆T (c) Heat taken by the gas (r) 3nR∆T (d) Work done on the gas (s) None of these

4. Match the following two columns. Column I (a) Isobaric expansion (b) Isochoric cooling (c) Adiabatic expansion (d) Isothermal expansion

Column II (p) W > ∆U (q) W < ∆U (r) Q = ∆U (s) Q < ∆U

300 — Waves and Thermodynamics 5. Heat taken by a gas in process a -b is 6 p0V 0. Match the following columns. p b

2p0 a

p0

V

2V0

V0

Column I (a) (b) (c) (d)

Column II

W ab ∆U ab Molar heat capacity in given process CV of gas

(p) (q) (r) (s)

2 p0V 0 4 p0V 0 2R None of these

Subjective Questions 1. Two moles of helium gas undergo a cyclic process as shown in

p

figure. Assuming the gas to be ideal, calculate the following 2 atm quantities in this process. (a) The net change in the heat energy. (b) The net work done. (c) The net change in internal energy.

1 atm

A

D 300 K

B C 400 K

T

2. 1.0 k-mol of a sample of helium gas is put through the cycle of operations shown in figure. BC is

an isothermal process and pA = 1.00 atm, V A = 22.4 m3 , pB = 2.00 atm. What are T A , TB and VC ? p

B

A

C V

3. The density (ρ) versus pressure ( p) graph of one mole of an ideal monoatomic gas undergoing a cyclic process is shown in figure. Molecular mass of gas is M. 2ρ0

2

(Density) ρ0 O

1 p0

3

(Pressure)

(a) Find work done in each process. (b) Find heat rejected by gas in one complete cycle. (c) Find the efficiency of the cycle.

2p0

Chapter 21

Laws of Thermodynamics — 301

4. An ideal gas goes through the cycle abc. For the complete cycle 800 J of heat flows out of the gas.

Process ab is at constant pressure and process bc is at constant volume . In process c-a, p ∝ V . States a and b have temperatures Ta = 200 K and Tb = 300 K. (a) Sketch the p -V diagram for the cycle. (b) What is the work done by the gas for the process ca?

5. A cylinder of ideal gas is closed by an 8 kg movable piston of area 60 cm 2 . The atmospheric pressure is 100 kPa. When the gas is heated from 30° C to 100° C, the piston rises 20 cm. The piston is then fastened in the place and the gas is cooled back to 30° C. If ∆Q1 is the heat added to the gas during heating and ∆Q2 is the heat lost during cooling, find the difference. 7  6. Three moles of an ideal gas C p = R at pressure p0 and temperature T0 is isothermally  2  expanded to twice its initial volume. It is then compressed at a constant pressure to its original volume. (a) Sketch p -V and p -T diagram for complete process. (b) Calculate net work done by the gas. (c) Calculate net heat supplied to the gas during complete process. (Write your answer in terms of gas constant = R )

7. Two moles of a gas ( γ = 5 / 3) are initially at temperature 27°C and occupy a volume of 20 litres. The gas is first expanded at constant pressure until the volume is doubled. Then, it is subjected to an adiabatic change until the temperature returns to its initial value. (a) Sketch the process on a p -V diagram. (b) What are final volume and pressure of the gas? (c) What is the work done by the gas?

8. An ideal monoatomic gas is confined by a spring loaded massless piston of cross-section 8.0 × 10−3 m 2. Initially, the gas is at 300 K and occupies a volume of 2.4 × 10−3 m3 and the spring is in its relaxed state. The gas is heated by an electric heater until the piston moves out slowly without friction by 0.1 m. Calculate

(a) the final temperature of the gas and (b) the heat supplied by the heater. The force constant of the spring is 8000 N/m, atmospheric pressure is 1.0 × 105 N/m2.The cylinder and the piston are thermally insulated.

7 9. An ideal diatomic gas  γ =  undergoes a process in which its internal energy relates to the  5 volume as U = α V , where α is a constant.

(a) Find the work performed by the gas to increase its internal energy by 100 J. (b) Find the molar specific heat of the gas.

10. For an ideal gas the molar heat capacity varies as C = CV + 3 aT 2. Find the equation of the process in the variables (T , V ) where a is a constant.

11. One mole of an ideal monatomic gas undergoes the process p = αT 1/ 2 , where α is a constant. (a) Find the work done by the gas if its temperature increases by 50 K. (b) Also, find the molar specific heat of the gas.

12. One mole of a gas is put under a weightless piston of a vertical cylinder at temperature T . The space over the piston opens into atmosphere. Initially, piston was in equilibrium. How much work should be performed by some external force to increase isothermally the volume under the piston to twice the volume? (Neglect friction of piston).

302 — Waves and Thermodynamics 13. An ideal monatomic gas undergoes a process where its pressure is inversely proportional to its temperature. (a) Calculate the molar specific heat for the process. (b) Find the work done by two moles of gas if the temperature changes from T1 to T2.

14. The volume of one mole of an ideal gas with the adiabatic exponent γ is changed according to a , where a is a constant. Find the amount of heat absorbed by the gas in the T process, if the temperature is increased by ∆T .

the relation V =

15. Two moles of a monatomic ideal gas undergo a cyclic process ABCDA as shown in figure. BCD is a semicircle. Find the efficiency of the cycle. p C

2p0 B

p0 p0 2

D A

V0

2V0

V

16. Pressure p, volume V and temperature T for a certain gas are related by p=

αT − βT 2 V

where, α and β are constants. Find the work done by the gas if the temperature changes from T1 to T2 while the pressure remains the constant. 5R 17. An ideal gas has a specific heat at constant pressure C p = . The gas is kept in a closed vessel 2 of volume V 0 at temperature T0 and pressure p0. An amount of 10 p0V 0 of heat is supplied to the gas. (a) Calculate the final pressure and temperature of the gas. (b) Show the process on p -V diagram.

18. Pressure versus temperature ( p -T ) graph of n moles of an ideal gas is shown in figure. Plot the corresponding: (a) density versus volume (ρ-V ) graph, (b) pressure versus volume ( p -V ) graph and (c) density versus pressure (ρ - p) graph.

p 4p0 2p0 p0

C

B

D A

T0

2T0

T

19. Three moles of an ideal gas being initially at a temperature Ti = 273 K were isothermally expanded 5 times its initial volume and then isochorically heated so that the pressure in the final state becomes equal to that in the initial state. The total heat supplied in the process is  Cp  80 kJ. Find γ  =  of the gas.  CV 

Answers Introductory Exercise 21.1 1. a = 80 J , b = 0 , c = 50 J , d = 40 J , e = 40 J , f = 40 J , g = 40 J

2. 327 J

Introductory Exercise 21.2 1. (a) − 6.8 × 10 4 J

(b) 1.78 × 10 5 J, out of gas

(c) No

2. (a) Positive (b) Into the system (c) In loop 1, into the system, In loop 2, out of the system 3. 2.67 × 10 −2 mol

4. W = 0, Q = ∆U = 1800 R

6. − 600 R

7.

5. (i) 2 p0V0

(ii) zero (iii)

3 p0V0 2

1 p0V0 2

Introductory Exercise 21.3 1. (a) 316 K 3. Process BC CA AB

2. Into gas

(b) 309.6 K Q

W

∆U

– – +

0 – +

− − + γ γ −1

4. (a) No (b) No (c) No

5. γ and

7. 1.18 MJ

8. 0.6 kJ, 1.0 kJ, 1.6

6. Q1 > Q2

Introductory Exercise 21.4 1. 2.72 × 10 5 cal, 72.72 % 4. 2 × 10 cal 6

3. (a) 25% (b) 6.25 × 10 5 J (c) 18.75 × 10 5 J

2. 60 %

5. 117° C and 52° C

6. Refrigerator A

7. 35 W

8. 18.18%

Exercises LEVEL 1 Assertion and Reason 1. (b)

2. (d)

3. (d)

4. (a)

5. (b)

6. (d)

7. (c)

8. (b)

9. (a)

10. (a or b)

6. (c)

7. (d)

8. (a)

9. (b)

10. (c)

Objective Questions 1. (a)

2. (c)

3. (c)

4. (a)

5. (d)

11 (d)

12. (c)

13. (c)

14. (c)

15. (d)

Subjective Questions 1. 2.67 × 10 −2 mol 3. (a) absorbs

(b) 7200 J

2. ∆U = (c) liberates, 7200 J

nR∆T for all processes γ −1

4. (a) 0.50 J

(b) 0.50 J

304 — Waves and Thermodynamics 5. (a) positive (b) I→ positive, II→ negative (c) into the system (d) I→ into the system II→ out of the system. 6. 1.68 × 103 W 7. (a) 316 K

8. 10 4 J

(b) 310 K

10. C = 3.1 R

9. (a) p0V0

(b)

11. 2.49 kJ

5 p0V0 , 3 p0V0 2 12. 113°C

(d)

25p0V0 8R

14. 2.75 × 10 4 J

13. 33600.2 J 15. (a) 169 J

p0V0 2

(c)

16. (a) −1.15 × 10 J 5

(b) 2087 J

17. (a) QABC = − 2P0V0

(b) ∆UABC = 0

(b) 2.55 × 10 J, out of gas 5

18. (a) AC

19. (a) – 150 J (b) – 400 J 21. (a) 241 K (b) 383 K (c) 1770 J 24. 166 J

(b) 150 J

(c) No

(c) 10 J

20. 11609.99287 J 22. (a) 200 J (b) 0.80°C

(d) 1770 J

LEVEL 2 Single Correct Option 1.(a)

2.(a)

3.(a)

4.(a)

5.(b)

6.(a)

7.(b)

8.(d)

9.(b)

10.(d)

11 (d)

12.(c)

13.(c)

14.(c)

15.(c)

16.(a)

17.(a)

18.(a)

19.(c)

20.(b)

21.(a)

22.(b)

23.(c)

24.(c)

25.(d)

26.(b)

27.(d)

More than One Correct Options 1.(b,d)

2.(a,c) 3.(a,b)

4.(c,d)

5.(a,b,c,d)

6.(a,b)

Comprehension Based Questions 1.(b)

2.(b)

3.(c)

4.(b)

5.(a)

Match the Columns 1. (a) → r

(b) → p

(c) → s

(d) → p

2. (a) → s

(b) → s

(c) → s

(d) → r

3. (a) → p

(b) → s

(c) → s

(d) → s

4. (a) → q

(b) → p,r

(c) → p

(d) → p

5. (a) → s

(b) → s

(c) → r

(d) → s

Subjective Questions 1. (a) 1153 J (b) 1153 J (c) zero 2. T A = 273 K, TB = 546 K, VC = 44.8 m3 3. (a) W12 =

− p0 M pM pM 3  ln (2), W 23 = 0 , W31 = 0 (b) 0  + ln 2  ρ0 ρ0 ρ0  2

4. (b) – 4000 J 6. (b) 3RT 0 ln(2) −

3a  2 −  T  2R 

2 (1 − ln 2) 5

5. 136 J 3 RT 0 2

8. (a) 800 K (b) 720 J 10. Ve

(c)

= constant

(c) 3RT 0 ln(2) −

21 RT 0 4

7. (b) 113.1 L, 0.44 × 10 5 N / m2 9. (a) 80 J

(b)

9R 2

11. (a) 207.75 J

(b) 2R

(c) 12479 J

Chapter 21 12. RT (1 − ln 2) 14.

13. (a)

(2 − γ ) R∆T

17. (a)

ρ

4p0

B,C

ρ0

2p0

A,D

C

B

p0

ρ 2ρ0

V0 2

V0

B

C

V

A D

p0 2p0

19. 1.4

23 23 p0 , T0 3 3

p

2ρ0

ρ0

7R (b) 4R (T 2 − T1) 2

15. 25.8%

(γ − 1)

16. α (T 2 − T1) − β (T 22 − T12 ) 18.

Laws of Thermodynamics — 305

p 4p0

D A

V0 2

V0

V

Calorimetary and Heat Transfer Chapter Contents 22.1 Calorimetry 22.2 Heat Transfer

308 — Waves and Thermodynamics

22.1 Calorimetry Heat transfer between two substances, due to a temperature difference (may be in same or different states) comes under the topic calorimetry. Let us discuss this topic pointwise.

Specific Heat (c) The amount of heat required to raise the temperature of unit mass of a substance by 1°C (or 1 K) is called its specific heat. Thus, c=

Q m∆θ

or

Q m∆T

…(i)

The SI unit of specific heat is J/kg-K. Because heat is so frequently measured in calories, the unit cal/g-°C is also used quite often.

Heat Required to Change The Temperature From Eq. (i), we can write Q = mc∆θ or

mc∆T

…(ii)

Thus Eq. (ii) is used when temperature of a substance changes without change in state. Note

(i) In general c is a function of temperature. But normally variation in c is small. So, it is assumed to be constant for wide range of temperature. For example “Specific heat of water is 1 cal / g -° C between 14.5 °C to 15.5 °C”. This implies that 1 cal of heat will be required to raise the temperature of 1 g of water from 14.5 °C to 15.5 °C. From 15.5 °C to 16.5 °C, a different amount of heat ( ≠ 1 cal ) will be required. But normally this variation is very small. So, it is assumed constant ( = 1 cal / g - °C ) from `0 °C to 100 °C. (ii) If c is given a function of temperature, then Eq. (ii) cannot be applied directly for calculation of Q. Rather integration will be used. Thus, θf

Tf

Q = m∫ cdT

or

m ∫ cd θ

(iii)

θi

Ti

In the above expression, c is a function of T ( or θ ). (iii) The specific heat of water is much larger than that of most other substances. Consequently, for the same amount of added heat, the temperature change of a given mass of water is generally less than that for the same mass of another substance. For this reason a large body of water moderates the climate of nearby land. In the winter, the water cools off more slowly than the surrounding land and tends to warm the land. In the summer, the opposite effect occurs as the water heats up more slowly than the land. (iv) Specific heat is also represented by s or S. So, in different problems we have taken all notations for it.

Latent Heat ( L ) The amount of heat required to change the state (or phase) of unit mass of a substance at constant temperature is called latent heat L. Thus, L=

Q m

…(iv)

Calorimetry and Heat Transfer — 309

Chapter 22

For a solid-liquid transition, the latent heat is known as the latent heat of fusion ( L f ) and for the liquid-gas transition, it is known as the latent heat of vaporization ( Lv ). For water at 1 atmosphere, latent heat of fusion is 80.0 cal/g. This simply means 80.0 cal of heat are required to melt 1.0 g of water or 80.0 cal heat is liberated when 1.0 g of water freezes at 0°C. Similarly, latent heat of vaporization for water at 1 atmosphere is 539 cal/g.

Heat Required to Change the State (or Phase) at Constant Temperature From Eq. (iv), we can see that …(v)

Q = mL So, this is the heat required to change the state (or phase) at constant temperature.

Extra Points to Remember ˜

In some problems, two or more than two substances at different temperatures are mixed. Heat is given by the hot bodies and taken by the cold bodies. Finally, all substances reach to a common equilibrium temperature. All such problems can be solved by a single equation, Heat given by hot bodies = Heat taken by cold bodies But note that, on both sides of this equation, we have to take the positive signs. To make them positive, always write ∆θ = θ Higher − θ Lower in the equation Q = mc∆θ when there is a change of temperature without change in state.

˜

V

Water equivalent of a container Normally, a liquid is heated in a container. So, some heat is wasted in heating the container also. Suppose water equivalent of a container is 10 g, then it implies that heat required to increase the temperature of this container is equal to heat required to increase the temperature of 10 g of water.

Example 22.1 How much heat is required to convert 8.0 g of ice at – 15°C to steam at 100°C? (Given , cice = 0.53 cal/g-°C, L f = 80 cal / g and L v = 539 cal/g, and cwater = 1 cal/g-°C ) Solution

Ice

Ice

Water

Water

–15°C

0°C

0°C

100°C

Q1

Q2

Q3

Steam Q4

100°C

Fig. 22.1

∴ Net heat required,

Q1 = mc ice (T f – Ti ) = ( 8.0) ( 0.53) [ 0 – (–15)] = 63.6 cal Q2 = mL f = ( 8) ( 80) = 640 cal Q3 = mc water (T f – Ti ) = ( 8.0) (1.0) [100 – 0] = 800 cal Q4 = mLv = ( 8.0) ( 539) = 4312 cal Q = Q1 + Q2 + Q3 + Q4 = 5815.6 cal

Ans.

310 — Waves and Thermodynamics V

Example 22.2 In the above problem if heat is supplied at a constant rate of q = 10 cal / min , then plot temperature versus time graph. Solution

T(°C)

D

100

E

B C t (min) 6.36 70.36 150.36 581.56 −15° A

Fig. 22.2

From A to B (i) Temperature of ice will increase from −15° C to 0°C. Q Total heat required (ii) tAB = = 1 Heat supplied per minute q 63.6 = 6.36 min 10 (iii) Between A and B we will get only ice. =

From B to C (i) Temperature of ( ice + water ) mixture will remain constant at 0 °C. Q 640 (ii) t BC = 2 = = 64 min q 10 ∴ t Total = t AB + t BC = 70.36 min (iii) Between B and C we will get both ice and water. From C to D (i) Temperature of water increases from 0°C to 100 °C Q 800 (ii) t CD = 3 = = 80 min q 10 ∴ t Total = t AB + t BC + t CD = 150.36 min (iii) Between C and D we will get only water. From D to E (i) Temperature of ( water + steam ) mixture will remain constant at 100 °C. Q 4312 (ii) t DE = 4 = = 431.2 min q 10 ∴ t Total = t AB + t BC + t CD + t DE = 581.56 min (iii) Between D and A we will get both water and steam. The corresponding graph is as shown in Fig. 22.2.

Chapter 22

Calorimetry and Heat Transfer — 311

Exercise AB and CD correspond to temperature change without change in state. BC and DE correspond to state change without change in temperature. In the given condition, prove that AB and CD are straight lines and slope of these lines is inversely proportional to specific heat in that state. Further, prove that lengths of lines BC and DE are proportional to latent heat in that state. V

Example 22.3 10 g of water at 70°C is mixed with 5 g of water at 30°C. Find the temperature of the mixture in equilibrium. Specific heat of water is 1 cal / g - °C Solution

or ∴ ∴ V

Let t ° C be the temperature of the mixture. From energy conservation, Heat given by 10 g of water = Heat taken by 5 g of water m1 c water | ∆t 1 | = m2 c water | ∆t 2 | (10) (1) ( 70 – t ) = 5 (1) ( t – 30) t = 56.67° C

Ans.

Example 22.4 The temperature of equal masses of three different liquids A, B and C are 12°C, 19°C and 28°C respectively. The temperature when A and B are mixed is 16°C and when B and C are mixed it is 23°C. What should be the temperature when A and C are mixed? Solution Let m be the mass of each liquid and s A , sB , sC specific heats of liquids A , B and C respectively. When A and B are mixed. The final temperature is 16° C. Heat gained by A = Heat lost by B ∴ i.e. ms A (16 − 12) = msB (19 − 16) 4 …(i) i.e. sB = s A 3 When B and C are mixed. Heat gained by B = Heat lost by C i.e. msB ( 23 − 19) = msC ( 28 − 23) 4 …(ii) i.e. sC = sB 5 4 4 16 From Eqs. (i) and (ii), SC = × S A = SA 5 3 15 When A and C are mixed, let the final temperature be θ Heat gained by A = Heat lost by C ∴ msA (θ − 12) = msC ( 28 − θ ) 16 i.e. θ − 12 = ( 28 − θ ) 15 By solving, we get 628 Ans. θ= = 20.26 ° C 31

312 — Waves and Thermodynamics V

Example 22.5 In a container of negligible mass 30 g of steam at 100°C is added to 200 g of water that has a temperature of 40°C. If no heat is lost to the surroundings, what is the final temperature of the system? Also find masses of water and steam in equilibrium. Take L v = 539 cal / g and cwater = 1 cal / g - ° C. Solution

Let Q be the heat required to convert 200 g of water at 40°C into 100°C, then

Q = mc∆T = ( 200) (1.0) (100 – 40) = 12000 cal Now, suppose m0 mass of steam converts into water to liberate this much amount of heat, then Q 12000 m0 = = = 22.26g L 539 Since, it is less than 30 g, the temperature of the mixture is 100°C. and V

Ans.

Mass of steam in the mixture = 30 – 22.26 = 7.74 g

Ans.

mass of water in the mixture = 200 + 22.26 = 222.26 g

Ans.

Example 22.6 In an insulated vessel, 0.05 kg steam at 373 K and 0.45 kg of ice at 253 K are mixed. Find the final temperature of the mixture (in kelvin). Given, L fusion = 80 cal / g = 336 J / g L vaporization = 540 cal / g = 2268 J / g sice = 2100 J / kg - K = 0.5 cal / g - K and (JEE 2006) swater = 4200 J / kg - K = 1 cal / g - K Solution

Q1

0.05 kg steam at 373 K → 0.05 kg water at 373 K Q2

0.05 kg water at 373 K → 0.05 kg water at 273 K Q3

0.45 kg ice at 253 K → 0.45 kg ice at 273 K Q4

0.45 kg ice at 273 K → 0.45 kg water at 273 K Q1 = ( 50) ( 540) = 27,000 cal = 27 k cal Q2 = ( 50) (1) (100) = 5000 cal = 5 k cal Q3 = ( 450) ( 0.5) ( 20) = 4500 cal = 4.5 k cal Q4 = ( 450) ( 80) = 36000 cal = 36 k cal Now, since Q1 + Q2 > Q3 but Q1 + Q2 < Q3 + Q4 ice will come to 273 K from 253 K, but whole ice will not melt. Therefore, temperature of the mixture is 273 K. V

Example 22.7 An ice cube of mass 0.1 kg at 0°C is placed in an isolated container which is at 227°C. The specific heat s of the container varies with temperature T according to the empirical relation s = A + BT , where A = 100 cal / kg - K and B = 2 × 10 −2 cal/kg-K 2 . If the final temperature of the container is 27°C, determine the mass of the container. (Latent heat of fusion for water = 8 × 104 cal/kg, specific heat of water = 103 cal / kg - K ). (JEE 2001)

Chapter 22

Calorimetry and Heat Transfer — 313

Let m be the mass of the container. Initial temperature of container, Solution

Ti = ( 227 + 273) = 500 K and final temperature of container, T f = ( 27 + 273) = 300 K Now, heat gained by the ice cube = heat lost by the container ( 0.1)( 8 × 104 ) + ( 0.1)(103 )( 27) = − m ∫

∴

300 500

( A + BT ) dT 300

 BT 2  10700 = − m  AT +  2  500 

or

After substituting the values of A and B and the proper limits, we get m = 0.495 kg

INTRODUCTORY EXERCISE

Ans.

22.1

(Take cice = 0.53 cal/g-°C, cwater = 1.0 cal/g-°C, (Lf )water = 80 cal/g and (Lv )water = 529 cal/g unless given in the question.) 1. 10 g ice at 0°C is converted into steam at 100°C. Find total heat required. (Lf = 80 cal/g, Sw = 1cal/g-°C, Lv = 540 cal/g) 2. Three liquids A, B and C of specific heats 1 cal/g -°C, 0.5 cal/g -°C and 0.25 cal/g -°C are at temperatures 20 °C, 40 °C and 60 °C respectively. Find temperature in equilibrium if they are mixed together. Their masses are equal.

3. Equal masses of ice (at 0°C) and water are in contact. Find the temperature of water needed to just melt the complete ice.

4. A lead bullet just melts when stopped by an obstacle. Assuming that 25 per cent of the heat is absorbed by the obstacle, find the velocity of the bullet if its initial temperature is 27°C. (Melting point of lead = 327°C, specific heat of lead = 0.03 cal/g-°C, latent heat of fusion of lead = 6 cal/g, J = 4.2 J/cal). (JEE 1981)

5. The temperature of 100 g of water is to be raised from 24°C to 90°C by adding steam to it. Calculate the mass of the steam required for this purpose.

(JEE 1996)

6. 15 g ice at 0°C is mixed with 10 g water at 40°C. Find the temperature of mixture. Also, find mass of water and ice in the mixture.

7. Three liquids P , Q and R are given 4 kg of P at 60 °C and 1 kg of R at 50 °C when mixed produce a

resultant temperature 55 °C. A mixture of 1 kg of P at 60 °C and 1 kg of Q at 50°C shows a temperature of 55 °C.What will be the resulting temperature when1 kg of Q at 60 °C is mixed with 1 kg of R at 50 °C?

22.2 Heat Transfer Heat can be transferred from one place to the other by any of three possible ways : conduction, convection and radiation. In the first two processes, a medium is necessary for the heat transfer. Radiation, however, does no have this restriction. This is also the fastest mode of heat transfer, in which heat is transferred from one place to the other in the form of electromagnetic radiation.

314 — Waves and Thermodynamics Conduction T1 > T2 Figure shows a rod whose ends are in thermal contact with a hot reservoir at temperature T1 and a cold reservoir at temperature T2 . T1 T2 Q (Cold) The sides of the rod are covered with insulating medium, so the (Hot) transport of heat is along the rod, not through the sides. The molecules at the hot reservoir have greater vibrational energy. This Fig. 22.3 energy is transferred by collisions to the atoms at the end face of the rod. These atoms in turn transfer energy to their neighbours further along the rod. Such transfer of heat through a substance in which heat is transported without direct mass transport is called conduction. Most metals use another, more effective mechanism to conduct heat. The free electrons, which move throughout the metal can rapidly carry energy from the hotter to cooler regions, so metals are generally good conductors of heat. The presence of ‘free’ electrons also causes most metals to be good electrical conductors. A metal rod at 5°C feels colder than a piece of wood at 5°C because heat can flow more easily from your hand into the metal. Heat transfer occurs only between regions that are at different temperatures, and the rate of heat flow dQ is .This rate is also called the heat current, denoted by H. Experiments show that the heat current dt dT is proportional to the cross-section area A of the rod and to the temperature gradient , which is the dx rate of change of temperature with distance along the bar. In general, dQ dT …(i) H= = – KA dt dx dQ dT is negative. The constant K, called a positive quantity since The negative sign is used to make dx dt the thermal conductivity is a measure of the ability of a material to conduct heat. A substance with a large thermal conductivity k is a good heat conductor. The value of k depends on the temperature, increasing slightly with increasing temperature, but K can be taken to be practically constant throughout a substance if the temperature difference between its ends is not too great. Let us apply Eq. (i) to a rod of length l and constant cross-sectional area A in which a steady state has been reached. In a steady state, the temperature at each point is constant in time. Hence, dT – = T1 – T2 dx Therefore, the heat ∆Q transferred in time ∆t is

 T – T2  ∆Q = KA  1  ∆t  l  Thermal Resistance ( R ) Eq. (ii) in differential form can be written as dQ ∆T ∆T =H = = dt l / KA R Here, ∆T = temperature difference (TD) and R =

l = thermal resistance of the rod. KA

…(ii)

…(iii)

Chapter 22

Calorimetry and Heat Transfer — 315

Extra Points to Remember ˜

Consider a section ab of a rod as shown in figure. Suppose Q1 heat enters into Q1 Q2 the section at ‘a’ and Q2 leaves at ‘b’, then Q2 < Q1. Part of the energy Q1 – Q2 is a b utilized in raising the temperature of section ab and the remaining is lost to Fig. 22.4 atmosphere through ab. If heat is continuously supplied from the left end of the rod, a stage comes when temperature of the section becomes constant. In that case, Q1 = Q2 if rod is insulated from the surroundings (or loss through ab is zero). This is called the steady state condition. Thus, in steady state temperature of different sections of the rod becomes constant (but not same). Hence, in the figure : Q

Q T1

T2

T3

T4

Insulated rod in steady state

Fig. 22.5

T1 = constant, T2 = constant etc. and T1 > T2 > T3 > T4

T T1

Now, a natural question arises, why the temperature of whole rod not becomes equal when heat is being continuously supplied? The answer is : there must be a temperature difference in the rod for the heat flow, same as we require a T4 potential difference across a resistance for the current flow through it. x

In steady state, the temperature varies linearly with distance along the rod if it is insulated. ˜

Fig. 22.6

Comparing equation number (iii), i.e. heat current dQ ∆T H= = dt R

 where, R = l     KA 

with the equation, of current flow through a resistance, dq ∆V = i = dt R

l    where, R =   σA 

We find the following similarities in heat flow through a rod and current flow through a resistance.

Table 22.1 S.No

Heat flow through a conducting rod

Current flow through a resistance

1.

Conducting rod

Electrical resistance

2.

Heat flows

Charge flows

3.

TD is required dQ Heat current H = = rate of heat flow dt ∆T TD = H= R R l R= KA

PD is required dq Electric current i = = rate of charge flow dt ∆V PD = i = R R l R= σA

K = thermal conductivity

σ = electrical conductivity

4. 5. 6. 7.

From the above table it is evident that flow of heat through rods in series and parallel is analogous to the flow of current through resistance in series and parallel. This analogy is of great importance in solving complicated problems of heat conduction.

316 — Waves and Thermodynamics Convection Although conduction does occur in liquids and gases also, heat is transported in these media mostly by convection. In this process, the actual motion of the material is responsible for the heat transfer. Familiar examples include hot-air and hot-water home heating systems, the cooling system of an automobile engine and the flow of blood in the body. You probably have warmed your hands by holding them over an open flame. In this situation, the air directly above the flame is heated and expands. As a result, the density of this air decreases and then air rises. When the movement results from differences in density, as with air around fire, it is referred to as natural convection. Air flow at a beach is an example of natural convection. When the heated substance is forced to move by a fan or pump, the process is called forced convection. If it were not for convection currents, it would be very difficult to boil water. As water is heated in a kettle, the heated water expands and rises to the top because its density is lowered. At the same time, the denser, cool water at the surface sinks to the bottom of the kettle and is heated. Heating a room by a radiator is an example of forced convection.

Radiation The third means of energy transfer is radiation which does not require a medium. The best known example of this process is the radiation from sun. All objects radiate energy continuously in the form of electromagnetic waves. Now, let us define few terms before studying the other topics.

Radiant Energy All bodies radiate energy in the form of electromagnetic waves by virtue of their temperature. This energy is called the radiant energy. Absorptive Power ‘a’ “It is defined as the ratio of the radiant energy absorbed by it in a given time to the total radiant energy incident on it in the same interval of time.” energy absorbed a= energy incident As a perfectly black body absorbs all radiations incident on it, the absorptive power of a perfectly black body is maximum and unity.

Spectral Absorptive Power ‘a λ ’ The absorptive power ‘a’ refers to radiations of all wavelengths (or the total energy) while the spectral absorptive power is the ratio of radiant energy absorbed by a surface to the radiant energy incident on it for a particular wavelength λ. It may have different values for different wavelengths for a given surface. Let us take an example, suppose a = 0.6, a λ = 0.4 for1000 Å and a λ = 0.7 for 2000 Å for a given surface. Then, it means that this surface will absorb only 60% of the total radiant energy incident on it. Similarly, it absorbs 40% of the energy incident on it corresponding to1000 Å and 70% corresponding to 2000 Å. The spectral absorptive power a λ is related to absorptive power a through the relation ∞

a = ∫ a λ dλ 0

Chapter 22

Calorimetry and Heat Transfer — 317

Emissive Power ‘e’ (Do not confuse it with the emissivity e which is different from it, although both have the same symbol e). “For a given surface it is defined as the radiant energy emitted per second per unit area of the surface.” It has the units of W/ m 2 or J/s-m 2 . For a black body e = σT 4 . Spectral Emissive Power ‘e λ ’ “It is emissive power for a particular wavelength λ.” Thus, ∞

e = ∫ eλ dλ 0

Stefan’s Law The rate at which an object radiates energy is proportional to the fourth power of its absolute temperature. This is known as the Stefan’s law and is expressed in equation form as P = σAeT 4 Here, P is the power in watts (J/s) radiated by the object, A is the surface area in m 2 , e is a fraction between 0 and 1 called the emissivity of the object and σ is a universal constant called Stefan’s constant, which has the value σ = 5.67 × 10 –8 W/ m 2 -K 4 Note Emissivity e is also sometimes denoted by er .

Perfectly Black Body A body that absorbs all the radiation incident upon it and has an emissivity equal to 1 is called a perfectly black body. A black body is also an ideal radiator. It implies that if a black body and an identical another body are kept at the same temperature, then the black body will radiate maximum power as is obvious from equation P = eAσT 4 also. Because e =1for a perfectly black body while for any other body e T0 , the net rate of heat transfer from the body to the surroundings is dQ  dT  4 4 = eAσ (T 4 – T04 ) or mc  –  = eAσ (T – T0 )  dt  dt Rate of cooling  dT  eAσ 4 (T – T04 ) – =  dt  mc

or

–

dT ∝ (T 4 – T04 ) dt

Newton’s Law of Cooling According to this law, if the temperature T of the body is not very different from that of the dT surroundings T0 , then rate of cooling – is proportional to the temperature difference between dt them. To prove it, let us assume that T = T0 + ∆T So that

 ∆T  T 4 = (T0 + ∆T ) 4 = T04 1 +  T0    4∆T  ≈ T04 1 +  T0  

∴

(T 4 – T04 ) = 4T03 ( ∆T )

or

(T 4 – T04 ) ∝ ∆T

4

(from binomial expansion)

(as T0 = constant)

320 — Waves and Thermodynamics Now, we have already shown that rate of cooling  dT  4 4 –  ∝ (T – T0 )  dt  and here we have shown that (T 4 – T04 ) ∝ ∆T if the temperature difference is small. Thus, rate of cooling dT – ∝ ∆T dt

or

dT = dθ or

as

–

dθ ∝ ∆θ dt

∆T = ∆θ

Variation of Temperature of a Body According to Newton’s Law Suppose a body has a temperature θ i at time t = 0.It is placed in an atmosphere whose temperature is θ 0 . We are interested in finding the temperature of the body at time t. Assuming Newton’s law of cooling to hold good or by assuming that the temperature difference is small. As per this law, Rate of cooling ∝ temperature difference  dθ   eAσ  3 or –  =  (4θ 0 ) (θ – θ 0 )  dt   mc 

Here,

∴

θ0 = constant

θi

θ

t=0

t=t

Fig. 22.9

 dθ  –  = α (θ – θ 0 )  dt   4eAσθ 30   α =   mc 

or

∴

θ0 = constant

t dθ = – α ∫ dt 0 i θ – θ0

(is a constant)

θ

∫θ

θi

θ = θ 0 + (θ i – θ 0 ) e – αt

θ

θ0

From this expression we see that θ = θ i at t = 0 and θ = θ 0 at t = ∞, i.e. temperature of the body varies exponentially with time from θ i to θ 0 ( < θ i ). The temperature versus time graph is as shown in Fig.22.10.

t

Fig. 22.10

Note If the body cools by radiation from θ 1 to θ 2 in time t , then taking the approximation  dθ  θ 1 – θ 2 –  =  dt  t

and

 θ + θ2  θ = θ av =  1   2 

 dθ  The equation  –  = α (θ – θ 0 ) becomes  dt  θ1 – θ2  θ + θ2  =α  1 – θ 0  2  t This form of the law helps in solving numerical problems related to Newton’s law of cooling.

Chapter 22

Calorimetry and Heat Transfer — 321

Wien’s Displacement Law At ordinary temperatures (below about 600°C) the thermal radiation emitted by a body is not visible, most of it is concentrated in wavelengths much longer than those of visible light. eλ 4000 K 3000 K 2000 K

4 1 2 3 Wavelength (mm) Power of black body radiation versus wavelength at three temperatures. Note that the amount of radiation emitted (the area under a curve) increase with increasing temperature. Fig. 22.11 0

Figure shows how the energy of a black body radiation varies with temperature and wavelength. As the temperature of the black body increases, two distinct behaviors are observed. The first effect is that the peak of the distribution shifts to shorter wavelengths. This shift is found to obey the following relationship called Wien’s displacement law. λ mT = b Here, b is a constant called Wien’s constant. The value of this constant for perfectly black body in SI unit is 2.898 × 10 –3 m-K. Thus, λm ∝

1 T

Here, λ m is the wavelength corresponding to the maximum spectral emissive power eλ . The second effect is that the total amount of energy the black body emits per unit area per unit time ( = σT 4 ) increases with fourth power of absolute temperature T. This is also known as the emissive power. We know ∞

e = ∫ eλ dλ = Area under eλ - λ graph = σT 4 0

Area ∝ T 4

or

⇒

eλ

A2 = (2) 4 A1 = 16 A1 eλ

A2 A1 λm

2T

T λ

λm 2

λ

Fig. 22.12

Thus, if the temperature of the black body is made two fold, λ m remains half while the area becomes 16 times.

322 — Waves and Thermodynamics V

Example 22.8 In which of the following process, convection does not take place primarily? (JEE 2005) (a) (b) (c) (d)

Sea and land breeze Boiling of water Warming of glass of bulb due to filament Heating air around a furnace Solution (c) Glass of bulb heats due to filament by radiation.

V

Example 22.9 A copper rod 2 m long has a circular cross-section of radius 1 cm. One end is kept at 100°C and the other at 0°C. The surface is insulated so that negligible heat is lost through the surface. In steady state, find (a) the thermal resistance of the bar (b) the thermal current H dT (c) the temperature gradient and dx (d) the temperature at a distance 25 cm from the hot end. Thermal conductivity of copper is 401 W/m-K. l l Solution (a) Thermal resistance, R = = KA K ( πr 2 ) R=

or

( 2) ( 401) ( π ) (10–2 ) 2

= 15.9 K/ W (b) Thermal current, H =

Ans.

∆T ∆θ 100 = = R R 15.9 H = 6.3 W

or

Ans.

(c) Temperature gradient =

0 – 100 = – 50 K/ m 2

= – 50° C/ m

Ans.

(d) Let θ be the temperature at 25 cm from the hot end, then θ°C

100°C

0°C

0.25 m 2.0 m

Fig. 22.13

or or

(θ – 100) = (temperature gradient) × (distance) θ – 100 = (– 50) (0.25) θ = 87.5° C

Ans.

Chapter 22 V

Calorimetry and Heat Transfer — 323

Example 22.10 Three rods made of the same material and having the same cross-section have been joined as shown in the figure. Each rod is of the same length. The left and right ends are kept at 0°C and 90°C, respectively. The temperature of junction of the three rods will be (JEE 2001) (b) 60° C (d) 20° C Solution Let θ be the temperature of the junction (say B). Thermal resistance of all the three rods is equal. Rate of heat flow through AB + Rate of heat flow through CB = Rate of heat flow through BD 90° − θ 90° − θ θ − 0 ∴ + = R R R Solving this equation, we get θ = 60° C ∴ The correct option is (b). Here, R = Thermal resistance

90°C

0°C

(a) 45° C (c) 30° C

90°C

Fig. 22.14 A 90°C

D

B

0°C

θ

90°C C

Fig. 22.15

Temperature difference (TD ) Note Rate of heat flow, H= Thermal resistance ( R ) l where, R= KA K = Thermal conductivity of the rod. This is similar to the current flow through a resistance (R) where current (i) = Rate of flow of charge Potential difference ( PD ) = Electrical resistance ( R ) l Here, R = where σ = Electrical conductivity. σA V

Example 22.11 Figure shows a copper rod joined to a steel rod. The rods have equal length and equal cross- sectional area. The free end of the copper rod is kept at 0° C and that of the steel rod is kept at 100° C. Find the temperature θ at the junction of the rods. Conductivity of copper = 390 W/m-° C and that of steel = 46 W/m-° C. θ

0° C Solution

Copper

Steel

H steel = H copper

∴ or or Solving we get, θ = 10.6° C

100 °C (two rods are in series)

( TD ) S ( TD )C = ( l / KA ) S ( l / KA )C K S ( TD ) S = K C ( TD )C 46 (100 − θ ) = 390 (θ − 0)

Note In the above problem heat flows from right to left or from higher temperature to lower temperature.

Ans.

324 — Waves and Thermodynamics V

Example 22.12 A spherical black body with a radius of 12 cm radiates 450 W power at 500 K. If the radius were halved and the temperature doubled, the power radiated in watt would be (JEE 1997) (a) 225 (b) 450 (c) 900 (d) 1800 4 Solution Power radiated = e r σT A or Power radiated ∝ (surface area) (T ) 4 . The radius is 1 halved, hence, surface area will become times. Temperature is doubled, therefore, T 4 becomes 4 16 times.  1 New power = ( 450)   (16) = 1800 W.  4 ∴ The correct option is (d).

Example 22.13 A cylinder of radius R made of a material of thermal conductivity K 1 is surrounded by a cylindrical shell of inner radius R and outer radius 2R made of a material of thermal conductivity K 2 . The two ends of the combined system are maintained at two different temperatures. There is no loss of heat across the cylindrical surface and the system is in steady state. The effective thermal conductivity of the system is (JEE 1998) (a) K 1 + K 2 (b) K 1 K 2/( K 1 + K 2 ) (c) ( K 1 + 3K 2 ) / 4 (d) ( 3K 1 + K 2 )/4 Solution Let R1 and R 2 be the thermal resistances of inner and outer portions. Since, temperature difference at both ends is same, the resistances are in parallel. Hence,

R

2R

V

Fig. 22.16

1 1 1 = + R R1 R 2 l KA 1 KA = R l R=

or ∴ ∴

K ( 4πR 2 ) K 1 ( πR 2 ) K 2 ( 3πR 2 ) = + l l l K + 3K 2 K= 1 4

∴ The correct option is (c).

Chapter 22 V

Calorimetry and Heat Transfer — 325

Example 22.14 A body cools in 10 minutes from 60°C to 40°C. What will be its temperature after next 10 minutes? The temperature of the surroundings is 10°C. Solution According to Newton’s law of cooling,  θ1 + θ 2    θ1 – θ 2    = α   – θ0   t   2   For the given conditions, 60 – 40  60 + 40  =α  – 10 10 2   Let θ be the temperature after next 10 minutes. Then, 40 – θ  40 + θ  =α  – 10 10  2  Solving Eqs. (i) and (ii), we get θ = 28° C

INTRODUCTORY EXERCISE

…(i)

…(ii)

Ans.

22.2

1. A rod is heated at one end as shown in figure. In steady state temperature of different sections becomes constant but not same. Why so?

Fig. 22.17

2. Show that the SI units of thermal conductivity are W/m-K. 3. Find SI units of thermal resistance. 4. Suppose a liquid in a container is heated at the top rather than at the bottom. What is the main process by which the rest of the liquid becomes hot?

5. Three rods each of same length and cross-section are joined in series. The thermal conductivity of the materials are K,2K and 3K respectively. If one end is kept at 200 °C and the other at 100 °C. What would be the temperature of the junctions in the steady state? Assume that no heat is lost due to radiation from the sides of the rods.

6. A rod CD of thermal resistance 5.0 K/W is joined at the middle of an identical rod AB as shown in figure. The ends A, B and D are maintained at 100°C and 25°C respectively. Find the heat current in CD. A 100°C

C

B 0°C

25°C

D Fig. 22.18

7. A liquid takes 5 minutes to cool from 80 °C to 50 °C . How much time will it take to cool from 60 °C to 30°C? The temperature of surroundings is 20 °C .

326 — Waves and Thermodynamics

Final Touch Points 1. Cooling by conduction or radiation (i) By conduction A body P of mass m and specific heat c Q Rod is connected to a large body Q (of specific heat infinite) P θ0 = constant l, K, A through a rod of length l, thermal conductivity K and area θi > θ 0 of cross-section A. Temperature of Q is θ 0( < θi ). This t = 0, θi temperature will remain constant as its specific heat is t = t, θ very high. Heat will flow from P to Q through the rod. If we neglect the loss of heat due to radiation then due to this heat transfer, temperature of P will decrease but temperature of Q will remain almost constant. At time t, suppose temperature of P becomes θ then due to temperature difference heat transfer through the rod. dQ TD θ − θ 0 …(i) =H = = dt R R l Here, R = KA Now, if we apply equation of calorimetry in P, then dQ  −d θ  …(ii) Q = mc ( − ∆θ ) or = mc    dt  dt Equating Eqs. (i) and (ii), we have −

dθ TD θ − θ0 = = = Rate of cooling dt mcR mcR

So, this is the rate of cooling by conduction. dθ or − ∝ TD dt (ii) By radiation

…(iii)

…(iv)

In article 22.2, we have already derived the expression of rate of cooling, −

dT eAσ 4 dT = (T − T04 ) or − ∝ (T 4 − T04 ) dt mc dt

…(v)

Further in Newton’s law of cooling, we have also seen that, if temperature difference between body and atmosphere is less than this rate of cooling, −

dT ∝ ∆T dt

or TD

…(vi)

In Newton’s law of cooling, we have also seen that temperature of the body falls exponentially, if dT dθ or − rate of cooling, − ∝ TD dt dt In Eq. (iv) of conduction and Eq. (vi) of radiation (special case when TD is small) dT dθ or − − ∝ TD dt dt So, in both cases temperature of the body will fall exponentially like, θ = θ 0 + (θi − θ 0 ) e − αt or T = T0 + (Ti − T0 ) e −αt 4eAσT03 1 In radiation, α = and in conduction α = mcR mc 2. For emissivity, we have used the term e or er . This is sometimes confused with emissive power e. Emissivity is unitless and always less than or equal to one. But emissive power has the units J/s -m 2 or watt/ m 2.

Solved Examples TYPED PROBLEMS Type 1. Based on calculation of thermal resistance

Concept (i) A thermal resistance of a conducting rod is calculated/required between two points or two surfaces (say a and b). The formula of thermal resistance is l R= KA Here, l is that dimension of conductor which is parallel to a and b and A is that cross-sectional area which is perpendicular to a and b. l can be applied (ii) From a to b if A is uniform at every point then direct formula R = KA otherwise we will have to use integration. V

Example 1 Thermal conductivity of the conductor shown in figure is K. Find thermal resistance between points a and b. a z y x b

Solution From a to b area of cross-section perpendicular to ab is uniform ( A = xy) and length along ab is z. Therefore, using the formula l , we have R= KA z Ans. R= Kxy V

Example 2 Thermal conductivity of inner core of radius r is K and of the outer one of radius 2r is 2K. Find equivalent value of thermal conductivity between its two ends. l r

2r

328 — Waves and Thermodynamics Solution The meaning of equivalent value of thermal conductivity means, if a single material of K e is used with same dimensions then value of thermal resistance should remain unchanged. ⇒

R1

R 2r

R2

Between two ends R1 and R2 are in parallel. Hence, 1 1 1 + = R1 R2 R l 1 KA . Therefore, will be given by . Using this in Eq. (i), we have KA R l K (πr 2) 2K [(π )(2r )2 − πr 2] K e (π )(2r )2 + = l l l Solving this equation, we get 7 Ke = K 4

…(i)

R is given by

V

Ans.

Example 3 A spherical body of radius ‘b’ has a concentric cavity of radius ‘a’ as shown. Thermal conductivity of the material is K. Find thermal resistance between inner surface P and outer surface Q. P

P a

b

Solution As we move from P to Q surface perpendicular to PQ is spherical and its size keeps on increasing (just like different layers of a spherical onion). So, first we will calculate thermal resistance of one layer at a distance r from centre and thickness dr by using the formula l . R= KA Q

P r a

b

Thermal resistance = dR

dr

In this formula, dimension of the layer along PQ is dr and the surface area perpendicular to PQ is 4πr 2

Calorimetry and Heat Transfer — 329

Chapter 22 dR =

∴

dr K (4πr 2)

Now, if we integrate dR from r = a to r = b, we will get the total thermal resistance between P and Q. Thus, b

b

a

a

R = ∫ dR = ∫

dr K (4πr 2)

Solving this expression, we get 1 4 πK

R= V

 1 1  −   a b

Ans.

Example 4 In the above example, if temperature of inner surface P is kept constant at θ 1 and of the outer surface Q at θ 2 ( < θ 1 ). Then, Find (a) rate of heat flow or heat current from inner surface to outer surface. (b) temperature θ at a distance r( a < r < b) from centre. Solution (a) θ2 θ1

H or

H

or

dQ dt

dQ Temperature difference = dt Thermal resistance (θ1 − θ 2)(4πK ) =  1 1  −   a b =

(θ1 − θ 2)(ab)(4πK ) (b − a )

(b)

Ans.

θ2 θ θ1 a

H1 P

H2 M Q

r b

In the figure, we can see that ∴

Heat current H 1 = Heat current H 2 (TD)PM (TD)MQ = RPM RMQ

…(i)

330 — Waves and Thermodynamics Using the result obtained in Example-3 of thermal resistance, we can find 1  1 1 RPM =  −  4 πK  a r  RMQ =

and

1  1 1  −  4 πK  r b 

Substituting the values in Eq. (i), we have θ1 − θ θ − θ2 = 1  1 1 1 1  −  −  4 πK  a r  4 πK  r

1  b

Solving this equation we can find θ.

Type 2. Mixed problems of calorimetry and conduction

Concept (i) In calorimetry, we have two equations, (when temperature changes without change in state) Q = mc∆θ and (when state changes without change in temperature) Q = mL The above two equations in differential form can be written as dQ  d θ  dm …(i) = mc  ±  or L      dt  dt dt dm is rate of conversion of mass from one state to another state. In the above equation dt (ii) In conduction, we have the equation, dQ TD or H = …(ii) dt R l where, R= KA dQ dQ In these types of problems of calorimetry is equated with of conduction. dt dt V

Example 5 One end of the rod of length l, thermal conductivity K and area of cross-section A is maintained at a constant temperature of 100 °C. At the other end large quantity of ice is kept at 0 °C. Due to temperature difference, heat flows from left end to right end of the rod. Due to this heat ice will start melting. Neglecting the radiation losses find the expression of rate of melting of ice. l, K, A

Ice at 0°C

100°C

Solution Putting,  dQ   dQ  =     dt  conduction  dt  calorimetry

Chapter 22

Calorimetry and Heat Transfer — 331

TD dm = L⋅ R dt dm TD = dt RL

∴ ⇒ So, this is the desired expression of

Ans.

dm . dt

In the above expression, TD = 100° C, R =

L = latent heat of fusion

and V

l KA

Example 6 B point of the rod shown in figure is maintained at 200 °C. At left end A, there is water at 100 °C and at right end C there is ice at 0 °C. Heat currents H 1 and H 2 will flow on both sides. Due to H 1 , water will convert into steam and due to H 2 ice will be melted. If latent heat of vaporization is 540 cal / g l and latent heat of fusion is 80 cal / g then neglecting the radiation losses find 1 so l2 that rate of melting of ice is two times the rate of conversion of water into steam. A

l1

B

l2

C Ice at 0°C

H1 200°C H2 Water at 100°C

Solution Using the relation of

dm derived in above example, dt dm TD (TD)KA = = dt RL lL

l   as R =   KA 

Given that,  dm  dm =2     dt  LHS  dt  RHS or

(TD)KA   (TD)KA  = 2   lL   lL  LHS RHS

K and A are same on both sides. Hence,  TD  TD =2     lL  RHS  lL  LHS Substituting the proper values, we have

∴

 100  200 =2  l2 × 80  l1 × 540  l1 80 4 = = l2 540 27

Ans.

332 — Waves and Thermodynamics Type 3. Mixed problems of Stefan’s law and Wien’s displacement law

Concept Stefan’s law Energy radiated per unit time = emissive power = e r σT 4 Energy radiated per unit time called power = e r σT 4 A Total energy radiated in time ' t ' = e r σT 4 At Above all three terms ∝ T 4 Wien’s displacement law 1 λm∝ T From Wien’s law normally we find the ratio of temperatures of two bodies and then this ratio is used in Stefan’s law. V

Example 7 Three discs, A, B and C having radii 2m, 4m and 6m respectively are coated with carbon black on their outer surfaces. The wavelengths corresponding to maximum intensity are 300 nm, 400 nm and 500 nm, respectively. The power radiated by them are Q A , Q B and QC respectively

(JEE 2004)

(a) QA is maximum (b) QB is maximum 4 Solution (b) Q ∝ AT and λ m T = constant.

(c) QC is maximum

(d) QA = QB = QC

Hence, Q∝ QA : QB : QC =

A (λ m )4

or Q ∝

r2 (λ m )4

(2)2 (4)2 (6)2 4 1 36 : : = : : = 0.05 : 0.0625 : 0.0576 (3)4 (4)4 (5)4 81 16 625

i.e. QB is maximum.

Type 4. Problems of cooling either by conduction or radiation

Concept In final touch points we have seen that if a body cools by conduction, then dθ TD …(i) rate of cooling, − = dt mcR Here, mc = C = heat capacity of the body This C is the heat required to raise the temperature of whole body by 1°C or 1K and R is the thermal resistance of conducting rod. So, Eq. (i) can also be written as dθ TD − = dt CR If the body cools by radiation, then dT eAσ …(ii) rate of cooling or − = (T 4 − T04 ) dt mc

Chapter 22

Calorimetry and Heat Transfer — 333

For small temperature difference, Newton’s law of cooling can be applied and dT …(iii) − ∝ TD dt Further, if rate of cooling ∝ TD as Eqs. (i) or (iii), temperature of the body falls exponentially. V

Example 8 Two spheres, one solid and other hollow are kept in atmosphere at same temperature. They are made of same material and their radii are also same. Which sphere will cool at a faster rate initially? Solution In the equation, −

dT eAσ 4 = (T − T04 ) dt mc

Only mass is different. Therefore, dT 1 ∝ dt m Mass of hollow sphere is less. So, hollow sphere will cool at a faster rate initially. −

V

Example 9 The ratio of specific heats of two spheres is 2 : 3, radii 1 : 2, emissivity 3 : 1 and density 1 : 1. Initially, they are kept at same temperatures in atmosphere. Which sphere will cool at a faster rate initially. Solution In the equation, −

dT eAσ 4 = (T − T04 ) dt mc A = surface area = 4πR2 4 m = Volume × density = πR3ρ 3

Substituting in the above equation, we have dT 3eσ dT e − = (T 4 − T04 ) or − ∝ dt Rρ c dt Rρ c 3 3  e  Now, = =    Rρc sphere-1 1 × 1 × 2 2 1 1  e  = =    Rρc sphere-2 2 × 1 × 3 6

and

 e    for sphere-1 is more.  Rρc So, first sphere will cool at faster rate initially. V

Example 10 Two metallic spheres S1 and S2 are made of the same material and have got identical surface finish. The mass of S1 is thrice that of S2 . Both the spheres are heated to the same high temperature and placed in the same room having lower temperature but are thermally insulated from each other. The ratio of the initial rate of cooling of S1 to that of S2 is (JEE 1995) (a)

1 3

(b)

1 3

(c)

3 1

 1 (d)    3

1/ 3

334 — Waves and Thermodynamics Solution −

dT eAσ 4 = (T − T 4o ) dt mc m1 = 3m2 1

∴

R1 = (3)3 R2 A R2 1  dT  −  ∝ ∝ 3 ∝  dt  m R R 1

(− dT /dt )1 R2  1 3 = =  (− dT /dt )2 R1  3

∴ ∴ V

The correct option is (d).

Example 11 Two identical conducting rods are first connected independently to two vessels, one containing water at 100°C and the other containing ice at 0°C. In the second case, the rods are joined end to end and connected to the same vessels. Let q1 and q2 gram per second be the rate of melting of ice in the two cases q respectively. The ratio 1 is (JEE 2004) q2 (a)

1 2

(b)

(c)

4 1

(d)

1 4

Temperature difference  dm =L   dt  Thermal resistance dm 1 1 or ∝ ⇒ q∝ dt Thermal resistance R R In the first case rods are in parallel and thermal resistance is while in second case rods are 2 in series and thermal resistance is 2R. q1 2R 4 = = q2 R / 2 1

Solution (c)

dQ  dm =L   dt  dt

2 1 or

Miscellaneous Examples V

Example 12 Two metal cubes with 3 cm-edges of copper and aluminium are arranged as shown in figure. Find (a) the total thermal current from one reservoir to the other 100°C (b) the ratio of the thermal current carried by the copper cube to that carried by the aluminium cube. Thermal conductivity of copper is 401 W/m-K and that of aluminium is 237 W/m-K. Solution (a) Thermal resistance of aluminium cube, R1 = or

R1 =

(3.0 × 10–2) (237) (3.0 × 10–2)2

= 0.14 K/W

l KA

Al 20°C Cu

Calorimetry and Heat Transfer — 335

Chapter 22

l KA (3.0 × 10–2) = 0.08 K/W R2 = (401) (3.0 × 10–2)2

and thermal resistance of copper cube, R2 = or

As these two resistance are in parallel, their equivalent resistance will be R1R2 R= R1 + R2 =

(0.14) (0.08) (0.14) + (0.08)

= 0.05 K/W Temperature difference Thermal current, H = ∴ Thermal resistance (100 – 20) = = 1.6 × 103 W 0.05 (b) In parallel thermal current distributes in the inverse ratio of resistance. Hence, H Cu RAl R 0.14 = = 1 = = 1.75 H Al RCu R2 0.08 V

Ans.

Ans.

Example 13 One end of a copper rod of length 1 m and area of cross-section 4.0 × 10 –4 m2 is maintained at 100°C. At the other end of the rod ice is kept at 0°C. Neglecting the loss of heat from the surroundings, find the mass of ice melted in 1 h. Given, K Cu = 401 W / m - K and L f = 3.35 × 105 J / kg . Solution Thermal resistance of the rod, 100°C

0°C H ,

l 1.0 R= = = 6.23 K/W KA (401) (4 × 10–4 ) ∴

Temperature difference Thermal resistance (100 – 0) = = 16 W 6.23

Heat current, H =

Heat transferred in 1 h, Q = Ht

Q  Q H =   t

= (16) (3600) = 57600 J Now, let m mass of ice melts in 1 h, then m= =

or

Q L

(Q = mL )

57600 3.35 × 105

= 0.172 kg = 172 g

Ans.

336 — Waves and Thermodynamics V

Example 14 Two bodies A and B have thermal emissivities of 0.01 and 0.81 respectively. The outer surface areas of the two bodies are same. The two bodies emit total radiant power at the same rate. The wavelength λ B corresponding to maximum spectral radiancy from B is shifted from the wavelength corresponding to maximum spectral radiancy in the radiation from A by 1.0 µm. If the temperature of A is 5802 K, calculate (a) the temperature of B, (b) wavelength λ B . PA = PB

Solution (a) ∴

eAσAATA4 = eB σABTB4 e  TB =  A   eB 

∴

1/ 4

TA

(as AA = AB )

Substituting the values,  0.01 TB =    0.81

1/ 4

(5802) = 1934 K

Ans.

(b) According to Wien’s displacement law,

∴ or Also, or or V

λA TA = λ B TB T   5802 λB =  A λ A =   λ  1934 A  TB  λ B = 3 λA λ B – λA = 1 µm  1 λ B –   λ B = 1 µm  3 λ B = 1.5 µm

Ans.

Example 15 5 g of water at 30°C and 5 g of ice at − 20°C are mixed together in a calorimeter. Find the final temperature of mixture. Water equivalent of calorimeter is negligible, specific heat of ice = 0.5 cal/g-°C and latent heat of ice = 80 cal/g. Solution

In this case heat is given by water and taken by ice

Heat available with water to cool from 30° C to 0° C = ms∆θ = 5 × 1 × 30 = 150 cal Heat required by 5 g ice to increase its temperature upto 0°C ms∆θ = 5 × 0.5 × 20 = 50 cal Out of 150 cal heat available, 50 cal is used for increasing temperature of ice from − 20° C to 0° C. The remaining heat 100 cal is used for melting the ice. If mass of ice melted is m g, then m × 80 = 100 ⇒ m = 1.25 g Thus, 1.25 g ice out of 5 g melts and mixture of ice and water is at 0° C.

Chapter 22 V

Calorimetry and Heat Transfer — 337

Example 16 A bullet of mass 10 g moving with a speed of 20 m/s hits an ice block of mass 990 g kept on a frictionless floor and gets stuck in it. How much ice will melt if 50% of the lost kinetic energy goes to ice? (Temperature of ice block = 0°C ). Solution

Velocity of bullet + ice block, (10 g ) × (20 m/s ) V = = 0.2 m/s 1000 g

(Pi = Pf )

1 1 mv2 − (m + M ) V 2 2 2 1 = [0.01 × (20)2 − 1 × (0.2)2] 2 1 = [4 − 0.04 ] = 1.98 J 2 1.98 Heat received by ice block = cal 4.2 × 2 Loss of KE =

∴

∴

= 0.24 cal (0.24 cal) Mass of ice melted = (80 cal / g ) = 0.003 g

V

Ans.

Example 17 At 1 atmospheric pressure, 1.000 g of water having a volume of 1.000 cm3 becomes 1671 cm3 of steam when boiled. The heat of vaporization of water at 1 atmosphere is 539 cal/g. What is the change in internal energy during the process? Solution

Heat spent during vaporisation, Q = mL = 1.000 × 539 = 539 cal Work done, W = P (Vv − Vl ) = 1.013 × 105 × (1671 − 1.000) × 10−6 169.2 = 169.2 J = cal 4.18 = 40.5 cal

∴ Change in internal energy, ∴

V

∆U = 539 cal − 40.5 cal = 498.5 cal

Example 18 At 1 atmospheric pressure, 1.000 g of water having a volume of 1.000 cm3 becomes 1.091 cm 3 of ice on freezing. The heat of fusion of water at 1 atmosphere is 80.0 cal/g. What is the change in internal energy during the process? Solution

Heat given out during freezing,

External work done

Q = − mL = − 1 × 80 = − 80 cal W = P (V ice − V water ) = 1.013 × 105 × (1.091 − 1.000) × 10−6

Ans.

338 — Waves and Thermodynamics = 9.22 × 10−3 J =

9.22 × 10−3 cal 4.18

= 0.0022 cal ∴ Change in internal energy, ∆U = Q − W = − 80 − 0.0022 = − 80.0022 cal V

Ans.

Example 19 Two plates each of area A, thickness L1 and L2 thermal conductivities K 1 and K 2 respectively are joined to form a single plate of thickness ( L1 + L2 ). If the temperatures of the free surfaces are T1 and T2 . Calculate

Heat flow

K1

K2

T2

T1 L1

L2

(a) rate of flow of heat (b) temperature of interface and (c) equivalent thermal conductivity.

Solution

(a) If the thermal resistance of the two plates are R1 and R2 respectively. Plates are

in series. RS = R1 + R2 = as and so

L1 L2 + AK 1 AK 2

L KA dQ ∆T H = = dt R (T1 − T2) A (T1 − T2) = = (R1 + R2)  L1 L2  K + K   1 2 R=

Ans.

(b) If T is the common temperature of interface, then as in series rate of flow of heat remains same, i.e. H = H 1 (= H 2) T1 − T2 T1 − T = R1 + R2 R1 T1R2 + T2R1 (R1 + R2)

i.e.

T=

or

 L2 L1  T1 K + T2 K   2 1 T=  L1 L2  K + K   1 2

L   as R =   KA 

Calorimetry and Heat Transfer — 339

Chapter 22

(c) If K is the equivalent conductivity of composite slab, i.e. slab of thickness L1 + L 2 and cross-sectional area A, then as in series (L1 + L 2) RS = R1 + R2 or = R1 + R2 AK eq K eq =

i.e.

V

L1 + L 2 L1 + L 2 = A (R1 + R2)  L1 L2  K + K   1 2

L    as R =  KA 

Example 20 One end of a rod of length 20 cm is inserted in a furnace at 800 K. The sides of the rod are covered with an insulating material and the other end emits radiation like a black body. The temperature of this end is 750 K in the steady state. The temperature of the surrounding air is 300 K. Assuming radiation to be the only important mode of energy transfer between the surroundings and the open end of the rod, find the thermal conductivity of the rod. Stefan constant σ = 6.0 × 10 −8 W/m 2 -K 4 . Take emissivity of the open end e = 1 Solution

Heat flowing through the rod per second in steady state,

750 K Air temp 300 K

Furnace 800 K 20 cm

dQ KAdθ = dt x Heat radiated from the open end of the rod per second in steady state, dQ = Aσ (T 4 − T04 ) dt

…(i)

…(ii)

From Eqs. (i) and (ii), Kdθ = σ (T 4 − T04 ) x K × 50 = 6.0 × 10−8 [(7.5)4 − (3)4 ] × 108 0.2 K = 74 W/m-K

or V

Ans.

Example 21 Three rods of material x and three rods of material y are connected as shown in figure. All the rods are of identical length and cross-sectional area. If the end A is maintained at 60°C and the junction E at 10°C, calculate temperature of junctions B, C and D. The thermal conductivity of x is 0.92 cal/cm-s°C and that of y is 0.46 cal/cm-s°C. C x 60°C A

y

x 10°C E

x B y

y D

340 — Waves and Thermodynamics Solution

Thermal resistance, R =

∴

l KA Rx K y = Ry K x

(as lx = ly and Ax = Ay )

0.46 0.92 1 = 2

=

So, if Rx = R then Ry = 2R CEDB forms a balanced Wheatstone bridge, i.e. TC = TD and no heat flows through CD. 1 1 1 ∴ = + RBE R + R 2R + 2R 4 R 3 The total resistance between A and E will be RBE =

or

RAE = RAB + RBE = 2R + ∴ Heat current between A and E is H =

(∆T )AE (60 – 10) 15 = = RAE (10 /3) R R

Now, if TB is the temperature at B, H AB =

(∆T )AB RAB

15 60 – TB = R 2R

or

TB = 30° C

or Further,

H AB = H BC + H BD 15 30 – TC 30 – TD = + R R 2R (30 – T ) 15 = (30 – T ) + 2

or or Solving this, we get or V

4 10 R= R 3 3

T = 20° C TC = TD = 20° C

Ans.

[TC = TD = T (say)]

Ans.

Example 22 A hollow sphere of glass whose external and internal radii are 11 cm and 9 cm respectively is completely filled with ice at 0°C and placed in a bath of boiling water. How long will it take for the ice to melt completely? Given that density of ice = 0.9 g / cm3 , latent heat of fusion of ice = 80 cal / g and thermal conductivity of glass = 0.002 cal / cm - s°C. Solution

In steady state, rate of heat flow 4πKr1r2∆T H = r2 – r1

(result is taken from Example 4)

Chapter 22 Substituting the values,

H =

Calorimetry and Heat Transfer — 341

(4) (π ) (0.002) (11) (9) (100 – 0) (11 – 9)

dQ = 124.4 cal/s dt dm This rate should be equal to, L dt  dm dQ /dt 124.4 = = 1.555 g/s ∴  =   dt  80 L or

m = ρ ice (4πr12) = (0.9) (4) (π ) (9)2 = 916 g ∴ Time taken for the ice to melt completely m 916 t= = = 589 s (dm /dt ) 1.555 Total mass of ice,

V

Ans.

Example 23 A point source of heat of power P is placed at the centre of a spherical shell of mean radius R. The material of the shell has thermal conductivity K. Calculate the thickness of the shell if temperature difference between the outer and inner surfaces of the shell in steady state is T. Solution Consider a concentric spherical shell of radius r and thickness dr as figure. In steady state, the rate of heat flow (heat current) through this shell will be ∆T (– dθ ) H = = dr R (k) (4πr 2) or

H = – (4πKr 2)

shown in l   R =   KA 

dθ dr

Here, negative sign is used because with increase in r, θ decreases

dθ r1

P dr r

r2 .

∴

∫r

1

This equation gives, In steady state, ∴ Thickness of shell,

4 πK θ 2 =– dθ H ∫θ1 r2 4πKr1r2 (θ1 – θ 2) H = (r2 – r1 )

r2 dr

H = P , r1r2 ≈ R2 and θ1 – θ 2 = T 4πKR2T r2 – r1 = P

Ans.

342 — Waves and Thermodynamics V

Example 24 A steam cylindrical pipe of radius 5 cm carries steam at 100°C. The pipe is covered by a jacket of insulating material 2 cm thick having a thermal conductivity 0.07 W/m-K. If the temperature at the outer wall of the pipe jacket is 20°C, how much heat is lost through the jacket per metre length in an hour? K = 0.07 W/m-K

7 cm

dr

r

5 cm

Solution

∴

Thermal resistance per metre length of an element at distance r of thickness dr is dr l   dR = R =   K (2πr ) KA 

Total resistance

R=∫

r2 = 7 cm r1 = 5 cm

dR

7.0 × 10 –2 m dr 1 ∫ × 10 –2 m r 5.0 2πK 1  7 ln   =  5 2πK 1 = ln (1.4) (2π ) (0.07)

=

= 0.765 K/W Temperature difference Heat current, H = Thermal resistance (100 – 20) = = 104.6 W 0.765 ∴

Heat lost in one hour = Heat current × time = (104.6) (3600) J = 3.76 × 105 J

Ans.

Exercises LEVEL 1 Take cice = 0.53 cal/g-°C, cwater = 1.0 cal/g-°C, ( L f )water = 80 cal/g and ( Lv )water = 529 cal/g unless given in the question.

Assertion and Reason Directions : (a) (b) (c) (d)

Choose the correct option.

If both Assertion and Reason are true and the Reason is correct explanation of the Assertion. If both Assertion and Reason are true but Reason is not the correct explanation of Assertion. If Assertion is true, but the Reason is false. If Assertion is false but the Reason is true.

1. Assertion : Specific heat of any substance remains constant at all temperatures. Reason :

It is given by

s=

1 dQ ⋅ m dT

2. Assertion : When temperature of a body is increased, in radiant energy, number of low wavelength photons get increased. Reason :

According to Wien’s displacement law λ m ∝

1 . T

3. Assertion : Warming a room by a heat blower is an example of forced convection. Reason :

Natural convection takes place due to gravity.

4. Assertion : A conducting rod is placed between boiling water and ice. If rod is broken into two equal parts and two parts are connected side by side, then rate of melting of ice will increase to four times. Reason : Thermal resistance will become four times.

5. Assertion : A normal body can radiate energy more than a perfectly black body. Reason :

A perfectly black body is always black in colour.

6. Assertion : According to Newton’s law, good conductors of electricity are also good conductors of heat. Reason : At a given temperature, eλ ∝ aλ for any body.

7. Assertion : Good conductors of electricity are also good conductors of heat due to large number of free electrons. Reason : It is easy to conduct heat from free electrons.

8. Assertion : Emissivity of any body (e) is always less than its absorptive power (a). Reason :

Both the quantities are dimensionless.

9. Assertion : Heat is supplied at constant rate from one end of a conducting rod. In steady state, temperature of all points of the rod becomes uniform. Reason : In steady state, temperature of rod does not increase.

344 — Waves and Thermodynamics 10. Assertion : A solid sphere and a hollow sphere of same material and same radius are kept at same temperatures in atmosphere. Rate of cooling of hollow sphere will be more. Reason : If all other conditions are same, then rate of cooling is inversely proportional to the mass of body.

Objective Questions 1. For an enclosure maintained at 2000 K, the maximum radiation occurs at wavelength λ m . If the temperature is raised to 3000 K, the peak will shift to (a) 0.5 λ m

(b) λ m

(c)

2 λm 3

(d)

3 λm 2

2. A substance cools from 75°C to 70°C in T1 minute, from 70°C to 65°C in T2 minute and from 65°C to 60°C in T3 minute, then

(a) T1 = T2 = T3 (c) T1 > T2 > T3

(b) T1 < T2 < T3 (d) T1 < T2 > T3

3. Two liquids are at temperatures 20°C and 40°C. When same mass of both of them is mixed, the temperature of the mixture is 32°C. What is the ratio of their specific heats? (b) 2/5 (d) 2/3

(a) 1/3 (c) 3/2

4. The specific heat of a metal at low temperatures varies according to S = aT 3 , where a is a constant and T is absolute temperature. The heat energy needed to raise unit mass of the metal from temperature T = 1 K to T = 2 K is (a) 3a

(b)

15a 4

(c)

2a 3

(d)

13a 4

5. The intensity of radiation emitted by the sun has its maximum value at a wavelength of 510 nm and that emitted by the North star has the maximum value at 350 nm. If these stars behave like black bodies, then the ratio of the surface temperatures of the sun and the north star is (a) 1.46 (c) 1.21

(b) 0.69 (d) 0.83

6. A solid material is supplied heat at a constant rate. The temperature of material is changing

Temperature

with heat input as shown in the figure. What does the slope of DE represent?

O

E C A

D

B Heat input

(a) Latent heat of liquid

(b) Latent heat of vaporization

(c) Heat capacity of vapour

(d) Inverse of heat capacity of vapour

7. Two ends of rods of length L and radius R of the same material are kept at the same temperature. Which of the following rods conducts the maximum heat? (a) L = 50 cm, R = 1 cm (c) L = 25 cm, R = 0.5 cm

(b) L = 100 cm, R = 2 cm (d) L = 75 cm, R = 1.5 cm

Chapter 22

Calorimetry and Heat Transfer — 345

8. 1 g of ice at 0°C is mixed with 1 g of steam at 100°C. After thermal equilibrium is achieved, the temperature of the mixture is (a) 100°C

(b) 55°C

(c) 75°C

(d) 0°C

9. A wall has two layers A and B each made of different materials. The layer A is 10 cm thick and B is 20 cm thick. The thermal conductivity of A is thrice that of B. Under thermal equilibrium temperature difference across the wall is 35°C. The difference of temperature across the layer A is (a) 30°C

(b) 14°C

(c) 8.75°C

(d) 5°C

10. A wall has two layers A and B each made of different materials. Both the layers have the same thickness. The thermal conductivity of material A is twice of B. Under thermal equilibrium the temperature difference across the layer B is 36°C. The temperature difference across layer A is (a) 6°C

(b) 12°C

(c) 18°C

(d) 24°C

11. The end of two rods of different materials with their thermal conductivities, area of cross-section and lengths all in the ratio 1 : 2 are maintained at the same temperature difference. If the rate of flow of heat in the first rod is 4 cal/s. Then, in the second rod rate of heat flow in cal/s will be (a) 1

(b) 2

(c) 8

(d) 16

Subjective Questions 1. A thin square steel plate 10 cm on a side is heated in a black smith’s forge to temperature of 800° C. If the emissivity is 0.60, what is the total rate of radiation of energy?

2. A lead bullet penetrates into a solid object and melts. Assuming that 50% of its kinetic energy was used to heat it, calculate the initial speed of the bullet. The initial temperature of the bullet is 27° C and its melting point is 327° C. Latent heat of fusion of lead = 2.5 × 104 J/ kg and specific heat capacity of lead = 125 J/ kg-K.

3. A ball is dropped on a floor from a height of 2.0 m . After the collision it rises upto a height of 1.5 m. Assume that 40% of the mechanical energy lost goes as thermal energy into the ball. Calculate the rise in the temperature of the ball in the collision. Specific heat of the ball is 800 J/K.

4. A nuclear power plant generates 500 MW of waste heat that must be carried away by water pumped from a lake. If the water temperature is to rise by 10°C, what is the required flow rate in kg/s?

5. The emissivity of tungsten is 0.4. A tungsten sphere with a radius of 4.0 cm is suspended within a large evacuated enclosure whose walls are at 300 K. What power input is required to maintain the sphere at a temperature of 3000 K if heat conduction along supports is neglected? Take, σ = 5.67 × 10–8 Wm 2 -K 4.

6. A pot with a steel bottom 1.2 cm thick rests on a hot stove. The area of the bottom of the pot is 0.150 m 2. The water inside the pot is at 100°C and 0.440 kg are evaporated every 5.0 minute. Find the temperature of the lower surface of the pot, which is in contact with the stove. Take Lv = 2.256 × 106 J/ kg and ksteel = 50.2 W/ m-K

7. A carpenter builds an outer house wall with a layer of wood 2.0 cm thick on the outside and a

layer of an insulation 3.5 cm thick as the inside wall surface. The wood has K = 0.08 W/ m-K and the insulation has K = 0.01 W/ m-K. The interior surface temperature is 19° C and the exterior surface temperature is – 10° C. (a) What is the temperature at the plane where the wood meets the insulation? (b) What is the rate of heat flow per square metre through this wall?

346 — Waves and Thermodynamics 8. A closely thermally insulated vessel contains 100 g of water at 0°C. If the air from this vessel is rapidly pumped out, intensive evaporation will produce cooling and as a result of this, water freeze. How much ice will be formed by this method? If latent heat of fusion is 80 cal/g and of evaporation 560 cal/g. [ Hint If m gram ice is formed, mL f = (100 – m ) Lv ]

9. In a container of negligible mass 140 g of ice initially at –15° C is added to 200 g of water that has a temperature of 40°C. If no heat is lost to the surroundings, what is the final temperature of the system and masses of water and ice in mixture?

10. A certain amount of ice is supplied heat at a constant rate for 7 minutes. For the first one minute the temperature rises uniformly with time. Then, it remains constant for the next 4 minutes and again the temperature rises at uniform rate for the last two minutes. Calculate the final temperature at the end of seven minutes. (Given, L of ice = 336 × 103 J/ kg and specific heat of water = 4200 J/ kg-K)

11. Four identical rods AB, CD , CF and DE are joined as shown in figure. The length, cross-sectional area and thermal conductivity of each rod are l , A and K respectively. The ends A, E and F are maintained at temperatures T1 , T2 and T3 respectively. Assuming no loss of heat to the atmosphere. Find the temperature at B, the mid-point of CD. T3

T2

F

C

E

B

D

T1 A

12. The ends of a copper rod of length 1 m and area of cross-section 1 cm 2 are maintained at 0° C and 100° C. At the centre of the rod there is a source of heat of power 25 W. Calculate the temperature gradient in the two halves of the rod in steady state. Thermal conductivity of copper is 400 Wm −1K –1 .

13. A copper sphere is suspended in an evacuated chamber maintained at 300 K. The sphere is maintained at a constant temperature of 500 K by heating it electrically . A total of 210 W of electric power is needed to do it. When the surface of the copper sphere is completely blackened, 700 W is needed to maintain the same temperature of the sphere. Calculate the emissivity of copper.

LEVEL 2 Single Correct Options 1. A cylindrical rod with one end in a steam chamber and the other end in ice results in melting of 0.1 g of ice per second. If the rod is replaced by another with half the length and double the radius of the first and if the thermal conductivity of material of second rod is 1/4 that of first, the rate at which ice melts in g/s will be (a) 0.4 (c) 0.2

(b) 0.05 (d) 0.1

Calorimetry and Heat Transfer — 347

Chapter 22

2. Two sheets of thickness d and 3d, are touching each other. The temperature just outside the thinner sheet is T1 and on the side of the thicker sheet is T3 . The interface temperature is T2. T1 , T2 and T3 are in arithmetic progression. The ratio of thermal conductivity of thinner sheet to thicker sheet is (a) 1 : 3 (c) 2 : 3

(b) 3 : 1 (d) 3 : 9

3. A long rod has one end at 0°C and other end at a high temperature. The coefficient of thermal conductivity varies with distance from the low temperature end as K = K 0(1 + ax ), where K 0 = 102 SI unit and a = 1 m −1. At what distance from the first end the temperature will be 100°C ? The area of cross-section is 1 cm 2 and rate of heat conduction is 1 W. (a) 2.7 m (c) 3 m

(b) 1.7 m (d) 1.5 m

4. Two rods are of same material and having same length and area. If heat ∆Q flows through them for 12 min when they are joined side by side. If now both the rods are joined in parallel, then the same amount of heat ∆Q will flow in (b) 3 min (d) 6 min

(a) 24 min (c) 12 min

5. Three rods of identical cross-sectional area and made from the same metal

A (T )

form the sides of an isosceles triangle ABC right angled at B as shown in the figure. The points A and B are maintained at temperatures T and 2 T respectively in the steady state. Assuming that only heat conduction takes place, temperature of point C will be (a) (c)

T 2 −1 3T 2+1

(b) (d)

T 2+1 T 3 ( 2 − 1)

90° B (√2T )

C

6. A kettle with 2 litre water at 27°C is heated by operating coil heater of power 1 kW. The heat is lost to the atmosphere at constant rate 160 J/s, when its lid is open. In how much time will water heated to 77°C with the lid open? (specific heat of water = 4.2 kJ/ ° C-kg) (a) 8 min 20 s (c) 14 min

(b) 6 min 2 s (d) 7 min

7. The temperature of the two outer surfaces of a composite slab consisting of two materials having coefficients of thermal conductivity K and 2K and thickness x and 4x respectively are AK (T2 − T1 ) T2 and T1(T2 > T1 ). The rate of heat transfer through the slab in steady state is f. x where, f is equal to

T2

(a) 1 (c) 2/3

x

4x

K

2K

(b) 1/2 (d) 1/3

T1

348 — Waves and Thermodynamics More than One Correct Options 1. A solid sphere and a hollow sphere of the same material and of equal radii are heated to the same temperature (a) (b) (c) (d)

both will emit equal amount of radiation per unit time in the beginning both will absorb equal amount of radiation per second from the surroundings in the beginning the initial rate of cooling will be the same for both the spheres the two spheres will have equal temperatures at any instant

2. Three identical conducting rods are connected as shown in figure. Given that θ a = 40° C, θ c = 30° C and θ d = 20° C . Choose the correct options. a

b

d

(a) (b) (c) (d)

c

temperature of junction b is 15°C temperature of junction b is 30°C heat will flow from c to b heat will flow from b to d

3. Two liquids of specific heat ratio 1 : 2 are at temperatures 2 θ and θ (a) if equal amounts of them are mixed, then temperature of mixture is 1.5 θ 4 (b) if equal amounts of them are mixed, then temperature of mixture is θ 3 (c) for their equal amounts, the ratio of heat capacities is 1 : 1 (d) for their equal amounts, the ratio of their heat capacities is 1 : 2

4. Two conducting rods when connected between two points at constant but different temperatures separately, the rate of heat flow through them is q1 and q2 (a) When they are connected in series, the net rate of heat flow will be q1 + q2 qq (b) When they are connected in series, the net rate of heat flow is 1 2 q1 + q2 (c) When they are connected in parallel, the net rate of heat flow is q1 + q2 qq (d) When they are connected in parallel, the net rate of heat flow is 1 2 q1 + q2

5. Choose the correct options. (a) Good absorbers of a particular wavelength are good emitters of same wavelength. This statement was given by Kirchhoff (b) At low temperature of a body the rate of cooling is directly proportional to temperature of the body. This statement was given by the Newton (c) Emissive power of a perfectly black body is 1 (d) Absorptive power of a perfectly black body is 1

Calorimetry and Heat Transfer — 349

Chapter 22

Match the Columns 1. Match the following two columns. Column I (a) (b) (c) (d)

Column II

Stefan’s constant Wien’s constant Emissive power Thermal resistance

(p) (q) (r) (s)

[ Lθ ] [ ML2 T–3 θ –2 ] [ MT–3 ] None of these

2. Heat is supplied to a substance in solid state at a constant rate. Its temperature varies with time as shown in figure. Match the following two columns. θ (°C)

f d

b

e

c

a t (s)

Column I (a) (b) (c) (d)

Column II

Slope of line ab Length of line bc Solid + liquid state Only liquid state

(p) (q) (r) (s)

de cd directly proportional to mass None of these

3. Six identical conducting rods are connected as shown in figure. In steady state temperature of point a is fixed at 100°C and temperature of e at − 80° C. Match the following two columns. c

Column I (a) (b) (c) (d)

Temperature of b Temperature of c Temperature of f Temperature of d

Column II (p) (q) (r) (s)

10°C 40°C – 20°C None of these

a

b

d

e

f

4. Three liquids A, B and C having same specific heats have masses m, 2m and 3m. Their temperatures are, θ , 2 θ and 3θ respectively. For temperature of mixture, match the following two columns. Column I (a) When A and B are mixed (b) When A and C are mixed (c) When B and C are mixed (d) When A , B and C all are mixed

Column II (p) 5 θ 2 (q) 5 θ 3 (r) 7 θ 3 (s) 13 θ 5

350 — Waves and Thermodynamics 5. Match the following two columns. Column I

Column II

(a) Specific heat

(p) watt

(b) Heat capacity

(q) J/kg-°C

(c) Heat current

(r) J/s

(d) Latent heat

(s) None of these

Subjective Questions 1. As a physicist, you put heat into a 500 g solid sample at the rate of 10.0 kJ/ min, while recording its temperature as a function of time. You plot your data and obtain the graph shown in figure. θ (°C) 50 40 30 20 10 –5

1

2

3

4

t (min)

(a) What is the latent heat of fusion for this solid? (b) What is the specific heat of solid state of the material?

2. A hot body placed in air is cooled according to Newton’s law of cooling, the rate of decrease of temperature being k times the temperature difference from the surroundings. Starting from t = 0, find the time in which the body will lose half the maximum temperature it can lose.

3. Three rods of copper, brass and steel are welded together to form a Y -shaped structure. The cross-sectional area of each rod is 4 cm 2. The end of copper rod is maintained at 100 °C and the ends of the brass and steel rods at 80 °C and 60 °C respectively. Assume that there is no loss of heat from the surfaces of the rods. The lengths of rods are : copper 46 cm, brass 13 cm and steel 12 cm. (a) What is the temperature of the junction point? (b) What is the heat current in the copper rod? K ( copper) = 0.92, K ( steel) = 0.12 and K ( brass) = 0.26 cal/ cm-s °C

4. Ice at 0°C is added to 200 g of water initially at 70°C in a vacuum flask. When 50 g of ice has been added and has all melted the temperature of the flask and contents is 40°C. When a further 80 g of ice has been added and has all melted the temperature of the whole becomes 10°C. Find the latent heat of fusion of ice.

5. A copper cube of mass 200 g slides down a rough inclined plane of inclination 37° at a constant speed. Assuming that the loss in mechanical energy goes into the copper block as thermal energy. Find the increase in temperature of the block as it slides down through 60 cm. Specific heat capacity of copper is equal to 420 J/ kg-K. (Take, g = 10 m/ s2)

6. A cylindrical block of length 0.4 m and area of cross-section 0.04 m 2 is placed coaxially on a thin metal disc of mass 0.4 kg and of the same cross-section. The upper face of the cylinder is

Chapter 22

Calorimetry and Heat Transfer — 351

maintained at a constant temperature of 400 K and the initial temperature of the disc is 300 K. If the thermal conductivity of the material of the cylinder is 10 watt/m-K and the specific heat of the material of the disc is 600 J/ kg -K, how long will it take for the temperature of the disc to increase to 350 K? Assume for purpose of calculation the thermal conductivity of the disc to be very high and the system to be thermally insulated except for the upper face of the cylinder.

7. A metallic cylindrical vessel whose inner and outer radii are r1 and r2 is filled with ice at 0° C. The mass of the ice in the cylinder is m. Circular portions of the cylinder is sealed with completely adiabatic walls. The vessel is kept in air. Temperature of the air is 50° C. How long will it take for the ice to melt completely. Thermal conductivity of the cylinder is K and its length is l. Latent heat of fusion is L.

8. An electric heater is placed inside a room of total wall area 137 m 2 to maintain the temperature inside at 20° C . The outside temperature is −10° C. The walls are made of three composite materials. The inner most layer is made of wood of thickness 2.5 cm the middle layer is of cement of thickness 1 cm and the exterior layer is of brick of thickness 2.5 cm. Find the power of electric heater assuming that there is no heat losses through the floor and ceiling . The thermal conductivities of wood, cement and brick are 0.125 W/ m °-C, 1.5 W/ m °-C and 1.0 W/ m °-C respectively.

9. A 2 m long wire of resistance 4 Ω and diameter 0.64 mm is coated with plastic insulation of thickness 0.66 mm. A current of 5 A flows through the wire. Find the temperature difference across the insulation in the steady state. Thermal conductivity of plastic is 0.16 × 10−2 cal/ s cm °-C.

10. Two chunks of metal with heat capacities C1 and C2 are interconnected by a rod of length l and cross-sectional area A and fairly low conductivity K . The whole system is thermally insulated from the environment. At a moment t = 0, the temperature difference between two chunks of metal equals ( ∆T )0. Assuming the heat capacity of the rod to be negligible, find the temperature difference between the chunks as a function of time .

11. A rod of length l with thermally insulated lateral surface consists of material whose heat conductivity coefficient varies with temperature as k = a / T , where a is a constant. The ends of the rod are kept at temperatures T1 and T2. Find the function T ( x ), where x is the distance from the end whose temperature is T1.

12. One end of a uniform brass rod 20 cm long and 10 cm 2 cross-sectional area is kept at 100° C. The other end is in perfect thermal contact with another rod of identical cross-section and length 10 cm. The free end of this rod is kept in melting ice and when the steady state has been reached, it is found that 360 g of ice melts per hour. Calculate the thermal conductivity of the rod, given that the thermal conductivity of brass is 0.25 cal/ s cm ° C and L = 80 cal/ g.

13. Heat flows radially outward through a spherical shell of outside radius R2 and inner radius R1. The temperature of inner surface of shell is θ1 and that of outer is θ 2. At what radial distance from centre of shell the temperature is just half way between θ1 and θ 2?

14. A layer of ice of thickness y is on the surface of a lake. The air is at a constant temperature – θ° C and the ice water interface is at 0°C. Show that the rate at which the thickness increases is given by dy Kθ = dt Lρy

where, K is the thermal conductivity of the ice, L the latent heat of fusion and ρ is the density of the ice.

Answers Introductory Exercise 22.1 1. 7200 cal 2. 31.43 °C 6. 0°C, mw = 15 g, mi = 10 g

3. 80°C 7. 52°C

4. 12.96 m/s

5. 12 g

Introductory Exercise 22.2 1. For heat flow to take place, there must be a temperature difference along the rod. 4. Conduction 5. 145.5°C, 118.2°C 6. 4 W 3. K W –1

7. 9 min

Exercises LEVEL 1 Assertion and Reason 1. (d)

2. (a)

3. (b)

4. (c)

5. (c)

6. (d)

7. (a)

8. (d)

9. (d)

10. (a)

Objective Questions 1.(c)

2.(b)

3.(d)

4.(b)

5.(b)

6.(d)

7.(b)

8.(a)

9.(d)

10.(c)

11.(c)

Subjective Questions 1. 900 W

2. 500 m/s

7. (a) −8.1° C (b) 7.7 W / m

2

10. 40°C

11.

3. 2.5 × 10 −3 °C

4. 1.2 × 10 4 kg/s 5. 3.7 × 10 4 W

8. 87.5 g

9. 0°C, mass of ice is 54 g and that of water is 286 g

3T1 + 2(T 2 + T3 )

12. 424°C/m, 212°C/m

7

6. 105°C

13. 0.3

LEVEL 2 Single Correct Option 1.(c)

2.(a)

3.(b)

4.(b)

5.(c)

6.(a)

7.(d)

More than One Correct Options 1.(a,b)

2.(b,d)

3.(b,d)

4.(b,c)

5.(a,d)

Match the Columns 1. (a) → s

(b) → p

(c) → r

(d) → s

2. (a) → s

(b) → r

4. (a) → q

(b) → p

3. (a) → q

(b) → p

(c) → p

(d) → r

5. (a) → q

(b) → s

(c) → p,r

(d) → s

(c) → s

(d) → q

(c) → s

(d) → r

Subjective Questions 1. (a) 40 kJ/kg (b) 1.33 kJ/kg-° C −3

5. 8.6 × 10 °C

6. 166 s

10. ∆T = (∆T )0e − αt , where α =

2. ln(2) / k

3. (a) 84°C (b) 1.28 cal/s

7. t = mL ln(r2 / r1) / 100 πKl KA(C1 + C 2 ) lC1C 2

T  11. T = T1  2   T1 

8. 17647 W

4. 90 cal/g

9. 2.23°C

x /l

12. 0.222 cal/cm-s-°C

13.

2R1 R 2 R1 + R 2

17. Wave Motion INTRODUCTORY EXERCISE

17.1

∂2 y = − ω 2a sin ωt ∂t 2 ∂2 y and =0 ∂x 2 ∂2 y ∂2 y Since, (constant) ≠ ∂t 2 ∂x 2 Hence, the given equation does not represent a wave equation. Coefficient of t c 2. Speed of wave = = Coefficient of x b

1.

3. The converse is not true means if the function can be represented in the form y = f (x ± vt ), it does not necessarily express a travelling wave. As the essential condition for a travelling wave is that the vibrating particle must have finite displacement value for all x and t. 4. (a) A has the dimensions of y. x t and are dimensionless. a T coefficient of t (b) v = coefficient of x x t (c) and are of same sign a T (d) ymax = A

INTRODUCTORY EXERCISE 1. Comparing the given equation with

17.2

y = sin (kx − ωt ) we find, (a) Amplitude A = 5 mm (b) Angular wave number k = 1 cm −1 2π (c) Wavelength, λ = = 2π cm k ω 60 (d) Frequency, ν = = Hz 2π 2π 30 = Hz π 1 π (e) Time period, T = = s ν 30 (f) Wave velocity, v = νλ = 60 cm s−1

2π 2π ,λ = ω k ∂y (b) Find and then substitute the given values of ∂t x and t. (c) and (d) Procedure is same.

2. (a) T =

INTRODUCTORY EXERCISE

17.3

1. (a) coefficient of x and t are of same sign.

314 2π ω = 10 m/s ⇒ λ = , f = 31.4 k 2π (c) vmax = ωA 2. (a) The velocity of the particle at x at time t is ∂y v= = (3.0 cm ) (−314 s−1 ) cos [(3.14 cm −1 ) ∂t x − (314 s−1 ) t ] (b) v =

= (−9.4 ms−1 )cos[(3.14cm −1 )x − (314 s−1 ) t ] The maximum velocity of a particle will be v = 9.4 ms−1 (b) The acceleration of the particle at x at time t is ∂v a= = − (9.4 ms−1 )(314 s−1 ) sin[(3.14 cm −1 ) ∂t x − (314 s−1 ) t ] = − (2952 ms−1 )sin[(3.14 cm −1 )x − (314 s−1 ) t ]. The acceleration of the particle at x = 6.0 cm at time t = 0.11 s is a = − (2952 ms−2 )sin[ 6π − 11π ] = 0 Coefficient of t 3. (a) Wave velocity = Coefficient of x 1/0.01 = = 5 m/s 1/0.05 Since, coefficient of t and coefficient of x are of same sign. Hence, wave is travelling in negative x-direction. or v = − 5 m/s (b) Velocity of particle, ∂y t   x vP = = 2cos  +   0.05 0.01 ∂t Substituting x = 0.2 and t = 0.3, we have vP = 2 cos(34 ) = 2(− 0.85) = − 1.7 m/s

Wave Motion — 355

Chapter 17 2π ∆x λ λ∆φ v (∆φ ) ∆x = = 2π 2πf (350) (π / 3) = (2π ) (500)

∆φ =

4. (a) ∴

6. (a)

1. fmax =

17.4

INTRODUCTORY EXERCISE

2. Suppose power of line source is P. Then,

INTRODUCTORY EXERCISE

17.5

1. The tension in the string is F = mg = 10 N. The mass per unit length is µ = 1.0 g cm −1 = 0.1 kg m −1. The wave velocity is, 10 N therefore, v = Fµ = = 10 ms−1. The 0.1 kg m −1 time taken by the pulse in travelling through 50 cm is, therefore 0.05 s. T 2. v = µ

at distance r, surface area is 2πrl. P P ∴ I = = S 2πrl 1 or I ∝ r Further, I ∝ A2 1 A∝ ∴ r 1 2 2 3. Energy density, u = ρω A 2 Total energy = (u) (V ) 1 = (ρV ) (2πf )2 A 2 2 = 2π 2 mf 2 A 2 = (2)(π )2 (0.08)(120)2 (0.16 × 10−3 )2

3. TCD = 3.2 g,TAB = 6.4 g

= 581 × 10− 3 J

T v= µ

4. v =

T , µ

5. v =

T T = ρS ρ (πd 2 / A )

m l

and

t=

4 × 20 (8920) (π ) (2.4 × 10− 3 )2

= 22 m / s

= 0.58 mJ 1 2 2 4. Power = ρω A Sv 2 But, ρS = µ

d v

4T = ρπd 2 =

17.6

P 1 1.0 1. I = = = W/m 2 4 πr2 4 π (1)2 4 π

λ min c fmin = λ max v 2. λ = for both f

µ=

…(i)

(c) From Eq. (i), we can see that v ∝ m. If m is doubled, then v will become 2 times. Hence, v from the relation λ = ⋅ λ will also become f 2 times. As f remains unchanged.

c

and

mg µ

1.5 × 9.8 = 16.3 m/ s 0.055 v 16.3 (b) λ = = = 0.136 m f 120

= π or 180°

INTRODUCTORY EXERCISE

T = µ

=

= 0.116 m ∆φ = (ω ) ∆t = (2πf ) (∆t ) = (2π ) (500)(10− 3 )

(b)

v=

and

v=

T µ

∴ (T = mg )

ρ Sv = µT 1 Hence, Power = (2πf )2 A 2 Tµ 2 = 2π 2 f 2 A 2 Tµ = (2) (π )2 (60)2 (0.06)2 80 × 5 × 10− 2 = 512W

356 — Waves and Thermodynamics 1 2

5. (a) As derived in above problem, power

6. P = ρω 2 A 2Sv

= 2π 2 f 2 A 2 Tµ = (2) (π )2 (200)2 (10− 3 )2 60 × (0.006) = 0.47 W

But,

ρS = µ

T T = µ ρS T T ∴ ρS = 2 or ρSv = v v Substituting in Eq. (i), we have 1 T P = (2πf )2 A 2 2 v 2 2 2 2π f A T = v (2) (π )2 (100) 2 (0.0005)2 (100) = 100 = 0.049 J = 49 mJ v=

and

1 (b) Energy density, u = ρω 2 A 2 2 Total energy = (u) (V ) 1  =  ρω 2 A 2 (lS ) 2  ρS = µ 1 ∴ Total energy = µl (2πf )2 A 2 2 = 2π 2 f 2 A 2µl But,

= (2) (π )2 (200)2 (10− 3 )2 (0.006) (2.0) = 9.4 × 10− 3 J = 9.4 mJ

…(i)

Exercise LEVEL 1 Assertion and Reason 3. If both waves ωt and kx are of opposite signs, then both are travelling in positive directions.

4. If f is doubled (which is source dependant), then λ will automatically become half, so that speed remains same. Because speed is only medium dependant.

5. Longitudinal or sound wave cannot travel in vacuum.

7. Electromagnetic wave which can travel with or without medium.

8. Just by observation we cannot say that µ 2 > µ 1 and hence v2 > v1. If their densities are different, then µ 2 may be less than µ 1 also.

9. At mean position kinetic energy is maximum and there is a maximum stretch in string. Because one side particles are moving up and the other side particles are moving down. Hence, potential energy is also maximum. ∂y 10. In vP = − v , ∂x ∂y (the slope) is also negative. vP is negative and ∂x Hence, the velocity v should also be negative.

Objective Questions

2π 2π = = 18 m k π /9 2π 2. ∆φ = ∆x = k∆x λ = (10π × 0.01) (10) = π 1 3. At t = 0, ymax = at x = 0 2 1 At t = 2 s, ymax = at x = 2 m 2 ∴ In 2 s, ymax has travelled 2 m in positive direction 2 ∴ v = + = + 1 m/s 3

1. λ =

4. Wave velocity and particle velocity are two different things.

5. ∆φ =

2π 2π φ(∆x ) (∆x ) = λ v =

6. v =

2π (25) (16 − 10) =π 300

ω T = k µ 2

∴

 v  as λ =  f 

 ω  30 T = µ   = (1.3 × 10−4 )    k  1 = 0.12 N

2

Chapter 17 2π 2π 2πN = = T (t / N ) t (2π ) (150) = 60 = (5π ) rad /s

7. Q ω =

Subjective Questions 1. (b) λ =

2π 2π = k 2π / 28

= 28 cm ω 2π /0.036 (c) f = = 2π 2π = 27.8 Hz (d) v = f λ = 778.4 cm/s = 7.8 m/s Since, ωt and kx have opposite signs. Hence, wave is travelling in positive direction. 2. (a) Put t = 0, x = 2cm 2π 2π = = 16 cm k (30π / 240) Coefficient of t (c) Wave velocity = Coefficient of x

(b) λ =

=

1 1/ 240

= 240 cm/s ω 30π (d) f = = = 15 Hz 2π 2π 3. At t = 0, y is maximum at x = 0. At t = 2s, y is maximum at x = 1m. Hence, in 2 s wave has travelled 2 m in positive x-direction 1 v = + m/s ∴ 2 = + 0.5 m/s

4. Since, coefficient of t and x are opposite signs, then wave is travelling along negative x-direction. Coefficient of t Further, speed of wave = Coefficient of x 2 = 2 m/s 1 Amplitude = maximum value of y 10 = =2m 5 ω 5. Wave speed, v = k and maximum particle speed, =

Wave Motion — 357

(vP )max = ωA From these two expressions, we can see that (vP )max = (kA ) V m 6. Mass per unit length, µ = l T Tl 500 × 2 v= = = µ m 0.06 = 129.1 m/s T as mass per unit length = ρS v = ρS

7. ∴

v=

0.98 9.8 × 10 3 × 10− 6

= 10 m/s

8. Since, wave is travelling along positive x-direction. Hence, coefficient of t and coefficient of x should have opposite signs. Further, Coefficient of t v= Coefficient of x Coefficient of t ∴ 2= Coefficient of x ∴ Coefficient of t = 2 (coefficient of x) = 2 × 1= 2 SI units 10 y= ∴ (x − 2t )2 + 2 λ 9. (a) = 2 cm 2 ⇒ λ = 4 cm v 40 f = = = 10 Hz λ 4 2π 5π  2π  (b) ∆φ = ∆x =   (2.5) =  4 λ 4 (c) ωt = θ or (2π / t ) = θ

θ π /3 = 2πf (2π ) (10) 1 = s 60 (d) At P, particle is at mean position. So, v = maximum velocity = ωA = 2πfA = (2π )(10)(2) = (40π ) cm/s = 125.7 cm/s = 1.26 m/s ∴

t=

358 — Waves and Thermodynamics  ∂y …(i) Further, vP = − (v )    ∂x  Sign of v1, the wave velocity is given positive. ∂y Sign of , slope of y - x graph is also positive. ∂x Hence, from Eq. (i) particle velocity is negative. ∴ vP = − 1.26 m/s Coefficient of t 10. (a) Wave velocity = Coefficient of x (1/ 0.01) = = 5 m/s (1/ 0.05) Since, coefficient of t and coefficient of x are of same sign. Hence, wave is travelling in negative x-direction. or v = − 5 m/s ∂y (b) Particle velocity, vP = ∂t t   x = 2 cos  +   0.05 0.01 Substituting x = 0.2 and t = 0.3, we have vP = 2 cos 34 = (2) (− 0.85) = − 1.7 m/s 12  v 11. (a) ω = 2πf = 2π   = 2π    0.4   λ = (60 π ) rad /s 2π 2π k= = = (5π ) m −1 λ 0.4 Since, wave is travelling along + ve x-direction, ωt and kx should have opposite sign. Further at t = 0, x = 0 the string has zero displacement and moving upward (in positive direction). Hence at x = 0, we should have A sin ω t not − A sin ωt. Therefore, the correct expression is y = A sin (ωt − kx ) or y = 0.05 sin (60πt − 5πx ) (b) Putting x = 0.2 m and t = 0.15 s in above equation we have, y = − 0.035 4 m = − 3.54 cm (c) In part (b), y = A / 2 A From y = to y = 0, time taken is 2 T 2π t= = 8 ω×8

=

2π (60π ) (8)

= 4.2 × 10− 3 s = 4.2 ms 2π 2π 12. (a)Q T = = = 0.02 s ω π /0.01 = 20 ms 2π 2π λ= = k π /2 = 4.0 cm

∂y ∂x π mm  t   x  − = −   cos π   2.0 cm 0.01 s  0.01 s 

(b) vP =

 x t   π m = − −   cos π   10 s   2.0 cm 0.01 s Put x = 1 cm and t = 0.01 s in above equation We get, vP = 0 (c) and (d) Putting the given values in the above equation, we get the answers. 2π 2π 13. (a) k = = = 0.157 rad /cm λ 40 1 1 T = = = 0.125 s f 8 2π 2π ω= = = 50.3 rad /s T 0.215 ω v = = 320 cm/s k (b) Amplitude is 15 cm. At t = 0, x = 0, y = + A. Hence, equation should be a cos equation. Further, wave is travelling in positive x-direction. Hence, ω t and kx should have opposite signs. 14. (a) Tx = (mL − x ) g = µ (L − x ) g where, µ = mass per unit length T vx = x = g (L − x ) ∴ µ dx (b) vx = − = g (L − x ) dt ∴

t

x=0

∫0 dt = − ∫x = L

g (L − x ) dx

Solving we get, t=2

L g

Chapter 17 15. (a) T cos dθ components are cancelled. T sin dθ

2. Q ω = 2πf = (2π ) (100)

components provide the necessary centripetal force to Pθ. dθ

= (200 π ) rad /s ω v = k

dθ dθ dθ Q

P

T

Wave Motion — 359

T

k=

∴

ω µ =ω v T

= (200π ) ∴

2T sin d θ = mPQ Rω

2

For small angles sin dθ ≈ dθ ∴

2T dθ = [µ (2R ) dθ ] Rω 2

Solving the equation, we get T = v = Rω µ L

M = ∫ kx dx =

16.

0

∴

k=

2M L2 T = µx

vx =

 = L  ∴

1

t

∫0 dt = L

or

kL2 2

 2 t=  3L 2 = 3

T = kx

T (2M / L2 ) x

T  − 1/ 2 dx =  x 2M  dt 2M L 1/ 2 x dx T ∫0 2M  3/ 2  L T  2ML T

= 2π m − 1 At zero displacement, ∂y ∂x  ∂y  T  (slope) v    ∂x   µ  ∴ | A| = = ω ω   35   (π / 20) −3  3.5 × 10  = 200π = 0.025 m 2π 2π 3. Q ω = = T 0.25 = 8π rad /s ω v= k ω 8π  50 ∴ k= = =   π m −1 v 0.48  3  vP = ω A = − v

y = A sin (ωt − kx ) 50   = A sin  8πt − πx   3 Put y = 3 cm, t = 1 s, x = 0.47 m showing we get A = 6 cm

4. v =

LEVEL 2 Single Correct Option 1. Q v = ∴

∴

ω k

3.5 × 10− 3 35

∴

T T = ρS ρ (πd 2 / 4 ) v∝

T d2

5. E ∝ ω 2 A 2 ω 600π = v 300 = 2π m − 1

k=

y = 0.04 sin (600 πt − 2πx )

Now, put t = 0.01 s and x = 0.75 m

or E is same ∴ or

E ∝ f 2 A2 fA = constant 1 A ∝ f

360 — Waves and Thermodynamics 6. The wave pulse is travelling along positive x-axis. Hence, at and bx should have opposite signs. Further, wave speed Coefficient of t v = Coefficient of x Coefficient of t ∴ 4= 1 ∴ Coefficient of t = 4 s− 1

More than One Correct Options Coefficient of t Coefficient of x 6.2 = = 3.1 m/s 2 2 A= = 0.1 m 20 3. y = A sin (πx + πt ) ∂y vP = = πA cos (πx + πt ) ∂t 2 ∂ y aP = 2 = − π 2 A sin (πx + πt ) ∂t Now, substitute t = 0 and given value of x. Since, ωt and kx are of same sign, hence the wave is travelling in negative x-direction. 4. Speed of wave Coefficient of t b = = =b Coefficient of x 1 2π 2π λ= = =a k (2π / a) Coefficient of t 5. Speed of wave = Coefficient of x 1/ b a = = 1/ a b 2π 2π λ= = =a k 2π / a 2π 2π T = = =b ω 2π / b

1. Q

v=

6. y - t graph is sine graph. Therefore, v - t graph is cos graph and a - t is − sine graph as ∂2 y ∂y and aP = 2 vP = ∂t ∂t

Match the Columns Coefficient of t b = Coefficient of x c (b) Maximum particle speed = ωA = (b) (a)

1. (a) Wave speed =

ω b = 2π 2π 2π 2π (d) λ = = k c ∂y vP = = (4 π ) cm/s cos [ πt + 2πx ] ∂t ∂2 y aP = 2 = (− 4 π 2 ) cm/s2 sin [ πt + 2πx ] ∂t Now substitute the values of t and x. ∂y For velocity, vP = − v ∂x ∂y = Slope of y - x graph. ∂x Sign of v is not given in the question. Hence, direction of vP cannot be determined. For particle acceleration, aP ∝ − y i.e. aP and y are away in opposite directions If y = 0, then aP = 0 1 Energy density = ρω 2 A 2 2 = Energy per unit volume Power = energy transfer per unit 1 Time = ρω 2 A 2Sv 2 Intensity = energy transfer per unit per unit area 1 P = ρω 2 A 2v = 2 S 1 Wave number = λ ∂y ∂2 y vP = , aP = 2 ∂t ∂t If ωt and kx are of same sign, then wave travel in negative x -direction. If they are of opposite signs, then wave travels in positive direction. (c) f =

2.

3.

4.

5.

Subjective Questions 1. (a) vP = − v 

dy   dx 

As vP and (slope)P are both positive, v must be negative. Hence, the wave is moving in negative x-axis. (b) y = A sin (ωt − kx + φ ) …(i) 2π π −1 k= = cm λ 2 A = 4 × 10−3 m = 0.4 cm

Chapter 17 dy = + ve dx ∴ vP = − v (slope) = + ve Further at t = 0, x = 0, y = + ve π φ= ∴ 4 Further, 20 3 = − v tan 60° At t = 0, x = 0, slope

∴

∴ ∴

v = − 20 cm/s v f = = 5 Hz λ ω = 2πf = 10π π π  y = (0.4 cm) sin 10πt + x +   2 4

λ=

= 0.02 cos 100(0.1t − x ) = 0.02 cos (10t − 100x ) m

Ans.

Energy carried per cycle P E = PT = = 2π 2 A 2 f µv f

Substituting the values, we have E = 1.6 × 10−5 J

2. (a) v = =

Ans.

64 12.5 × 103 × 0.8 × 10−6 Ans.

(b) ω = 2πf = 2π (20) = 40π rad/s ω 40π π −1 k= = = m v 80 2  π   ∴ y = (1.0 cm ) cos (40πs−1 )t −  m−1 x  2    Ans. (c) Substituting x = 0.5 m and t = 0.05 s, we get 1 Ans. y= cm 2 (d) Particle velocity at time t. ∂y vP = ∂t  π   = − (40π cm/s) sin (40π s−1 )t −  m −1 x  2    Substituting x = 0.5 m and t = 0.05 s, we get Ans. vP = 89 cm/s

3. k eff = 2k = 1.0 N/m 1 k eff 10 −1 = s m 2π 2π v = 0.1 m/s, A = 0.02 m f =

Ans.

The distance between two successive maxima 2π Ans. =λ= = 0.0628 m 100 4. (a) Dimensions of A and Y are same. Similarly, dimensions of a and x are same. (b) As the wave is travelling towards positive x-axis, there should be negative sign between term of x and term of t. Further, speed of wave Coefficient of t v= Coefficient of x ∴ Coefficient of t = (v ) × coefficient of x

5. From the given figure, we can see that

T ρS

= 80 m/s

v 0.1 2π = m = f  10  100    2π 

 2π  y = A cos   (vt − x )  λ

(c) P = 2π 2 A 2 f 2 µv ∴

Wave Motion — 361

(a) Amplitude, A = 1.0 mm (b) Wavelength, λ = 4 cm 2π (c) Wave number, k = = 1.57 cm −1 ≈ 1.6 cm −1 λ v 20 (d) Frequency, f = = = 5 Hz λ 4 T 1 6. v = or v ∝ ρS ρ ∴

v1 ρ2 = v2 ρ1

∴

ρ 1  v2  1  1 =  =  =  2 ρ 2  v1  4

2

7.

v1 = = v2 =

2

T1 µ1 4.8 = 20 m/s 12 . × 10−2 T2 7.5 = µ2 1.2 × 10−2

= 25 m/s Pulses will meet when xA = xB or 20t = 25(t − 0.02) ∴ t = 0.1 s and xA or xB = 20 × 0.1 = 2 m

362 — Waves and Thermodynamics 1 2

8. (a) P = ρω 2 A 2Sv or

A=

or 1 2P ω ρSv

…(i)

Here, ρS = µ = mass per unit length 6 × 10−3 = kg/m 8 2πv ω = 2πf = λ 2π × 30 = 0.2 Substituting these values in Eq. (i), we have 0.2 2 × 50 × 8 A= 2π × 30 6 × 10−3 × 30 = 0.0707 m = 7.07 cm (b) P ∝ vω 2 or

P ∝ v (v 2 )

or

P∝v

∴

m 2 x ω ∫ x dx x=L L  x 2 L2  m − T = ω2  −  L 2  2 T

− ∫ dT = 0

mω 2 2 (L − x 2 ) 2L T v= µ

T =

or

mω 2 2 (L − x 2 ) 2L m/L

=

=ω

dx ω = L2 − x 2 dt 2

or ∴

∫0 dt =

2 ω

∫0

∴

t=

2 ω

 −1  x   sin  L   0

t

3

When wave speed is doubled, then power will become eight times. m 9. − dT = (dm)xω 2 =  dx xω 2 L 

L2 − x 2 2

=

L

dx L − x2 2

2 π π ⋅ = ω 2 2ω

L

Ans.

18. Superposition of Waves INTRODUCTORY EXERCISE

1.

I max  I 1 / I 2 + 1  = I min  I 1 / I 2 − 1

2. (a) (b)

18.1

2

Amax A1 + A2 = Amin A1 − A2 I max  Amax  =  I min  Amin 

3. Path difference of difference of

π . 2

2

λ is equivalent to a phase 4

INTRODUCTORY EXERCISE 18.3 1. Wall will be a node (displacement). Therefore, shortest distance from the wall at which air particles have maximum amplitude of vibration (displacement antinode) should be λ /4. v 330 Here, λ= = = 0.5 m f 660 0.5 ∴ Desired distance is = 0.125 m. 4 2. 5 f − 2 f = 54 ∴

3 f = 54 f = 18 Hz f ∝ T

∴

f1 T = 1 f2 T2

3.

√2 A0

3A0

220 = 260

or 45°

√2 A0

Anet = 5 A0 I ∝ A2 ⇒

I net = 25 I 0

INTRODUCTORY EXERCISE

1. It is either 0 or π. 2. (a)

v=

18.2

ω 40π = k π /3

= 120 cm/s (b) Distance between adjacent nodes λ π π = = = 2 k (π / 3) = 3 cm ∂y πx (c) vP = = (− 200 π ) sin sin 40πt ∂t 3 Now, substitute the given values of x and t. 3. In this case, node points will also oscillate, but standing waves are formed. 2π 2π 4. λ = = = 15.7 m k 0.4 ω 200 f = = = 31.8 Hz 2π 2π v = f λ = 500 m/s

2.2g (2.2 + M ) g

Solving we get, M = 0.873 kg f1 250 5 4. (a) = = f2 300 6 So, f1 is 5th harmonic and f2 is 6th harmonic. v 2.5 T / µ (b) f1 = (5) = = 2l l 2 2 µ f1 l (0.036) (250)2 (1)2 ∴ T = = 2 (2.5)2 (2.5) = 360 N T /µ  T /µ  5. 6   = 2  2l  2 l  2  1 l1 ∴ =3 l2

INTRODUCTORY EXERCISE 18.4 1. At mean position, energy is in the form of kinetic energy.

2. (a) Assuming tension to be same v∝

µ RHS 4 = 2vLHS = 20 cm/s

µ RHS = ∴

vRHS

1 µ

364 — Waves and Thermodynamics  v − v1  (b) Ar =  2  Ai  v 2 + v1 

 v − v1  2 −3 Ar =  2  Ai = × 10 m  v2 + v1  3

Ar 20 − 10 1 = = Ai 20 + 10 3

∴

 2v2  8 −3 At =   Ai = × 10 m  v1 + v2  3

2 × 20 4 At  2v2  = =  = Ai  v1 + v2  20 + 10 5

In second medium, speed becomes two times. Therefore, λ also becomes two times. So, k remains one-half, value of ω will remain unchanged. Further, second medium is rarer medium (v2 > v1 ). Hence, there is no change in phase angle anywhere. 2 yr = × 10−3 cos π (2.0x + 50t ) ∴ 3 8 and yt = × 10−3 cos π (x − 50t ) 3

3. Speed of wave in first medium is, Coefficient of t Coefficient of x 50 = = 25 m/s 2 v2 = 50 m/s Ai = 2 × 10−3 m v1 =

Exercises LEVEL 1

9.

A

A

Assertion and Reason A

1. At x = 0, y = y1 + y2 = A sin ωt + A cos ωt. So, it is neither a node nor an antinode. 2. They are called stationary because net energy transfers from any section is zero, if amplitudes of constituent waves are zero.  2v2  3. At =   Ai  v 1 + v2 

A

Resultant amplitude is A. So, intensity will remain same. 10. Phase difference may be different but it should remain constant with time.

Objective Questions 2.

6 mm

If v1 > v2 or 2 is rarer, then At > Ai

6 mm A ⇒

4. N 1 A1 N 2 A2 N 3 A3 N 4 3λ l= 2

4 mm

A = (8)2 + (6)2 = 10 mm

3. (i) Two waves must travel in opposite directions.

v 5. f = η    2l 

As η increases, frequency increases. Hence, λ decreases. 6. Amplitudes may be different also.

7.

(ii) At x = 0, y = y1 + y2 should be zero at all times. 6. f1 : f2 : f3 = 1: 2 : 3 λ=

λ 8

Energy lying between

λ 4

λ λ and will be more. 8 4

v f

or

λ∝

1 f

1 1 : 2 3 8. All frequencies are integral multiples of 35 Hz. v 9. 3   = 300  2l  ∴

X=0

8 mm

12 mm

∴

λ 1 : λ 2: λ 3 = 1:

v = 200 l = (200) (1) = 200 m/s

Chapter 18 10. These are multiples of 30 Hz. Hence, fundamental frequency f0 = 30 Hz v Now, f0 = 2l ∴ v = 2 f0 l = 2 × 30 × 0.8 = 48 m/s  2π  11. Q ∆φ =   (∆x )  λ

Superposition of Waves — 365

(b) Since, reflected waves come in the same medium, we can say that P ∝ A2 2

2 1 Pr  Ar   − A /3 =  =  =  A  9 Pi  Ai 

∴ Fraction of power transmitted 1 8 =1− = 9 9

3. A = (10)2 + (20)2 + 2(10) (20) cos 60° = 26.46 cm

 2π   2π  =  (∆x ) =   (∆x )  vT   v/ f 

20 A

  2π =  (16 − 10)  300 × 0.04 

60° φ

=π

12. v =

tan φ =

ω T = k ρS 2

 ω  30 ∴ T = ρS   = (8000 × 10−6 )    k  1 = 7.2 N

Subjective Questions 1. AR = 4 2 cm = 5.66 cm 4

2

10

20 sin 60° 10 + 20 cos 60°

= 0.866 φ = 40.89° = 0.714 rad 4. Find y1 and y2 at given values of x and t and then simply add them to get net value of y. T 16 5. v= = = 20 m/s µ (0.4 × 10−3 )/(10−2 ) ∴

d 2l 0.4 m = = v v 20 = 0.02 s (b) At half the time pulse is at other end and it gets a phase difference of π. Hence, the shape is as shown below.

(a) t = AR

4

v 2. v2 = 1 2  v − v1  (a) Ar =  2  A  v1 + v2   (v / 2) − v1  = 1 A  v1 + v1 / 2  A =− 3  2v2  At =  A  v 1 + v2   2 × v1 / 2  = A  v1 + v1 / 2  2 = A 3

6. (a) For fixed end. 1 cm/s O 2 cm 2 cm 1 cm/s

By the superposition of these two pulses we will get the resultant. But only to the left of point O, where string is actually present. (b) For free end O 2 cm 2 cm s

366 — Waves and Thermodynamics 7. y = y1 + y2 = (6.0 cm ) sin (πx ) cos (0.6πt ) …(i) Ax = 6.0 cm sin (πx ) In parts (a), (b) and (c) substitute the given values of x and find the displacement amplitude at these locations. (d) At antinodes Ax = maximum = ± 6.0 cm π 3π 5π , ∴ πx = , 2 2 2 or x = 0.5 cm, 1.5 cm, 2.5 cm 8. (a) Distance between successive antinodes λ π π = = = = 2 cm 2 k (π / 2)

12. Fundamental frequency = (490 − 420) Hz f0 = 70 Hz T /µ v f0 = = 2l 2l T /µ (450)/ 0.005 l= = 2 f0 2 × 70

∴

(b) Amax = 2 A = (2π ) cm If x = 0 is taken as node, we shall taken sin equation for Ax ∴ Ax = Amax = sin kx = (2π cm ) sin kx π Put k = and x = 0.5 2 T 20 9. v= = = 47.14 m/s µ 9 × 10− 3 v 47.14 f1 = = 2l 2 × 30 = 0.786 Hz Next three frequencies are 2 f1, 3 f1 and 4 f1 T /µ v 10. (a) f = = 2l 2l ∴ T = 4 l2 f 2 µ  1.2 × 10− 3  = (4 ) (0.7)2 (220)2   0.7   = 163 N (b) f3 = 3 f1 = 3 × 220 = 660 Hz 11. Fundamental frequency, T /µ v f = = 2l 2l (50)/(0.1 × 10− 3 /10− 2 ) 2 × 0.6 = 58.93 Hz Let, n th harmonic is the highest frequency, then (58.93) n = 20000 ∴ n = 339.38 Hence, 339 is the highest frequency. ∴ fmax = (339) (58.93) Hz = 19977 Hz

But, ∴

= 2.142 m v 13. λ = = 0.5 m f 1

2

3

4

 λ l = 4  =1m  2

14. (a) f4 = 4 f0 f4 400 = = 100 Hz 4 4 (b) f7 = 7 f0 = 700 Hz 1 15. Fundamental frequency ∝ l 1 1 1 f1 : f2 : f3 = : : l1 l2 l3 ∴

f0 =

= 1: 2 : 3 1 1 1 l1 : l2 : l3 = : : = 6 : 3 : 2 1 2 3 6  6 l1 =   (1) m = m  11 11

∴

3  3 l2 =   (1) m = m  11 11 2  2 l3 =   (1) m = m  11 11

16. f ∝ ∴

1 l f1 l2 = f2 l1

or

 f  l2 =  1  l1  f2 

 124  =  (90 cm ) = 60 cm  186 

=

17. (a)

λ = 15 cm 2 ∴ λ = 30 cm 2π  π  k= =   cm − 1  15 λ ω=

2π 2π − 1 = s T 0.075

Superposition of Waves — 367

Chapter 18 Since, x = 0 is a node we will write sin equation. ∴ Ax = Amax sin kx and y = Ax sin ωt ω (2π / 0.075) (b) v = = k (π /15) = 400 cm/s = 4 m/ s (c) x = 0

T = …(i)

1 = 0.126 s f

v = fλ = 1470 cm/s vmax = ω Amax

(e)

= (50) (5.60) = 280 cm/s (f) Frequency and hence ω will become ∴

x=0 P

ω′ = 50 × k=

x

8 = 133 rad /s 3 k ∝ω

or

Hence, k will become

15 cm

x = 7.5 − 3 = 4.5 cm From Eq. (i), π  Ax = (0.85 cm ) sin  × 4.5  15  = 0.688 cm v 48 18. (a) f1 = = 2l 3 = 16 Hz v λ1 = = 3m f1

λ = l or λ = 2l = 1.6 m 2 v = f λ = 60 × 1.6 = 96 m/s T (b) v = µ

20. (a)

(c)

f3 = 3 f1 and hence λ 3 =

λ1 3

 0.04  2 T = µv 2 =   (96)  0.8  = 461 N vmax = ω Amax

(c) Forth harmonic means f4 . λ1 4

19. (a)

8 times. 3

 8 k ′ =   (0.034 ) = 0.0907 rad /cm  3

∴

(b) Second overtone means f3 .

f4 = 4 f1 and λ 4 =

ω v

8 times. 3

amax

= (2πf ) Amax = (2π ) (60) (0.003) = 1.13 m/s = ω 2 Amax = (2πf )2 Amax = (2π × 60)2 (0.003) = 426.4 m/s2

Third harmonic 5.6 (b) A = = 2.8 cm 2 (c) k = 0.034 cm −1 2π = 184.7 cm k 3λ l= = 277 cm 2 (d) λ = 184.7 cm ω 50 f = = = 7.96 Hz 2π 2π

21. (a) y = y1 + y2 π  = 0.2 sin (x – 3.0t ) + 0.2 sin  x – 3.0t +   2 = A sin (x – 3.0t + θ )

λ=

A2

0.2

A

θ

A1 0.2

368 — Waves and Thermodynamics A = (0.2)2 + (0.2)2

Here,

θ=

and

nλ 1 =l 2

24. (a)

π 4

π  y = 0.28 sin  x – 3.0t +  Ans.  4

∴

(100/ 4 × 10− 2 ) 2×2

= 12.5 Hz  v fb = 3   = 3 fa = 37.5 Hz  2l  (ii) ω = 2πf ∴ ω a = 2πfa = 25π ω b = 2πfb = 75π T 100 v= = µ 4 × 10− 2 = 50 m/s ω k= v ωa 25π π ka = = = v 50 2 ω b 75π 3π kb = = = v 50 2

∴

23. (a) l = λ 1 / 4 ∴ ∴

∴ (b)

or

λ 1 = 4 l = 16 m , l = 3λ 2 / 4 4l λ2 = = 5.33 m 3 l = 5λ 3 / 4 4l λ3 = = 3.2 m 5 T v= µ =

Now, f1 =

400 = 100 m/s (0.16/4 ) v 100 = = 6.25 Hz λ 1 16

Similarly, f2 and f3.

…(i)

(n + 1) λ 2 =l 2 (n + 1) (0.48) =l 2

l 0.24 From Eqs. (i) and (ii) we have, n 8 = n+1 9

Ans.

(n + 1) =

or

T /µ v fa = = 2l 2l =

l 0.27

Similarly,

0.32 = (0.2)2 + (0.2)2 + (2) (0.2) (0.2) cos φ

22. (i)

n=

or

(b) Since, the amplitude of the resulting wave is 0.32 m and A = 0.2 m, we have Solving this, we get φ = ± 1.29 rad

n (0.54 ) =l 2

∴

…(ii)

(b) From Eq. (i), l = (0.27) n = 0.27 × 8 = 2.16 m

λ = l (Fundamental) 2 ∴ λ = 2l = 4.32 m 25. From fixed end there will be a phase change of π. Further, wave will start travelling in opposite direction. Hence, ωt and kx both will now become position. From free end there is no change in phase. d 4+4 26. t1 = 1 = =8s v 1 After one reflection from fixed boundary wave is inverted. d 4 + 10 + 16 t2 = 2 = = 20 s v 1 After two times reflection from fixed boundaries wave pulse will again become upright. 27. Let us plot at t = 3 s (c)

Y (cm) 3 1

X (cm) 4 5 6 7 8 9

y = y1 + y2

Net Y (cm) 4 1

X (cm) 4 5 6 7 8 9

Similarly, we can draw at other times.

Chapter 18

LEVEL 2 Single Correct option 1. Q

f ∝ T f1 T = 1 f2 T2

∴

3 T + 2.5 = 2 T Solving, we get T = 2 N T /µ T /ρS v 2. f = = = 2l 2l l

7. Q

∴

∴

T / ρ (πd 2 / 4 ) l T f ∝ dl  f1 T  d  l  =  1   2  2 f2  T2   d1   l1  =

or ∴

Superposition of Waves — 369 f2 = 2 f1 = 100 Hz f3 = 3 f1 = 150 Hz v n1 = = 25 Hz 4l n2 = 3n1 = 75 Hz n3 = 5n1 = 125 Hz 1 f ∝ l 1 1 1 l1 : l2 : l3 = : : f1 f2 f3 1 1 1 = : : = 12 : 4 : 3 1 3 4   12 l1 =   (114 )  12 + 4 + 3

∴

= 72 cm   4 l2 =   (114 )  12 + 4 + 3 = 24 cm   3 l3 =   (114 )  12 + 4 + 3

1  3  2     =3 2 2  1  1 1 k 3. f ∝ ⇒ l= l f =

Now, ∴ ∴

l = l1 + l2+ l3 k k k = + + f0 f1 f2 1 1 1 = + + f0 f1 f2

= 18 cm

8. f ∝ T f T W = 1 = f /2 T2 W − V1

k f3 1 f3

4. y1 + y2 = 2 A sin (ωt − kx ) = y4

2= (say)

4 ρw 3 f W Vρg = = f /3 W − V2 Vρg − V ρ l g

Solving ρ =

Now, y4 and y3 produce standing waves where, (Amplitude of constituent wave) Amax = 2 = 2 (2 A ) = 4 A 5. Tension in the string will be given by YA∆l YA  ∆l 1  T = = =   as  η l η l Now,

9= ∴

f ∝v T /ρA f 1 v1 = = f 2 v2 Y /ρ =

ρl =

ρ ρ − ρl 8 32 ρ= ρω 9 27

ρl ρw 32 = = 1.18 27

∴ Relative density of liquid =

T 1 = YA η

T /µ v 6. f1 = = 2l 2l 100/ 0.01 = = 50 Hz 2

vρg ρ = Vρg − Vρ w g ρ − ρw

9.

λ = 1.5 m 2 ∴

λ = 3.0 m 2π  2π  − 1 =k = =  m λ  3

370 — Waves and Thermodynamics Let us take antinode at x = 0, then 4mm

–X

X=0

2mm

∴

f5 5 f1 5 = = f2 2 f1 2

 2π  (2 mm ) = (4 mm ) cos   x  3

π  2π    x =  3 3

or x = 0.5 m The asked distance is 2x or 1.0 m. 10. f = 3Hz ∴

ω = 2πf = (6π ) rad /s ω 6π k= = = (2π ) rad /m v 3 y = A sin (kx − ωt ) = A sin (2πx − 6πt ) Putting y=± A and x = 3 , we get π (2π ) (3) − 6π t = 2 11 6πt = π ∴ 2 11 ∴ t= s 12 11. ∆φ 1 = π / 2  2π  ∆φ 2 =   (∆x )  λ

14. I ∝ A 2 I r = 0.64 I i Ar = 0.8 Ai = 0.8A Reflected from a denser medium. Hence, a phase change of π will occur. Reflected wave will travel in opposite direction. Hence, ωt and kx will have positive signs. v v 15. η   =    2l  B  2l  A ∴

 T /ρS   v  =  η=  2l  B  2l  A or

= 3π 5π ∴ ∆φ net = ∆φ 2 − ∆φ 1 = 2 12. Fundamental frequency, f0 = 450 − 400 = 50 Hz

∴

l=

T /µ v = 2l 2l T /µ 490/ 01 . = 2 f0 2 × 50

= 0.7 m

 4T /ρπ   4T /ρπ   =  η  2ld  B  2ld  A

…(i)

Given, TB = 2TA , lB = 2lA dB = 2 dA and ρB = 2 ρA Putting in Eq. (i), we get η=4 η = 4 means 4th harmonic or 3rd overtone.

More than One Correct Options 1. f ∝ T

 2π  =   (1.5λ )  λ

f0 =

2 f5 5 2 = × 480 5 = 192 Hz

f2 =

∴

X

Ax = Amax cos kx ∴

13.

f2 T = 2 f1 T1 ∴

f1 + 15 1.21 T1 = f1 T1

= 1.1 Solving we get f1 = 150 Hz v∝ T ∴

v2 T = 2 v1 T1

or

 1.21 T1  v2 =   v1 T1   = 1.1 v1

Superposition of Waves — 371

Chapter 18 Hence, increase in v is 10% λ =l 2 ∴ λ = 2l ∴ Fundamental wavelength = 2λ is unchanged.  2v2  4. At =   Ai  v1 + v2 

3. Ax = − 1.6 sin ax ∴ x = 0 is a node Second antinode is at a distance. λ λ 3λ 3  2π  x= + = =   4 2 4 4 k  But, k = a

5.

x=

∴

If v2 > v1 , At > Ai

3π 2a

Match the Columns λ  2π  l = 4 = 2λ = 2    k  2

1. v ∝

µ 2 = 9µ 1 v Hence, v2 = 1 3 So let v1 = 3 units, then v2 = 1 unit  v − v1  (a) A1 =  2  Ai  v 1 + v2 

4π = k 6. Two identical waves should travel in opposite directions. 7. y = y1 + y2 = (2 A cos kx ) sin ωt = Ax sin ωt Here, Ax = 2 A cos kx At x = 0, Ax is maximum or 2A. So, it is an antinode. Next antinode will occur at λ x = , λ … etc. 2 π 2π or x= , … etc. k k

and ∴

A1 v − v1 = 2 A2 2 v2

3−1 =1 2

=

v1 =3 v2 1 (c) I = ρω 2 A 2v 2 ρ is not given, so we cannot find I 1 / I 2. 1 (d) P = ρω 2 A 2Sv 2 But, Sρ = µ 1 P = ω 2 A 2µv ∴ 2 or (as ω → same) P ∝ A 2µv

1. Reflected and incident rays are in the same medium. Hence, I ∝ A2 I r has become 64% or 0.64 times of I i Ar = 0.8 Ai = 0.8 A

2. y = yi + yr

2

P1  A1   µ 1   v1  =      P2  A2   µ 2   v2 

= A sin (ax + bt + π / 2) + 0.8 A sin (ax − bt + π / 2) π   = 0.8 A sin  ax + bt +   2  π   2   π  + 0.2 A sin  ax + bt +   2

 2v2  A2 =   Ai  v1 + v2 

(b)

Comprehension Based Questions

∴

1 µ

1  1 = (1)2   (3) =  9 3

+ 0.8 A sin (ax − bt +

or

P2 =3 P1

2. (i)

= − 1.6 sin ax sin bt + 0.2A cos (bt + ax ) ∴ c = 0.2

(ii)

372 — Waves and Thermodynamics (a) Second overtone mode = f3 = 3 f1 Fifth harmonic mode = f5 = 5 f1 So, the ratio is 3/ 5, v 1 or λ ∝ (d) λ = f f λ3 f5 5 ∴ = = λ5 f3 3

So, percentage of energy transmitted will be Ans. (100 − 51)% or 49%.

3. Amplitude at a distance x is A = a sin kx First node can be obtained at x = 0, and the second at x = π/k At position x, mass of the element PQ is dm = (ρS )dx Its amplitude is A = a sin kx Hence mechanical energy stored in this element is

 2v2   Ai  v1 + v2 

3. (d) At = 

If v2 > v1, i.e. 2 is rarer then At > Ai . φ φ 4. A = 2 A0 cos and I = 4 I 0 cos2 2 2 5. Second overtone frequency means f3 = 3 f1 = 210 Hz ∴ f1 = 70 Hz = fundamental frequency Third overtone frequency = f4 = 4 f1 Second harmonic frequency = f2 = 2 f1

Subjective Questions 1. (a) v1 = T /µ 1 1 F /µ 1 , 2 v3 = F / (µ 1 / 4 ) = 2 F /µ 1 L L L + + ∴ t = t1 + t2 + t3 = v1 v2 v3

N

N

x=0

1 (ρSA 2ω 2 ) dx 2 1 = (ρSa2ω 2 sin 2 kx ) dx 2 Therefore, total energy stored between two adjacent nodes will be

Solving this, we get E= Ans

2. Let ai and ar be the amplitudes of incident and

1 (dm) A 2ω 2 2

dE =

or

E=∫

µ1 F

7L 2

p k

dE =

(energy of particle in SHM)

v2 = F / 4µ 1 =

=

x=

dx

4. l =

λ 2

x = π/k x=0

πSρω 2a2 4k

λ = 2l, k =

or

dE

Ans.

2π π = λ l

reflected waves. a

at

ai

x=0

Hence,

ai + ar =6 ai − ar ar 5 = ai 7 2

Now,

x=l

The amplitude at a distance x from x = 0is given by

ar

Then,

λ/2

2 Er  ar   5 =  =   7 Ei  ai 

= 0.51 or percentage of energy reflected is E 100 × r = 51%. Ei

A = a sin kx (given)

Total mechanical energy at x of length dx is 1 dE = (dm) A 2ω 2 2 1 = (µdx )(a sin kx )2 (2πf )2 2 or

Here,

dE = 2π 2 µf 2 a2 sin 2 kx dx T    v2  µ  π and k = f = 2= l λ (4 l 2 )

…(i)

Chapter 18

i.e. the amplitude of transmitted wave will be 2.0 cm. The expressions of Ar and At are derived as below.

Substituting these values in Eq. (i) and integrating it from x = 0 to x = l , we get total energy of string. E=

π 2a2T 4l

Ans.

Derivation

5. Tension, T = 80 N Amplitude of incident wave, Ai = 3.5 cm Mass per unit length of wire PQ is 0.06 1 m1 = = kg/m 4.8 80 and mass per unit length of wire QR is 0.2 1 m2 = = kg/m 2.56 12.8 Q

P

R

l1 = 4.8 m l2 = 2.56 m Mass = 0.06 kg Mass = 0.2 kg

(a) Speed of wave in wire PQ is 80 v1 = T /m1 = = 80 m/s 180 / and speed of wave in wire QR is v2 = T /m2 =

Superposition of Waves — 373

80 = 32 m/s 1/12.8

∴ Time taken by the wave pulse to reach from P to R is 4.8 2.56  4.8 2.56 t= + = +  s  80 v1 v2 32  = 0.14 s

Ans.

(b) The expressions for reflected and transmitted amplitudes (Ar and At ) in terms of v1 , v2 and Ai are as follows : v − v1 Ar = 2 Ai v2 + v1 2v2 and At = Ai v 1 + v2 Substituting the values, we get  32 − 80  Ar =   (3.5) = − 1.5 cm  32 + 80 i.e. the amplitude of reflected wave will be 1.5 cm. Negative sign of Ar indicates that there will be a phase change of π in reflected wave.  2 × 32  Similarly, At =   (3.5) = 2.0 cm  32 + 80

Suppose the incident wave of amplitude Ai and angular frequency ω is travelling in positive x-direction with velocity v1, then we can write …(i) yi = Ai sin ω [ t − x/v1 ] In reflected as well as transmitted wave, ω will not change, therefore, we can write …(ii) yr = Ar sin ω [ t + x/v1 ] and …(iii) yt = At sin ω [ t − x/v2 ] Now as wave is continuous, so at the boundary (x = 0). Continuity of displacement requires yi + yr = yi for x = 0 Substituting from Eqs. (i), (ii) and (iii) in the above, we get …(iv) Ai + Ar = At Also at the boundary, slope of wave will be continuous, i.e. ∂yi ∂yr ∂yt + = [ for x = 0 ] δx δx ∂x v  …(v) which gives Ai − Ar =  1  At  v2  Solving Eqs. (iv) and (v) for Ar and At , we get the required equations, i.e. v −v 2v2 Ar = 2 1 Ai and At = Ai v2 + v1 v2 + v1

6. When A is a node Suppose n1 and n2 are the complete loops formed on left and right side of point A. Then, f1 = f2 or or

v  n1  1  = n2  2L

 v2     2L

µ1 1 2 3 n1  v2  = , , , K ,etc =  = µ2 3 6 9 n2  v1   1   as v ∝   µ

∴ Possible frequencies are v1  v  3v , 2  1  , 1 , K , etc. 2L  2L 2L

 T  v1 =  µ 

374 — Waves and Thermodynamics 1 2L

or

T 1 , µ L

T 3 , µ 2L

8π   = 0.02 sin  πt −   3

T , K , etc. 2µ

When A is an antinode Suppose n1 and n2 are complete loops on left and right side of point A, λ λ n1 1 + 1 = L 2 4 v1  n1 1  or f1 =  +  L 2 4

y = y1 + y2 = 0.06 sin πt cos 4 π − 0.06 cos πt sin 4 π 8π 8π + 0.02 sin πt cos − 0.02 cos πt sin 3 3 Ans. = 0.05 sin πt − 0.0173 cos πt

9. Resultant amplitude

λ λ n2 2 + 2 = L 2 4 v2  n2 1  f2 =  +  L 2 4

or

⇒

n1 = 3, n2 = 10, …

…

2

 2 A  A Ar =   +    3  2 5 = A 6 A/2 3 tan φ = − =− 2A/3 4

etc.

…

Therefore, the possible frequencies are v 1  1 1  v 1  2 1  v1  3 1   +  ,  +  ,  +  , K , etc. L  2 4 L  2 4 L  2 4

∴

3 4L L = 4

T 5 , µ 4L λ 4

T 7 , µ 4L

Ans.

Ans.

10. Speed of longitudinal waves in the rod, v=

P

In the next higher mode, there will be total 6 loops and the desired frequency is  6   (100) = 300 Hz  2

2

φ = − tan −1 (3/ 4 )

or

T , ..., etc. µ

L=λ

8. Q k =

Ar

A/2

n1 = 1, n2 = 4 n1 = 2, n2 = 7

7.

2A /3

φ

2n1 + 1 1 = 2n2 + 1 3

For

or

A

A /2

f1 = f2,

Substituting we get,

A/3

Ans.

ω π = v 3

π   y1 = 0.06 (πt − kx ) = 0.06 sin  πt − × 12   3 = 0.06 sin (πt − 4 π ) Similarly, π   y2 = 0.02 sin (πt − kx′ ) = 0.02 sin  πt − × 8   3

Y 1.6 × 1011 = = 8000 m/s ρ 2500

At the clamped position nodes will be formed. Between the clamps integer number of loops will be formed. Hence, λ n1 = 80 2 or …(i) n1λ = 160 R

P

Q

S

Between P and R, P is a fixed end and R is the free end. It means the number of loops between P and λ R will be odd multiple of . Then, 4 (2n2 − 1) λ =5 2 2 or …(ii) (2n2 − 1) λ = 20

Chapter 18 Also between Q and S, (2n3 − 1)λ = 60 From Eqs. (i) and (ii), we get n1 160 = =8 2n2 − 1 20 and from Eqs. (i) and (iii). n1 160 8 = = 2n3 − 1 60 3

…(iii) …(iv)

…(v)

For minimum frequency n1 , n2 and n3 should be least from Eqs. (iv) and (v). We get, n1 = 8, n2 = 1, n3 = 2 20 λ= = 20 cm [from Eq. (ii)] 2n2 − 1 ∴

fmin

= 0.2 m v 8000 = = λ 0.2 = 40 kHz

Next higher frequency corresponds to n1 = 24 , n2 = 2 and n3 = 5 f = 120 kHz

Ans.

Ans.

11. (a) Distance between two nodes is λ/2 or π/k. The volume of string between two nodes is π V = s k

…(i)

Superposition of Waves — 375

Energy density (energy per unit volume) of each wave will be 1 u1 = ρω 2 (8)2 = 32 ρω 2 2 1 and u2 = ρω 2 (6)2 = 18 ρω 2 2 ∴ Total mechanical energy between two consecutive nodes will be E = (u1 + u2 ) V π = 50 ρω 2S k (b) y = y1 + y2 = 8 sin (ωt − kx ) + 6 sin (ωt + kx ) = 2 sin (ωt − kx ) + {6 sin (ωt + kx ) + 6 sin (ωt + kx )} = 2 sin (ωt − kx ) + 12 cos kx sin ωt Thus, the resultant wave will be a sum of standing wave and a travelling wave. Energy crossing through a node per second = power or travelling wave 1 ∴ P = ρω 2 (2)2 Sv 2 1 2  ω = ρω (4 )(S )    k 2 =

2ρω 3S k

19. Sound Waves 19.1

INTRODUCTORY EXERCISE

2. Q v ∝ T

1. ∆pmax = BAk

∴

∆p ∆pmax B = max = Ak A (2π / λ ) (∆pmax ) λ = 2πA (14 ) (0.35) = (2π ) (5.5 × 10− 6 )

2. λ min =

v fmax λ max

 270   (332) v− 3° C =   273  = 330.17 m/s  273 + 30   (332) v30° C =  273  

= 1.4 × 105 N/m 2 1450 = 20000 = 0.0725 m = 7.25 cm v 1450 = = = 72.5 m fmin 20

= 349.77 m/s The difference in these two speeds is approximately 19.6 m/s. B 3. Q v = fλ = ρ ∴

3. (a) Displacement is zero when pressure is maximum. (b) ∆pmax = BAk

= 3.6 × 109 N/m 2

4.

4. In the above problem, we have found that ∆pmax 2πfρv ω    as v = and 2πf = ω   k

INTRODUCTORY EXERCISE

19.2

1. v ∝ T

= 819° C

2

v  or T2 =  2  T1 = (2)2 (273)  v1 

= 1092 K

INTRODUCTORY EXERCISE

19.3

1. (a) ∆pmax = BAk

Now substituting the value, we have (12) (8.18) = 1.04 × 10− 5 m A= (129 . ) (2700)2

v2 T = 2 v1 T1

(7/ 5) (8.31) (273) (32 × 10− 3 )

= 315 m/s

= 3.63 × 10− 6 m

(∆pmax )k = ρω 2

γRT M

v= =

10 = 3 (2π ) (10 ) (1.29) (340)

∴

B = ρ ( f λ )2 = (900) (250 × 8)2

 ω = (ρv 2 ) A   = 2πfAρv  v ∆p ∴ A = max 2πfρv

A=

 T  v 2 =  2  v1  T1 

 ω = (ρv 2 ) ( A )    v = (2πfA ρv ) = (2π ) (300) (6.0 × 10− 3 )(1.2) (344 ) = 4.67 Pa v (∆p)2max (b) I = 2B =

(∆p)2max 2ρv

=

(4.67)2 (2) (1.2) (344 )

= 0.0264 W/m 2 = 2.64 × 10− 2 W/m 2

(as B = ρv 2)

Sound Waves — 377

Chapter 19 I L = log10    I 0

(c)

 2.64 × 10− 2   = 10 log10  − 12   10 = 104 dB I 2. L2 − L1 = 10 log10 1 I2 Given, L2 − L1 = 9 dB Solving the equation, we get I1 = 7.9 I2 1 3. I ∝ 2 r 2

2 I 1  r2   300 =  =  = 100  30  I 2  r1  I Now, L1 − L2 = 10 log10 1 I2 I1 Substituting = 100 I2

∴

We get, L1 − L2 = 20 dB v (∆p)2max 4. (a) I = 2B (∆p)2max = 2ρv For finest sound, I =

(2 × 10− 5 )2 2 × 1.29 × 345

= 4.49 × 10− 13 W / m 2 I L = 10 log10    I 0  4.49 × 10− 13   = 10 log10  10−12   = − 3.48 dB Same formulae can be applied for loudest sound. (b) (∆p)max = BAk  ω = (ρv 2 ) ( A )    v = 2πfρv ∴

A=

(∆p)max 2πfρv

For finest sound, A=

2 × 10− 5 (2π ) (500) (1.29) (345)

= 1.43 × 10− 11 m

INTRODUCTORY EXERCISE

19.4

v 1. ∆x = λ /2 = 2f v 330 ∴ f = = = 1375 Hz 2∆x 2(0.12) 2π 2. (a) ∆φ = (∆x ) λ λ∆φ v (∆φ ) ∆x = = ∴ 2π ( f ) (2π ) (350) (π / 3) = (500) (2π ) = 0.1166 m = 11.7 cm  2π  (b) ∆φ =   ∆t = (2πf ) ∆t T  = (2π ) (500) (10− 3 ) = π or

180°

INTRODUCTORY EXERCISE 19.5 1. Length of the organ pipe is same in both the cases. v 2l and frequency of third harmonic of closed pipe will be  v f2 = 3    4 l

Fundamental frequency of open pipe is f1 =

Given that, f2 = f1 + 100 or f2 − f1 = 100 3  v   1  v  or   −     = 100 4  l   2  l  v ⇒ = 100 Hz 4l v or f1 = 200 Hz ∴ 2l Therefore, fundamental frequency of the open pipe is 200Hz. 2. First harmonic of closed pipe = Third harmonic of open pipe  v l1 1 v = =3  ⇒ ∴ ∴ l2 6 4 l1  2l2 

378 — Waves and Thermodynamics v = 512 Hz 4l v fo = = 2 fc = 1024 Hz 2l v 4. (a) f1 = 4l v 345 ∴ l= = 4 f1 4 × 220

3. fc =

= 0.392 m  v   v (b) 5   = 3    4 lc   2l0  l0 =

∴

6  6 lc =   (0.392)  5 5

= 0.47 m

INTRODUCTORY EXERCISE

19.6

1. Frequency of first will decrease by loading wax over it. Beat frequency is increasing. Hence, f2 > f1 or f2 − f1 = 4 ∴ f1 = f2 − 4 = 256 − 4 = 252 Hz 2. By putting wax on first tuning fork, its frequency will decrease. Beat frequency is also decreasing. Hence, f1 > f2 ∴ f1 − f2 = 3 or f1 = 3 + f2 = 3 + 384 = 387 Hz

INTRODUCTORY EXERCISE 19.7 1. Source is moving towards the observer

5. x=0 (i)

(ii)

 v   330  f′= f   = 450    v − vs   330 − 33

(iii)

3λ 4 λ l 0.8 ∴ = = 4 3 3 = 0.267 m Displacement antinode is at λ x = = 0.267 m 4 λ and x = 3 = 0.8 m 4 (b) Pressure antinode is displacement node: In third figure, Pressure antinode or displacement node is at λ x = 0, x = 2 and x=λ 400 5 6. (a) = 560 7 Since, these are odd harmonics (5 and 7). Hence, pipe is closed. (c) Given frequencies are integer multiples of 80 Hz. Hence, fundamental frequency is 80 Hz. v = 80 4l v 344 ∴ l= = 320 320 = 1.075 m (a) In second figure, l =

f ′ = 500 Hz  v  f1 = f    v − vs 

2.

 340   340 f1 = f   = f   306  340 − 34  and ∴

 340   340 f2 = f   = f    323  340 − 17 f1 323 19 = = f2 306 18 v + v0   v   v + vA  5.5 = 5    v 

3. Using the formula, f ′ = f  

We get, and

 v + vB  6.0 = 5    v 

v = speed of sound vA = speed of train A vB = speed of train B v Solving Eqs. (i) and (ii), we get B = 2 vA Here,

4. Observer is stationary and source is moving.  v  During approach, f1 = f    v − vs 

…(i) …(ii)

Chapter 19  320  = 1000    320 − 20

Sound Waves — 379

 320  = 1000   = 941.18 Hz  320 + 20  f − f2  |% change in frequency| =  1  × 100  f1 

= 1066.67 Hz  v  During recede, f2 = f    v + vs 

≈ 12%

Exercise LEVEL 1

10. Wavelength remains same.

Assertion and Reason 1. Closed pipe Frequencies are,

v 3v 5v , , 4l 4l 4l

Open pipe Frequencies are v  v  v ,2 ,3     2l  2l 2l They are never same. 

v   ⇒ f ′ > f but f ′ = constant.  v − vs 

2. f ′ = f 

3. Change in pressure is maximum at a point where displacement is zero. I 5. L = 10 log10 I0

Objective Questions 1. Sound waves cannot travel in vacuum. 3. f0 =

v 2l

and

∴

4. v = ∴

Increase in the value of L is not linear with I. I ∆L = L2 − L1 = 10 log 2 I1

∴

γRT T ∝ M M TO 2 TN 2 = MO2 MN2

(γ = 1.4 for both)

 MO2   TN TO 2 =   2  MN2 

∆L = 3dB if I 2 = 2I 1 6. It is independent of pressure as long as temperature remains constant.

7. fA − fB = 4 Hz

v 4l f0 = 2 fc

fc =

 32 =   (273 + 15)  28 = 329 K = 56° C

5.

When A is loaded with wax fA will decrease. So, fA − fB will be less than 4 Hz till fA > fB . But fB − fA may be greater than 4 Hz when fB becomes greater than fA. 450 3 8. = 750 5 Successive harmonics are odd 3 and 5. Hence, it is a closed pipe. v of an open pipe and 9. f0 = 2 (l + 1.2 r) v for a closed pipe. f0 = 4 (l + 0.6 r)

Third overtone

6. f ∝ v and v ∝ T 7. In air speed of sound is less, hence air is denser medium. v v 8. = 2l0 4 lc

⇒ lc : l0 = 1 : 2

380 — Waves and Thermodynamics 9. f ∝ v ∝ T ∴

(T → tension)

f1 T = 1 f2 T2

 f  ⇒ T2 = T1  2   f1 

n = n1 − n2 =

18.

2

∴

2

 256 = (10)   = 6.4 kg  320

10.

But beat frequency between A and B is also decreasing. ∴

S

Both S and S′ are moving toward observer. Hence, fS = fS ′ or fb = 0 10 v v 11. fb = = f1 − f2 = − 3 1 1.01 Solving we get, v = 337 m/s  v   v  12. f ′ = f   = f   = 0.5 f  v + vs   v + v

13. I R = I max = ( I 1 + I 2 )2 I 1 = I 2 = I 0, then IR = 4I0

If

λ = 52 − 17 = 35 cm 2 ∴ λ = 70 cm = 0.7 m v = f λ = 500 × 0.7 = 350 m/s   v 15. f ′ = f    v ± vs cos θ 

At θ = 90° ; f ′ = f ∴ n1 = 0 16. Fundamental frequency, v 340 f0 = = = 85 Hz 4l 4 × 1

5 m/s

fb = fS ′ − fS

S'

5 m/s 5 m/s

 355 − 5  = f − fS = 180 − 180    355 + 5 = 5 Hz

∴

fA > fB fA − fB = 5 Hz

…(i)

fB = fA − 5 = 345 Hz

Now, fB ~ fC = 4 Hz fC is either 341 Hz or 349 Hz. If it is 341 Hz, then beat frequency with A will be 9 Hz. If it is 349 Hz, then beat frequency will be 1 Hz. If wax is loaded on A, its frequency will decrease. To produce 6 beats/s with C it should become either 347 Hz (if fC = 341 Hz) or it should become 343 Hz (if fC = 349 Hz).

f′ T′ 101 = = = 1.0049 f T 100

Six frequencies can be produced below 1 kHz. Those six frequencies are, f0, 3 f0, 5 f0, 7 f0, 9 f0 and 11 f0. As 13 f0 = 1105 Hz > 1000 Hz or 1 kHz. 17. There is no relative motion between O and S ′. S

or

If it becomes 347 Hz, then only it produces 2 beats/s with B, which is given in the question. Ans. ∴ fC = 341 Hz λ 20. = 122 − 40 = 82 cm 2 ∴ Next resonance length = 122 cm + 82 cm = 204 cm 21. f ∝ T

14.

O

nλ 1λ 2 λ2 − λ1

19. By putting wax on A. Its frequency will decrease.

S'

O

v=

v v − λ1 λ2

∴

22. λ =

f ′ = (1.0049) (200) ≈ 201 Hz fb = f ′ − f = 1 Hz v 340 = = 1 m = 100 cm f 340

λ = 25 cm 4 Air column lengths required are, λ 3λ 5λ etc. , , 4 4 4 or 25 cm, 75 cm, 125 cm etc. Maximum we can take 75 cm. ∴ Minimum water length = 120 − 75 = 45 cm

Chapter 19 23. Pressure node means displacement antinode 7λ = 105 cm 4 λ = 15 cm 4

Sound Waves — 381

L2 − 10 = 10 log10 (16) L2 = 22 dB 2π 2π 30. λ = = = 4m k π /2 or or

l=

Displacement antinodes are at a distance λ 3λ 5λ 7λ and from closed end or at a , , 4 4 4 4 distance of 15 cm, 45 cm, 75 cm and 105 cm. 24. Number of moles ∝ Volume n1 M O 2 + n2M H 2 M = n1 + n2 (1) (32) + (1) (2) = = 17 2 γRT 1 Now, v= ∝ M M v1 M2 2 = = v2 M1 17 

v   = constant. But f1 > f  v − vs 

25. f1 = f 

 v  f2 = f   = constant, but f2 < f  v + vs  1 26. f ∝ l Both frequencies will becomes half. Hence, fb = f1 − f2 will also become half.  320   320  27. fb = 243   − 243   = 6 Hz  320 − 4   320 + 4 

31. λ =

v 330 = = 0.55 m = 55 cm f 600

d

The desired distance, d =

Other frequencies are 3 f1 : 5 f1 etc. or 240 Hz, 400 Hz etc. Open pipe Fundamental frequency is v 320 f1 = = = 100 Hz 2l 2 × 1.6 Other frequencies are 2 f1 : 3 f1 : 4 f1 etc. or 200 Hz, 300 Hz and 400 Hz etc. So, then resonate at 400 Hz. 29. Resultant amplitude will become 4 time. Therefore, resultant intensity is 16 times I L2 − L1 = 10 log10 1 I2

λ = 13.75 cm 4

v v − λ1 λ2 332 332 = − ≈13 Hz 0.49 0.5 33. ( fb )A = fB − fA

32. fb = f1 − f2 =

 300 + 30 = 300   − 300 = 30 Hz  300  ( f b )B = f A − f B  300  = 300   − 300 = 33.33 Hz  300 − 30 v + v0  34. fa = f   

v

 v0 f = a −1 v f  v − v0  fr = f    v  v0 f =1− r v f

∴

28. Closed pipe Fundamental frequency is v 320 f1 = = = 80 Hz 4l 4 × 1

5λ = 5 m 4

∴

…(i)

…(ii)

From Eqs. (i) and (ii), we get f + fr f = a 2

Subjective Questions 1. Speed of sound wave, v=

B = ρ

Wavelength, λ =

(2 × 109 ) = 1414 m/s 103 v = 5.84 m f

Ans.

382 — Waves and Thermodynamics  v ± vm   = f  v ± vm 

2. f ′ = f 

4.

∴

γRT M 1.40 × 8.31 × 300 = = 1321 m/s 2 × 10− 3

v=

3.

L1 = 10 log10 = 10 log10 L2 = 10 log10

5.

B = ρ ( f λ )2 = 1.33 × 1010 N/m 2 v=

(b)

 10− 6   − 12  = 60 dB  10   I 2    I 0

∴

l Y = t ρ

2   1.5  l Y = ρ   = (6400)   −4  t  3.9 × 10 

 v − vw − v0  f′ = f    v − v w + vS 

Y T F (as T = F and µ = ρA) = 30 = 30 ρ µ ρA F Y = ∴ A 900 10. Equal volume of different gases contains equal number of moles at STP. ∴ n∝V n M + n2M 2 M = 1 1 n1 + n2 =

6. v=

S

d2

d1

2d1 = v t1 vt 332 × 3/ 2 ∴ d1 = 1 = 2 2 = 249 m vt 332 × 5/ 2 Similarly, d2 = 2 = 2 2 = 415 m ∴ Total distance = d1 + d2 = 664 m Next echo he will hear after time, 5 3 t = t2 + t1 = + = 4 s 2 2 γp = ρ

(5/3) × 0.76 × 13.6 × 103 × 9.8 0.179

where,

p = hρg

∴

v = 972 m/s

2

= 9.47 × 1010 N/m 2

9.

 340 − 5 − 20  = 300    340 − 5 + 10  = 274 Hz

7. v =

B ρ

= (1300) (400 × 8)2

 I 1    I 0

 10− 9  = 10 log10  − 12  = 30 dB  10  L1 = 2L2

v = fλ =

8. (a)

(2) (2) + (1) (28) = 10.67 2+1 γRT M

∴

v2 T M = 2⋅ 1 v1 T1 M 2

or

 T M  v 2 =  2 ⋅ 1  v1  T1 M 2   300 2   (1300) = × 273 10.67  

11.

≈ 591 m/s I L1 = 10 log10    I 0  I  100 = 10 log10  − 12   10  Solving we get, I = 10− 2 W/m 2 P Now, I= 4 πr 2 ∴ P = I (4 πr2 ) = (10− 2 )(4 π )(40)2 = 201 W

Sound Waves — 383

Chapter 19 I L = 10 log10    I 0

12. (a)

 I  60 = 10 log10  − 12   10  Solving we get, I = 10− 6 W/m 2 (b) I = ∴

P S

P = (I ) (S ) = (10− 6 ) (120 × 10− 4 ) = 1.2 × 10− 8 W

I  13. (a) L2 − L1 = 10 log10  2   I 1 I  ∴ 13 = 10 log10  2   I 1 I2 Solving we get, = 20 I1 (b) From the above equation we can see that we do not need L1 and L2 separately. P 5 14. (a) I = = 4 πr2 4 π (20)2 = 9.95 × 10− 4 W/m 2 1 (b) I = ρω 2 A 2v 2 ∴ A= =

I L = 10 log10    I 0

16.

 I  ∴ 102 = 10 log10  − 12   10  Solving we get, I = 1.59 × 10− 2 W/m 2 P I = 4 πr 2 ∴ P = (I ) (4 πr2 ) = (1.59 × 10− 2 ) (4 π ) (20)2 = 80 W 1 2 2 17. I = ρω A v 2 = 2ρπ 2 f 2 A 2v

= (2) (1.29) (π ) (300) (0.2 × 10− 3 )2 (330) L = 10 log10 (I / I 0 )

where, I 0 = 10− 12 W/m 2 Substituting the value we get, L =134.4 dB

18. (a) v =

2I ρ (2πf )2 v 2 × 9.95 × 10− 4 1.29 (2π × 300)2 (330) ∴

I    I 0 − 12

60 = 10 log10 (I /10

−6

2 × 10 1.29 (2π × 800)2 (330)

= 1.36 × 10− 8 m = 13.6 nm

2.18 × 109 1000

= 1476 m/s v 1476 λ= = = 0.43 m f 3400 1 I = ρω 2 A 2v 2 2I A= ρ (2πf )2 v

)

Solving, we get I = 10− 6 W/m 2

=

B = ρ

=

Now, using the result derived in above problem. 2I A= ρ (2πf )2 v

2

= 30.27 W/m 2

= 1.15 × 10− 6 m

15. L = 10 log10

(as ω = 2πf ) 2

2 × 3 × 10− 6 (1000) (2π × 3400)2 (1476)

= 9.44 × 10− 11 m (b)

v=

γp = ρ

(1.4 ) (105 ) (1.2)

= 341.56 m/s v 341.56 λ= = = 0.1 m f 3400 A=

2I ρ (2πf )2 v

384 — Waves and Thermodynamics =

2 × 3.0 × 10− 6 1.2 × (2π × 3400)2 (341.56)

= 5.66 × 10− 9 m Aair 5.66 × 10− 9 = = 60 Awater 9.44 × 10− 11 1 I = ρω 2 A 2v 2 ρ and v are less in air. So, for same intensity A should be large. B v (∆p)2max 19. I = ⇒ v= ρ 2B (c)

∴

B = ρv 2

∴

I = =

ω 1297 = = 3.93 rad /s v 330 Now, x = 0 (the open end) is an antinode. Hence, we can write the equation, y = A cos kx sin ωt λ 23. = (84 − 50) cm = 34 cm 2 λ Next length = 84 + = 118 cm 2 λ = 68 cm or 0.68 m v = fλ = (512) (0.68) = 348.16 m/s  v v  24. 2   =  2   2l1   4 l2  k=

(∆p)2max 2ρv

T /µ  v2  =  l1  4 l2 

or

(6 × 10−5 )2 2 × 1.29 × 330

v l  T = µ  2 1  4 l2 

∴

= 4.2 × 10− 12 W/m 2 L = 10 log10

 4 × 10− 3   340 × 40   =   0.4   4 × 100 

I    I 0

 4.2 × 10− 12   = 10 log10  − 12  10 

∴

v 344 = = 382.2 Hz 2l 2 × 0.45

First two overtones are 2 f1 and 3 f1. v 344 (b) Fundamental, f1 = = = 191.1 Hz 4 l 4 × 0.45 First two overtones are 3 f1 and 5 f1. λ 21. = (45 − 15) cm 2 ∴ λ = 60 cm v = fλ = (500) (0.6) = 300 m/s Lowest frequency when open from both ends is 300 = 250 Hz 2 × 0.6 v 22. Fundamental frequency = 2l 330 f = 2 × 0.8 =

f = 206.25 Hz ω = 2πf = 1297 rad /s

2

= 11.56 N v f = 4 (l + 0.6 π )

25.

= 6.23 dB

20. (a) Fundamental, f1 =

2

v = 4 f (l + 0.6 π ) = 4 × 480 (0.16 + 0.6 × 0.025) = 336 m/s v 26. (a) f = 4l v 345 ∴ l= = 4 f 4 × 220 = 0.392 m  v   v 5  =3   4 lc   2l0 

(b)

6 6 lc = (0.392) 5 5 = 0.470 m v1 v v 27. = 2 ⇒ 1 = 0.4 2 (0.8l ) 4 l v2 v 340 28. (a) λ = = = 1.13 m f 300 ∴

v ⋅ 2l

l0 =

(b) Ahead  v   340  f′= f   = 300    v − vs   340 − 30 = 329 Hz

Chapter 19

  0.32 = 0.625    0.32 + 0.245

v 340 = = 1.03 m f ′ 329

λ′ =

= 0.354 Hz v 0.32 λ= = = 0.904 m f ′ 0.354

Behind  v   340  f′ = f   = 300    v + vs   340 + 30 = 275.67 Hz v 340 λ′ = = f ′ 275.67

 v + v0    v − vs 

31. (a) f ′ = f 

 340 + 18 = 262    340 − 30

= 1.23 m 29. (a) Possible fundamental frequencies of retuned string is (440 ± 15 . ) Hz. (b) f ∝ T

= 302 Hz  v − v0  (b) f ′ = f    v + vs 

f1 T = 1 f2 T2

∴

 340 − 18  = 262   = 228 Hz  340 + 30

2

∴

 f  T2 =  2  T1  f1 

 441.5 (i) T2 =    440 

2

T1 = 1.0068 T1 T − T1 % change = 2 × 100 T1 = + 0.68% 2  438.5 (ii) T2 =   T1 = 0.9932 T1  440  T − T1 × 100 % change = 2 T1 = − 0.68%

30. (a) λ = 0.12 Surface wave speed, v = 0.32 m/s v 0.32 f′ = = = 2.66 Hz λ 0.12 1 1 f = = = 0.625Hz T 1.6  v  Now, using f ′ = f    v − vs  Here, vs = velocity of source or velocity of duck  0.32  ∴ 2.66 = 0.625    0.32 − vs  Solving we get, vs = 0.245 m/s (b) Behind the duck  v  f′ = f    v + vs 

Sound Waves — 385

32.

1 m/s O

S′

S

1 m/s

fb = fS − fS ′  340   340  4= f  − f    340 − 1  340 + 1 1   4 = f 1 −   340

or

−1

1   − f 1 +   340

−1

Applying Binomial, we get 1  1    4 = f 1 +  − f 1 −    340 340 f or f = 680 Hz = 170 7 33. Beat frequency between P and Q is or 3.5 Hz. 2 On loading P with wax, its frequency will decrease and beat frequency is increasing. Q

5 m/s 5 m/s

∴ or Given, or

Q′

fQ > fP fQ − fP = 3.5 fQ′ − fQ = 5  332 + 5 fQ   − fQ = 5  332 − 5 

Solving this equation, we get fQ = 163.5 Hz Now, from Eq. (i) we have fP = 160 Hz

…(i)

386 — Waves and Thermodynamics 34. fb = f1 − f2

=

 340   340  2 = 680   − 680    340 − vs   340 + vs  v   = 680 1 − s   340

−1

v   − 680 1 + s   340

= 0.7 Since, γ θC , therefore only reflection will take place. …(i) 2. ∆pmax = BAk

44. In front of locomotive,  v  f′ = f    v − vs   344  = 500   = 548 Hz  344 − 30 ∴

λ′ =

v= ∴

v 344 = = 0.628 m f ′ 548

Behind the locomotive,  v  f′ = f    v + vs 

∴

∴

 344  = 500   = 460 Hz  344 + 30 v 344 λ′ = = = 0.748 cm f ′ 460

the frequencies are odd multiple of 85 Hz, the pipe must be closed at one end. (b) Now, the fundamental frequency is the lowest, i.e. 85 Hz. v 85 = ∴ 4l 340 Ans. ⇒ l= =1 m 4 × 85

46. Let the frequency of the first fork be f1 and that of second be f2. We then have,

We also see that f1 > f2 ∴ f1 − f2 = 8 f1 33 and = f2 32

…(i) …(ii)

f2 = 256 Hz

= 0.025 m = 2.5 cm  2v2  At =   Ai  v1 + v2 

3.

At 2 × 100 2 = = Ai 200 + 100 3

∴

4. fb = f1 − f2  340   340  10 = 1700   − 1700    340 + v2   340 + v1  −1

v   − 1700 1 + 1   340

v1 − v2 = 2 m/s

5. Velocity of source after 1 s, Ans.

−1

Using Binomial therefore, we get v  v    10 = 1700 1 − 2  − 1700 1 − 1    340 340  1700 = (v1 − v2 )    340  ∴

f1 = 264 Hz

l

(0.01 × 105 ) 1.3 × (200)2 × (1/1.3)

v   = 1700 1 + 2   340

Solving Eqs. (i) and (ii), we get and

B = ρv 2 3λ l= 2 2l λ= 3 2π 2π 3π k= = = λ 2l / 3 l 3π 1 −1 = = m 3.9 π 13 .

=

or

v f1 = 4 × 32 v f2 = 4 × 33

B ρ

Substituting in Eq. (i), we have ∆p ∆p A = max = max Bk ρv 2k

45. (a) The given frequencies are in the ratio 5 : 7 : 9. As

and

Sound Waves — 387

vs = gt = 10 × 1 = 10 m/s

388 — Waves and Thermodynamics A1 = 2 and

 v + v0   v − v0  ∆f = f   − f    v − vs   v + vs 

2

I max  A1 + A2  9 =  = I min  A1 − A2  1

 300 + 2   300 − 2  = 150   − 150    300 − 10  300 + 10 7λ 6. L = 4

11. l =

= 12 Hz

4L 7 2π k= λ 2π 7π = = 4 L / 7 2L

∴

x=0

7. ∆xmax = 3m = λ v 330 = = 110 Hz. For frequencies less λ 3 than 100 Hz, λ will be more than 3m and no maximum will be obtained.

VS

O

60°

 300 + 10  = 500    300 + 10 − 20 = 534 Hz 10 m/s

20 m/s S2

S1

S'1 10 m/s

fb = fS ′ − fS1

10.

∴

and

λ∝v v∝ T

∴

λ∝ T

∴

λ2 T = 2 λ1 T1

∴

 T  λ2 =  2 λ1  T1 

 340 + 20   340 + 20  = 500   − 500    340 − 10   340 + 10  ≈ 31 Hz ω (400π ) f1 = 1 = = 200 Hz 2π 2π ω 404 π f2 = 2 = = 202 Hz 2π 2π fb = f2 − f1 = 2 Hz

(as f = constant) (T = temperature)

 16 + 273   (45.33) =  51 + 273  = 42.8 cm

12. Q ∴

E

 v + w cos 60°  f′= f    v + w cos 60° − vs 

9.

4 l 4 (34 ) = = 45.33 cm 3 3 v λ= f

λ=

f =

W = 20 m/s N

8.

3λ 4

∴ λ=

x = 0 is a node  7π   L ∴ Ax = a sin kx = a sin     = a  2L   7 

∴

A2 = 1

f1 = f2  330 − v   330 + v  176   = 165    330   330 − 22

Solving this equation we get, v = 22 m/s n1M 1 + n2M 2 13. M = n1 + n2 (2) (32) + (3) (48) = = 41.6 5 v or f ∝ v f = 4l γRT But, v= M 1 or v∝ M 1 f ∝ ∴ M f2 M1 ∴ = f1 M2 or

 M1   f1 f2 =   M2   32   (200) = 175.4 Hz =  48 

Sound Waves — 389

Chapter 19  v + v0  f = f0    v  10t   (as v0 = gt) = 103 1 +   v  Hence, f versus t graph is a straight line of slope 104 . v 104 100 = slope = ∴ v 3 ∴ v = 300 m / s ∂y 15. Q vp = − v   , where v = + ve.  ∂x  ∂y or slope is positive. At E , ∂x Hence, vp is negative. ∂y At D , or slope is zero. Hence, vp is zero. ∂x

(b) First overtone frequency of closed pipe = 3 f1 and first overtone frequency of open pipe 5  = 2 f2 = 2  f1 = 2.5 f1 4 

14.

More than One Correct Options 1. Q

 v f =n   4 l

∴

l=

(n = 1, 3, 5…)

or

(d) Tenth harmonic of closed pipe = 10 f1 Eighth harmonic of open pipe 5  = 8 f2 = 8  f1 = 10 f1 4 

5. f =

> f0 but f is constant. During receding.  v  f = f0    v − vs  < f0 but f is again constant.

Comprehension Based Questions 1.

or

f2 = fundamental frequency of open pipe. Given, 5 f1 = 4 f2 5 or f2 = f1 4 ∴ f2 > f1

v

8−v

S Man

From conservation of linear momentum, 50 (8 − v ) = 150 v ∴ v = 2 m/s 8 − v = 6 m/s  v + v0  f1 = f0    v − vs 

v∝ T

v2 ∝ T v v (d) f0 = or 2l 4l 1 ⇒ f0 ∝ l 4. Let f1 = fundamental frequency of closed pipe and

O

Plank

γRT γR (t + 273) = M M v 2 ∝ (t + 273)

(c) v = T /µ ∴

γ RT / M v = 4 (l + 0.6r ) 4 (l + 0.6 r)

(d) v does not depend on pressure if temperature is kept constant. 6. During approach,  v  f = f0    v − vs 

 330  nv =n  4f  4 × 264 

= (0.3125 n) m = (31.25 n) cm Now, keep on substituting n = 1, 3, etc. 2. (a) Velocity of sound wave in air independent of pressure if T = constant (b) v =

(c) Fifteenth harmonic of closed pipe = 15 f1 Twelfth harmonic of open pipe 5  = 12 f2 = 12  f1 = 15 f1 4 

 330 + 2 332 = f0  f0 =  330 − 6  324

2.

6 m/s S

O

2 m/s

 v − v0  f2 = f0    v + vs   330 − 2   328 = f0  =  f0  330 + 6  336

3. f1 > f0 but f1 = constant. Similarly, f2 < f0 but f2 = constant

390 — Waves and Thermodynamics 4. I ∝ A 2

Match the Columns v 1. (a) f = 2l v v f fc = = = 4 (2l ) 8l 4 (b) Second overtone frequency = 5 fc = 5 ( f / 4 ) = 1.25 f (c) Third harmonic frequency = 3 fc = 3 ( f / 4 ) = 0.75 f (d) First overtone frequency = 3 fc = 3 ( f / 4 ) = 0.75 f O2

O1

2.

T

v 4

or

A∝ I

Due to a point source, P (P = power of source) I = 4 πr 2 1 1 or I ∝ 2 and A ∝ r r Due to a line source P 1 or I ∝ I = 2πl r 1 ∴ A∝ r

5. Q k = π 2π =2m k 5λ 5 l= = (2) 4 4 = 2.5 m Distance of displacement node from the closed end λ or = λ 2 = 1 m or 2m λ=

O3 v 4

T'

(a) fb = fT ′ − fT  v   v  = f   − f    v − v/4  v + v/4 8 = f 15  v + v/4 (b) fb = fT ′ − fT = f   − f  v − v/4  2 = f 3 (c) T and T ′ both are approaching towards O3. ∴ fT = fT ′ or fb = 0 (d) Not T and T ′ both are receding from O3. Hence, again fT = fT ′ or fb = 0 3. Let f1 = frequency of tuning fork and f2 = frequency of streamed wire Given, f1 > f2 ∴ fb = f1 − f2 (a) By loading wax on turning fork f1 will decrease. Hence, f1 − f2 may be less than fb or f2 − f1 may be greater than fb . (b) If prongs are filed, f1 will increase. Hence, f1 − f2 > fb (c) If tension in stretched wire is increased, f2 will increase. Hence, f1 − f2 may be less than fb or f2 − f1 may be greater than fb . (d) If tension in stretched wire is decreased, f2 will decrease. Hence, f1 − f2 > fb

Pressure node means displacement antinode. Its distance from closed end λ 3λ 5λ or or = 4 4 4 = 0.5 m,1.5 m and 2.5 m

Subjective Questions 1. By definition sound level = 10 log or ⇒

I = 60 I0

I = 106 I0 I = 10−12 × 106 = 1 µW/m 2

Power entering the room = 1 × 10−6 × 2 = 2 µW Energy collected in a day = 2 × 10−6 × 86400 = 0.173 J

Ans.

2. (a) At a distance r from a point source of power P, the intensity of the sound is

Sound Waves — 391

Chapter 19 P 0.8 = 4 πr2 (4 π ) (1.5)2

I = or

5. (a) λ =

I = 2.83 × 10−2 W /m2

S1

…(i)

d sin θ =

(∆p)m = 2 × 2.83 × 10−2 × 1.29 × 340 = 4.98 N/m 2

= 3.0 × 10−6 m

Ans.

3. (a) Fundamental frequency when the pipe is open at both ends is v 340 = 2l 2 × 0.6

= 283.33 Hz

Ans.

(b) Suppose the hole is uncovered at a length l from the mouthpiece, the fundamental frequency will be v f1 = 2l v 340 l= = ∴ 2 f1 2 × 330 = 0.515 m = 51.5 cm

Ans.

Note Opening holes in the side effectively shortens the length of the resonance column, thus increasing the frequency.

4. 2 H 2 + d 2 / 4 − d = nλ Now, 2 (H + h)2 + d 2 / 4 − d = nλ +

…(i) λ 2

Solving these two equations, we get λ = 2 4 (H + h)2 + d 2 − 2 4 H 2 + d 2

…(ii)

λ 2

λ θ = sin − 1    2d   0.57 = sin − 1    4 

Ans.

(b) Pressure oscillation amplitude (∆p)m and displacement oscillation amplitude A are related by the equation (∆p)m = BAk ω Substituting B = ρv 2 , k = v and ω = 2πf We get , (∆p)m = 2πAρvf (∆p)m 4.98 A= = ∴ 2πρvf (2π ) (1.29) (340) (600)

d = 2m

S2

∴

From Eqs. (i) and (ii),

θ

d

Further, the intensity of sound in terms of (∆p)m , ρ and v is given by (∆p)2m …(ii) I = 2ρv

f1 =

v 340 = = 0.57 m f 600

= 8.14 °

 λ or θ = sin − 1    d

(b) d sin θ = λ

 0.57 = sin − 1    2  = 16.5° (c) d sin θ = nλ ∴

nmax =

d λ

(When sin θ = 1)

2 = 3.5 0.57 ∴ Maximum 3 maximas can be heard corresponding to n = 1, 2 and 3 (beyond θ = 0°) 2π 2π λ 6. ∆φ 1 =   ∆x =   = π  λ  λ 2 =

(a) ∆φ net = ∆φ 1 + ∆φ 2 =π+0 =π So, they will interfere destructively and I net = 0 (b) No interference will take place. Hence, I net = I 1 + I 2 = 2I 0 (c) ∆φ net = ∆φ 1 ± ∆φ 2 (as ∆φ 2 = π ) = 2π or 0 In both conditions, they interfere constructively. Hence, I = I max = 4 I 0 2π  2πf  7. (a) ∆φ = (∆x ) =   (∆x )  v  λ (2π ) (170) = (11 − 8) 340 = 3π = φ (say ) θ …(i) I = 4 I 0 cos2 2

392 — Waves and Thermodynamics Substituting value of φ, we get I =0 (b) Now, φ = 3π ± π = 4π or 2π Hence, from Eq. (i) I = 4I0 I Now, L1 − L2 = 10 log10 1 I2

10. Given length of pipe, l = 3 m

 4I  L1 − 60 = 10 log10  0  = 6  I0 

∴

∴ L1 = 66 dB (c) Sources are incoherent. Hence, I = I 0 + I 0 = 2I 0 I L1 − L2 = 10 log10 1 I2 2I 0 = 10 log10 I0 ∴ or

L1 − 60 = 10 log10 2 = 3 L1 = 63 dB P 8. (a) I = 4 πr 2 ∴

I1 =

The longitudinal oscillations of an air column can be viewed as oscillations of particle displacement or pressure wave or density wave. Pressure variation is related to particle displacement as ∂y p (x , t ) = − B (B = Bulk modulus) ∂x  3 BAπ   3πx   3πv  =  sin   sin   t  l   l   l 

1 × 10− 3 = 19.9 × 10− 6 W/m 2 4 π (2)2

= 19.9 µW/m 2 I2 =

1 × 10− 3 = 8.84 × 10− 6 W/m 2 4 π (3)2

The amplitude of pressure variation is 3 BAπ B or B = ρv 2 pmax = ⇒ v= l ρ 3 ρv 2 Aπ p l or A = max2 pmax = ∴ l 3 ρv π

= 8.84 µW/m 2 (b) I p = I max = ( I 1 +

I 2 )2

= 55.3 µW/ m 2

Here, pmax = 1% or p0 = 103 N/m 2

(c) I p = I min = ( I 1 − I 2 )2

Substituting the values, (103 )(3) A= = 0.0028 m (3)(103 . )(332)2 π

= 2.2 µW/ m 2 (d) I p = I 1 + I 2 = 28.7 µW/ m 2

9. vs = v0 = v Let u be the speed of sound. Then,  u + v0 cos θ  f′ = f    u + vs cos θ 

or

 u + v cos θ  = f    u + v cos θ  f′ = f

x=0

Third harmonic implies that  λ 3  =l  2 l 2l 2 × 3 or λ= = =2m 3 3 The angular frequency is ω = 2πf . x=l 2πv (2π )(332) or = = λ 2 ω = 332π rad/s The particle displacement y(x , t ) can be written as y (x , t ) = A cos kx sin ωt 2π 2π 3π k= = = λ (2l/3) l 3πv ω  and ω = kv = v =   l k  3πx   3πv  y (x , t ) = A cos  ∴  ⋅ sin   t  l   l 

vs

θ

S

v0 θ O

Ans.

or Ans. A = 0.28 cm According to definition of Bulk modulus ( B ) − dP …(i) B= (dV /V ) Mass m or V = Volume = Density ρ m Vdρ or dV = − 2 dρ = − ρ ρ dV dρ or =− V ρ

Chapter 19 Substituting in Eq. (i), we get ρ (dP ) dρ = B or amplitude of density oscillation is ρ p dρ max = pmax = max B v2 =

(b) vs = 30 m/min

1 ρ  = 2 B v 

103 = 9 × 10−3 kg/m 3 (332)2

wave

I    I 0

∴

L = 60 dB Hence, I = (106 )I 0 = 10−6 W/m 2 Power P Intensity, I = = Area 4 πr 2 ⇒ P = I (4 πr2 )

N

y = a sin (kx − ωt ) k = 15π and ω = 6000π ω Velocity of the sound, v = k 6000π = = 400 ms−1 as v = 15π

Hence, ρ =

P = (10−6 )(4 π )(500)2 = 3.14 W  30  (1.0 × 103 )    100 ∴ Time, t = 3.14 t = 95.5 s 12. Let n1 harmonic of the chamber containing H2 is equal to n2 harmonic of the chamber containing O2. Then, A

Ans.

14. (a) Comparing with the equation of a travelling

I 0 = 10−12 W/m 2

where,

Spacing between the pies will be 300 − 30 or 13.5 m = 20  v   300  and f′ = f   = (20)    v − vs   300 − 30 = 22.22 min −1

11. Sound level (in dB) L = 10 log10

Sound Waves — 393

A

B ρ

B 1.6 × 105 = = 1 kg/m 3 v2 (400)2

Ans.

(b) Pressure amplitude, p0 = BAk p 24π Hence, A = 0 = Bk 1.6 × 105 × 15π = 10−5 m = 10 µm

Ans.

(c) Intensity received by the person, W W I = = 4 πR 2 4 π (10)2 Also,

I =

p02 W (24 π )2 ⇒ = 2 2ρv 2 × 1 × 400 4 π (10)

W = 288π 3 W

Ans.

15. (a) Path difference and hence phase difference at P 0.5 m

∴ ∴

from both the sources is 0°, whether θ = 45° , or θ = 60°. So, both the wave will interfere constructively. Or maximum intensity will be obtained. From I max = ( I 1 + I 2 )2 we have

0.5 m

 vH   vO  n1  2  = n2  2   4l   4l  n1 vO 2 300 3 = = = n2 vH 2 1100 11  vH   1100  fmin = 3  2  = 3    4 × 0.5  4l  = 1650 Hz

I0 = ( I +

Ans.

13. This problem is a Doppler’s effect analogy. (a) Here,

f = 20 min −1

v = 300 m/min vs = 0 and v0 = 0 300 Spacing between the pies = = 15 m Ans. 20 and f ′ = f = 20 min −1

I )2 = 4 I (I 1 = I 2 = I , say )

∴ I = I 0/4 When one source is switched off, no interference will be obtained. Intensity will be due to a single source or I 0/4. (b) At θ = 60° , also maximum intensity or I 0 will be observed.

16. (a) Frequency of second harmonic in pipe A = frequency of third harmonic in pipe B v  v  ∴ 2 A =3 B   2lA   4 lB 

394 — Waves and Thermodynamics γ ARTA MA

(b) Velocity of sound in air is

or

vA 3 = vB 4

or

γA γB

∴

M A γ A  16  5/3  16 =  =    M B γ B  9   7/5  9 

or

M A  25  16 400 =    = M B  21  9  189

⇒

γ B RTB MB

=

MB 3 = MA 4

Air speed va = 5 m/s

3 4 Source

(as TA = TB )

va =

=

Substituting

γA MB . γB MA

25 189 3 × = 21 400 4

B = ρ

 344 − 5  = 105   Hz  344 − 5 − 10  f2 = 1.0304 × 105 Hz

v = 200 Hz 4l Decreasing the tension in the string decrease the beat frequency. Hence, the first overtone frequency of the string should be 208 Hz (not 192 Hz) 1 T 208 = ∴ l µ f1 =

Ans.

2.088 × 109 = 1445 m/s 103

∴

Frequency of sound in water will be v 1445 Hz f0 = w = λ w 14.45 × 10−3

T = µ ⋅ l 2 (208)  2.5 × 10−3   (0.25)2 (208)2 =  0.25 

f0 = 105 Hz

= 27.04 N

(a) Frequency of sound detected by receiver (observer) at rest would be source

Ans.

19. (a) Wavelength of sound ahead of the source is λ′ =

observer(At rest)

v − vs 332 − 32 = = 0.3 m f 1000

Ans.

 v + v0   332 + 64  (b) f ′ = f   = 1000   v − v   332 − 32  s

vs = 10m/s

vr = 2 m/s

= 1320 Hz

 v w + vr  f1 = f0    v w + v r − vs 

Ans.

(c) Speed of reflected wave will remain 332 m/s. Ans. (d) Wavelength of reflected wave. v − v0 332 − 64 λ′′ = = f′ 1320

 1445 + 2  = 105   Hz  1445 + 2 − 10 f1 = 1.0069 × 105 Hz

Ans.

18. Frequency of fundamental mode of closed pipe,

17. Velocity of sound in water is vw =

(1.4)(8.31)(20 + 273) 28.8 × 10−3

∴ Frequency of sound detected by receiver (observer) in air would be  va − vw  f2 = f0    v a − vw − vs 

(as TA = TB )

M B 189 from part (a), we get = M A 400 fA = fB

γRT = M

= 344 m/s Frequency does not depend on the medium. Therefore, frequency in air is also f0 = 105 Hz.

(b) Ratio of fundamental frequency in pipe A and in pipe B is fA vA/2lA vA (as lA = lB ) = = fB vB /2lB vB γ ARTA MA = γ B RTB MB

Observer at (rest)

vs = 10 m/s

Ans.

= 0.2 m

Ans.

20. Thermometry, Thermal Expansion and Kinetic Theory of Gases INTRODUCTORY EXERCISE

20.1

T − 0 TF − 32 1. (a) C = 100 180 Putting TF = 0, we get TC = − 17.8° C T − 0 TF − 32 TK − 273 TF − 32 (b) C or = = 100 180 100 180 Putting TK = 0, we get TF = − 459.67° F TC − 0 TF − 32 = 100 180 Putting TF = 2TC , we get TC = 160° C T (b) Putting TF = C , we get TC = − 24.6° C 2 TC 100 3. = 52 − 5 99 − 5

2. (a)

2. Density of water will increase by increasing the temperature from 0 to 4°C, then it will decrease. Fraction of volume immersed is given by ρ fi = s ρl ρ l is first increased the decreased first decreased the increased. Or percentage of volume above water level will first increased then decreased. ρ ρ′ 3. fi = s ⇒ fi = s ρl ρl ′ fi′  ρ′s   ρ l   1 + γ l ∆T   1 + γ 2 ∆T  =   = =  fi  ρ s   ρ′l   1 + γ s ∆T   1 + γ 1 ∆T  1 Now, ≈ (1 − γ 1 ∆T ) 1 + γ 1 ∆T fi ′ = (1 + γ 2∆T ) (1 − γ 1∆T ) ∴ fi

∴

∴ TC = 50° C

Neglecting γ 1 γ 2 (∆T )2 term, we get fi′ ≈ (γ 2 − γ 1 ) ∆T fi

Correct reading (°C) 100 TC 0

5

52 99

Wrong reading (°C)

T − 0 TF − 32 Now, using the equation, C = 100 180 Putting TC = 50° C , we get TF = 122° F T − 0 TF − 32 TK − 273 TF − 32 4. C or = = 100 180 100 180 Putting TF = TK , We get, TF or TK = 574.25 TC − 0 TF − 32 5. = 100 180 Putting TC = TF , we get TC or TF = − 40° C or − 40° F

INTRODUCTORY EXERCISE

20.2

1 2 Temperature is decreased. Hence, l or T will decrease. So, it will gain time. ∆T ∆T Time gained = (as T ′ ≈ T ) t≈ t T′ T 1 1 = α∆θt = (1.2 × 10− 5 ) (30) (24 × 60 × 60) 2 2 = 15.55 s

1. ∆T = T α∆θ

4. Steel Brass

Since, α B > α Fe On cooling brass will contract more. 5. ∆d = d (α B − α Fe ) ∆θ ∆d ∴ ∆θ = d (α B − α Fe ) 0.01 = = 20.83° C 60 × 0.8 × 10− 5 ∴ New temperature = (30 + 20.83) = 50.83° C

6. (a) ∆l = l α∆θ = (88.42) (2.4 × 10− 5 ) (35 − 5) = 0.064 cm (b) At higher temperature, it measures less. Hence, l = l′ + ∆l = (88.42) + (0.064 ) = 88.484 cm ∆l − l α ∆θ 7. × 100 = × 100 l l = − (α∆θ ) × 100 = − (1.2 × 10− 5 ) (35) × 100 = − 0.042%

396 — Waves and Thermodynamics INTRODUCTORY EXERCISE

1.

20.3

 m pV = nRT =   RT M ∴

 VM  T =  p  mR 

7 1 = 28 4 11 1 = n2 = number of moles of CO2 = 44 4 n M + n2M 2 M = 1 1 n1 + n2

3. n1 = number of moles of nitrogen =

 1  1   (28) +   (44 )  4  4 = = 36  1  1   +   4  4 RM ρ= RT (1.01× 105 ) (36 × 10− 3 ) = (8.31) (290) = 1.5 kg/m 3

pV = nRT ∴

temperature T is given by  f  U = n  RT  2  f = degrees of freedom = 5 for O2 and 3 for Ar Hence, U = U O 2 + U Ar 5  3  = 2  RT  + 4  RT  2  2  where,

= 11RT

2. Average translational kinetic energy of an ideal 3 kT which depends on 2 temperature only. Therefore, if temperature is same, translational kinetic energy of O2 and N2 both will be equal. f 3 K K 3. T = = T = T fR 2 K R K O 3 ∴ KT = K O 2 gas molecule is

INTRODUCTORY EXERCISE

pV (ρgh) V n= = RT RT (13.6 × 103 ) (9.8) (10− 6 )(250 × 10− 6 ) = 8.31 × 300

1. vrms =

2. vav

= 1.33 × 10− 8 = (1.33 × 10 ) (6.02 × 10 ) = 8 × 10

15

3RT = M

20.5

3 × 8.31 × 373.15 2 × 10− 3

= 2156 m/s = 2.15 km/s 500 + 600 + 700 + 800 + 900 = = 700 m/s 5

∴ Number of molecules = (n) N A −8

20.4

1. Internal energy of n moles of an ideal gas at

= 12 atm

4.

Slope of p2 is more. Hence, p2 < p1. 1 6. p = (nRT ) V 1 i.e. p versus graph is a straight line passing V through origin of slope nRT . 7. pV and T both are constants.

INTRODUCTORY EXERCISE

 n  T  p2 =  2   2  p1  n1   T1   1  87 + 273 =    (20 atm )  2  27 + 273

Now,

nRT  nR  =  T p  p

For given mass, V - T graph is a straight line passing through origin having nR 1 slope = ∝ p p

At constant volume T - p graph is a straight line of VM 1 slope, = ∝ mR m Slope of m1 is less. Hence, m1 is greater. nRT or p ∝ nT 2. p = V p2 n2T2 ∴ = p1 n1T1 or

5. V =

(500)2 + (600)2 + (700)2

23

vrms =

+ (800)2 + (900)2 5

= 714 m/s

Chapter 20

Thermometry, Thermal Expansion and Kinetic Theory of Gases — 397 f 3 kT = kT 2 2  3 =   (1.38 × 10− 23 ) (300)  2

4. Helium gas is monoatomic. So, its degree of freedom f = 3. Average kinetic energy of 1 molecule of gas f 3 = kT = kT 2 2 3 = (1.38 × 10− 23 ) (300) 2 = 6.21 × 10−21 J

5. (a) vrms =

= 6.21 × 10− 21 J 3RT M 2 mv rms (4 × 10− 3 ) (103 )2 T = = 3R 3 × 8.31

vrms =

6. ∴

3RT M

For He gas : vrms =

=

= 160 K

3 × 8.31× 300 4 × 10− 3

7. Suppose n1 molecules have v1 velocity and n2 molecules have v2 velocity. Then, n v + n2 v2 vav = 1 1 n2 + n2

= 1368 m/s 3 × 8.31 × 300 For Ne gas : vrms = 20.2 × 10− 3 = 609 m/s (b) Each gas is monoatomic for which degree of freedom f = 3. Hence, average kinetic energy of one atom

vrms =

But,

n1 v12 + n2 v22 n1 + n2

Now, vrms ≥ vav because v1 and v2 may be in opposite directions also.

Exercises LEVEL 1

Objective Questions

Assertion and Reason

(1)2 + (0)2 + (2)2 + (3)4 4 = 3.5 m/s

2. vrms =

1. Straight line passing through origin represents isochoric process. 3. In monoatomic gas, there is no rotational degree of freedom. 4. Temperature of the gas is associated with random or disordered motion of the gas molecules. By the motion of container total kinetic energy of gas molecules will increase. But temperature will not increase. 5. Internal energy is equally distributed in all degrees of freedom. For example KT 3 = in diatomic gas KR 2 As

fT = 3 and fR = 2

7. Density of water is maximum at 4°C. Hence, volume is minimum at 4°C. 8. U ∝ T ∝ pV

9. p =

1  mN  2   vrms 3 V  mN Total mass of gas = = density of gas V Volume of gas

3. T1 = 273 + 27 = 300 K T2 = 273 + 927 = 1200 K vrms ∝ T T has become four times. Therefore, vrms will become two times. 4. Hydrogen and oxygen both are diatomic gases. Therefore, average kinetic energy per molecule is f kT 2 5 or (as f = 5) kT 2 5. E ∝ T T2 E2 2E = = =2 ∴ T1 E1 E ∴

T2 = 2T1 = 2 (273 + 10) = 566 K = 293° C

7. If p = constant, then V ∝ T

398 — Waves and Thermodynamics 8. It means rod is compressed from its natural length by ∆l.

6. The temperature on the platinum scale is

= (− αdθ ) = − (12 × 10− 6 ) (50)

Rt − R0 × 100° R100 − R0 86 − 80 = × 100° C 90 − 80

= − 6 × 10− 4

= 60° C

t=

∆l l∆α∆θ ∴ Strain = − =− l l

7. Let the relation between the thermometer reading

9. ∆l1 = ∆l2 ∴

l1 α 1 ∆θ = l2 α 2 ∆θ l1 α 2 9 or = = l2 α 1 11 19 In option (a), ratio is . 11 nRT mR (t + 273) 10. V = = p Mp m ratio is same. Therefore, slope and intercept of p straight line between V and t should also remain same.

and centigrade be y = ax + b Given, at x = 100, y = 80 and at x = 0, y = 10 ∴ 80 = 100 a + b, 10 = b ⇒ a = 0.7 Now, we have to find x when y = 59 ∴ 59 = 0.7x + b ⇒ x = 70 Ans. ∴ The answer is 70°C. 8. If T be the corresponding temperature, 3RT 3R (300) = MO MH M  T = (300)  O   MH 

∴

Subjective Questions

= 4800 K

1. For Q.Nos. 1 and 2 Apply the formula, TC − 0 TK − 273 TF − 32 TR − 0 = = = 180 80 100 100 3. In the above equation, putting TF = TC , We get TF = − 40° F or TC = − 40° C T − 32 4. Initially TC =  F  (100)  180   68.2 − 32 =  (100)  180  = 2011 . °C TC = 20.11 + 40 = 60.11° C

Finally

 180 TF = 32 +  T  100 C

∴

 180 = 32 +   (60.11)  100 0°C 100° TC Thermometer

0° 20°

32° 80°

TC 100 or = 32 − 20 80 − 20

TC = 20° C

Ans.

9. Mass of 6.02 × 10 molecules is 17 g or 0.017 kg. 23

∴ Mass of one NH3 molecule 0.017 = kg 6.02 × 1023 = 2.82 × 10− 26 kg n1 + n2 n1 n2 10. = + γ −1 γ1 − 1 γ2 − 1 3+ 2 3 2 = + ∴ γ − 1 1.67 − 1 1.4 − 1 5 = 9.48 γ −1 Solving, we get γ = 1.53 γp 11. v = ρ ∴

= 140.2° F

5.

Ans.

ρv 2 1.3 × (330)2 = = 1.4 p 1.013 × 105 2 γ =1+ ( f = degree of freedom ) f 2 1.4 = 1 + f γ=

Solving, we get f =5

Chapter 20

Thermometry, Thermal Expansion and Kinetic Theory of Gases — 399

12. 4 g hydrogen = 2 moles hydrogen

f has become

1 11.2 litre He at STP = mole of He 2 RT p = pH + pHe = (nH + nHe ) V 1 8.31 × (300 K)  = 2 +   2 (20 × 10−3 )m3 = 3.12 × 105 N/m 2

∴

Ans.

∴

V ∝T Vi V f = Ti T f Tf =

Vf Vi

3 times and V has become 3 times. 2 5 3 × 3 or 7.5 times. ∴ U will become × 3 2 3RT T 17. vrms = ∝ M M vrms is same. Hence, T T =     MH MO become

13. Since, pressure is constant. ∴

2

or

TH 2

Ti

T f = 2Ti = 600K f ∴ ∆U = nR∆T 2 3 = R (600 − 300) = 450R 2 14. From energy conservation principle, Ei = E f or E1 + E2 = E

n1T1 + n2T2 n1 + n2

18. (i) ∴

Ans.

1 15. Number of gram moles = = n 18 Avogadro number, N A = 6.02 × 1023 / g-mol ∴ Total number of molecules, N = nN A = 3.34 × 1022 ∴ Number of molecules per cm 2 N (R → in cm) = 4 πR 2 3.34 × 1022 = = 6500 (4 π ) (6400 × 105 )2

16. pV = nRT ∴

∴ or

= 20 K = − 253° C 3RT = 11.2 × 103 m/s M (11.2 × 103 )2 M T = 3R (11.2 × 103 )2 × 2 × 10− 3 = 3 × 8.31

Ans.

3  3  3  ∴ n1  RT1 + n2  RT2 = (n1 + n2 )  RT  2  2  2  T =

2

 MH2   TO =   2  MO2   2 =   (273 + 47)  32

⇒

∴

5 time (3 for He and 5 for O2). p has 3

nRT = pV f U = (nRT ) ( f = degree of freedom ) 2 f U = ( pV ) 2 U ∝ fpV

= 10059 K (ii) Escape velocity from moon = 2gm Rm = 1.6 × 2 × 1750 × 103 = 2366 m/s Substituting in Eq. (i), we have (2366)2 × 2 × 10− 3 T = 3 × 8.31 = 449 K 19. At constant volume, p∝T p1 p2 = T1 T2

∴

p T2 =  2  T1  p1 

∴

 160 =  (273) = 546 K  80   6.5 − 2.5  Rt − T0   × 100 =   × 100  R100 − R0   3.5 − 2.5

20. t = 

= 400°

…(i)

400 — Waves and Thermodynamics 26. n1 = n2

21. At constant volume, p∝T p2 T2 = p1 T1

∴

p or T2 =  2  T1  p1 

 45 + 75 =  (273 + 30)  5 + 75  = 450 K = 450 − 273 = 177° C

∆l = (α∆θ ) l Stress = Y × strain = γα∆θ ∴ Tension = ( A ) (stress) = YAα∆θ Substituting the value, we get T = YAα∆θ = (2.0 × 1011 ) (2 × 10− 6 ) (12 . × 10− 5 ) (40)

22. Strain =

= 192 N 23. Change in weight = upthrust on 100% volume ∆W ′ F ′  1 + γ s ∆θ  ∴ = =  ∆W F  1 + γ l ∆θ  (50 − 45.1) 1 + (12 × 10− 6 ) (75) = (50 − 45) 1 + γ l (75)

or

Solving this equation, we get γ l = 3.1 × 10− 4 per ° C

24. (a) n =

−2

pV (1.52 × 10 ) (10 ) = RT (8.31) (29815 . ) 6

= 6.135 pM (1.52 × 106 ) (2 × 10−3 ) (b) ρ = = RT 8.31 × 298.15 = 1.24 kg/m 3 (c) ρ ∝ M ∴ ∴

(if P and T are same) ρO 2 ρH 2

=

MO2 MH2

 MO2   ρH ρ O 2 =   2  MH2   32 =   (1.23)  2 = 19.68 kg/m 3

25. If T = constant, then ∴

p1V1 = p2V2 V   70 p2 = p1  1  = (1 atm ) =    6  V2  = 11.7 atm

∴ ∴

p1V1 p2V2 = RT1 RT2  p  T   1   270 V2 =  1   2  V1 =     (500)  0.5  300  p2   T1  = 900 m 3

27. p = constant V ∝T h∝T hf h = i T f Ti

∴ or ∴

 Tf  hf =   hi  Ti 

or

 273 + 100 =  (4 ) = 5.09 cm  273 + 20  0.025 28. n1 = = 8.93 × 10− 4 28 0.04 n2 = = 0.01 4 n1 ∴ = 0.089 n2 In equilibrium p and T are same. ∴ pV = nRT ∴ V ∝ n or L ∝ n L1 n1 = = 0.089 ∴ L2 n2

29. n = n1 + n2 pV pV pV = 1 1+ 2 2 RT RT RT p1 V1 + p2 V2 ∴ p= V (0.11) (138 . ) + (0.16) (0.69) = 011 . + 0.16 = 0.97 MPa

30. (n1 + n2 )1 = (n1 + n2 ) f ∴ ∴

piV1 piV2 pV1 pV2 + = + RTi RTi RT1 RT2 P (V1 + V2 ) P= Ti (V1 / T1 + V2 /T2 ) (1 atm ) (600) =  400 200 (293)  +   373 273 = 1.13 atm

Chapter 20 31. V =

Thermometry, Thermal Expansion and Kinetic Theory of Gases — 401

nRT (1) (8.31) (27315 . ) = p 1.013 × 105

36. (a) Average translational kinetic energy of any type

= 0.0224 m = 22.4 L 3

32. n1 = n2 ∴

V  T  p2 =  1   2  p1  V2   T1   0.75  273 + 157 =  (1.5 × 105 )    0.48  273 + 27 

p1V1 p2V2 = RT1 RT2

or

= 3.36 × 105 Pa 1.4 = 0.05 28 30% of the gas dissociates into atoms. ∴ n1′ = 2 (0.3) (0.05) + (0.7) (0.05) = 0.065 0.4 = 0.1 n2 = number of moles of He = 4 ∴ n = n1′ + n2 = 0.165 nRT (0.165) (8.31) (1500) Now, P = = V (5 × 10− 3 )

33. n1 = number of moles of N 2 =

= 4.1 × 10 N/m 5

of molecules is 3 3 kT = (1.38 × 10− 23 ) (300) 2 2 = 6.21 × 10− 21 J 2 (b) energy is associated with rotation. Hence, it 5 is diatomic gas. 5  Q = nCV ∆T = (1)  R (1) 2  5 = × 8.31 = 20.8 J 3 n1C p1 + n2C p 2 37. Cp = n1 + n2 2.5R + 3.5R = = 3R 1+ 1 n1 CV1 + n2 CV2 CV = n1 + n2 1.5R + 2.5R = 2R 2 Cp 3 γ= = = 1.5 CV 2 =

2

34. In diatomic gas rotational degree of freedom is two. ∴ Rotational KE of one molecule 1 f I ω 2 = kT = kT 2 2 2kT ∴ ω= I =

38. pV − b = constant (as f = 2)

2 × 1.38 × 10− 23 × 300 8.28 × 10− 38 × 10− 9

= 1013 rad /s

35. (a) C p = (1 + f / 2) R 29 = (1 + f / 2) 8.31 Solving, we get f =5 (b) pT = constant ∴ p ( pV ) = constant or pV 1/ 2 = constant C in the process pV x = constant is given by R f R C = CV + = R+ 1− x 2 1− x ∴

 f 1  29 = 8.31  +   2 1 − 1/ 2

Solving, we get

f =3

C =

R R + γ − 1 1 − (− b)

C = 0 , if γ − 1 = − (1 + b) ∴ b=−γ 39. pV − 1 = constant R C = CV + ∴ 1− x x= − 1

Here ∴

C = CV +

2K 3k pV 3kpV Now, n = = RT 2KR

40. K =

3 kT 2

R 2

=

⇒ T =

3 (200 × 103 ) (0.1 × 10− 3 ) (1.38 × 10− 23 ) 2 × 6 × 10− 21 × 8.31

= 8.3 × 10− 3 mol Now, N = nN A

= (8.3 × 10− 3 ) (6.02 × 1023 )

= 5 × 1021

402 — Waves and Thermodynamics 41. vrms =

3RT = M

3 × 8.31 × (273 + 57) 46 × 10− 3

= 423 m/s vrms

(c) vrms =

vrms

= 484 m/s

∆p = 2 mvrms

45. Let mass of nitrogen = (m) g. Then, mass of

Mass of one gas molecule, 46 m= g 6.02 × 1023 = 7.64 × 10−26 kg Let n molecules strike per second per unit area. Then, F ∆P / ∆t Pressure = = = 2 mn vrms A A Pressure 2 × 1.013 × 105 ∴ n= = (2m) vrms 2 × 7.64 × 10− 26 × 423 = 3.1 × 1027 3 3  pV  kT = k   2 2  nR  3 pV = k 2 (M / mN A ) R 3kPV m N A = 2MR

42. KT =

(3 × 138 . × 10− 23 ) (100 × 103 ) =

(2 × 10− 6 ) (8.0 × 10− 23 ) (6.02 × 1023 ) 2 × (50 × 10− 3 ) (8.31)

= 4.8 × 10−22 J  T′   573   v0 =   v0 v′0 =   T  293 

= 1.40 v0 (b) vrms does not depend on pressure, as long as temperature remains constant. (c) vrms =

3RT M

or

vrms ∝

3 kT 2 3 = (1.38 × 10− 23 ) (300) 2 = 6.21 × 10− 21 J

44. (a) KT =

oxygen = (100 − m) g. Number of moles of m nitrogen, n1 = and number of moles of oxygen 28  100 − m n2 =    32  For air ρ=

pM p  m1 + m2  =   RT RT  n1 + n2 

 1.01 × 105  100 × 10− 3   8.31 × 273  (m / 28) + (100 − m) / 32  Solving this equation, we get m = 76.5 g This is also percentage of N2 by mass on air as total mass we have taken is 100 g. 46. n1 = n2 p1 V1 p2 V2 ∴ = RT1 RT2 pV T ( p + ρgh) V1 T2 ∴ V2 = 1 1 2 = 0 p2 T1 p0 T1 ∴

1.284 =

=

(1.01 × 105 + 103 × 10 × 40) (20) (273 + 20) 1.01 × 105 (273 + 4 )

≈ 105 cm 3

43. (a) vrms ∝ T ∴

3 RT 2 3 = × 8.31 × 300 2 = 3740 J 3RT 3 × 8.31 × 300 = M 32 × 10− 3

(b) KT in one mole =

1 M

Q n∆T Q and heat capacity C p = ∆T ∴ C p = nC P Similarly, Cv = nCV Now, C p − Cv = n (C P − CV ) − nR C p − Cv 29.1 ∴ n= = = 3.5 R 8.31 5  (b) C p = nC P = n  R 2 

47. (a) Molar heat capacity C p =

= (3.5) (2.5) (8.31) = 72.75 J/ K

Chapter 20

Thermometry, Thermal Expansion and Kinetic Theory of Gases — 403

3  Cv = nCV = n  R 2  = (3.5) (1.5) (8.31) = 43.65 J/K 7  (c) C p = nC P = n  R 2  = (3.5) (3.5) (8.31) = 101.8 J/ K 5  Cv = nCV = n  R 2  = (3.5) (2.5) (8.31) = 72.75 J/K 48. (a) As discussed in the above problem, Cv = nCV C Cv ∴ n= v = CV (3/ 2)R 35 = = 2.81 1.5 × 8.31 (b) Internal energy,  f  U = n  RT  2  f = degree of freedom = 3  3 U = (2.81)   (8.31) (273)  2

∴

= 9562 J = 9.56 kJ 5  5 (c) C P = R =   (8.31)  2 2 = 20.8 J/mol-K

LEVEL 2

3. T = mV + C

or

∴ p - V graph is as shown in graph (a).

4. p2V = constant 2

 nRT  ∴   V = constant  V  ∴

p1V1 p2V2 p (V1 + V2 ) ∴ + = RT1 RT2 RT T1 T2 p (V1 + V2 ) Solving, we get T = p1 V1 T2 + p2 V2 T1

2. ∆l = lα∆θ ∴

α=

∆l 0.008 = = 8 × 10− 6 per °C l ∆θ (10) (100)

γ = 3α = 2.4 × 10− 5 per ° C ∆V = Vγ ∆θ = (100) (2.4 × 10− 5 ) (100) ∴

= 0.24 cc V ′ = V + ∆V = 100.24 cc

T2 ∝V T ∝ V

or

V is made three times. So, T will become 3 times. pV 5. n = RT 1 1 or m ∝ ∴ n∝ T T m1 T2 273 + 37 = = m2 T1 273 + 7 310 = = 1.1 280 8RT 6. vav = πM T and M are same. Hence, average speed of O2 molecules in both vessels will remain same or v1. pV 7. n = RT pVN A ∴ N = nN A = RT (1013 . × 105 × 10− 13 ) (10− 6 ) (6.02 × 1023 ) = (8.31) (300)

Single Correct Option 1. (n1 + n2 )i = (n1 + n2 ) f

pV = mV + C nR C  p = nR  m +   V

∴

= 2.4 × 106

8. vrms =

3RT = M

3 × 8.31 × 273 28 × 10− 3

= 493 m/s Now, increase in gravitation PE = decrease in kinetic energy (mg h) 1 2gh 2 2 or ∴ = m vrms = vrms h 2 1 + (h/ R ) 1+ R 2 × 9.81 × h or = (493)2 1 + (h / 6400 × 103 ) Solving this equation, we get h ≈ 12000 m or 12 km

404 — Waves and Thermodynamics 9. U ∝ T According to given graph, p∝V ∴ p∝T Hence, V = constant 10. 28 g of N2 mean 1 g-mol Now, n =

pV RT

 pf  =  ni  pi 

nf

or

n∝

p T

 Ti   1  273 + 57 11  =    T   2  273 + 27 = 20  f

11  11 or nf =   ni = mol  20 20 11 9 ∆n = ni − nf = 1 − = g - mol 20 20 9 63 ∆m = × 28 g = g 20 5

11. 4 g H2 means 2 g-moles and 8 g He means 2 g-moles. Now, M =

n1M 1 + n2 M 2 (2) (2) + (2) (4 ) = n1 + n2 2+ 2

= 3 g/mol PM (1.013 × 105 ) (3 × 10− 3 ) ρ= = = 0.13 kg/m 3 RT (8.31) (273) 1 12. ρ = V Given, ∴ or ∴

p ∝ Kρ 1 ρ ∝ V pV = constant T = constant

More than One Correct Options 1. p ∝ ρ as per process. If ρ becomes half, then p will becomes 2 times. Further, ∴

ρ∝

1 V

p2 = constant (1/V )

( pV ) p = constant (as pV ∝ T ) Tp = constant 1 ∴ p∝ T or P - T graph is a rectangular hyperbola. or Hence,

1 V If volume becomes 4 times than p will remain half T V ∝ p

2. p ∝

∴

Vp2 = constant

∴

T  2   p = constant  P

pT = constant 1 or p∝ T i.e. p - T graph is a rectangular hyperbola. If p is halved than T will becomes two times. 1 4. ρ ∝ V V is doubled so ρ will remain half. U ∝ T ∝ pV According to given graph, p ∝ V ∴ U ∝ T ∝ p2 or V 2 or

V is doubled, soU and T both will become four times. p∝V T T  ∝V ∴  as p ∝   V V ∴

T ∝V2

or T - V graph is a parabola passing through origin. nRT mRT  mR  6. V = = = T p Mp  Mp i.e. V - T graph is a straight line passing through origin of slope  mR  m =  ⇒ Slope ∝ Mp p   Hence, slope depends on both m and p. nRT (m /V )RT 7. (a) p = = V M So, given m in the question is mass of gas per unit volume. (b) m = total mass of gas (c) m = mass of one molecule of gas.

Match the Columns n fT (2)(3) RT RT = = 3RT 2 2 n f RT (2)(2) RT (b) K R = R = = 2RT 2 2

1. (a) KT =

Chapter 20

Thermometry, Thermal Expansion and Kinetic Theory of Gases — 405

(c) Potential energy = 0 (d) U = K R + KT = 5 RT 1 2. U ∝ T and ρ ∝ V According to given graph, 1 U ∝ρ ⇒ T ∝ V 1 1 or p ∝ 2 ∴ pV ∝ V V Density is increasing. So, volume is decreasing. Hence, pressure will increase. U is increasing. So, temperature will also increase. T 1 Now, ∝ 2 V V T Volume is decreasing. Hence, will increase. V γ RT 3. Speed of sound = M

4. (a) Density of water is maximum at 4°C. From 0°C to 4° C, density will first increase, then decrease. (b) and (c) : Same logic as given in option (a). l (d) T = 2π or T ∝ l g As temperature is increased, length will increase. So, time period will also increase. 5. (a) p = constant. Hence p - T graph is as shown below.

i.e. T - V graph is a straight line passing through origin. pM (d) ρ = RT 1 If p = constant, then ρ ∝ T i.e. ρ -T graph is rectangular hyperbola.

Subjective Questions 1. pV = nRT pdV = nRdT dV nR = , dV = VγdT dT P dV / dT nR 1 γ= = = ∴ V PV T R R x , 2. In the process pV = constant C = + γ −1 1− x Given, C = R and γ = 7/5 5 Substituting these values, we get x = 3 1 5/ 3 Now, pV = constant or p ∝ (V )5/ 3 By increasing volume to two times pressure will decrease (2)5/ 3 times. vrms ∝ T

vrms ∝

or vrms will become

pV

(2) times (2)5/ 3

vrms will become (2)−1/ 3 times 1 times. or (2)1/ 3 or

p

Now, T

pM RT If p = constant, then 1 1 or ρ ∝ ρ∝ T U

(b) ρ =

(as U ∝ T )

U

P

(c) If p = constant, then T ∝V

or

p ∝ (no. of collisions) (vrms ) 1 1 ∝ (no. of collisions) (2)5/ 3 (2)1/ 3

or number of collisions will decrease (2)4/ 3 times. Ans. 3. p1V = n1RT p2V = n2RT

(T = 273 K)  m1 − m2  ∴ ( p1 − p2 )V = (n1 − n2 )RT =   RT  M  ∆m …(i) (∆p)V = RT M In the initial condition (at STP) RT p …(ii) = 0 M ρ (T = 273 K)

406 — Waves and Thermodynamics ∴ ρ 0 = ρ 50 (1 + 50γ ) Substituting in Eq. (i), we have % of fraction submerged  1 + 0.3 × 10−5   1 + 50γ s   × 100 =  × 100 =  −5  1 + 50γ l   1 + 8 × 10 

From Eqs. (i) and (ii), we get ρV∆p 1.2 × 0.4 × 0.24 ∆m = = P0 1.0 = 0.1152 kg = 115.2 g

Ans.

4. p2 = pressure at depth x p1

= 99.99 % p2

x

ρ 100gh100 = ρ 0gh0

7.

x ∆h

p1V1 = p2V2 p1 ( Ah) = p2 ( A )(h − ∆h) p (h − ∆h) …(i) ∴ p1 = 2 h Initially, mg = ρgxA mg Now, p2 = p0 + ρgx = p0 + A Substituting in Eq. (i), we have mg   ∆h  Ans. p1 =  p0 +  1 −   A  h RT 5. p = (n = 1) V R or …(i) p = (T0 + αV 2 ) V For minimum attainable pressure dp − RT0 = 0 or + αR = 0 dV V2 T or V = 0 α d2p At this volume we can see that is positive or p dV 2 is minimum. From Eq. (i) RT0 pmin = + αR T0/α T0/α or

= 2R α T0

Ans.

(ρ )50 =

ρ0 1 + 50γ

 1  39.2   = 40  1 + 100γ  Solving this equation we get γ = 2.0 × 10−4 per °C

Ans.

8. Let ∆l be the change in length. (let ∆ls > ∆l > ∆la ) Strain in steel =

∆ls − ∆l l0

and strain in aluminium = In equilibrium,

∆l − ∆la l0

2Fs = Fa

 ∆l − ∆l   ∆l − ∆la  2 s Ys A =    Ya A  l0   l0  or 2(l0α sθ − ∆l )Ys = (∆l − l0α aθ )Ya Solving this equation, we get  2α Y + α aYa  ∆l =  s s  l0θ  2Ys + Ya  ∴ Total length   α Y + 2α sYs   = l0 + ∆l = l0 1 +  a a  θ  2Ys + Ya   

Ans.

9. From ∆l = lα∆θ we have, 0.05 = 25α A (100) ∴

α A = 0.00002 per ° C 0.04 = 40 α B (100) α B = 0.00001 per ° C

In third case let l is the length of rod A. Then length of rod B will be (50 − l ) cm.

6. At 50°C, density of solid = density of liquid. ρ  At 0°C, fraction submerged  s  × 100  ρ l  0° C

Ans. ρ 100 h or = 0 ρ0 h100

…(i)

∆l = ∆l1 + ∆l2 or 0.03 = l (0.00002) (50) + (50 − l ) (0.00001) (50) Solving we get, l = 10 cm and 50 − l = 40 cm Ans.

21. Laws of Thermodynamics INTRODUCTORY EXERCISE

6. VT = constant

21.1

V ( pV ) = constant pV 2 = constant

(as T ∝ pV )

1. ∆U will be same along all paths. Then, we can

or ∴

2. ∆U = Q − W

Comparing with pV x = constant, we have

apply Q = W + ∆U for all. = 254 − (−73) = 327 J

INTRODUCTORY EXERCISE

21.2

1. (a) Work done by gas under constant pressure is W = p (V f + Vi ) = (1.7 × 105 )(0.8 − 1.2) = − 6.8 × 104 J (b)

Q = W + ∆U

x=2 nR∆T W = 1− x (2)(R )(300) = 1− 2 = − 600 R 7. pV 2 = K = p0V02 K p= 2 ∴ V W = ∫ pdV = ∫

= − 6.8 × 104 − 1.1 × 105 or | Q | = 1.78 × 105 J

2V

W 2 = − ve as cycle is anti-clockwise Since, |W 1 | > |W 2 | ∴ Net work done by the system is positive. (b) In cyclic process, Qnet = W net Since, W net is positive. So, Qnet is also positive. So, heat flows into the system. (c) W 1 = Q1 = + ve , so into the system. W 2 = Q2 = − ve , so out of the system. 3. Helium is monoatomic. So, degree of freedom f = 3. nf 3n U = RT = RT 2 2 2U 2 × 100 or n= = 3RT 3 × 8.31 × 300

4. In isochoric process,

Ans.

K K − V0 2V0 K = 2V0 =

Substituting, K = p0V02 , we have 1 W = p0 V0 2

INTRODUCTORY EXERCISE

= 1800 R 5. W = Area under p-V diagram Further p ∝ V along AB. Therefore, pB = 2 pA = 2 p0

21.3

1. (a) Q = nCV ∆T 3  200 = (1)  R (T f − 300) 2  or 200 = 1.5 × 8.31(T f − 300) Solving we get, T f = 316 K (b) Q = nC p ∆T

W =0 3  Q = ∆U = nCV ∆T = 4  R (300) 2 

K dV V2

 K 0 = −  V  V0

As Q is negative. So, net heat flow is out of the gas. 2. (a) W 1 = + ve, as cycle is clockwise

= 2.67 × 10−2 mol

2V0

V0

(Table 21.3)

5  200 = (1)  R (T f − 300) 2  or 200 = 2.5 × 8.31 (T f − 300) Solving we get, T f = 309.6 K 2. p = constant ∴

T ∝V

408 — Waves and Thermodynamics As the gas expands V increases. So, T also increases. Hence, ∆T is positive. Therefore, in the expression, Q = nCp ∆T Q is positive. 3. AB V is increasing, so W = + ve . Product of pV and therefore T and therefore U is increasing. So, ∆U is also + ve. Q = W + ∆U ∴ Q is also positive. BC V = constant ∴ ∴ CA

W =0 Q = ∆U = − ve

U is a state function. Hence, ∆U depends only on the initial and final positions. Therefore, ∆U 1 = ∆U 2 But, W1 > W2 as the area under 1 is greater than area under 2. Hence, Q1 > Q2

7. W = ∫

2

1

Solving we get, W = 1.18 × 10 6 J

8. W = nR ∆T

Therefore, C = Cp . ∆Q nCp ∆T Cp Now, = = =γ ∆U nCV ∆T CV ∆Q ∆Q and = ∆W ∆Q – ∆U nCp ∆T = nCp ∆T – nCV ∆T

=

C p – CV

=

INTRODUCTORY EXERCISE

1. Given,

as, ∴

21.4

Q1 = 106 cal T1 = (827 + 273) = 1100 K T2 = (27 + 273) = 300 K Q2 T2 (in Carnot engine) = Q1 T1 Q2 =

T2  300  6 ⋅ Q1 =   (10 )  1100 T1

= 2.72 × 105 cal

C p /CV

Ans.

Efficiency of the cycle,  T η = 1 − 2  × 100  T1 

C p /CV – 1

or

6. Q1 = W 1 + ∆U 1 and Q2 = W 2 + ∆U 2

300   η = 1 −  × 100  1100 = 72.72%

p

Ans.

2. T1 = 2100 K, T2 = 700 K, ηactual = 40 %

1 A

(in isobaric process)

= (1)(8.31)(72) = 600 J = 0.6 kJ ∆U = Q − W = 1.0 kJ C p n C p ∆T Q γ= = = CV n CV ∆T ∆U 1.6 = = 1.6 1.0

and

γ γ –1

p

1

= 1.18 MJ

Wgas = 0 as on the right hand side there is vacuum. ∴ ∆U = Q − W = 0 ∴ ∆T is also zero, as U ∝T 5. In an isobaric process p = constant.

=

2

pdV = ∫ (αV 2 ) dV = ∫ (5 × 1.01 × 105 ) V 2dV

V is decreasing. ∴ W = − ve ∆U is also negative. ∴ Q = W + ∆U is also negative. 4. Q = 0 as the box is insulated.

Cp

Vf

Vi

B

A

W1

2

Maximum possible efficiency of Carnot engine is T 700 ηmax = 1 − 2 = 1 − T1 2100

B

W2 V

V

= 0.666 = 66.6%

Chapter 21 Actual efficiency as the percentage of the maximum possible efficiency is ηactual 40 × 100 = × 100 ηmax 66.6 = 60% T2 375 3. (a) η = 1 − = 1 − = 0.25 = 25% T1 500

Ans.

βA > βB Therefore, refrigerator A is better. 7. Here, T1 = 25 + 273 = 298 K T2 = − 10 + 273 = 263 K Q2 = 263 J/s Coefficient of performance is given by Q2 T2 β= = Q1 − Q2 T1 − T2 Q1 T1 = ∴ Q2 T2 T 298 ∴ Q1 = 1 × Q2 = × 263 T2 263

= 6.25 × 105 J (c) Q2 = Q1 − W = (25 − 6.25) × 105 = 18.75 × 105 J 4. T1 = 627 + 273 = 900 K T2 = 27 + 273 = 300 K Q1 = 3 × 106 cal

= 298 Js−1

 T  Q  η = 1 − 2  = 1 − 2   T1   Q1 

∴

Average power consumed, = 298 − 263 = 35 J s−1 = 35 W W 8. η = net × 100 ΣQ+ ve

Q2 T2 = Q1 T1 T Q2 = 2 × Q1 T1

W net = Area under the cycle pV = 0 0 2 ΣQ+ve = QABC = W ABC + ∆U ABC = (Area under ABC) + nCV ∆T 5  3  =  p0V0 + n  R (TC − TA ) 4  2 

3 × 106 × 300 = 106 cal 900 Work done by the engine, W = Q1 − Q2 = 3 × 106 − 106 cal =

5. η1 = 1 −

T2 T1

or

= 2 × 106 cal 1 T = 1− 2 6 T1

...(i)

p

In the second case, the temperature of the sink is reduced by 65°C. Hence, T − 65 η2 = 1 − 2 T1 1 T2 − 65 ...(ii) or = 1− 3 T1 Solving Eqs. (i) and (ii), we get T1 = 390 K = 117 ° C, T2 = 325 K = 52 ° C 6. Coefficient of performance is given by T2 β= T1 − T2 (−10 + 273) βA = (27 + 273) − (−10 + 273) = 7.1

(−27 + 273) = 5.6 (17 + 273) − (−27 + 273)

βB =

(b) W = ηQ1 = 0.25 × 25 × 105

∴

Laws of Thermodynamics — 409

3p 0 2

B A

p0 p0 2

D

V0 2

V0

3V 0 2

V

5 3 p0V0 + ( pC VC − pAVA ) 4 2 5 3  3p V pV  = p0V0 +  0 0 − 0 0  4 2 2 2  11 = p0V0 4 ( p0V0 / 2) η= × 100 = 18.18 % (11p0V0 / 4 ) =

∴

C

Exercises Vf   1  = nRT ln    2  Vi 

LEVEL 1

2. Wgas = nRT ln 

Assertion and Reason 1. Q = 0, ∆U = − W In expansion, W is positive. So, ∆U is negative. Hence, T also decreases.

2.

p

p

V

V (ii)

(i)

In both figures, Vi = V f but W 1 = Hatched area W2 = 0 3. First law is just energy conservation law, which can be applied for any system. 5. Q − W = ∆U is state function. So, it will remain constant for any two thermodynamic states. 6. For given temperatures T1 and T2 efficiency of a heat engine can’t be greater than efficiency of Carnot engine. 7. pT = constant ∴

T    T = constant V 

or

V ∝T2

If temperature is increased, then volume will also increase. So, work done is positive. 8. Q = 0 as chamber is adiabatic W = 0 as expansion is taking place in vacuum. ∆U = Q − W = 0 ∴ ∴ U or T = constant or pV = constant 1 ∴ p∝ V

Objective Questions 1. p∝V2 T ∝ V or T ∝ V 3 ⇒ V ∝ T 1/ 3 V T is increasing. So, V will also increase. Hence, work done will be positive.

∴

= − nRT ln 2 ∴ Work done on the gas = + nRT ln 2 3. ∆U is same p. Q = W + ∆U W 1 = + ve,W 2 = 0 and W 3 = − ve 4. In adiabatic compression temperature increases. Hence, pV increases, as T ∝ pV 5. W = − π (Radius)2 = − π (Radius) (Radius)  p − p1   V2 − V1  =−π 2     2   2 

6.

p 1 2 3 V

1 → Isobaric 2 → Isothermal 3 → Adiabatic Minimum area is under graph-3. 7. W 1 > W 2 as area under graph-a is more. Hence, ∆Q1 > ∆Q2 , as ∆Q = W + ∆U and ∆U 1 = ∆U 2  T 8. η = 1 − 2  × 100  T1  T2 = 300 K and T1 = 600 K η = 50% n RT 9. p = V V = constant, but temperature will decrease with time. So, p will decrease. 10. In adiabatic compression, internal energy of gas increases. So, temperature increases. ∴

γ

γ

11. pT 1 − γ = constant or p ∝ T γ − 1 ∴

α=

γ 1.4 = = 3.5 γ − 1 1.4 − 1

Chapter 21

4. Qnet = W net = area under the cycle

R 1− x 5 R 5 R = R+ = R− 2 1− x 2 x −1

12. C = CV +

= (2 × 105 )(2.5 × 10−6 ) = 0.5 J

5. (a) W I = + ve (as cycle is clockwise)

C is negative if 2 5 ∴ x < 1.4 If x lies between 1 and 1.4, then also C is negative. n C + n2CV2 13. CV = 1 V1 n1 + n2 x −1
|W II | ∴ W net = + ve (c) Qnet = W net = + ve (d) In loop-I, QI = W I = + ve In loop-II, QII = W II = − ve 6. Work done per cycle = area under the cycle 1 = (3 − 2)(10−3 )(30 − 10)(1.01 × 105 ) 2 = 1010 J Total work done per second 100 = 1010 × 60 = 1683 J/s 3  7. (a) Q = nCV ∆T = (1)  R (T f − Ti ) 2 

∴ 200 = 2.5 × 8.31 (T f − 300) ∴ T f ≈ 310 K 8. V = constant

= 2.67 × 10−2 mol ∆U = nCV ∆T nR∆T = γ −1

Laws of Thermodynamics — 411

R ) γ −1

pV = nRT , V∆p = nR (∆T ) Q = nCV ∆T 5 5  = n  R ∆T = (V∆p) 2  2 5 −3 = × (10 × 10 )(4 × 105 ) 2 = 104 J

9. (a) W = + (Area of the cycle) 1 (2V0 − V0 )(3 p0 − p0 ) 2 = p0V0 (b) QCA = nC p ∆T =+

Ans.

412 — Waves and Thermodynamics 5  = n  R (TA − TC ) 2  5 ( pAVA − pC VC ) 2 5 5 = ( p0V0 − 2 p0V0 ) = − p0V0 2 2 QAB = n CV ∆T 3  = n  R (TB − TA ) 2  3 = ( pBVB − pAVA ) 2 = 3 p0V0 (c) W BC = + (Area under the graph) = 2 p0V0 ∆U BC = n CV ∆T 3  = n  R (TC − TB ) 2  =

3 3 = ( pCVC − pBVB ) = − p0V0 2 2 p0V0 ∴ QBC = W BC + ∆U BC = 2 (d) p - V equation along path BC is  2p  p = −  0  V + 5 p0  V0  or

 2p  pV = −  0  V 2 + 5 p0V  V0 

or

 2p  RT = −  0  V 2 + 5 p0V  V0 

 2 p0 2  …(i) 5 p0V − V ⋅ V    0 For T to be maximum, dT =0 dV 4p 5 p0 − 0 V = 0 ∴ V0 5 ∴ V = V0 4 So at this volume, temperature is maximum. Substituting this value of V in Eq. (i), we get 25 p0V0 Tmax = 8R 10. First process V = constant ∴ p∝T Pressure becomes three times. ∴

T =

1 R

Therefore, temperature also becomes three times. (T f = 3Ti ) Second process p = constant ∴ V ∝T Volume is doubled. So, temperature also becomes two times. (T f ′ = 6Ti ) Q Q + Q2 Now, C = = 1 n ∆T n ∆T nCV ∆T1 + n C p ∆T2 = n ∆T 5  7   R (3Ti − Ti ) +  R (6Ti − 3Ti ) 2  2  = (6Ti − Ti ) = 3.1 R

11. First Process V = constant ∴ T ∝p Pressure becomes half. So, temperature also becomes half. Q1 = n1CV ∆T = 2CV (150 − 300) = − 300 CV Second Process p = constant Q2 = nCp ∆T = 2 C p (300 − 150) = 300 C p Q = Q1 + Q2 = 300(Cp − CV ) = 300 R = 300 × 8.31 = 2493 J 3 12. ∆U = Q − W = nCV ∆T = n  R (T f − Ti ) 2  3  ∴ 1200 − 2100 = 5  × 8.31 (T f − 400) 2  ∴

T f = 385 K ≈ 113 ° C

m 2 = = 2.0002 × 10−3 m 3 ρ i 999.9 m 2 Vf = = = 2 × 10−3 m 3 ρ f 1000

13. Vi =

∆U = Q − W = ms ∆θ − pa ∆V

= (2)(4200)(4 ) − (10)5 (−0.002 × 10−3 )

= 33600.2 J

Chapter 21 14.

m 0.01 = = 10−5 m 3 ρ i 1000 m 0.01 Vf = = = 0.0167 m 3 ρf 0.6

Laws of Thermodynamics — 413

18. (a) W AC is less than W ABC as area under graph is

Vi =

less. p

A 5 m3

= 27530 J 15. (a) W = p0 ∆V = (1.013 × 10 )(1671 − 1) × 10 = 169 J (b) Q = mL = (10−3 )(2.256 × 106 ) = 2256 J ∴ ∆U = Q − W = 2087 J 16. (a) W = p∆V = (2.30 × 105 )(1.2 − 1.7) = − 1.15 × 105 J (b) Q = W + ∆U = (−1.15 × 105 ) + (−1.40 × 105 ) = − 2.55 × 10 J

17. (a) and (b) ( pV )A = ( pV )C = 3 p0V0 TA = TC or ∆T = 0 ∆U ABC = 0 QABC = W ABC = Area under the graph Ans. = − 2 p0V0

i.e. Heat is released during the process. (c) Process AB ρ - p graph will be a straight line parallel to ρ - axis because pressure is constant. Further, V is decreasing. Therefore, ρ will increase. Process BC V = constant ⇒ ρ = constant Therefore, ρ- p graph is a straight line parallel to p - axis. ρ - p graph is as shown below. ρ

ρ0

B

V

∴

∴

U B = 20 J ∆U = Q − W AB UB −UA = Q − 0 20 − 10 = Q Q = 10 J

∴ (QAB + QBC ) − (W AB + W BC ) = QAC − W AC ∴ (250 + QBC ) − (500 + 0) = 300 − 700 Solving, we get QBC = − 150 J (b) QCDA = (WCD + W DA ) + (U A − UC )

20.

= (−800 + 0) − (UC − U A ) = − 800 − (300 − 700) = − 400 J m 1.0 V = = ρ 8.92 × 103 = 1.12 × 10−4 m 3

C

∆V = Vγ ∆θ = (1.12 × 10−4 )(2.1 × 10−5 )(30)

A 2p0

p

Ans.

19. (a)UC − U A along two paths should be same.

γ = 3α = 2.1 × 10−5 per ° C

p0

Ans.

(c) From A to B,

5

2ρ0

15 m3

(b) For process A to C, Q = 200 J Work done W AC = area under AC 1 = (8 + 4 ) × 10 2 = 60 J From first law of thermodynamics, ∆U = Q − W AC UC − U A = 200 − 60 ∴ UC = U A + 140 = 10 + 140 = 150 J

−6

5

C

4 Pa

− 105 (0.0167 − 10−5 )

∴ ∴

B

8 Pa

∆U = Q − W = ms∆θ + mL − p0∆V = (0.01 × 4200 × 100) + (0.01)(2.5 × 10)6

= 7.056 × 10−8 m 3

414 — Waves and Thermodynamics W = p∆V

23. Along the process CD apply, −8

= (1.01 × 10 )(7.056 × 10 ) 5

= 7.13 × 10−3 J Q = mc ∆Q = (1)(387)(30) ∆U = Q − W = 11609.99287 J pV (2 × 105 )(1.0 × 10−2 ) 21. (a) T = = nR (1)(8.31)

V  + nRTC ln  D  + pD (VA − VD )  VC 

TV γ − 1 = constant

∴

∴ T f V fγ − 1 = TiViγ − 1 V  T f = Ti  i  Vf 

γ−1

 1.0 × 10−2  = (240.7)  −3   5.0 × 10  ≈ 383 K (c) Work done on the gas = ∆U

5/ 3− 1

p∆V = nR∆T Work done under constant pressure is W = p∆V = nR∆T = (0.2)(8.31)(100) = 166.2 J

LEVEL 2 Single Correct Option 2V

1. W = ∫ pdV V

=∫

= nCV ∆T 3  = (1)  R (T f − Ti ) 2  = 1.5 × 8.31(383 − 240.7) ≈ 1770 J (d) ∆U ≈ 1770 J 22. (a) From conservation of linear momentum, 0.01 × 200 = (2 + 0.01) v ∴ v ≈ 1 m/s Mechanical energy dissipated in collision, = Ki − K f 1 1 = × (0.01)(200)2 − (2.01)(1)2 2 2 ≈ 199 J m 2000 + 10 (b) n = = = 10.05 M 200 Using, Q = nCV ∆T Q 199 ∆T = = ∴ nCV 10.05 × 3 × 8.31 = 0.80° C

nRTA = pAVA nRTC = pC VC

Further, and 24. pV = nRT

= 240.7 K (b) In adiabatic process,

∴

pV = constant and find VC Similarly along path AB, again apply, pV = constant and find VB . Now, W net = W AB + W BC + WCD + W DA V  = nRTA ln  B  + pB (VC − VB )  VA 

2V 

V

n RT    dV  V − b

 2V − b = nRT ln    V−b  Put n = 1,  2V − b W = RT ln    V−b 

∴

2. For AB

V = constant

∴

W =0

For BC

Vf  W = n RT2 ln    Vi  V  = RT2 ln  2   V1 

For CA

V ∝T

∴

p = constant W = nR∆T = R (T1 − T2 ) 3. Q Q = n Cp ∆T 7 7  RT0 = (10)  R (T f − T0 ) 2  2 T f = 1.1 T0

Chapter 21 If p = constant, then V ∝ T If temperature becomes 1.1 times, then volume will also become 1.1 times. 4. W net = area under the cycle

9. T ∝ U TC = TD = T0 = 300 K TA = TB = 2T0 = 600 K Qnet = QAB + QBC + QCD + QDA Vf  = nRTA ln   + nCV (TC − TB )  Vi 

= 2 p0V0 Q+ve = QABC = W ABC + ∆U ABC = area under the graph + n CV ∆T 3  = (3 p0V0 ) + n  R (TC − TA ) 2  = (3 p0V0 ) +

Vf  + nRTC ln   + nCV (TA − TD )  Vi  = (1)(R )(600) ln (2) + (1)(CV )(−300)  1 + (1)(R )(300)ln   + (1)(CV )(300)  2

3 ( pCVC − pAVA ) 2

= 10.5 p0V0 W 4 η = net = Q+ve 21 = area under the graph + ∆U 3p V = − p0V0 − 0 0 2 5 = − p0V0 2 = constant

In the process, pV x = constant R C = CV + 1− x 3 R Here, C = R + = 2R 2 1+ 1

∴

p = constant Qnet = W net = W 12 + W 23 + W 31

∴

W 23 = Qnet − W 12

11. QA = QB

3

∴

2 Vf

V

V = constant ∴ T ∝p If p is decreasing, then T will also decrease. 8. TV γ − 1 = constant T1  V2  =  T2  V1  L  =  2  L1 

γ−1

5/ 3 − 1

n− 1

= constant

pV = constant

 R  W =  ∆T  1 − x x = −1 R R ∴ W = ∆T = (T2 − T1 ) 2 2 14. ∆U = Q − W = 450 J = nCV ∆T Here,

3  = n  R ∆T 2  L  =  2  L1 

(as T ∝ pV )

n

In the process, pV x = constant

In process 3 − 1

∴

( pV )V

Bulk modulus in the process pV γ = constant is γp. Hence, in the given process Bulk modulus is np. 13. pV −1 = constant

1

Vi

nCp ∆TA = nCV ∆TB Cp ∆TB = ⋅ ∆TA = γ∆TA CV = (1.4 )(30) = 42 K TV n − 1 = constant

12. ∴

p

(as W 31 = 0)

= − 300 − nR∆T = − 300 − 2 × 8.31 × 300 = − 5286 J

∴

Q = n C∆T = (1)(2R )(2T0 − T0 ) = 2RT0

7.

= 300 R ln 2 T ∝V

10. 1-2

5. Q = W + ∆U

6. pV −1

Laws of Thermodynamics — 415

2/ 3

∴

 300 n ∆T =    R 

416 — Waves and Thermodynamics Q n ∆T 600 = = 2R 300/R

C =

Now,

20. In second process, p ∝ V i.e. p-V graph is straight line passing through origin. p

15. W = p∆V = p0 (2V0 − V0 ) = p0V0 ∆U = U f − U i = 2 p0 (2V0 ) − 2 p0V0 = 2 p0V0 ∴ Q = W + ∆U = 3 p0V0 Q 16. C = n ∆T or C ∝Q ∆U is same. But W 1 < W 2 as area under p-V graph is less. Hence, Q1 < Q2 C 1 < C 2. C1 W1 Further, T ∝ pV ( pV )2 > ( pV )1 ∴ T2 > T1 21. dQ = − dU or dU + dW = − dU or ∴ or

17. Qnet = W net QAB = QCD = 0

5  QBC = n Cp ∆T = (1)  R (4T − T ) 2 

= 7.5RT QDA = n Cp ∆T 5  = (1)  R (3T − 5T ) 2  = − 5RT W net = Qnet = 7.5RT − 5RT = 2. 5RT 18. Uρ = constant ∴

∴ or ∴

 1 T   = constant V  T ∝V p = constant ∆U ∆U = W Q − ∆U n CV ∆T = n C p ∆T − n CV ∆T

1 1 3 = = γ −1 5 −1 2 3 19. W = Area under F -x graph =

= 17.5 J Now, Q = W + ∆U = 20 J

2dU + dW = 0 2 [ n CV dT ] + pdV = 0   R   2 n   dT  + pdV = 0   γ − 1 

or

2nRdT  nRT  +  dV = 0  V  γ −1

or

 2  dT dV + =0    γ − 1 T V

Integrating we get,  2    ln (T ) + ln (V ) = ln (C )  γ − 1 Solving we get TV

γ−1 2

= constant

22. In a cyclic process, Qnet = W net ∴ QAB + QBC + QCA = area under the graph 1 ∴ 600 + 200 + QCA = (3 × 10−4 )(5 × 105 ) 2 = 75 J ∴ QCA = − 725 J = WCA + ∆UCA = − 725 = (− Area under the graph ) + ∆UCA 1  = −  × 11 × 105 (3 × 10−4 ) + ∆UCA 2  Solving we get, ∆UCA = − 560 J (lying on same isotherm ) 23. T2 = T3 ∴

∆U 1 = ∆U 2 W = Hatched area 1 = ( pi + pf )(V f − Vi ) 2

Chapter 21 =

1 ( piV f − piVi + pf V f − pf Vi ) 2 p

26. Pressure becomes half. So, temperature is doubled.  1  as T ∝   p

∆U = n CV ∆T pV Ti = 0 0 nR 2 p0V0 Tf = nR

pf pi Vi

Vf

V

Now, pi Vi and pf V f are same for both processes. Further pi and Vi , area also same for both processes. For 1 → 3 V f is more and pf is less. Hence, W will be more. Now, Q = W + ∆U W 13 > W 12 (as ∆U is same) ∴ Q13 > Q12 24. ∆U net = 0 ∴ ∴

∆U 1 + ∆U 2 + ∆U 3 = 0 Q1 + (Q2 − W 2 ) + ∆U 3 = 0 (200 kJ) + (−100 + 50) kJ + ∆U 3 = 0 ∴ ∆U 3 = − 150 kJ In adiabatic process, W = − ∆U ∴ W 3 = − ∆U 3 = 150 kJ 25. QBC = 0 QCD = n Cp ∆T 5  = n  R (TD − TC ) 2  5 = ( pDVD − pC VC ) 2 5 = (105 − 2 × 105 ) 2 = − 25 × 104 J QDA = n CV ∆T 3  = n  R (TA − TD ) 2  3 = ( pAVA − pDVD ) 2 3 = (2.4 × 105 − 105 ) 2 = 21 × 104 J Now,

Laws of Thermodynamics — 417

W net = Qnet = (9 − 25 + 21) × 104 J = 5 × 104 J

p0V0 nR  3   p0V0  3 p0V0 ∴ ∆U = n  R  =  2   nR  2  pi  27. W = nRT0 ln    pf  ∴

∴

∆T = T f − Ti =

W BC = 2 W AB  pC   p / 2 nRT0 ln   = 2nRT0 ln  0   p0 / 2  p0  Solving this equation we get, p pC = 0 8

More than One Correct Options 1. p p

A

p 2

B V

V 2V

Now, see the hint of Q-No 9 (d) of subjective questions for Level 1. 2. Temperature is increased. So, internal energy will also increase. ∆U = + ve ∴ Further, pV 2 = constant ∴

T  2   V = constant or V 

V ∝

1 T

Temperature is increased. So, volume will decrease and work done will be negative. In the process pV x = constant, molar heat capacity is given by R C = CV + 1− x Here, x = 2 ∴ C = CV − R

418 — Waves and Thermodynamics CV of any gas is greater than R. So, C is positive. Hence, from the equation, Q = nC∆T Q is positive if T is increased. or ∆T is positive. 3. ab (as V = constant) W =0 Q1 = ∆U 1 = nCV ∆T 3  = (2)  R (2T0 − T0 ) = 3RT0 2  bc (as T = constant) ∆U = 0 Vf  Q2 = W 2 = nRT ln   ∴  Vi 

cd

da

= (2)(R )(2T0 )ln (2) = 4 RT0 ln (2) W =0 Q3 = ∆U 3 = nCV ∆T 3  = (2)  R (T0 − 2T0 ) 2  = − 3RT0 ∆U = 0 Vf  Q4 = W 4 = nRT ln    Vi   1 = (2)(R )(T0 ) ln    2

= − 2RT0 ln(2) Now in complete cycle, ∆U net = 0 Qnet = W net = 2RT0 ln(2) = + ve 4. ab ρ = constant ∴ ∴

V = constant W =0 Q = ∆U ∆U is positive, as U is increasing. Hence, Q is also positive. 1 bc ∝T ρ ∝U ⇒ ∴ V ρ is decreasing, so V is increasing. Hence, work done is positive. 1 Further, ∝ T (T ∝ pV ) V ∴ pV 2 = constant In the process, pV x = constant, Molar heat capacity is given by R C = CV + 1− x

Here, x = 2 ∴ C = CV − R For any of the gas, CV ≠ R. ∴ C ≠0 ∴ Q = nC∆T ≠ 0 as ∆U ≠ 0 and ∆T ≠ 0 ca ρ is increasing. Hence, V is decreasing. So, work done is negative.

5.

Q1 = nCV ∆T

6. 1-2

Q2 = nCp ∆T Q3 = 0 C p > CV Q2 > Q1 > Q3 V = constant

∴

∴ p∝T T is increasing. So, p will also increase. 2-3 V ∝T ∴ p = constant. V and T both are increasing. Process 4-1 is reverse of 2-3.

Comprehension Based Questions pV 2

1. Qnet = W net = area under the cycle = 2.

CV 1 5 = = Cp γ 3

3. In adiabatic process, γ −1

p1 − γT γ = constant

or T ∝ p

γ

γ−1 γ

∴

TB  pB  =  TA  pA 

∴

 2 TB = (1000)    3  2 = (1000)    3

5/ 3 − 1 5/ 3

0.4

= 850 K

4. In adiabatic process, W AB = − ∆U AB

(as Q = 0 )

= nCV (TA − TB ) 3  = (1) R (TA − TB ) 2   3 25 = (1)  ×  (1000 − 850) 2 3 = 1875 J

Chapter 21 5. W BC = 0 ∴ ∴

(as V = constant) T pC TC Q

(as V = constant) ∝p = pB / 2 = TB / 2 = 425 K = ∆U = nCV (∆T ) 3  = (1)  R (TC − TB ) 2  3 25 × × (425 − 850) 2 3 = 5312.5 J =

Match the Columns 1. (a) W = nR∆T = (2)(R )(T ) = 2RT (b) ∆U = nCV ∆T 3  = (2)  R (T ) = 3RT 2  (c) W = − ∆U = − 3RT (d) ∆U = 3RT (in all processes)

2. For both processes, (as V ∝ T ) p = constant ∴ C = C p for same gases. For W , ∆U and Q we will require number of moles also. or pV 1/ 2 = constant 3. p2V = constant x Comparing with pV = constant, we get 1 x= 2 ∴ Molar heat capacity, R C = CV + 1− x 3  =  R + 2R = 3.5 R 2  Q = nC∆T = n (3.5R ) (∆T ) = 3.5 nR∆T ∆U = nCV ∆T 3  = n  R ∆T = 1.5 nR∆T 2  ∴ W by the gas = Q − ∆U = 2nR∆T and work done on the gas = − 2nR∆T 4. (a) Q = nCp ∆T ∆U = nCV ∆T W = Q − ∆U = nR∆T W < ∆U as CV > R

Laws of Thermodynamics — 419

(b) W = 0 ∆U = − ve Q = ∆U = − ve

∴ (c) Q = 0 ∴ (d) ∆T = 0 ⇒

(as V is increasing) W = + ve ∆U = − W = − ve ∆U = 0

Vf  Q = W = nRT ln   = + ve  Vi 

and

5. (a) W ab = area under p-V graph 3 p0V0 = 1.5 p0V0 2 (b) ∆U ab = Qab − W ab 3p V = 6 p0V0 − 0 0 = 4.5PV 0 0 2 Q W + ∆U (c) C = = n∆ T n∆ T 6 p0V0 = 2R = pV  pV n  b b − a a  nR nR  ∆U (d) CV = n∆T 4.5p0V0 = = 1.5R  pbVb paVa  − n   nR nR  =

Subjective Questions 1. In a cyclic process, ∆U = 0 Qnet = W net QAB = n Cp ∆T 5  = (2)  R (400 − 300) 2  = 500R p QBC = n R TB ln  i   pf   2 = (2)(R )(400) ln    1 = 800R ln (2) QCD = nCp ∆T = − 500R p QDA = n RTD ln  i   pf 

420 — Waves and Thermodynamics  1 = (2)(R )(300) ln    2 = − 600R ln (2) ∴

Qnet = W net = (200R ) ln (2) = (200)(8.31)(0.693) ≈ 1153 J

pV (1.013 × 105 )(22.4) 2. TA = A A = nR (103 )(8.31) = 273 K Along path AB V = constant ∴ T ∝p p is doubled. So, T is also doubled. ∴ TB = 2TA = 546 K. Further, ∴ ∴

Substituting the values like above we get, 1.5 p0 M ∆U 2 = ρ0 pM W 3 = Q 2 − ∆U 2 = 0 ρ0 For Process 3-1 Density is constant. Hence, volume is constant. ∴ W3 = 0 ∴ Q3 = ∆U 3 = nCV ∆T 3  = (1)  R (T1 − T3 ) 2  =

=−

TB = TC pB VB = pC VC ( p )(V ) (2)(22.4 ) VC = B A = ( pA ) 1

1 ∝p V ∴ Process is isothermal. ∆U 1 = 0 ∴

p Q1 = W 1 = nRT ln  i   pf  pM PM  1 ) = 0 ln   (as n = 1and RT =  2 ρ0 ρ =−

p0 M ln (2) ρ0

For Process 2-3 Q2 = Qp = nCp ∆T 5  = (1) R (T3 − T2 ) 2  =

5  p3 M p2 M  −   2  ρ3 ρ2 

=

5  2 p0M 2 p0M  −   2  ρ0 2ρ 0 

= 2.5

p0 M ρ0

∆U 2 = nCV ∆T

1.5p0 M ρ0

(b) ΣQ− ve = | Q1 + Q3 | p M 3  = 0  + ln 2  ρ0  2  −p M  (+ p0 M /ρ 0 ) +  0 ln 2  ρ0  (2.5 p0 M /ρ 0 ) 2 = (1 − ln 2) 5

= 44.8 m 3

3. For Process 1 - 2 ρ ∝ p

3  p0 M 2 p0 M  −   2  ρ0 ρ0 

W net (c) η = = ΣQ+ ve

4. (a)

p c

1.5p0 p0

a

b

V0

1.5V0

V

W net = Qnet in a cycle. Since, Qnet is negative (flows out of the gas). Hence, W net should also be negative. Or, cycle should be anti-clockwise as shown in figure. From a to b p = constant ∴ V ∝T T has become 1.5 times. Therefore, V will also become 1.5 times. From c to a p∝V 1 times. Therefore, V has become 1.5 1 times. p will also become 1.5

Chapter 21 In a cycle | Qnet | = |W net | = Area of cycle 1 800 = (0.5 p0 )(0.5V0 ) ∴ 2 ∴ p0V0 = 6400 J W ca = − Area under the graph 1 = − (2.5 p0 )(0.5V0 ) 2 = − 0.625p0 V0 = − 0.625 × 6400 = − 4000 J 5. First process is isobaric. ∴ ∆Q1 = nCV ∆T + p∆V Second process is isochoric ∴ ∆Q2 = nCV ∆T mg   ∆Q1 − ∆Q2 = p∆V = p0 + [ Ax ]  A  = ( p0 A + mg )x

−4

= [10 × 60 × 10 5

(c) QTotal

7. (a)

(p0, V0)

(b) p0 =

40

113.1

V (L)

nRT0 2 × 8.31 × 300 = V0 20 × 10–3 = 2.5 × 105 N/m 2

A

p1 = p0 = 2.5 × 105 N/m 2 V1 = 2V = 40 × 10−3 m 3 Process AB :

V ∝T

∴ Process BC

T1 = 2T0 = 600 K

Using T1V1γ − 1 = T2V2γ − 1, we get

B C V

B T

and (b) For the process AB, p0V0 pB (2V0 ) = T0 T0 ∴

(p1, V1) B C (p , V ) 2 2

20

+ 8 × 10 ](0.2)

A

A

0.44

p

C

p (× 105 N/m2)

2.5

Ans.

p

3 RT0 2 21 = 3RT0 ln (2) − RT0 4

∴ WTotal = 3RT0 ln (2) −

= 136 J

6. (a)

Laws of Thermodynamics — 421

p0 2 ∆U = 0 Q = W + ∆U V = nRT ln B VA

= 0.44 × 105 N/m 2

Ans.

Ans.

(c) WTotal = W 1 + W 2 = p0 (V1 − V0 ) +

pB =

= 3RT0 ln 2 For the process BC, 2V0 V0 = T0 TC T ∴ TC = 0 2 3 T  W = nR∆T = 3R  0 − T0 = − RT0 2  2 21 Q = nCp ∆T = − RT0 4

V2 = 2 2V1 = 113.1 × 10−3 m 3 nRT2 (2)(8.31)(300) p2 = = V2 113.1 × 10−3

nR (T1 − T2 ) γ −1

Substituting the values, we get WTotal = 12479 J p1V1 p2V2 8. (a) = T1 T2  pV T  T2 =  2 2 1   p1V1 

Here, p1 = 1.0 × 105 N/m 2 V1 = 2.4 × 10−3 m 3

Ans.

…(i)

422 — Waves and Thermodynamics T1 = 300 K kx p2 = p1 + A = 1.0 × 105 +

∴

8000 × 0.1 8 × 10−3

W = Q − ∆U = n(C − CV )∆T W C − CV = ∆U CV  C − CV  W =  ∆U  CV   9/ 2 − 5/ 2 =  (100) = 80 J  5/ 2 

∴

= 2.0 × 105 N/m 2 V2 = V1 + Ax = 2.4 × 10−3 + 8 × 10−3 × 0.1 = 3.2 × 10–3 m 3 Substituting in Eq. (i), we get T2 = 800 K

10. dQ = dU + dW CdT = CV dT + pdV (CV + 3aT 2 ) dT = CV dT + pdV

Ans.

(b) Heat supplied by the heater, Q = W + ∆U Here, ∆U = nCV ∆T  p1V1   3  =   R (800 − 300)  RT1   2  =

(1.0 × 105 )(2.4 × 10–3 )(1.5)(500) (300)

= 600 J 1 W = kx 2 + p1∆V 2 1 = × (8000)(0.1)2 + (1.0 × 105 )(0.1)(8 × 10–3 ) 2 = (40 + 80) J = 120 J Ans. ∴ Q = 600 + 120 = 720 J

 RT  3aT 2 dT = pdV =   dV  V 

∴

dV  3a   TdT =  R V

∴

Integrating, we get  3aT 2    = ln V − ln C  2R  V = Ce

11. (a) p = αT

3aT 2 2R

TV −1/ 2 = constant

or

pV 1/ 2 = constant

Comparing with pV x = constant, we have 1 x= 2 ∴ Molar specific heat R R C = + γ −1 1− x R R = + (γ = 7/ 5) 7/ 5 − 1 1 − 1/ 2 5 9R Ans. = R + 2R = 2 2 Q = nC∆T ∆U = nCV ∆T

3aT 2 2R

= constant

pT −1/ 2 = constant

or

p1/ 2V −1/ 2 = constant pV −1 = constant ∴

x = –1  R  ∆W =   (∆T )  1 − x

As U ∝ T ∴ T ∝ V 1/ 2 or

−

or Ve

1/ 2

U ∝ V

9.

Ans.

 8.31 =  (50)  2  = 207.75 J R 3 R (b) C = CV + = R + = 2R 1− x 2 2

Ans. Ans.

12. F + pA = p0 A F = ( p0 − p) A

F

2V

W = ∫ ( p0 − p) A dx V 2V

2V

= ∫ p0 dV − ∫ pdV V

p0 A p

V

 dV  = p0V − ∫ RT   V  V  2V

= p0V − RT ln (2) = RT − RT ln (2) = RT (1 − ln 2)

Ans.

Chapter 21

Laws of Thermodynamics — 423 1  p0   + p0 (2V0 − V0 )  2 2 3 = − p0V0 4 3  = nCV ∆T = (2)  R (TB − TA ) 2  3    n = 2, CV = R  2   p0V0 p0V0  = 3R  −   2R 4R 

1 or pT = constant T ∴ p ( pV )) = constant or pV 1/ 2 = constant

∆W DA = −

In the process, pV x = constant,

∆U AB

13. (a) p ∝

Molar heat capacity is C = CV +

R 1− x

3 R R+ 1 2 1− 2 3 7 = R + 2R = R 2 2 (b) W = Q − ∆U = nC∆T − nCV ∆T = n(C − CV )∆T 3  7 = 2 R − R (T2 − T1 )  2 2  = 4 R (T2 − T1 ) a 14. V = T or VT = constant or V ( pV ) = constant ∴ pV 2 = constant or

C =

= Ans.

Ans.

 2 − γ R R + = R γ − 1 1 − 2  γ − 1

C =

Now,

Q = nC∆T  2 − γ = (1)   R∆T  γ − 1  2 − γ =  R∆T  γ − 1

15. Process AB is isochoric (V = constant). Hence, ∆W AB = 0 ∆W BCD = p0V0 +

π V  ( p0 )  0   2 2

π  =  + 1 p0V0 4 

3  ∆U BCD = nCV ∆T = (2)  R (TD − TB ) 2 

Hence, ∆QBCD = ∆U BCD + ∆W BCD  π 5 =  +  p0V0  4 2 ∆UDA = nCV ∆T 3  = (2)  R (TA − TD ) 2 

x=2

∴

pV   T =   nR 

pV  3  2p V = (3R )  0 0 − 0 0  = p0V0  2R 2R  2

In the process pV x = constant, molar heat capacity is R R C = + γ −1 1− x Here,

3 p0V0 = ∆QAB 4

Ans.

2p V   pV = (3R )  0 0 − 0 0   4R 2R  9 = − p0V0 4 ∴ ∆QDA = ∆U DA + ∆W DA 9 3 = − p0V0 − p0V0 4 4 = − 3 p0V0 Net work done is, 3 π W net =  + 1 −  p0V0 4 4 = 1.04 p0V0 and heat absorbed is Qab = ∆Q+ ve  3 π 5 =  + +  p0V0 = 4.03 p0V0  4 4 2 Hence, efficiency of the cycle is W η = net × 100 Qab 1.04 p0V0 = × 100 4.03 p0V0 = 25.8%

Ans.

424 — Waves and Thermodynamics 16. p =

αT − βT 2 V

(p = constant)

V =

Hence

αT − βT 2 p

 α − 2βT  dV =   dT  p 

or

W = ∫ pdV = ∫

T2

T1

 α − 2β T  p  dT  p 

W = [αT − βT 2 ]TT12

or

= α (T2 − T1 ) − β (T22 − T12 )

Ans.

17. (a) First law of thermodynamics for the given process from state 1 to state 2 Q12 − W 12 = U 2 − U 1 Here,

Q12 = + 10 p0V0 joule

W 12 = 0 (Volume remains constant) U 2 − U 1 = nCV (T2 − T1 ) nCV (T2 − T1 ) = 10 p0V0 For an ideal gas, p0V0 = nRT0 and ∴

∴

C p − CV = R CV = Cp − R 5R 3R = −R= 2 2

or p - V graph will be a rectangular hyperbola with increasing p and decreasing V . 1 ρ ∝ . Hence, ρ -V graph is also a rectangular V hyperbola with decreasing V and hence increasing ρ. pM   ρ∝ p ρ=  RT  Hence, ρ - p graph will be a straight line passing through origin, with increasing ρ and p. Process B -C is an isochoric process, because p - T graph is a straight line passing through origin i.e. V = constant Hence, p - V graph will be a straight line parallel to p - axis with increasing p. Since, V = constant hence ρ will also be constant Hence ρ -V graph will be a dot. ρ- p graph will be a straight line parallel to p -axis with increasing p, because ρ = constant Process C-D is inverse of A - B and D - A is inverse of B - C . Different values of p, V , T and ρ in tabular form are shown below p

V

T

ρ

A

p0

V0

T0

ρ0

 3R  n   (T2 − T0 ) = 10 nRT0  2

B

2 p0

V0 2

T0

2ρ0

23 T0 3 As p ∝ T for constant volume 23 p2 = p0 3

C

4 p0

V0 2

2T0

2ρ0

D

2 p0

V0

2T0

ρ0

T2 =

(b)

Here,

T  V0 = nR  0   P0 

and

ρ0 =

P0M RT0

The corresponding graphs are as follows

2

ρ

p

2ρ0

B,C

1 V

18. Process A -B is an isothermal process i.e. Hence,

T = constant 1 p∝ V

ρ0

A,D

V0 2

V0

V

Chapter 21

19. First process

p C

4p0

∴

2p0

D

B

T = constant 1 p∝ V

V is made 5 times. Therefore, p will become A

p0 V0 2

V0

V

ρ B

2ρ0 ρ0

Laws of Thermodynamics — 425

C

Second process V = constant ⇒ T ∝ p Pressure again becomes first times. So, temperature will also become 5 times. Q1 + Q2 = 80 × 103 Vf  ∴ nRT ln   + nCV ∆T = 80 × 103  Vi 

A

 8.31  ∴ (3)(8.31)(273)ln(5) + (3)   (5 × 273 − 273)  γ − 1

D

p0 2p0

1 th. 5

4p0

p

= 80 × 103 Solving we get, γ = 1.4

22. Calorimetry and Heat Transfer INTRODUCTORY EXERCISE

1.

2.

3.

4.

5.

22.1

Q = mLf + msw ∆θ + mLf = m (Lf + sw ∆θ + Lf ) = 10 [ 80 + 1 × 100 + 540 ] = 7200 cal Let θ (> 60° ) is the mixture temperature in equilibrium. Heat given = Heat taken 0 = m(1)(θ − 20) + m(0.5)(θ − 40) + m (0.25)(θ − 60) Solving the equation, we get θ = 31.43° C mL = ms (θ − 0° ) L 80 ∴ θ= = = 80° C s 1 75% heat is retained by bullet 3 1  mv 2 = ms∆θ + mL  4  2 (8s∆θ + 8L) or v= 3 Substituting the values, we have (8 × 0.03 × 4.2 × 300) + (8 × 6 × 4.2) v= 3 = 12.96 m/s Let m be the mass of the steam required to raise the temperature of 100 g of water from 24°C to 90°C. Heat lost by steam = Heat gained by water ∴ m (L + s∆θ 1 ) = 100 s∆θ 2 (100)(s)(∆θ 2 ) or m= L + s(∆θ 1 ) Here, s = specific heat of water = 1 cal/g-°C, L = latent heat of vaporization = 540 cal/g ∆θ 1 = (100 − 90) = 10° C and ∆θ 2 = (90 − 24 ) = 66° C Substituting the values, we have (100)(1)(66) m= = 12 g (540) + (1)(10)

∴ m = 12 g 6. Heat liberated by 10 g water (at 40° C) when it converts into water at 0° C

Q = m1 s ∆θ = 10 × 1 × 40 = 400 cal Mass of ice melted by this heat, Q 400 m2 = = = 5 g < 15 g L 80 Therefore, whole ice is not melted. Temperature of mixture is 0°C. Mass of water = m1 + m2 = 15 g Mass of ice = 15 − m2 = 10 g 7. Similar to Example 22.4

INTRODUCTORY EXERCISE

22.2

KA (θ 1 − θ 2 ) t l Ql J-m ∴ K = = A (θ 1 − θ 2 ) t m 2 K - s W (as J / s = W) = m-K l m 3. R= = KA (W / m -K) m 2 K = = KW −1 W 4. In convection, liquid is heated from the bottom. Q=

2.

5.

200°C K H1

∴

2K H2

θ1

θ2

3K H3 100°C

H1 = H2 = H3 200 − θ 1 θ 1 − θ 2 θ 2 − 100 = = (l / KA ) (l / 2KA ) (l / 3KA )

Solving these equations we get, θ 1 = 145.5° C and θ 2 = 118.2° C

6.

R/2 100°C

H1

R/2

θ H2

H3

R 25°C

H1 + H2 = H3 100 − θ 25 − θ θ − 0° + = (R / 2) R R/2

0°C

Chapter 22 Solving we get, θ = 45° C TD 45° C − 25° C HCD = = R 5 = 4W 7. Let us start with the assumption that   ∆θ ∝ Temperature difference ∆

Calorimetry and Heat Transfer — 427 ∴ and

 80 − 50  80 + 50  − 20   =α    5   2  60 − 30  60 + 30  − 20   =α    t   2

Solving these two equations, we get t = 9 min

Exercises LEVEL 1

5. λ m ∝

Assertion and Reason

4.

5. 6. 9.

10.

6.

7.

8.

Objective Questions 1. λ m ∝

1 T

(λ m )2 T1 2000 2 = = = (λ m )1 T2 3000 3 2 ∴ (λ m )2 = (λ m )1 3 2. At higher temperature radiation is more. So, cooling is fast. 3. Q1 = Q2 ∴

∴

ms1 (32 − 20) = ms2 (40 − 32) s1 8 2 = = s2 12 3

∴

2

4. Q = ∫ dQ = ∫ msdT 1

2

= ∫ (1)(aT 3 ) dT = 1

15a 4

(λ m )1 T2 T1 (λ m )2 or = = (λ m )2 T1 T2 (λ m )1 350 = = 0.69 510 dQ = CdT dT 1 ∴ = = Slope of T - Q graph. dQ C l l l Thermal resistance = = ∝ KA K (πR 2 ) R 2 l The rod for which 2 is minimum will conduct R maximum heat. Heat taken by 1 g ice in transformation from ice at 0°C to water at 100°C is Q = mL + ms∆θ = (1) (80) + (1) (1) (100) = 180 cal Mass of steam condensed to give this much heat is Q 180 1 m= = = g L 540 3 This is less than 1g or total mass of steam. Therefore, whole steam is not condensed and mixture temperature is 100°C. H A = HTotal (TD)A TD ∴ = RA RA + RB ∴

1. Specific heat is a function of temperature. But for most of the substances variation with temperature is almost negligible. R R1 = 2R and R2 = 2 1 Thermal resistance becomes th. Therefore heat 4 current becomes four times. If temperature of a normal body is more than a perfectly black body, it can radiate more energy. This law was given by Kirchhoff. In steady state, temperature of different sections (perpendicular to the direction of heat flow) becomes constant but not same. dT eAσ 4 dT 1 − = (T − T04 ) or − ∝ dt mc dt m

1 T

9.

∴

 RA  (TD)A =   TD  RA + RB    1 =  TD  1 + RB / RA 

...(i)

RB  lB   K A A   20 =    =   (3) = 6 RA  lA   K B A   10 

428 — Waves and Thermodynamics Substituting in Eq. (i), we get  1  (TD)A =   × 35 = 5° C  1 + 6

5. P = er σ A (T 4 − T04 ) = (0.4 )(5.67 × 10−8 )(4 π )(4 × 10−2 )2[(3000)4 − (300)4 ]

10. Thermal resistance, l 1 R= ∝ ⇒ KA = 2 KB KA K R ∴ RA = B 2 So, let RA = R then RB = 2R Now, HA = HB R  (TD)A (TD)B or (TD)A =  A  (TD)B ∴ =  RB  RA RB  R =   (36° C) = 18° C  2R 

11.

R=

 1  1 1 = (2)     =  2  2 2 Therefore, thermal resistance of 2nd rod is half. Hence, rate of heat flow will be twice.

Subjective Questions 1. Total rate of radiation of energy = er σ T 4 A = (0.6)(5.67 × 10−8 )(800 + 273)4 (0.1)2 (2) ≈ 900 W 1 1 2 2.  mv  = ms∆θ + mL  2 2 v = 2 s∆θ + L ∴ = 2 (125) (327 − 27) + 2.5 × 104 = 500 m/s

3. 0.4 [ mg ∆ h ] = ms ∆θ 0.4 g∆h 0.4 × 9.8 × 0.5 = s 800 −3 = 2.5 × 10 ° C

∆θ =

4. P = 

dm  s ⋅ ∆θ  dt 

∴

dm dQ TD θ − 100 = =H =  =  dt  dt R (l / KA )

6. L  ∴

 Ll   dm θ = 100 +      KA   dt   2.256 × 106 × 1.2 × 10−2   0.44  = 100 +     50.2 × 0.15    60 × 5 = 105° C

7. (a) H 1 = H 2 θ

l KA R2  l2   K 1 A1  =   ⋅  R1  l1   K 2 A2 

∴

= 3.7 × 104 W

dm P 500 × 106 = = dt s∆θ (4200) × 10 = 11904 kg/s = 1.2 × 104 kg/s

–10°C

H2

H1

R2

R1

Wood

19°C

Insulation

19 − θ θ + 10 = ∴ R1 R 19 − θ θ + 10 or = (l1 / K 1 A ) (l2 / K 2 A ) K 1 (19 − θ ) K 2 (θ + 10) = ∴ l1 l2 0.01 (19 − θ ) 0.08 (θ + 10) = ∴ 3.5 2.0 Solving, we get θ = − 8.1° C 19 − θ (b) H = H 1 = (l1 / K 1 A ) K 1 A (19 − θ ) = l1 (0.01) (19 + 8.1) (1) = 3.5 × 10−2

TD  H =   R

= 7.7 W / m 2

8. Hint is already given in the question 9. Let mixture is water at θ°C (where 0° C< θ < 40° C) Heat given by water = Heat taken by ice ∴ (200) (1) (40 − θ ) = (140) (0.53) (15) + (140) (80) + (140) (1) (θ − 0) Solving we get, θ = − 12.7° C

Chapter 22 Since, θ < 0° C and we have assumed the mixture to be water whose temperature can't be less than 0° C. Hence, mixture temperature θ = 0° C. Heat given by water in reaching upto 0° C is, θ = (200) (1) (40 − 0) = 8000 cal. Let m mass of ice melts by this heat, then 8000 = (140) (0.53) (15) + (m) (80) Solving we get m = 86 g ∴ Mass of water = 200 + 86 = 286 g Mass of ice = 140 − 86 = 54 g J . 10. Let heat is supplied at a constant rate of (α ) min In 4 minutes, (when temperature remains constant) ice will be melting. ∴ } mL = (4 ) α α L 336 × 103 ∴ = = m 4 4 = 84 × 103 J/min - kg Now in last two minutes, Q = ms ∆θ ∴ (α ) (2) = (m) (4200) (θ − 0° C) α we get, Substituting the value of m θ = 40° C

11.

T3

T2 R H2

R R/2

R/2

T

Calorimetry and Heat Transfer — 429

Solving this equation, we get θ ≈ 206° C Now, temperature gradient on θ − 100 LHS = = 212° C/m 0.5 θ−0 and on RHS = = 424 ° C/m 0.5 13. Net power = er σ (T 4 − T04 ) A In first case, 210 = er σ [(500)4 − (300)4 ] A

...(i)

In second case, 700 = σ [(500)4 − (300)4 ] A

...(ii)

Dividing Eq. (i) by Eq. (ii), we get 210 er = = 0.3 700

LEVEL 2 Single Correct Option 1. R =

1 times. 4 ∴ R will become half. Hence, heat current will become two times. Therefore, rate of melting of ice will also become two times or 0.2 g/s. l is halved, A is four times and K is

2. K1 T1

H3

d

R T1

∴

H1 + H2 = H3  T3 − T   T2 − T  T − T1   +  =  3R / 2   3R / 2  R

12.

T =

H1

∴

3T1 + 2 (T2 + T3 ) 7

θ

100°C

25 W

0°C H2

H 1 + H 2 = 25 ∴

T2

T3 K2

H1

∴

l KA

θ − 100 θ−0 + = 25 0.5/ (400 × 10−4 ) 0.5/ (400 × 10−4 )

3a

H1 = H2 T1 − T2 T − T3 = 2 (d / K 1 A ) (3d / K 2 A ) K 1  T2 − T3  1 = ⋅ K 2  T1 − T2  3

But T1 : T2 and T3 are in AP ∴ T2 − T3 = T1 − T2 K1 1 ∴ = K2 3 TD 3. H = R TD 100 − 0 R= = ∴ H 1 = 100 kW −1

430 — Waves and Thermodynamics Now, or

x

x

0

0K

R = ∫ dR = ∫ 100 = ∫

x 0

dx 0 (1 + ax ) A

More than One Correct Options 1. Total radiation per second is given by P = er σT 4 A

dx 102 (1 + x ) (10−4 )

Here, er : σ : T and A all are same. Hence, P will be same. Same is the case with absorption per second.  dT  Now, P = ms  −   dt 

Solving this equation we get, x = 1.7 m 4. Let R = thermal resistance of each rod. In first case, Rnet = R + R = 2R R In second case, Rnet = 2

(in series)

In second case thermal resistance has become

1 th. 4

So, heat current will become 4 times. So time taken to flow same amount of heat will be reduced 1 to th. 4

5.

P 1  dT  or Rate of cooling  −  = ∝  dt  ms m Mass of hollow sphere is less. So, its initial rate of cooling will be more.

(in parallel)

2.

40°C H1 θ

T A

H2 20°C

H2

B √2T

C

H1

H1 = H2 2T − TC T −T = C (l / KA ) ( 2l / KA )

∴

Solving we get, TC =

3T 2+1

6. Net power available per second, P = 1000 − 160 = 840 J/s Heat required, Q = ms∆θ = (2) (4200) (77 − 27) = 420000 J Q ∴ Time required = = 500 s = 8 min 20 s P 3X X 4X 7. Rnet = R1 + R2 =   + =  KA  2KA KA Now,

H =

(T2 − T1 )  KA (T2 − T1 )   1 =   3 (3 X / KA )  x 1 f = 3 =

∴

dQ TD = dt Rnet

H3 30°C

H1 = H2 + H3 40 − θ θ − 20 θ − 30 ∴ = + R R R where, R = thermal resistance of each rod Solving this equation, we get θ = 30° C Heat flows from higher temperature to lower temperature. 3. ms (2θ − θ′ ) = m (2s) (θ′ − θ ) Solving this equation we get, 4 θ′ = θ 3 Further, heat capacity C = ms or C ∝ s C 1 s1 1 ∴ = = C 2 s2 2 TD 4. q1 = R1 TD ∴ R1 = q1 TD Similarly, R2 = q2 In series, qs =

TD TD = R1 + R2 TD + TD q1 q2

(as m is same)

Chapter 22 =

q1 q2 q1 + q2

 1  TD In parallel, qp = = TD    Rnet  Rnet  1 1 = TD  +   R1 R2  q   q = TD  1 + 2  = q1 + q2  TD TD

5. (b) If temperature difference is small, the rate of cooling is proportional to TD.

Match the Columns 1. (a) E = σT 4 = energy radiated per unit surface are per unit time by a black body ∴

σ=

2 −2  E   ML T  = 2 4  4 T   L T θ 

= [ MT −3 θ −4 ] b = λ mT

(b)

∴ [ b ] = [ Lθ ] (c) Emissive power is energy radiated per unit time per unit surface area. ∴ (d) H = ∴

 ML2 T −2  −3 [E ] =  2  = [ MT ]  L T  d Q TD = dt R TD θ R= = (d Q / dt ) [ ML2T −2 /T ]

= [ M −1 L−2 T 3 θ ] dQ dθ 2. (a) = ms dt dt  d θ  dQ / dt ∴   =  dt  ms 1 ∴ Slope of θ - t graph ∝ m (b) QTotal = mL or

 dQ    (time) = mL  dt 

or time ∝ m or length of line bc ∝ m (d) ab : only solid state bc : solid + liquid state cd : only liquid state de : liquid + gaseous state ef : only gaseous state.

Calorimetry and Heat Transfer — 431 3. If R = thermal resistance of one rod c H/2 a 100°C H

H/2 d

b

e H –80°C

H/2

H/2 f

Then, Rnet = 3R 100 − (−80) 60 H = = ∴ 3R R 60 100 − θ b (a) ab H = = R R ∴ θ b = 40° C H 30 40 − θ c (b) bc = = 2 R R ∴ θC = 10° C (c) Same as above (d) de 60 θ d − (−80) H = = R R ∴ θ d = − 20° C 4. (a) m (s)(θ′ − θ ) = 2m (s) (2θ − θ′ ) Solving we get, 5 θ 3 (b) ms (θ′ − θ ) = 3m (s) (3θ − θ′ ) θ′ =

Solving we get, 5 θ 2 (c) 2m (s) (θ′ − 2θ ) = 3m (s) (3θ − θ′ ) θ′ =

Solving we get, θ′ =

13 θ 5

(d) ms (θ′ − θ ) + 2 ms (θ′ − 2θ ) + 3ms (θ′ − 3θ ) =0 Solving we get, θ′ =

7 θ 3

5. Unit of heat capacity is J / ° C. Unit of latent heat is J/kg.

Subjective Questions 1. (a) Between t = 1 min to t = 3 min, there is no rise in the temperature of substance. Therefore, solid melts in this time.

432 — Waves and Thermodynamics Q Ht 10 × 2 = = m m 0.5 = 40 kJ/kg Q (b) From Q = ms∆T or s = m∆T Specific heat in solid state 10 × 1 s= = 133 . kJ/kg-° C 0.5 × 15

4. Let heat capacity of flask is C and latent heat of

L=

2. Let T0 be the temperature of surrounding and T be the temperature of hot body at some instant. Then, dT − = K (T − T0 ) dt T t dT or ∫Tm T − T0 = − K ∫0 dT

fusion of ice is L. Then, C (70 − 40) + 200 × 1 × (70 − 40) = 50 L + 50 × 1 × (40 − 0) or …(i) 3C − 5L = − 400 Further, C (40 − 10) + 250 × 1 × (40 − 10) = 80L + 80 × 1 × (10 − 0) or …(ii) 3C − 8L = − 670 Solving Eq. (i) and (ii), we have Ans. L = 90 cal/g

5. ms∆θ = work done against friction = (µmg cos θ ) d (but µmg cos θ = mg sin θ ) (µg cos θ )d (g sin θ )d ∆θ = = s s  3 (10)  (0.6)  5 = 420

(Tm = temperature at t = 0) Solving this equation, we get (i) T = T0 + (Tm − T0 ) e−Kt Maximum temperature it can lose is (Tm − T0 ) From Eq. (i), T − T0 = (Tm − T0 )e− Kt

= 8.57 × 10−3 ° C ≈ 8.6 × 10−3 ° C

Given that

T − T0 T − T0 = m = (Tm − T0 )e− Kt 2 Solving this equation we get, ln (2) t= K 3. Let θ be the temperature of junction. Then, 100°C

Ans.

dQ dθ KA (θ 1 − θ 2 ) = mc = dt dt l

6. Using

Area = 0.4 m2 Temp = 400 K

Cylinder

0.4 m

80°C H2

H1

B

C

Disc

θ H3

Mass = 0.4 kg c = 600 J/kg-K

S

We have, 60°C

H1 + H2 = H3 100 − θ 80 − θ θ − 60 ∴ + = (46/ 0.92 A ) (13/ 0.26 A ) (12/ 012 . A) ∴

Solving this equation we have, θ = 84 ° C (b) Heat current in copper rod, 100 − θ H1 = (46/0.92 × 4 ) = 128 . cal/s

∴

 dθ  KA (400 − θ ) mc   =  dt  0.4 dθ (10)(0.04)(400 − θ ) (0.4)(600) = dt 0.4  dθ  1 dt   =  400 − θ  240 t

350

∫0dt = 240 ∫300

dθ 400 − θ

Solving this, we get Ans.

t ≈ 166 s

Ans.

Chapter 22

Calorimetry and Heat Transfer — 433

10. Let T1 be the temperature of C 1 and T2 the

7. Consider a differential cylinder

temperature of C 2 at some instant of time. Further let T be the temperature difference at that instant.

dr

C1

C2 H

T1

T2

r

Then,

H = ∴ or

T KA  dT  C 1  − 1 = H = = (T )  dt  l/KA l

dQ dθ dθ = KA = (2πKrl ) dt dr dr r 2 dr 50 H = dθ 2πKl ∫r 1 r ∫0 r  H ln  2  = 50 2πKl  r1 

∴

H =

∴

∴

2.5 × 10−2 1.0 × 10−2 2.5 × 10−2 + + 0.125 × 137 1.5 × 137 1.0 × 137

= 0.0017 ° C-s /J Temperature difference Now, heat current H = Net thermal resistance 30 = = 17647 W 0.0017 l 9. Thermal resistance of plastic coating  Rt =   KA  t Rt = (d = diameter) K (πd )l (0.16 × 10

Now, ∴

+

dT2 KA = dt lC 2

Further,

−

dT dT dT =− 1 + 2 dt dt dt KA (C 1 + C 2 ) = T l C 1C 1

R = R1 + R2 + R3 =

=

and Ans.

l 8. Three thermal resistances are in series.  R =   KA  ∴

dT dT dT =− 1 + 2 dt dt dt dT1 KA − = (T ) dt lC 1 −

∴

100πKl ln (r2 / r1 )

Ht = mL mL mL ln (r2 / r1 ) t= = H 100πKl

Now,

(Q T = T1 − T2 ) dT T KA   and C 2  + 2  = H = = (T )  dt  l/KA l

−2

dT KA (C 1 + C 2 ) =− T l C 1C 2

T

∫∆T

or

0

Solving, we get T = ∆T0e− αt α=

where,

11.

= 0.0223 ° C-s /J TD i Re = Rt

H dx

T1

H =

T2

TD (− dT ) = R (dx )/KA

 dT   a =−    dx   T ∴

T2

∫T

1

−

dT H = T aA

 A = constant …(i) 

l

∫0dx

 T  Hl ln  1  =  T2  aA

TD = i 2ReRt = (5)2 (4 )(0.0223) = 2.23° C

KA (C 1 + C 2 ) lC 1C 2

T

x

0.06 × 10−3 × 4.18 × 102 )(π )(0.64 × 10−3 )(2) 2

Ans.

t

∫0 dt

or

H =

aA ln (T1/T2 ) l

434 — Waves and Thermodynamics Substituting in Eq. (i), we have aA  T1   dT  aA ln   = −    dx  T  T2  l T  ln  1  T dT  T2  x or ∫T1 T = − l ∫0 dx

1

or

−

−

x l

x l

or

T  T1  =  T1  T2 

or

T  T = T1  1   T2 

x l

x

T  l = T1  2   T1 

Ans.

12. L 

dm TD  =  dt  R1 + R2

80 × 360 = 3600

− dθ  dθ  13. H = =  −  (4 πKr2 ) dr/K (4 πr2 )  dr  H r dr

dθ = −

r dr R1R2 (θ 1 − θ 2 ) ∫ 2 R R2 − R1 1 r

1 1  R1R2 θ1 + θ2 (θ 1 − θ 2 ) −  − θ1 = R2 − R1 2  r R1   1 1 θ1 − θ2 R1R2 = (θ 1 − θ 2 )  −   R1 r 2 R2 − R1

1 1 R2 − R1 R1 + R2 = − = r R1 2R1R2 2R1R2

100

20 10 + 0.25 × 10 K × 10 Solving this equation, we get K = 0.222 cal/cm-s-° C

∫

 (θ1 + θ 2 )      2 θ1

1 1 R2 − R1 − = R1 r 2R1R2

ice at 0°C R2

R1

or

or or

H

100°C

4 πK (R1R2 )(θ 1 − θ 2 ) R2 − R1

Substituting this value of H in Eq. (i), we have R1R2 dθ 2 (θ 1 − θ 2 ) = − (r ) R2 − R1 dr ∴

−

H =

∴

T  T  x ln   = − ln (T1/ T2 ) = ln  1   T1   T2  l

∴

dr 4 πK θ 2 dθ =− H ∫θ1 r2 R2 − R1 4 πK = (θ 1 − θ 2 ) R1R2 H R2

∫R

or

or Ans. …(i)

r=

2R1R2 R1 + R2

Ans.

14. See the extra points just before solved examples. Growth of ice on ponds. We have already derived that 1 ρL 2 t= y 2 Kθ ∴

dt ρLy = dy K θ

∴

dy K θ = dt Lρy

JEE Main and Advanced Previous Years’ Questions (2018-13) JEE Main 1. Two moles of an ideal monoatomic gas occupies a volume V at 27°C. The gas expands adiabatically to a volume 2 V . Calculate (i) the final temperature of the gas and (ii) change in its internal energy. (2018)

(a) (i) 189 K (ii) 2.7 kJ (c) (i) 189 K (ii) −2.7 kJ

(b) (i) 195 K (ii) −2.7 kJ (d) (i) 195 K (ii) 2.7 kJ

2. A silver atom in a solid oscillates in simple harmonic motion in some direction with a frequency of 1012 per second. What is the force constant of the bonds connecting one atom with the other? (Take, molecular weight of silver = 108 and Avogadro number = 6.02 × 1023 g mol −1 ) (2018) (a) 6.4 N/m (c) 2.2 N/m

(b) 7.1 N/m (d) 5.5 N/m

3. A granite rod of 60 cm length is clamped at its middle point and is set into longitudinal vibrations. The density of granite is 2.7 × 103 kg/m 3 and its Young’s modulus is 9.27 × 1010 Pa. What will be the fundamental frequency of the longitudinal vibrations? (2018) (a) 5 kHz (c) 10 kHz

(b) 2.5 kHz (d) 7.5 kHz

4. An observer is moving with half the speed of light towards a stationary microwave source emitting waves at frequency 10 GHz. What is the frequency of the microwave measured by the observer? (speed of light = 3 × 108 ms −1 ) (2017) (a) 12.1 GHz (c) 15.3 GHz

(b) 17.3 GHz (d) 10.1 GHz

5. Cp and CV are specific heats at constant pressure and constant volume, respectively. It is observed that

Cp − CV = a for hydrogen gas Cp − CV = b for nitrogen gas. The correct relation between a and b is (2017) (a) a = b (c) a = 28b

(b) a = 14b 1 (d) a = b 14

6. A copper ball of mass 100 g is at a temperature T . It is dropped in a copper calorimeter of mass 100 g, filled with 170 g of water at room temperature. Subsequently, the temperature of the system is found to be 75°C. T is (Given, room temperature = 30°C, specific heat of copper = 01 (2017) . cal/g°C) (a) 885°C (c) 825°C

(b) 1250°C (d) 800°C

7. An external pressure p is applied on a cube at 0°C so that it is equally compressed from all sides. K is the bulk modulus of the material of the cube and α is its coefficient of linear expansion. Suppose we want to bring the cube to its original size by heating. The temperature should be raised by (2017) (a)

p αK

(c) 3pKα

3α pK p (d) 3αK (b)

8. The temperature of an open room of volume 30 m 3 increases from 17°C to 27°C due to the sunshine. The atmospheric pressure in the room remains 1 × 105 Pa. If n i and n f are the number of molecules in the room before and after heating, then (2017) n f − n i will be (a) 138 . × 1023 (c) −2.5 × 1025

(b) 2.5 × 1025 (d) −161 . × 1023

2

Waves & Thermodynamics 9. A uniform string of length 20 m is suspended from a rigid support. A short wave pulse is introduced at its lowest end. It starts moving up the string. The time taken to reach the support is (Take, (2016) g = 10 ms−2) (a) 2 π 2 s

13. n moles of an ideal gas undergoes a process A and B as shown in the figure. The maximum temperature of the gas during the process will be (2016) p

(b) 2 s

A

2 p0

(c) 2 2 s

(d) 2 s p0

10. A pipe open at both ends has a fundamental frequency f in air. The pipe is dipped vertically in water, so that half of it is in water. The fundamental frequency of the air column is now (2016) f 2 (c) 2f

3f 4 (d) f

(a)

(b)

11. A pendulum clock loses 12 s a day if the temperature is 40°C and gains 4 s a day if the temperature is 20°C. The temperature at which the clock will show correct time, and the coefficient of linear expansion α of the metal of the pendulum shaft are, respectively. (2016) (a) 25°C, α = 1.85 × 10−5 / ° C (b) 60°C, α = 1.85 × 10−4 / ° C (c) 30°C, α = 1.85 × 10−3 / ° C (d) 55°C, α = 1.85 × 10−2 / ° C

12. An ideal gas undergoes a quasi static, reversible process in which its molar heat capacity C remains constant. If during this process the relation of pressure p and volume V is given by pV n = constant, then n is given by (Here, Cp and CV are molar specific heat at constant pressure and constant volume, respectively) (2016) (a) n = (b) n = (c) n = (d) n =

Cp CV C − Cp C − CV Cp − C C − CV C − CV C − Cp

B

2V0

V0

(a)

9 p0V0 4 nR

(b)

3 p0V0 2 nR

(c)

9 p0V0 2 nR

V

(d)

9p0V0 nR

14. A train is moving on a straight track with

speed 20 ms −1 . It is blowing its whistle at the frequency of 1000 Hz. The percentage change in the frequency heard by a person standing near the track as the train passes him is close to (speed of sound = 320 ms −1 ) (2015)

(a) 12%

(b) 6%

(c) 18%

(d) 24%

15. Consider a spherical shell of radius R at temperature T. The black body radiation inside it can be considered as an ideal gas of photons with internal energy per unit U volume u = ∝ T 4 and pressure V 1 U  p =   . If the shell now undergoes an 3 V  adiabatic expansion, the relation between T and R is (2015) (a)T ∝ e − R (b)T ∝

1 R

(c)T ∝ e − 3R (d) T ∝

1 R3

16. Consider an ideal gas confined in an isolated closed chamber. As the gas undergoes an adiabatic expansion, the average time of collision between molecules increases as V q, where V is the  C  volume of the gas. The value of q is  γ = p  CV   (2015)

3γ + 5 (a) 6

γ+1 (b) 2

3γ − 5 (c) 6

γ −1 (d) 2

3

Previous Years’ Questions (2018-13) 17. A solid body of constant heat capacity 1 J/°C is being heated by keeping it in contact with reservoirs in two ways (i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies same amount of heat. (ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir supplies same amount of heat. In both the cases, body is brought from initial temperature 100°C to final temperature 200°C. Entropy change of the body in the two cases respectively, is

21. One mole of diatomic ideal gas undergoes a cyclic process ABC as shown in figure. The process BC is adiabatic. The temperatures at A, B and C are 400 K, 800 K and 600 K, respectively. Choose the correct statement. (2014) p B 800 K

(2015)

600 K C

A 400 K

(a) ln2, ln2 (c) 2 ln 2, 8 ln 2

18. A pipe of length 85 cm is closed from one end. Find the number of possible natural oscillations of air column in the pipe whose frequencies lie below 1250 Hz. The velocity of sound in air is 340 m/s. (2014) (a) 12

(b) 8

(c) 6

(d) 4

19. Parallel rays of light of intensity

I = 912 Wm−2 are incident on a spherical black body kept in surroundings of temperature 300 K. Take Stefan constant σ = 5.7 × 10−8 Wm −2 K −4 and assume that the energy exchange with the surroundings is only through radiation. The final steady state temperature of the black body is close to (2014) (a) 330 K

(b) 660 K

(c) 990 K

(d) 1550

20. Three rods of copper, brass and steel are welded together to form a Y-shaped structure. Area of cross-section of each rod is 4 cm 2. End of copper rod is maintained at 100°C whereas ends of brass and steel are kept at 0°C. Lengths of the copper, brass and steel rods are 46, 13 and 12 cm respectively. The rods are thermally insulated from surroundings except at ends. Thermal conductivities of copper, brass and steel are 0.92, 0.26 and 0.12 in CGS units, respectively. Rate of heat flow through copper rod is (2014) (a) 1.2 cal/s (c) 4.8 cal/s

V

(b) ln2, 2 ln 2 (d) ln2, 4 ln 2

(b) 2.4 cal/s (d) 6.0 cal/s

(a) The change in internal energy in whole cyclic process is 250 R (b) The change in internal energy in the process CA is 700 R (c) The change in internal energy in the process AB is −350 R (d) The change in internal energy in the process BC is −500 R

22. A sonometer wire of length 1.5 m is made of steel. The tension in it produces an elastic strain of 1%. What is the fundamental frequency of steel if density and elasticity of steel are 7.7 × 103 kg/m 3 and 2.2 × 1011 N/m 2 respectively? (2013)

(a) 188.5 Hz (b) 178.2 Hz (c) 200.5 Hz (d) 770 Hz

23. If a piece of metal is heated to temperature θ and then allowed to cool in a room which is at temperature θ 0. The graph between the temperature T of the metal and time t will be closed to (2013) T (a) T O

(b) t

O

T

t

T

(c) θ 0 O

θ0

(d)

t

θ0 O

t

4

Waves & Thermodynamics

24. The shown p-V diagram represents the thermodynamic cycle of an engine, operating with an ideal monoatomic gas. The amount of heat, extracted from the source in a single cycle is (2013) B

2p0 p0

C D

A

2V0

V0

25. An ideal gas enclosed in a vertical cylindrical container supports a freely moving piston of mass M. The piston and the cylinder have equal cross-sectional area A. When the piston is in equilibrium, the volume of the gas is V 0 and its pressure is p0. The piston is slightly displaced from the equilibrium position and released. Assuming that the system is completely isolated from its surrounding, the piston executes a simple harmonic motion with frequency (2013) (a)

(a) p0V0

13 (b)   p0V0  2

1 Aγ p0 2 π V0 M

11 (c)   p0V0 2

(d) 4 p0V0

(c)

1 2π

A 2 γp0 MV0

(b)

1 V0 Mp0 2 π A2 γ

(d)

1 2π

MV0 Aγ p0

Answer with Explanations 1. (c) For adiabatic process relation of temperature and volume is, T2 V2γ − 1 = T1 V1γ − 1 ⇒ T2 (2 V )2 / 3 = 300( V )2 / 3 [γ =

5 for monoatomic gases] 3

≈ − 2.7 kJ T2 ≈ 189 K, ∆U ≈ 2.7 kJ

2. (b) For a harmonic oscillator,

∴

⇒

3. (a)

Y ρ

v λ 1 Y f = 2L ρ

So, frequency f =

300 T2 = 2 / 3 ≈ 189 K ⇒ 2 Also, in adiabatic process, ∆Q = 0, ∆U = − ∆W − nR( ∆T ) 3 25 or ∆U = = −2 × × ( 300 − 189) γ −1 2 3

T = 2π

∴Wave speed, v =

m 1 , where k = force constant and T = k f k = 4 π 2f 2m 2 108 × 10 −3 22 = 4 ×   × (1012 )2 ×  7  6.02 × 10 23 k = 7.1 N / m λ 4

⇒

=

1 2 × 60 × 10 −2

927 . × 1010 2.7 × 10 3

≈ 5000 Hz f = 5 kHz

4. (b) As observer is moving with relativistic speed; formula

∆f vradial , does not apply here. = f c

Relativistic doppler’s formula is 1 + v /c fobserved = factual ⋅   1− v /c Here, So, ∴

1/ 2

v 1 = c 2 3/2 fobserved = factual    1/ 2 

1/ 2

fobserved = 10 × 3 = 17.3 GHz

5. (b) By Mayor’s relation, for 1 g mole of a gas, L

From vibration mode, λ = L ⇒ λ = 2L 2

Cp − CV = R So, when n gram moles are given, R Cp − CV = n

5

Previous Years’ Questions (2018-13) As per given question, R a = C p − C V = ; for H 2 2 R ; for N 2 b = Cp − CV = 28 a = 14b

6. (a) Heat gained (water + calorimeter) = Heat lost by copper ball ⇒

mw s w ∆T + mc sc ∆T = mBs B∆T

⇒

170 × 1 × 45 + 100 × 01 . × 45 = 100 × 01 . × (T − 75) ∴ T = 885° C ∆V p p 7. (d) K = = ⇒ V K ( − ∆V / V ) pV p pV ⇒ − ∆V = = V( 3α ) ∆T ⇒ ∆T = ⇒ K 3αK K N RT NA pVNA pVNA We have, nf − ni = − RTf RTi

8. (c) From pV = nRT =

⇒

= − 2.5 × 10 25 ∆n = − 2.5 × 10 25

9. (c) At distance x from the bottom v =

⇒

∫

L 0

 mgx     L  T = = µ  m    L

⇒

dx gx ⇒ = dt

 x1 / 2 t g ∫ dt ⇒  0  (1 / 2 ) 2 L t = ⇒t =2 g

x −1 / 2dx =

x g

  = g ⋅t 0  20 = 2 2s 10

v 2l

Now, after half filled with water it becomes a closed pipe l of length . 2 Fundamental frequency of this closed pipe, v v f′ = = =f 4 ( l /2 ) 2 l

In the process pV n = constant, molar heat capacity is given by R R R C = + = CV + γ − 1 1− n 1− n C − CV R C −CV = ⇒ 1− n = p C − CV 1− n  C − C V  (C − C V ) − (C p − C V ) C − C p ⇒ n = 1−  p =  = C −CV C −CV  C −CV 

13. (a) p-V equation for path AB

p  p p = −  0  V + 3 p0 ⇒ pV = 3 p0 V − 0 V 2 V0  V0 

or

T =

1  pV p 2 =  3 p0 V − 0 V  nR nR  V0 

For maximum temperature, dT 2 p0 V =0 = 0 ⇒ 3 p0 − V0 dV 3 p 3 p0 V = V0 and p = 3 p0 − 0 = ⇒ 2 V0 2 Therefore, at these values :  3 p0   3V0      2   2  9 p0 V0 ∴ Tmax = = nR 4nR

14. (a) Observer is stationary and source is moving.  v  During approach, f1 = f    v − vs

L 11. (a) T0 = 2 π g T ′ = T0 + ∆T = 2 π

From Eq. (i), α ( ∆θ)t ∆T α∆θ = ⇒ ∆t = T0 2 2 α ( 40 − 25)(24 × 3600) 12 = ⇒ α = 1.85 × 10 −5 / ° C 2

L

10. (d) Fundamental frequency of open pipe. f =

⇒ 3 (θ − 20) = 40 − θ ⇒ 4θ = 100 ⇒ θ = 25° C Time gained or lost is given by  ∆T  ∆t ∆T =  t t ≈ T0  T0 + ∆T 

12. (b) ∆Q = ∆U + ∆W

10 5 × 30 1 1  × 6.02 × 10 23 .  − nf − ni =   300 290  8.3

∴

1 L (1 + α ∆θ)  L 2 = 2 π (1 + α ∆θ) g g  α ∆θ  ≈ T0  1 +   2  α ∆θT0 …(i) ∴ ∆T = T ′−T0 = 2 α ∆θ1T0 ∆T1 12 40 − θ or = = ⇒ ∆T2 α ∆θ 2T0 4 θ − 20

=2 π

L + ∆L ⇒ T ′ = T0 + ∆T g

 320  = 1000   = 1066.67 Hz  320 − 20 

6

Waves & Thermodynamics  v  f2 = f    v + vs

During recede,

So, we have 6 possibilities.

 320  = 1000   = 941.18 Hz  320 + 20   f − f2  |% change in frequency| =  1  × 100 ≈ 12%  f1 

15. (b) Given,

U ∝ T4 V U = αT 4 V

...(i)

It is also given that, P =

Alternate method In closed organ pipe, fundamental node

nR 0T 1 1 U  = (αT 4 )   ⇒ V 3 V 3

l = 0.85 =

(R 0 = Gas constant) or

VT 3 =

3nR 0 = constant α

4 ∴  πR 3  T 3 = constant or RT = constant 3  ∴

T ∝

16. (b) Average time between two collisions is given by 1 2 πnvrms d 2

...(i)

N Here, n = number of molecules per unit volume = V 3RT and vrms = M Substituting these values in Eq.(i) we have, V τ ∝ T

...(ii)

For adiabatic process, TV γ − 1 = constant V Substituting in Eq. (ii), we have τ ∝  1   γ −1  V  or

τ ∝V

 γ −1  1+   2 

or

1 + γ    2 

τ ∝ V

17. (a) Entropy is a state functions. Therefore in both cases answer should be same.

18. (c) For closed organ pipe =

(2 n + 1)v [n = 0, 1, 2 ...... ] 4l

(2 n + 1)v < 1250 4l 4 × 0.85 (2 n + 1) < 1250 × 340 (2 n + 1) < 12.5 2 n < 1150 . ⇒ n < 525 . So, n = 0, 1, 2, 3 ,...,5

λ = 0.85 4 ⇒ λ = 4 × 0.85 c As we know, ν = λ 340 = 100 Hz ⇒ 4 × 0.85 i.e.

1 R τ =

λ 4

∴ Possible frequencies = 100 Hz, 300 Hz, 500 Hz, 700 Hz, 900 Hz, 1100 Hz below 1250 Hz.

19. (a) In steady state

S =4πR2

I πR2 Incident

Radiation

Energy incident per second = Energy radiated per second ∴ IπR 2 = σ (T 4 − T04 ) 4 πR 2 ⇒ I = σ (T 4 − T04 ) 4 ⇒

T 4 − T04 = 40 × 10 8

⇒ T − 81 × 10 8 = 40 × 10 8 ⇒ T 4 = 121 × 10 8 4

⇒

T ≈ 330 K

20. (c) In thermal conduction, it is found that in steady state the heat current is directly proportional to the area of cross-section A which is proportional to the change in temperature (T1 − T2 ). ∆Q KA(T1 − T2 ) Then, = ∆t x

7

Previous Years’ Questions (2018-13) According to thermal conductivity, we get 100°C l1 K1

dQ2 dt l2

K2

23. (c) According to Newton’s cooling law, option (c) is correct answer.

24. (b) Heat is extracted from the source means heat is dQ1 dt

K3 dQ3 l3 dt

given to the system (or gas) or Q is positive. This is positive only along the path ABC. Heat supplied ∴ Q ABC = ∆U ABC + W ABC = nC V (Tf − Ti ) + Area under p-V graph 3 = n R  (TC − TA ) + 2 p0 V0 2  3 ( nRTC − nRTA ) + 2 p0 V0 2 3 = ( pC VC − pA VA ) + 2 p0 V0 2 3 = ( 4 p0 V0 − p0 V0 ) + 2 p0 V0 2 13 = p0 V0 2

0°C

0°C

=

dQ 1 dQ 2 dQ 3 = + dt dt dt . (T − 0) 0.92 (100 − T ) 026 . (T − 0) 012 + = 12 46 13 ⇒ T = 40° C 0.92 × 4 (100 − 40) dQ1 ∴ = = 4.8 cal/s dt 40 i.e.

21. (d) According to first law of thermodynamics, we get (i) Change in internal energy from A to B i.e. ∆U AB ∆U AB = nC V (TB − TA ) 5R = 1× ( 800 − 400) = 1000 R 2 (ii) Change in internal energy from B to C

25. (c) In equilibrium, p0 A = Mg when slightly displaced downwards,

∆U BC = nC V (TC − TB ) 5R = 1× ( 600 − 800) 2

x0

= − 500 R (iii) ∆Uisothermal = 0 (iv) Change in internal energy from C to A i.e. ∆UCA ∆UCA = nC V (TA − TC )

p  dp = − γ  0  dV  V0   As in adiabatic process, dp = − γ p     dV V

= − 500 R

22. (b) Fundamental frequency of sonometer wire v 1 f = = 2l 2l

T 1 = µ 2l

∴

Here, µ = mass per unit length of wire.

⇒ ⇒

T Y∆l 1 Y∆l = ⇒ f = A l 2 l ld ∆l = 0.01 l = 1.5 m, l d = 7.7 × 10 3 kg / m3

⇒ Y = 2.2 × 1011 N / m2 After substituting the values we get, f ≈ 178.2 Hz

Tl A∆l

Restoring force, F = (dp) A  γ p0  =−  ( A ) ( Ax )  V0 

T Ad

Also, Young’s modulus of elasticity Y =

…(i)

F ∝− x Therefore, motion is simple harmonic comparing with F = − kx we have γ p0 A 2 k= V0 ∴

f =

1 2π

k m

=

1 2π

γ p0 A 2 MV0

JEE Advanced 1. One mole of a monoatomic ideal gas undergoes a cyclic process as shown in the figure (where, V is the volume and T is the temperature). Which of the statements below is (are) true ? (More than One Correct Option, 2018)

cylinders in the steady state is 200K, then (Numerical Value, 2018) K 1 / K 2 = .......... . Insulating material

T1

K1

K2

T2

L

T

L

II

4. In an experiment to measure the speed of III

I IV

V

(a) Process I is an isochoric process (b) In process II, gas absorbs heat (c) In process IV, gas releases heat (d) Processes I and III are not isobaric

2. Two men are walking along a horizontal straight line in the same direction. The main in front walks at a speed 1.0ms −1 and the man behind walks at a speed 2.0 ms −1. A third man is standing at a height 12 m above the same horizontal line such that all three men are in a vertical plane. The two walking men are blowing identical whistles which emit a sound of frequency 1430 Hz. The speed of sound in air 330 ms −1. At the instant, when the moving men are 10 m apart, the stationary man is equidistant from them. The frequency of beats in Hz, heard by the stationary man at this instant, is ............. . (Numerical Value, 2018)

3. Two conducting cylinders of equal length but different radii are connected in series between two heat baths kept at temperatures T1 = 300K and T2 = 100K, as shown in the figure. The radius of the bigger cylinder is twice that of the smaller one and the thermal conductivities of the materials of the smaller and the larger cylinders are K 1 and K 2, respectively. If the temperature at the junction of the

sound by a resonating air column, a tuning fork of frequency 500 Hz is used. The length of the air column is varied by changing the level of water in the resonance tube. Two successive resonances are heard at air columns of length 50.7 cm and 83.9 cm. Which of the following statements is (are) true?(More than One Correct Option, 2018) (a) The speed of sound determined from this experiment is 332 m s −1 (b) The end correction in this experiment is 0.9 cm (c) The wavelength of the sound wave is 66.4 cm (d) The resonance at 50.7 cm corresponds to the fundamental harmonic

5. One mole of a monoatomic ideal gas undergoes an adiabatic expansion in which its volume becomes eight times its initial value. If the initial temperature of the gas is 100 K and the universal gas constant R = 8.0 j mol −1 K −1 , the decrease in its internal energy in joule, is ............ . (Numerical Value, 2018)

6. One mole of a

p

II 3p0 monoatomic ideal gas undergoes four IV III I thermodynamic p0 processes as shown schematically in the V V0 3V0 pV-diagram below. Among these four processes, one is isobaric, one is isochoric, one is isothermal and one is adiabatic. Match the processes mentioned in List-l with the corresponding statements in List-II. (Matching Type Question, 2018)

9

Previous Years’ Questions (2018-13) List-I

List-II

7. Which of the following options is the only

P. In process I

1.

Work done by the gas is zero

Q. In process II

2.

Temperature of the remains unchanged

gas

(a) (II) (iii) (S) (c) (III) (iii) (P)

R. In process III

3.

No heat is exchanged between the gas and its surroundings

S. In process IV

4.

Work done by the gas is 6 p0 V0

(b) P → 1; Q → 3; R → 2; S → 4 (c) P → 3; Q → 4; R → 1; S → 2 (d) P → 3; Q → 4; R → 2; S → 1

Directions (Q.Nos. 7-8) Matching the information given in the three columns of the following table. An ideal gas is undergoing a cyclic thermodynamic process in different ways as shown in the corresponding p-V diagrams in column 3 of the table. Consider only the path from state 1 to state 2. W denotes the corresponding work done on the system. The equations and plots in the table have standards notations and used in thermodynamic processes. Here γ is the ratio of heat capacities at constant pressure and constant volume. The number of moles in the gas is n. (Matching Type, 2017) Column 1

1 γ −1 ( p2 V2 − p1V1 )

Column 2

(i)

Isothermal

p

(Q)

p

1

2

1 2 V

(III) W1→ 2 = 0

(iii) Isobaric

(R)

p 1 2

V

(iv) Adiabatic (IV) W 1→ 2 = − nRT V  ln  2   V1 

8. Which one of the following options is the correct combination? (a) (II) (iv) (P) (c) (II) (iv) (R)

(b) (III) (ii) (S) (d) (IV) (ii) (S)

(S)

p

1

end of a uniform rope of constant mass per unit length. The top end of the rope is attached to a fixed rigid support at O. A transverse wave pulse (Pulse 1) of wavelength λ 0 is produced at point O on the rope. The pulse takes time TOA to reach point A. If the wave pulse of wavelength λ 0 is produced at point A (Pulse 2) without disturbing the position of M it takes time TAO to reach point O. Which of the following options is/are correct? (More than One Correct Option, 2017) O

Pulse 1

Pulse 2 A M

(a) The timeTAO = TOA (b) The wavelength of Pulse 1 becomes longer when it reaches point A (c) The velocity of any pulse along the rope is independent of its frequency and wavelength (d) The velocities of the two pulses (Pulse 1 and Pulse 2) are the same at the mid-point of rope

Column 3

(P)

V

(II) W1→ 2 (ii) Isochoric = − pV2 + pV1

(b) (II) (iii) (P) (d) (II) (iv) (R)

9. A block M hangs vertically at the bottom

(a) P → 4; Q → 3; R → 1; S → 2

(I) W 1→ 2 =

correct representation of a process in which ∆U = ∆Q − p∆V ?

2

V

10. A stationary source emits sound of frequency f0 = 492 Hz. The sound is reflected by a large car approaching the source with a speed of 2 ms −1. The reflected signal is received by the source and superposed with the original. What will be the beat frequency of the resulting signal in Hz? (Given that the speed of sound in air is 330 ms −1 and the car reflects the sound at the frequency it has received). (Single Integer Type, 2017)

10

Waves & Thermodynamics

11. Two loudspeakers M and N are located 20 m apart and emit sound at frequencies 118 Hz and 121 Hz, respectively. A car in initially at a point P, 1800 m away from the midpoint Q of the line MN and moves towards Q constantly at 60 km/h along the perpendicular bisector of MN. It crosses Q and eventually reaches a point R, 1800 m away from Q. Let v( t ) represent the beat frequency measured by a person sitting in the car at time t. Let ν P , νQ and ν R be the beat frequencies measured at locations P, Q and R respectively. The speed of sound in air is 330 ms−1. Which of the following statement(s) is (are) true regarding the sound heard by the person? (More than One Correct Option, 2016)

(a) The plot below represents schematically the variation of beat frequency with time ν(t) P Q

νQ

P 3V 5 = constant. Consider another thermodynamic process that brings the system from the same initial state to the same final state in two steps : an isobaric expansion at Pi followed by an isochoric (isovolumetric) process at volume V f . The amount of heat supplied to the system in the two-step process is approximately (Single Correct Option, 2016)

(a) 112 J

(b) 294 J

(c) 588 J

(d) 813 J

13. A water cooler of storage capacity 120 litres can cool water at a constant rate of P watts. In a closed circulation system (as shown schematically in the figure), the water from the cooler is used to cool an external device that generates constantly 3 kW of heat (thermal load). The temperature of water fed into the device cannot exceed 30°C and the entire stored 120 litres of water is initially cooled to 10°C. The entire system is thermally insulated. The minimum value of P (in watts) for which the device can be operated for 3 hours is (Single Correct Option, 2016) Cooler

Hot

Device

R t

(b) The rate of change in beat frequency is maximum when the car passes through Q (c) νP + νR = 2 νQ (d) The plot below represents schematically the variations of beat frequency with time ν(t)

(Specific heat of water is 4.2 kJ kg−1K −1 and the density of water is 1000 kg m −3 ) (a) 1600

P νQ

Cold

(b) 2067

(c) 2533

(d) 3933

14. A metal is heated in a furnace where a

Q R

t

12. A gas is enclosed in a cylinder with a movable frictionless piston. Its initial thermodynamic state at pressure Pi = 105 Pa and volume V i = 10−3 m3 changes to a final state at Pf = (1/ 32) × 105 Pa and V f = 8 × 10−3 m3 in an adiabatic quasi-static process, such that

sensor is kept above the metal surface to read the power radiated ( P ) by the metal. The sensor has a scale that displays log2 ( P / P0 ), where P0 is a constant. When the metal surface is at a temperature of 487°C, the sensor shows a value 1. Assume that the emissivity of the metallic surface remains constant. What is the value displayed by the sensor when the temperature of the metal surface is raised to 2767°C? (Single Integer Type, 2016)

Previous Years’ Questions (2018-13)

11

15. The ends Q and R of two thin wires, PQ

Ignoring the friction between the piston and the cylinder, the correct statements is/are (More than One Correct Option, 2015)

and RS, are soldered (joined) together. Initially, each of the wire has a length of 1 m 10°C. Now, the end P is maintained at 10°C, while the end S is heated and maintained at 400°C. The system is thermally insulated from its surroundings. If the thermal conductivity of wire PQ is twice that of the wire RS and the coefficient of linear thermal expansion of PQ is 1.2 × 10−5 K −1, the change in length of the wire PQ is (Single Correct Option, 2016)

(a) 0.78 mm (c) 1.56 mm

(b) 0.90 mm (d) 2.34 mm

16. A container of fixed volume has a mixture of one mole of hydrogen and one mole of helium in equilibrium at temperature T . Assuming the gases are ideal, the correct statements is/are (a) The average energy per mole of the gas mixture is 2RT (b) The ratio of speed of sound in the gas 6 mixture to that in helium gas is 5 (c) The ratio of the rms speed of helium atoms to 1 that of hydrogen molecules is 2 (d) The ratio of the rms speed of helium atoms to 1 that of hydrogen molecules is 2 (More than One Correct Option, 2015)

17. An ideal monoatomic gas is confined in a horizontal cylinder by a spring loaded piston (as shown in the figure). Initially the gas is at temperature T1, pressure P1 and volume V1 and the spring is in its relaxed state. The gas is then heated very slowly to temperature T2, pressure P2 and volume V 2. During this process the piston moves out by a distance x.

(a) IfV2 = 2V1 and T2 = 3T1, then the energy 1 stored in the spring is PV 1 1 4 (b) IfV2 = 2V1 and T2 = 3T1, then the change in internal energy is 3PV 1 1 (c) IfV2 = 3V1 and T2 = 4T1, then the work done by 7 the gas is PV 1 1 3 (d) IfV2 = 3V1 and T2 = 4T1, then the heat 17 supplied to the gas is PV 1 1 6

18. Two spherical stars A and B emit blackbody radiation. The radius of A is 400 times that of B and A emits 104 times λ  the power emitted from B. The ratio  A  λB of their wavelengths λ A and λ B at which the peaks occur in their respective radiation curves is (Single Integer Type, 2015)

19. Four harmonic waves of equal frequencies and equal intensities I 0 have phase π 2π and π. When they are angles 0, , 3 3 superposed, the intensity of the resulting wave is nI 0. The value of n is (Single Integer Type, 2015)

20. Two vehicles, each moving with speed u on the same horizontal straight road, are approaching each other. Wind blows along the road with velocity w. One of these vehicles blows a whistle of frequency f1. An observer in the other vehicle hears the frequency of the whistle to be f2. The speed of sound in still air is v. The correct statement(s) is (are) (More than One Correct Option, 2014)

(a) If the wind blows from the observer to the source, f2 > f1. (b) If the wind blows from the source to the observer, f2 > f1 (c) If the wind blows from the observer to the source, f2 < f1 (d) If the wind blows from the source to the observer, f2 < f1

12

Waves & Thermodynamics

21. One end of a taut string of length 3 m along the x-axis is fixed at x = 0. The speed of the waves in the string is 100 ms −1. The other end of the string is vibrating in the y-direction so that stationary waves are set up in the string. The possible waveform(s) of these stationary wave is (are) (More than One Correct Option, 2014)

πx 50 πt cos 6 3 πx 100 πt (b) y (t ) = A sin cos 3 3 5 πx 250 πt (c) y (t ) = A sin cos 2 3 5 πx (d) y (t ) = A sin cos 250 πt 2 (a) y (t ) = A sin

and W bf = 100 J respectively. The heat supplied to the system along the path iaf, ib and bf are Qiaf , Qib and Qbf respectively. If the internal energy of the system in the state b is U b = 200 J and Qiaf = 500 J, the ratio Qbf / Qib is (Single Integer Type, 2014)

24. Parallel rays of light of intensity

I = 912 Wm −2 are incident on a spherical black body kept in surroundings of temperature 300 K. Take Stefan constant σ = 5.7 × 10−8 Wm −2K −4 and assume that the energy exchange with the surroundings is only through radiation. The final steady state temperature of the black body is close to (Single Correct Option 2014)

22. A student is performing an experiment using a resonance column and a tuning fork of frequency 244 s −1. He is told that the air in the tube has been replaced by another gas (assume that the column remains filled with the gas). If the minimum height at which resonance occurs is (0.350 ± 0.005) m, the gas in the tube is (Useful information : 167RT = 640 J1/ 2 mole−1/ 2 ; 140RT = 590 J1/ 2 mole−1/ 2. The molar masses M in grams are given in the options. Take the value of 10/ M for each gas as given there.)

(a) 330 K (c) 990 K

(b) 660 K (d) 1550 K

Passage (Q. Nos. 25-26) In the figure a container is shown to have a movable (without friction) piston on top. The container and the piston are all made of perfectly insulating material allowing no heat transfer between outside and inside the container. The container is divided into two compartments by a rigid partition made of a thermally conducting material that allows slow transfer of heat.

(Single Correct Option, 2014)

(a) Neon (M = 20, 10 / 20 = 7 / 10) (b) Nitrogen (M = 28, 10 / 28 = 3 / 5) (c) Oxygen (M = 32, 10 / 32 = 9 / 16) (d) Argon (M = 36, 10 / 36 = 17 / 32)

23. A thermodynamic

p

system is taken a f from an initial state i with internal energy i b U i = 100 J to the V final state f along two different paths iaf and ibf, as schematically shown in the figure. The work done by the system along the paths af, ib and bf are W af = 200 J, W ib = 50 J

The lower compartment of the container is filled with 2 moles of an ideal monoatomic gas at 700 K and the upper compartment is filled with 2 moles of an ideal diatomic gas at 400 K. The heat capacities per mole of an 3 5 ideal monoatomic gas are CV = R, C p = R, 2 2 and those for an ideal diatomic gas are 5 7 CV = R, C p = R. 2 2 (Passage Type, 2014)

13

Previous Years’ Questions (2018-13) 25. Consider the partition to be rigidly fixed so that it does not move. When equilibrium is achieved, the final temperature of the gases will be (a) 550 K

(b) 525 K

(c) 513 K

(d) 490 K

p 32p0

F

p0

E

26. Now consider the partition to be free to move without friction so that the pressure of gases in both compartments is the same. Then total work done by the gases till the time they achieve equilibrium will be (a) 250 R

(b) 200 R

(c) 100 R

(d) −100 R

27. Two rectangular blocks, having indentical dimensions, can be arranged either in configuration I or in configuration II as shown in the figure. One of the blocks has thermal conductivity K and the other 2 K . The temperature difference between the ends along the x-axis is the same in both the configurations. It takes 9 s to transport a certain amount of heat from the hot end to the cold end in the configuration I. The time to transport the same amount of heat in the configuration II is (Single Correct Option, 2013) Configuration II Configuration I K

(a) 2.0 s

2K

2K K

(b) 3.0 s

(c) 4.5 s

X

(d) 6.0 s

28. Two non-reactive monoatomic ideal gases have their atomic masses in the ratio 2 : 3. The ratio of their partial pressures, when enclosed in a vessel kept at a constant temperature, is 4 : 3. The ratio of their densities is (Single Correct Option, 2013) (a) 1 : 4

(b) 1 : 2

(c) 6 : 9

(d) 8 : 9

29. One mole of a monatomic ideal gas is taken along two cyclic processes E → F → G → E and E → F → H → E as shown in the p-V diagram. The processes involved are purely isochoric, isobaric, isothermal or adiabatic.

G

H

V

V0

Match the paths in Column I with the magnitudes of the work done in Column II and select the correct answer using the codes given below the lists. (Matching Type, 2013)

Column I

Column II

P. G → E

1.

160 p0 V0 ln 2

Q. G → H

2.

36 p0 V0

R. F → H

3.

24 p0 V0

S. F → G

4.

31 p0 V0

Codes P Q R S (a) 2 3 1 4 (c) 4 3 2 1

P (b) 1 (d) 2

Q R S 2 4 3 3 4 1

30. The figure below shows the variation of specific heat capacity (C ) of a C solid as a function of temperature (T ). 100 200 300 400 500 The temperature is T(K) increased continuously from 0 to 500 K at a constant rate. lgnoring any volume change, the following statement(s) is (are) correct to reasonable approximation. (More than One Correct Option, 2013)

(a) the rate at which heat is absorbed in the range 0-100 K varies linearly with temperatureT (b) heat absorbed in increasing the temperature from 0-100 K is less than the heat required for increasing the temperature from 400-500 K (c) there is no change in the rate of heat absorbtion in the range 400-500 K (d) the rate of heat absorption increases in the range 200-300 K

14

Waves & Thermodynamics

31. A horizontal stretched string, fixed at two ends, is vibrating in its fifth harmonic according to the equation, y( x , t ) = ( 0.01 m) [sin( 62.8 m −1 )x ] cos[( 628 s−1 )t ]. Assuming π = 314 . , the correct statement(s) is (are)

(a) the number of nodes is 5 (b) the length of the string is 0.25 m (c) the maximum displacement of the mid-point of the string from its equilibrium position is 0.01 m (d) the fundamental frequency is 100 Hz (More than One Correct Option, 2013)

Answer with Explanations 1. (b,c,d) (b) Process-II is isothermal expansion,

(c)

∆U = 0 , W > 0 ∆Q = W > 0 Process-IV is isothermal compression, ∆U = 0 , W < 0

4. (a, c) Let n th harmonic is corresponding to 50.7 cm and ( n + 1)th harmonic is corresponding 83.9 cm.

∆Q = W < 0 (d) Process-I and III are not isobaric because in isobaric process T ∝ V hence, T-V graph should be a straight line passing through origin.

2. (5 Hz) Observer (stationary) 13m cos θ =

13m 12m A

θ

θ 2m/s

5m

5m

5 13

B 1m/s

  330 fA = 1430   330 − 2 cosθ       1 2 cos θ  = 1430  ≈ 1430 1 + 2 cos θ  330   1 −  330   [from binomial expansion]

λ ∴ Their difference is . 2 ∴

λ = (83.9 − 50.7) cm 2

or

λ = 66.4 cm

λ = 16.6 cm ∴ 4 Length corresponding to fundamental mode must be λ close to and 50.7 cm must be an odd multiple of 4 this length. 16.6 × 3 = 49.8 cm. Therefore, 50.7 is 3rd harmonic. If end correction is e, then 3λ e + 507 . = 4 e = 49.8 − 507 . = − 0.9 cm ∴ Speed of sound, v = fλ ⇒ v = 500 × 66.4 cm/s = 332 m/s 5 3 T - V equation in adiabatic process is

5. (900) Given, n = 1, γ =

  330 fB = 1430    330 + 1cosθ  cosθ  ≈ 1430 1 − 330  

∴

3 cos θ  Beat frequency = fA − fB = 1430  330  = 13 cos θ 5 = 13   = 5.00 Hz  13 

⇒

TV γ − 1 = constant

3. (4) Rate of heat flow will be same, ∴

300 − 200 200 − 100 = R1 R2

∴

R1 = R 2

⇒

K1 A = 2 =4 K2 A1

⇒

 as H = dQ = T ⋅ D     dt R 

L1 L2 = K1 A1 K 2 A 2

⇒

T1V1

γ −1

= T2V2

γ −1

V  T2 = T1  1   V2 

γ −1

2

1 3 = 100 ×    8

T2 = 25 K 3 C V = R for monoatomic gas 2 3R  ∴ ∆U = nC V ∆T = n ×   (T − T )  2  2 1 = 1×

3 × 8 × (25 − 100) = − 900 J 2

∴ Decrease in internal energy = 900 J

15

Previous Years’ Questions (2018-13) dp  dp  = γ      dV  isothermal  dV  adiabatic

6. (c) List-I

(P) Process I

⇒ Adiabatic ⇒ Q = 0

(Q) Process II

⇒ Isobaric

∴

W = p∆V = 3 p0[3V0 − V0 ] = 6 p0V0

(R) Process III ⇒ Isochoric ⇒ W = 0 (S) Process (IV) ⇒ Isothermal ⇒ Temperature = Constant

7. (b) ∆U = ∆Q − p∆V ∆U + p∆V = ∆Q As ∆U ≠ 0, W ≠ 0, ∆Q ≠ 0. The process represents, isobaric process Wgas = − p( ∆V ) = − p( V2 − V1 ) = − pV2 + pV1 Graph ‘P’ satisfies isobaric process.

8. (b) Work done in isochoric process is zero. W12 = 0 as ∆V = 0 Graph ‘S’ represents isochoric process. T , so speed at any 9. (a,c,d or a,c) v = µ position will be same for both pulses, therefore time taken by both pulses will be same. v λf = v ⇒ λ= f since when pulse 1 reaches at A tension and hence speed decreases therefore λ decreases.

10. (6)

Car vc=2m/s

11. (b,c,d) Speed of car, 500 m/s 3 At a point S, between P and Q  C + v cos θ  ν ′M = ν M  ;   C  C + v cos θ  ′ νN = νN     C v cos θ   ⇒ ∆v = ( v N − v M )  1 +   C  v = 60 km / h =

Similarly, between Q and R v cos θ  ∆ν = ( ν N − ν M )  1 −   C  dθ d ( ∆ν) v = ± ( ν N − ν M ) sinθ dt C dt θ ≈ 0° at P and R as they are large distance apart. ⇒ Slope of graph is zero. at Q, θ = 90° dθ is maximum sinθ is maximum also value of dt dθ v as = , where v is its velocity and r is the length of dt r the line joining P and S. and r is minimum at Q. ⇒ Slope is maximum at Q. V At P, ν P = ∆ν = ( ν N − ν M ) 1 +   C V (θ ≈ 0°) At R, ν R = ∆ν = ( ν N − ν M ) 1 −   C (θ ≈ 0°) At Q, νQ = ∆ν = ( ν N − ν M ) (θ = 90° ) From these equations, we can see that ν P + ν R = 2 νQ

12. (c) In the first process : pi Viγ = pf Vfγ p

 v + vC  f1 = f0   (v = speed of sound)  v  Frequency of reflected sound as observed at the source  v   v + vC  f2 = f1   = f0    v − vC   v − vC  Beat frequency = f2 − f0  v + vC   2 vC  = f0  − 1 = f0    v − vC   v − vC  2 ×2 = 492 × = 6 Hz 328 ⇒

λ ∝ v ∝T

Vi

pi

Frequency observed at car

Vf

pf V

⇒

pi  Vf  =  pf  Vi 

γ

⇒ 32 = 8 γ

⇒ γ =

For the two step process W = pi( Vf − Vi )

= 10 5(7 × 10 −3 )

= 7 × 10 2 J

5 3

…(i)

16

Waves & Thermodynamics f ( pf Vf − pi Vi ) 2 1 1 2 2 =  × 10 − 10   γ − 1 4

∆U =

∆U = −

2T − 20 = 400 − T or 3T = 420 T = 140°

or ∴

140° Q, R

P

3 3 9 ⋅ × 10 2 = − × 10 2 J 2 4 8

10°C

Q = 7 × 10 2 − =

9 × 10 2 8

47 × 10 2 J = 588 J 8

13. (b) Heat generated in device in 3 h = time × power = 3 × 3600 × 3 × 10 3

Tx = (130 x + 10° ) Change in length dx is suppose dy Then, dy = αdx (Tx − 10) 1

0

0

1

  αx 2 ∆y =  × 130 2 0  . × 10 −5 × 65 ∆y = 12

Heat absorbed by coolant = Pt = 324 × 10 5 − 120 × 1 × 4.2 ×10 3 × 20 J

∆y = 78.0 × 10 −5 m = 0.78 mm

Pt = ( 325 − 100.8) × 10 5 J = 2232 . × 10 5 J

16. (a, b, d) (a) Total internal energy

2232 . × 10 5 = 2067 W 3600 p 14. (9) log 2 1 = 1, p0 p Therefore, 1 = 2 p0 P=

f1 f nRT + 2 nRT 2 2 U 1 (Uave )per mole = = [5RT + 3RT ] = 2 RT 2n 4 n C + n2C p 2 (b) γ mix = 1 p1 n2C v1 + n2C v 2 U =

According to Stefan's law, p ∝ T 4

p2 = 2 × 44 p0

p log 2 2 = log 2 [2 × 4 4 ] = log 2 2 + log 2 4 4 p0 = 1 + log 2 2 8 = 1 + 8 = 9

15. (a)

7R 5R + (1) 2 2 = 3 = 5R 3R 2 (1) + (1) 2 2 n1M1 + n2M 2 M1 + M 2 2 + 4 = = = =3 n1 + n2 2 2 (1)

4

p2  T2  2767 + 273  4 =   =   =4  487 + 273  p1  T1 

10°C

∆y

∫ dy = ∫ αdx (130 x + 10 − 10)

= 324 × 10 J

P

Junction

Temperature of junction is 140°C Temperature at a distance x from end P is

Heat used to heat water = ms ∆θ = 120 × 1 × 42 . × 10 3 × 20 J

p2 p = 2 = 44 ⇒ p1 2 p0

dx

x

5

⇒

400°C

K

Q − W = ∆U ⇒

S

Mmix

Speed of sound V =

γRT M

⇒

γ M

V ∝

S

Q, R 2K

K

1m

1m

400°C

Rate of heat flow from P to Q dQ 2 KA (T − 10) = dt 1 Rate of heat flow from Q to S dQ KA ( 4000 − T ) = dt 1 At steady state, state rate of heat flow is same 2 KA (T − 10) = KA ( 400 − T ) ∴ 1

(d) ⇒

Vmix = VHe

γ mix M × He = γ He Mmix

Vrms =

3RT M

Vrms ∝

1 VHe , = M VH

3 /2 4 × = 5/ 3 3

MH = MHe

6 5

2 1 = 4 2

17. (a, b, c) Note This question can be solved if right hand side chamber is assumed open, so that its pressure remains constant even if the piston shifts towards right.

17

Previous Years’ Questions (2018-13) =

p0 p1= p0

7 p1V1 3 4 +  p1 ⋅ 3V1 − p1V1   3 2 3

7 p1V1 9 41p1V1 + p1 V1 = 3 2 6 3 Note ∆U = ( p2 V2 − p1 V1 ) has been obtained in 2 part (b). =

x p0 = p1 p kx

18. (2) Power, P = ( σT 4 A ) = σT 4 ( 4 πR 2 ) P ∝ T4R2

or,

…(i)

According to Wien’s law, pV = nRT T p∝ ⇒ V Temperature is made three times and volume is doubled 3 ⇒ p2 = p1 2 V − V1 2 V1 − V1 V1 ∆V Further x = = 2 = = A A A A 3 p1 kx p1 A p2 = = p1 + ⇒ kx = 2 2 A Energy of spring 1 2 p1 A pV kx = x= 11 2 4 4 3   (b) ∆U = nc v ∆ T = n  R  ∆T 2  3 = ( p2 V2 − p1 V1 ) 2 (a)

= (c) p2 = ⇒ ⇒ ⇒

3 2

1 T (λ is the wavelength at which peak occurs) ∴ Eq. (i) will become, λ∝

P∝

1/ 4

R2  λ∝   P 

⇒

λ A  RA  =  λ B  RB 

1/ 2

4 kx p1 = p1 + 3 A p1 A kx = 3 ∆V 2 V1 x= = A A Wgas = ( p0 ∆V + Wspring ) = ( p1 Ax +

1 kx ⋅ x ) 2

1/ 4

p1V1 7 p1V1 = 3 3

(d) ∆Q = W + ∆U 7 p1V1 3 = + ( p2V2 − p1V1 ) 3 2

=2

19. (3) Let individual amplitudes are A0 each. Amplitudes can be added by vector method. A3

A2 60°

60°

60°

A1

A1 = A 2 = A 3 = A 4 = A 0 Resultant of A1 and A 4 is zero. Resultant of A 2 and A 3 is A=

A 02 + A 02 + 2 A 0 A 0 cos 60° =

This is also the net resultant. Now,

I ∝ A2

∴ Net intensity will become 3I0. ∴ Answer is 3. P S

2V 1 p A 2V = +  p1 A ⋅ 1 + ⋅ 1 ⋅ 1   A 2 3 A  = 2 p1V1 +

1/ 4

 PB     PA 

1 = [400]1/ 2  4   10 

A4

p2 =

λ4

or

  3    2 p1  (2 V1 ) − p1V1  = 3 p1V1  

4 p1 3

R2

θ θ M

N 10 m

18 m

3 A0

18

Waves & Thermodynamics

20. (a,b) When wind blows from S to O

So

 v + w + u f2 = f1   v + w −u or f2 > f1 when wind blows from O to S  v − w + u f2 = f1   ⇒ f2 > f1 v − w −u

So,

= 500 J − 200 J = 300 J = Uf − Ui U f = U iaf + U i = 300 J + 100 J = 400 J ∆U ib = U b − U i = 200 J − 100 J = 100 J

21. (a,c,d) There should be a node at x = 0 and antinode at

Q ib = ∆U ib + W ib

x = 3 m. Also,

ω = 100 m / s. k ∴ y = 0 at x = 0 and y = ± A at x = 3 m. Only (a), (c) and (d) satisfy the condition. λ 22. (d) Minimum length = 4 ⇒ λ = 4l Now, v = f λ = (244) × 4 × l as l = 0.350 ± 0.005 ⇒ v lies between 336.7 m/s to 346.5 m/s γ RT Now, v = , here M is molecular mass in M × 10 −3

∆U iaf = Q iaf − W iaf

= 100 J + 50 J = 150 J

v =

Q ibf = ∆U ibf + W ibf = ∆U iaf + W ibf = 300 J + 150 J = 450 J So, the required ratio

Q bf Q ibf − Q ib = Q ib Q ib =

450 − 150 =2 150

24. (a) In steady state

gram

I

= 100 γRT ×

10 ⋅ M

S =4πR2 πR2 Incident

Radiation

For monoatomic gas, γ = 1.67 10 ⇒ v = 640 × M

Energy incident per second = Energy radiated per second

For diatomic gas,

∴

IπR 2 = σ (T 4 − T04 ) 4 πR 2

⇒

I = σ (T 4 − T04 ) 4

γ = 1.4 ⇒ ∴

10 M

v = 590 × vNe = 640 ×

7 = 448 m/s 10

v Ar = 640 ×

17 = 340 m/s 32

vO 2 = 590 ×

9 = 331.8 m/s 16

vN 2 = 590 ×

3 = 354 m/s 5

∴ Only possible answer is Argon.

23. (2)

W ibf = W ib + W bf W iaf

= 50 J + 100 J = 150 J = W ia + W af

= 0 + 200 J = 200 J Q iaf = 500 J

⇒ ⇒ ⇒ ⇒

T 4 − T04 = 40 × 10 8 T 4 − 81 × 10 8 = 40 × 10 8 T 4 = 121 × 10 8 T ≈ 330 K

25. (d) Let final equilibrium temperature of gases is T Heat rejected by gas by lower compartment = nC V ∆T 3 = 2 × R (700 − T ) 2 Heat received by the gas in above compartment = nC p ∆T 7 = 2 × R (T − 400) 2 Equating the two, we get 2100 − 3T = 7T − 2800 ⇒ Τ = 490 K

19

Previous Years’ Questions (2018-13) 26. (d) ∆W1 + ∆U1 = ∆Q1

V  nRT ln  f  = 32 p0V0 ln  Vi 

...(i)

∆W 2 + ∆U 2 = ∆Q 2

...(ii)

∆Q1 + ∆Q 2 = 0

= 32 p0V0 ln 2 5

∴ ( nC p ∆T )1 + ( nC p ∆T )2 = 0 But n1 = n2 = 2 ∴

= 160 p0 V0 ln 2 In G → E, ∆W = p0∆V = p0( 31V0 ) = 31 p0V0 In G → H work done is less than 31 p0V0 i .e.,24 p0V0 ln F → H work done is 36 p0V0

5 7 R (T − 700) + R (T − 400) = 0 2 2

Solving, we get T = 525 K Now, from equations (i) and (ii), we get ∆W1 + ∆W 2 = − ∆U1 − ∆U 2 as ∴

∆Q1 + ∆Q 2 = 0 ∆W1 + ∆W 2 = − [( nC V ∆T )1 + ( nC V ∆T )2 ] 3 = − 2 × R × ( 525 − 700) 2  +2 ×

5 R × ( 525 − 400) 2 

= −100R

27. (a) R I = R1 + R 2 = 

l   l  3 l   +  =    KA   2 KA  2  KA  1 1 1 = + RII R1 R2

KA 2 KA + l l l R or = I RII = 3KA 4.5 Since thermal resistance R II is 4.5 times less than thermal resistance R I . =

t II =

∴

or

F

32T0

T0

E

Iso

atic

iab Ad

p0

H V0

R = rate of absortion of heat =

the

dQ ∝C dt

(i) in 0 − 100 K C increase, so R increases but not linearly (ii) ∆Q = mC∆T as C is more in (400 K − 500 K) then ( 0 − 100 K ) so heat is increasing (iii) C remains constant so there no change in R from (400 K − 500 K) (iv) C is increases so R increases in range (200 K − 300 K )

31. (b,c)

l

Number of nodes = 6 From the given equation, we can see that 2π k= = 62.8 m−1 λ 2π m = 0.1 m ∴ λ= 62.8 5λ l= = 0.25 m 2

ρ1  p1   M1   2   4  8 =   =   = ρ2  p2   M 2   3   3  9

p 32p0

Q = mCT dQ dT = mc dt dt

P

pM RT ρ ∝ pM

Isochoric

29. (c)

30. (b, c, d)

tI 9 s =2s = 4.5 4.5

28. (d) ρ = ∴

 32 V0     V0 

The mid-point of the string is P, an antinode ∴maximum displacement = 0.01 m

rm

ω = 2 πf = 628 s −1

al

Isobaric

∴

G 32V0

V

In F → G work done in isothermal process is

f =

628 = 100 Hz 2π

But this is fifth harmonic frequency. f ∴ Fundamental frequency f0 = = 20 Hz. 5

Understanding Physics

JEE Main & Advanced

ELECTRICITY AND MAGNETISM

Understanding Physics

JEE Main & Advanced

ELECTRICITY AND MAGNETISM

DC PANDEY [B.Tech, M.Tech, Pantnagar, ID 15722]

ARIHANT PRAKASHAN (Series), MEERUT

Understanding Physics

JEE Main & Advanced

ARIHANT PRAKASHAN (Series), MEERUT All Rights Reserved

© SARITA PANDEY No part of this publication may be re-produced, stored in a retrieval system or by any means, electronic, mechanical, photocopying, recording, scanning, web or otherwise without the written permission of the publisher. Arihant has obtained all the information in this book from the sources believed to be reliable and true. However, Arihant or its editors or authors or illustrators don’t take any responsibility for the absolute accuracy of any information published and the damage or loss suffered thereupon. All disputes subject to Meerut (UP) jurisdiction only.

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Understanding Physics

JEE Main & Advanced

PREFACE The overwhelming response to the previous editions of this book gives me an immense feeling of satisfaction and I take this an opportunity to thank all the teachers and the whole student community who have found this book really beneficial. In the present scenario of ever-changing syllabus and the test pattern of JEE Main & Advanced, the NEW EDITION of this book is an effort to cater all the difficulties being faced by the students during their preparation of JEE Main & Advanced. The exercises in this book have been divided into two sections viz., JEE Main & Advanced. Almost all types and levels of questions are included in this book. My aim is to present the students a fully comprehensive textbook which will help and guide them for all types of examinations. An attempt has been made to remove all the printing errors that had crept in the previous editions. I am extremely thankful to (Dr.) Mrs. Sarita Pandey, Mr. Anoop Dhyani, Nisar Ahmad for their endless efforts during the project. Comments and criticism from readers will be highly appreciated and incorporated in the subsequent editions.

DC Pandey

Understanding Physics

JEE Main & Advanced

CONTENTS 23. CURRENT ELECTRICITY 23.1 23.2 23.3 23.4 23.5 23.6

Introduction Electric Current Electric Currents in Conductors Drift Velocity and Relaxation Time Resistance of a Wire Temperature Dependence of Resistance 23.7 Ohm’s Law

24. ELECTROSTATICS 24.1 Introduction 24.2 Electric Charge 24.3 Conductor and Insulators 24.4 Charging of a Body 24.5 Coulomb’s Law 24.6 Electric Field 24.7 Electric Potential Energy 24.8 Electric Potential 24.9 Relation Between Electric Field and Potential

25. CAPACITORS 25.1 Capacitance 25.2 Energy Stored in a Charged Capacitor 25.3 Capacitors 25.4 Mechanical Force on a Charged Conductor 25.5 Capacitors in Series and Parallel

1-108 23.8 The Battery and the Electromotive Force 23.9 Direct Current Circuits, Kirchhoff ’s Laws 23.10 Heating Effects of Current 23.11 Grouping of Cells 23.12 Electrical Measuring Instruments 23.13 Colour Codes for Resistors

109-231 24.10 Equipotential Surfaces 24.11 Electric Dipole 24.12 Gauss’s Law 24.13 Properties of a Conductor 24.14 Electric Field and Potential Due To Charged Spherical Shell or Solid Conducting Sphere 24.15 Electric Field and Potential Due to a Solid Sphere of Charge

233-333 25.6 Two Laws in Capacitors 25.7 Energy Density 25.8 C-R Circuits 25.9 Methods of Finding Equivalent Resistance and Capacitance

Understanding Physics

JEE Main & Advanced 26. MAGNETICS 26.1 Introduction 26.2 Magnetic Force on a Moving Charge(Fm) 26.3 Path of a Charged Particle in Uniform Magnetic Field 26.4 Magnetic Force on a Current Carrying Conductor 26.5 Magnetic Dipole 26.6 Magnetic Dipole in Uniform Magnetic Field 26.7 Biot Savart Law 26.8 Applications of Biot Savart Law 26.9 Ampere's Circuital Law 26.10 Force Between Parallel Current

335-454 26.11 26.12 26.13 26.14 26.15 26.16 26.17 26.18

Carrying Wires Magnetic Poles and Bar Magnets Earth’s Magnetism Vibration Magnetometer Magnetic Induction and Magnetic Materials Some Important Terms used in Magnetism Properties of Magnetic Materials Explanation of Paramagnetism, Diamagnetism and Ferromagnetism Moving Coil Galvanometer

27. ELECTROMAGNETIC INDUCTION 27.1 Introduction 27.2 Magnetic Field Lines and Magnetic Flux 27.3 Faraday’s Law 27.4 Lenz’s Law 27.5 Motional Electromotive Force

27.6 Self-Inductance and Inductors 27.7 Mutual Inductance 27.8 Growth and Decay of Current in an LR Circuit 27.9 Oscillations in L-C Circuit 27.10 Induced Electric Field

28. ALTERNATING CURRENT 28.1 Introduction 28.2 Alternating Currents and Phasors 28.3 Current and Potential Relations 28.4 Phasor Algebra

455-560

561-607 28.5 Series L-R Circuit 28.6 Series C-R Circuit 28.7 Series L-C-R Circuit 28.8 Power in an AC Circuit

Hints & Solutions JEE Main & Advanced Previous Years' Questions (2018-13)

609-731 1-36

Understanding Physics

JEE Main & Advanced

SYLLABUS JEE Main ELECTROSTATICS Electric charges Conservation of charge, Coulomb’s law-forces between two point charges, forces between multiple charges; Superposition principle and continuous charge distribution. Electric field Electric field due to a point charge, Electric field lines, Electric dipole, Electric field due to a dipole, Torque on a dipole in a uniform electric field. Electric flux, Gauss’s law and its applications to find field due to infinitely long uniformly charged straight wire, Uniformly charged infinite plane sheet and uniformly charged thin spherical shell. Electric potential and its calculation for a point charge, Electric dipole and system of charges; Equipotential surfaces, Electrical potential energy of a system of two point charges in an electrostatic field. Conductors and insulators, Dielectrics and electric polarization, Capacitor, Combination of Capacitors in series and in parallel, Capacitance of a parallel plate capacitor with and without dielectric medium between the plates, Energy stored in a capacitor.

CURRRENT ELECTRICITY Electric current, Drift velocity, Ohm’s law, Electrical resistance, Resistances of different materials, V-I characteristics of Ohmic and non-ohmic conductors, Electrical energy and power, Electrical resistivity, Colour code for resistors; Series and parallel combinations of resistors; Temperature dependence of resistance. Electric cell and its Internal resistance, Potential difference and emf of a cell, combination of cells in series and in parallel. Kirchhoff ’s laws and their applications. Wheatstone bridge, Meter bridge. Potentiometer – principle and its applications.

Understanding Physics

JEE Main & Advanced

MAGNETIC EFFECTS OF CURRENT AND MAGNETISM Biot-Savart law and its application to current carrying circular loop. Ampere’s law and its applications to infinitely long current carrying straight wire and solenoid. Force on a moving charge in uniform magnetic and electric fields. Cyclotron. Force on a current-carrying conductor in a uniform magnetic field. Force between two parallel current-carrying conductors-definition of ampere. Torque experienced by a current loop in uniform magnetic field; Moving coil galvanometer, its current sensitivity and conversion to ammeter and voltmeter. Current loop as a magnetic dipole and its magnetic dipole moment. Bar magnet as an equivalent solenoid, magnetic field lines; Earth’s magnetic field and magnetic elements. Para-, dia- and ferro- magnetic substances. Magnetic susceptibility and permeability, Hysteresis, Electromagnets and permanent magnets.

ELECTROMAGNETIC INDUCTION AND ALTERNATING CURRENTS Electromagnetic induction; Faraday’s law, Induced emf and current; Lenz’s law, Eddy currents. Self and mutual inductance. Alternating currents, Peak and rms value of alternating current/voltage; Reactance and impedance; LCR series circuit, Resonance; Quality factor, power in AC circuits, Wattless current. AC generator and transformer.

Understanding Physics

JEE Main & Advanced

JEE Advanced GENERAL Verification of Ohm’s law using voltmeter and ammeter. Specific resistance of the material of a wire using meter bridge and post office box.

ELECTRICITY AND MAGNETISM Coulomb’s law, Electric field and potential, Electrical potential energy of a system of point charges and of electrical dipoles in a uniform electrostatic field, Electric field lines, Flux of electric field, Gauss’s law and its application in simple cases, such as, to find field due to infinitely long straight wire, Uniformly charged infinite plane sheet and uniformly charged thin spherical shell. Capacitance, Parallel plate capacitor with and without dielectrics, Capacitors in series and parallel, Energy stored in a capacitor. Electric current, Ohm’s law, Series and parallel arrangements of resistances and cells, Kirchhoff ’s laws and simple applications, Heating effect of current. Biot-Savart’s law and Ampere’s law, Magnetic field near a current-carrying straight wire, Along the axis of a circular coil and inside a long straight solenoid, Force on a moving charge and on a current-carrying wire in a uniform magnetic field. Magnetic moment of a current loop, Effect of a uniform magnetic field on a current loop, Moving coil galvanometer, Voltmeter, Ammeter and their conversions.

ELECTROMAGNETIC INDUCTION Faraday’s law, Lenz’s law, Self and mutual inductance, RC, LR and LC circuits with DC and AC sources.

Current Electricity Chapter Contents 23.1

Introduction

23.2

Electric current

23.3

Electric currents in conductors

23.4

Drift velocity and Relaxation time

23.5

Resistance of a wire

23.6

Temperature dependence of resistance

23.7

Ohm's law

23.8

The battery and the electromotive force

23.9

Direct current circuits, Kirchhoff's laws

23.10 Heating effects of current 23.11 Grouping of cells 23.12 Electrical measuring instruments 23.13 Colour codes for resistors

2 — Electricity and Magnetism

23.1 Introduction An electrical circuit consists of some active and passive elements. The active elements such as a battery or a cell, supply electric energy to the circuit. On the contrary, passive elements consume or store the electric energy. The basic passive elements are resistor, capacitor and inductor. A resistor opposes the flow of current through it and if some current is passed by maintaining a potential difference across it, some energy is dissipated in the form of heat. A capacitor is a device which stores energy in the form of electric potential energy. It opposes the variations in voltage. An inductor opposes the variations in current. It does not oppose the steady current through it. Fundamentally, electric circuits are a means for conveying energy from one place to another. As charged particles move within a circuit, electric potential energy is transferred from a source (such as a battery or a cell) to a device in which that energy is either stored or converted to another form, like sound in a stereo system or heat and light in a toaster or light bulb. Electric circuits are useful because they allow energy to be transported without any moving parts (other than the moving charged particles themselves). In this chapter, we will study the basic properties of electric currents. We’ll study the properties of batteries and how they cause current and energy transfer in a circuit. In this analysis, we will use the concepts of current, potential difference, resistance and electromotive force.

23.2 Electric Current Flow of charge is called electric current. The direction of electric current is in the direction of flow of positive charge or in the opposite direction of flow of negative charge. Current is defined quantitatively in terms of the rate at which net charge passes through a cross-section area of the conductor. dq dq or i = Thus, I= dt dt We can have the following two concepts of current, as in the case of velocity, instantaneous current and average current. dq Instantaneous current = = current at any point of time and dt q Average current = t Hence-forth unless otherwise referred to, current would signify instantaneous current. By convention, the direction of the current is assumed to be that in which positive charge moves. In the SI system, the unit of current is ampere (A). 1 A = 1 C/s Household currents are of the order of few amperes.

Flow of Charge If current is passing through a wire then it implies that a charge is flowing through that wire. Further, dq …(i) i= ⇒ dq = idt dt

Chapter 23

Current Electricity — 3

Now, three cases are possible : If given current is constant, then from Eq. (i) we can see that flow of charge can be obtained directly by multiplying that constant current with the given time interval. Or,

Case 1

∆q = i × ∆t Case 2

If given current is a function of time, then charge flow can be obtained by integration. Or, tf

∆q = ∫ i dt ti

Case 3

If current versus time is given, then flow of charge can be obtained by the area under the

graph. ∆q = area under i - t graph

Extra Points to Remember ˜

˜

˜

˜

˜

The current is the same for all cross-sections of a conductor of non-uniform cross-section. Similar to the water flow, charge flows faster where the conductor is smaller in cross-section and slower where the conductor is larger in cross-section, so that charge rate remains unchanged. Electric current is very similar to water current, consider a water tank kept at some height and a pipe is connected to the water tank. The rate of flow of water through the pipe depends on the height of the tank. As the level of water in the tank falls, the rate of flow of water through the pipe also gets reduced. Just as the flow of water depends on the height of the tank or the level of water in the tank, the flow of current through a wire depends on the potential difference between the end points of the wire. As the potential difference is changed, the current will change. For example, during the discharging of a capacitor potential difference and hence, the current in the circuit decreases with time. To maintain a constant current in a circuit a constant potential difference will have to be maintained and for this a battery is used which maintains a constant potential difference in a circuit. Though conventionally a direction is associated with current (opposite to i1 the motion of electrons), it is not a vector as the direction merely represents the sense of charge flow and not a true direction. Further, current does not θ i = i1+ i2 obey the law of parallelogram of vectors, i.e. if two currents i1 and i 2 reach i2 a point we always have i = i1 + i 2 whatever be the angle between i1 and i 2 . According to its magnitude and direction, current is usually divided into two Fig. 23.1 types : (i) Direct current (DC) If the magnitude and direction of current does not vary with time, it is said to be direct current (DC). Cell, battery or DC dynamo are its sources. (ii) Alternating current (AC) If a current is periodic (with constant amplitude) and has half cycle positive and half negative, it is said to be alternating current (AC). AC dynamo is the source of it. If a charge q revolves in a circle with frequency f, the equivalent current, i =qf

˜

˜

In a conductor, normally current flow or charge flow is due to flow of free electrons. Charge is quantised. The quantum of charge is e. The charge on any body will be some integral multiple of e, i.e. q = ± ne where, n = 1, 2, 3K

4 — Electricity and Magnetism V

Example 23.1 In a given time of 10 s, 40 electrons pass from right to left. In the same interval of time 40 protons also pass from left to right. Is the average current zero? If not, then find the value of average current. Solution No, the average current is not zero. Direction of current is the direction of motion of positive charge or in the opposite direction of motion of negative charge. So, both currents are from left to right and both currents will be added. ∴ I av = I electron + I proton q q = 1 + 2 t1 t2 =

40 e 40 e + 10 10

( q = ne )

= 8e = 8 × 1.6 × 10−19 A = 1.28 × 10−18 A

Ans.

V

Example 23.2 A constant current of 4 A passes through a wire for 8 s. Find total charge flowing through that wire in the given time interval. Solution Since, i = constant ∴ ∆q = i × ∆ t =4×8 = 32 C

V

Example 23.3 A wire carries a current of 2.0 A. What is the charge that has flowed through its cross-section in 1.0 s ? How many electrons does this correspond to? q Solution Q i = t Ans. ∴ q = it = ( 2.0 A) (1.0 s ) = 2.0 C ∴

q = ne q 2.0 n= = e 1.6 × 10–19 = 1.25 × 1019

V

Ans.

Example 23.4 The current in a wire varies with time according to the relation i = ( 3.0 A) + ( 2.0 A / s) t (a) How many coulombs of charge pass a cross-section of the wire in the time interval between t = 0 and t = 4.0 s? (b) What constant current would transport the same charge in the same time interval?

Chapter 23 Solution

i=

(a)

dq dt 4

q

∫ 0 dq = ∫ 0

∴

Current Electricity — 5

idt

4

q = ∫ ( 3 + 2t ) dt

∴

0

4

= [ 3t + t 2 ] 0 = [12 + 16] = 28 C (b) i = V

q 28 = = 7A t 4

Ans. Ans.

Example 23.5 Current passing through a wire decreases linearly from 10 A to 0 in 4 s. Find total charge flowing through the wire in the given time interval. i (A) Solution Current versus time graph is as shown in figure. Area under this graph will give us net charge flow. 10 Hence, ∆q = Area 1 = × base × height t (s) 4 2 Fig. 23.2 1 = × 4 × 10 2 Ans. = 20 C

INTRODUCTORY EXERCISE

23.1

1. How many electrons per second pass through a section of wire carrying a current of 0.7 A? 2. A current of 3.6 A flows through an automobile headlight. How many coulombs of charge flow through the headlight in 3.0 h?

3. A current of 7.5 A is maintained in wire for 45 s. In this time, (a) how much charge and (b) how many electrons flow through the wire?

4. In the Bohr model, the electron of a hydrogen atom moves in a circular orbit of radius 5.3 × 10−11 m with a speed of 2.2 × 106 m /s. Determine its frequency f and the current I in the orbit.

5. The current through a wire depends on time as, i = (10 + 4t ) Here, i is in ampere and t in seconds. Find the charge crossed through a section in time interval between t = 0 to t = 10 s.

6. In an electrolyte, the positive ions move from left to right and the negative ions from right to left. Is there a net current? If yes, in what direction?

6 — Electricity and Magnetism

23.3 Electric Currents in Conductors Conductors are those materials which can conduct electricity. Conductors can be broadly classified into two groups : (i) Solid conductors (ii) Electrolyte conductors Normally in atoms and molecules, the negatively charged electrons and the positively charged nuclei are bound to each other and are thus not free to move. In solid conductors (notably metals), some of the electrons (called free electrons) are free to move within the bulk materials. In these conductors, current flow takes place due to these free electrons. Positive ions in these conductors are almost fixed. They do not move. So, they do not contribute in the current. In electrolyte solutions however, both positive and negative ions can move. In our following discussions, we will focus only on solid conductors so that the current is carried by the negatively charged free electrons in the background of fixed positive ions.

Theory of Current Flow through Solid Conductors At room temperature, the free electrons in a conductor move randomly with speeds of the order of 10 5 m /s.Since, the motion of the electrons is random, there is no net charge flow in any direction. For any imaginary plane passing through the conductor, the number of electrons crossing the plane in one direction is equal to the number crossing it in the other direction. Therefore, net current is zero from any section. –

– –

Fig. 23.3

When a constant potential difference V is applied between the ends of the conductor as shown in Fig. 23.4, an electric field E is produced inside the conductor. The conduction electrons within the conductor are then subjected to a force – eEand move overall in the direction of increasing potential. i –

E

+

V

vd

– –

– –

–

+

V

– – –

–

Fig. 23.4

However, this force does not cause the electrons to move faster and faster. Instead, a conduction electron accelerates through a very small distance (about 5 × 10 −8 m) and then collides with fixed ions or atoms of the conductor. Each collision transfers some of the electron’s kinetic energy to the ions (or atoms). Because of the collision, electron moves slowly along the conductor or we can say that it acquires a drift velocity vd in the direction opposite to E (in addition to its random motion.) The drift motion of free electrons produce an electric current in the opposite direction of this motion or in the direction of electric field (from higher potential to lower potential). It is interesting to note

Chapter 23

Current Electricity — 7

that the magnitude of the drift velocity is of the order of 10 – 4 m /s or about 10 9 times smaller than the average speed of the electrons of their random (or thermal) motion. The above discussion can be summarized as follows : 1. Free electrons inside a solid conductor can have two motions : (i) random or thermal motion (speed of the order of 10 5 m /s) (ii) drift motion (speed of the order of 10 – 4 m /s) 2. Net current due to random (or thermal motion) is zero from any section, whereas net current due to drift motion is non-zero. 3. In the absence of any electric field (or a potential difference across the conductor) free electrons have only random motion. Hence, net current from any section is zero. 4. In the presence of an electric field (or a potential difference across the conductor) free electrons have both motions (random and drift). Therefore, current is non-zero due to drift motion. 5. Drift motion of free electrons is opposite to the electric field. Therefore, direction of current is in the direction of electric field from higher potential to lower potential. V

Example 23.6 Electric field inside a conductor is always zero. Is this statement true or false? False. Under electrostatic conditions when there is no charge flow (or no current) in the conductor, electric field is zero. If current is non-zero, then electric field is also non-zero. Because the drift motion (of free electrons) which produces a net current starts only due to electric force on them. Solution

INTRODUCTORY EXERCISE

23.2

1. All points of a conductor are always at same potential. Is this statement true or false?

23.4 Drift Velocity and Relaxation Time As discussed before, in the presence of electric field, the free electrons experience an electric force of magnitude. (as q = e ) F = qE or eE This will produce an acceleration of magnitude, F eE a= = ( m = mass of electron ) m m Direction of force (and acceleration) is opposite to the direction of electric field. After accelerating to some distance an electron will suffer collisions with the heavy fixed ions. The collisions of the electrons do not occur at regular intervals but at random times. Relaxation time τ is the average time between two successive collisions. Its value is of the order of 10 −14 second. After every collision, let us assume that drift motion velocity of electron becomes zero. Then, it accelerates for a time interval τ, then again it collides and its drift motion velocity becomes zero and so on.

8 — Electricity and Magnetism If v d is the average constant velocity (called drift velocity) in the direction of drift motion, then relation between v d and τ is given by eEτ vd = m

Extra Points to Remember ˜

In some standard books of Indian authors, the relation is given as eEτ vd = 2m Initially, I was also convinced with this expression. But later on after consulting many more literatures in eEτ this. I found that vd = is correct. But at this stage it is very difficult for me to give its correct proof. m Because the correct proof requires a knowledge of high level of physics which is difficult to understand for a class XII student.

Current and Drift Velocity Consider a cylindrical conductor of cross-sectional area A in which an electric field E exists. Drift velocity of free electrons is v d and n is number of free electrons per unit volume (called free electron density). i E

vd vd ∆t

Fig. 23.5

Consider a length v d ∆t of the conductor. The volume of this portion is Av d ∆t. Number of free electrons in this volume = (free electron density ) × (volume) = (n) ( Av d ∆t ) = nAv d ∆t All these electrons cross the area A in time ∆t. Thus, the charge crossing this area in time ∆t is ∴ or or

∆q = ( nAv d ∆t ) ( e) ∆q i= = neAv d ∆t i = neAv d

Thus, this is the relation between current and drift velocity.

Chapter 23

Current Electricity — 9

Current Density Current per unit area (taken normal to the current), i / A is called current density and is denoted by j. The SI units of current density are A/ m 2 . Current density is a vector quantity j directed along E. j=

i A

But i = neAv d , therefore j = nev d

Extra Points to Remember ˜

˜

Drift velocity of electrons in a conductor is of the order of 10−4 m/s, then question arises in everybody’s mind that why a bulb glows instantly when switched on? Reason is : when we close the circuit, electric field is set up in the entire closed circuit instantly (with the speed of light). Due to this electric field, the free electrons instantly get drift velocity in the entire circuit and a current is established in the circuit instantly. The current so set up does not wait for the electrons to flow from one end of the conductor to the other end. If a current i is flowing through a wire of non-uniform cross-section, then current will remain constant at all cross-sections. But drift velocity and current density are inversely proportional to the area of cross-section. This is because i = neAvd

or

1 i or vd ∝ neA A i 1 or j ∝ j = A A

1

2 i

i

vd =

Further,

Fig. 23.6

So, in the figure i1 = i 2 = i but, (vd )2 > (vd )1 and j 2 > j1 because A 2 < A 1

Note Later we can also prove that electric field at 2 is also more than electric field at 1 . V

Example 23.7 An electron beam has an aperture of 1.0 mm 2 . A total of 6.0 × 1016 electrons go through any perpendicular cross-section per second. Find (a) the current and (b) the current density in the beam. Solution

(a) The current is given by i=

q ne = t t

Substituting the values we have, i=

( 6.0 × 1016 ) (16 . × 10−19 ) 1

= 9.6 × 10−3 A

Ans.

(b) The current density is j=

i 9.6 × 10−3 = A 10−6

= 9.6 × 103 A/ m 2

Ans.

10 — Electricity and Magnetism V

Example 23.8 Calculate the drift speed of the electrons when 1 A of current exists in a copper wire of cross-section 2 mm2 . The number of free electrons in 1 cm3 of copper is 8.5 × 1022 . Solution

n = free electron density, = 8.5 × 1022 per cm 3

(Given)

= ( 8.5 × 10 ) (10 ) per m 22

6

3

= 8.5 × 1028 per m 3 From i = neAv d , we get vd =

i neA

Substituting the values in SI units we have, vd =

1 ( 8.5 × 10 )(1.6 × 10−19 )( 2 × 10−6 ) 28

= 3.6 × 10−5 m/s

INTRODUCTORY EXERCISE

Answer

23.3

1. When a wire carries a current of 1.20 A, the drift velocity is 1.20 × 10−4 m /s. What is the drift velocity when the current is 6.00 A? 2. Find the velocity of charge leading to 1 A current which flows in a copper conductor of cross-section 1cm 2 and length 10 km. Free electron density of copper is 8.5 × 1028/ m 3. How long will it take the electric charge to travel from one end of the conductor to the other?

23.5 Resistance of a Wire Resistance of a wire is always required between two points or two surfaces (say P and Q). l P

Q A

Fig. 23.7

Here, l = length of wire, A = area of cross-section Now, R ∝l 1 and R∝ A Combining Eqs. (i) and (ii), we get l R =ρ A

…(i) …(ii)

…(iii)

Chapter 23

Current Electricity — 11

Here, ρ is called resistivity of the material of the wire. This depends on number of free electrons present in the material. With increase in number of free electrons the value of ρ decreases. Note

1 , where σ = conductivity. σ (ii) SI units of resistivity are Ω-m (ohm-metre). (iii) SI units of conductivity are ( Ω-m) −1 . (iv) In Eq. (iii), l is that dimension of conductor which is parallel to P and Q and A is that cross-sectional area, which is perpendicular to P and Q. (i) ρ =

V

Example 23.9 Two copper wires of the same length have got different diameters, (a) which wire has greater resistance? (b) greater specific resistance? l 1 Solution (a) For a given wire, R = ρ , i.e. R ∝ A A So, the thinner wire will have greater resistance. (b) Specific resistance (ρ ) is a material property. It does not depend on l or A. So, both the wires will have same specific resistance.

V

Example 23.10 A wire has a resistance R. What will be its resistance if it is stretched to double its length? Solution

Let V be the volume of wire, then V = Al V A= l

∴ Substituting this in R = ρ

l , we have A

R =ρ

l2 V

So, for given volume and material (i.e.V and ρ are constants) R ∝ l2 When l is doubled, resistance will become four times, or the new resistance will be 4R. V

Example 23.11 The dimensions of a conductor of specific resistance ρ are shown below. Find the resistance of the conductor across AB, CD and EF. A

b E

D

a F

c

C

B

Fig. 23.8

12 — Electricity and Magnetism l A Resistance across AB, CD and EF in tabular form is shown below.

Solution

R =ρ

Table 23.1

V

I

A

B

AB

c

a×b

CD

b

a×c

EF

a

b×c

c ab b ρ ac a ρ bc ρ

Example 23.12 A copper wire is stretched to make it 0.1% longer. What is the (JEE 1978) percentage change in its resistance? ρl l Solution ( V = volume of wire) R =ρ = A V /l = ∴

ρl 2 V

R ∝ l2

( ρ and V = constant)

For small percentage change % change R = 2 (% change in l ) = 2 ( 01 . %) = 0.2% Since R ∝ l 2 , with increase in the value of l, resistance will also increase.

INTRODUCTORY EXERCISE

23.4

1. In household wiring, copper wire 2.05 mm in diameter is often used. Find the resistance of a 35.0 m long wire. Specific resistance of copper is 1.72 × 10−8 Ω- m.

2. The product of resistivity and conductivity of a conductor is constant. Is this statement true or false?

3. You need to produce a set of cylindrical copper wires 3.50 m long that will have a resistance of 0.125 Ω each. What will be the mass of each of these wires? Specific resistance of copper = 1.72 × 10–8 Ω - m, density of copper = 8.9 × 103 kg/m 3.

4. Consider a thin square sheet of side L and thickness t, made of a material of resistivity ρ. The resistance between two opposite faces, shown by the shaded areas in the figure is (JEE 2010) (a) directly proportional to L t (b) directly proportional to t (c) independent of L (d) independent of t

L

Fig. 23.9

Chapter 23

Current Electricity — 13

23.6 Temperature Dependence of Resistance If we increase the temperature of any material, the following two effects can be observed : (i) Numbers of free electrons increase. Due to this effect conductivity of the material increases. So, resistivity or resistance decreases. (ii) The ions of the material vibrate with greater amplitude and the collision between electrons and ions become more frequent. Due to this effect resistivity or resistance of the material increases.

In Conductors There are already a large number of free electrons. So, with increase in temperature effect-(i) is not so dominant as effect-(ii). Hence, resistivity or resistance of conductors increase with increase in temperature. Over a small temperature range (upto 100°C), the resistivity of a metal (or conductors) can be represented approximately by the equation, ρ(T ) = ρ 0 [1 + α (T – T0 )]

…(i)

where, ρ 0 is the resistivity at a reference temperature T0 (often taken as 0°C or 20°C) and ρ (T ) is the resistivity at temperature T, which may be higher or lower than T0 . The factor α is called the temperature coefficient of resistivity. The resistance of a given conductor depends on its length and area of cross-section besides the resistivity. As temperature changes, the length and area also change. But these changes are quite small and the factor l / A may be treated as constant. R ∝ ρ and hence,

Then,

R (T ) = R 0 [1 + α (T – T0 )]

…(ii)

In this equation, R (T ) is the resistance at temperature T and R 0 is the resistance at temperature T0 , often taken to be 0°C or 20°C. The temperature coefficient of resistance α is the same constant that l appears in Eq. (i), if the dimensions l and A in equation R = ρ do not change with temperature. A

In Semiconductors At room temperature, numbers of free electrons in semiconductors (like silicon, germanium etc.) are very less. So, with increase in temperature, effect-(i) is very dominant. Hence, resistivity or resistance of semiconductors decreases with increase in temperature or we can say that temperature coefficient of resistivity α for semiconductors is negative. V

Example 23.13 The resistance of a thin silver wire is 1.0 Ω at 20°C. The wire is placed in a liquid bath and its resistance rises to 1.2 Ω. What is the temperature of the bath? α for silver is 3.8 × 10 –3 /°C. Solution

Here,

R (T ) = R 0 [1 + α (T − T0 )]

R (T ) = 1.2 Ω,

R 0 = 1.0 Ω, α = 3.8 × 10–3 / ° C and T0 = 20° C

Substituting the values, we have or Solving this, we get

1.2 = 1.0 [1 + 3.8 × 10–3 (T – 20)] 3.8 × 10–3 (T – 20) = 0.2 T = 72.6° C

Ans.

14 — Electricity and Magnetism V

Example 23.14 Read the following statements carefully

(JEE 1993)

Y : The resistivity of semiconductor decreases with increase of temperature. Z : In a conducting solid, the rate of collisions between free electrons and ions increases with increase of temperature. Select the correct statement (s) from the following (a) Y is true but Z is false (b) Y is false but Z is true (c) Both Y and Z are true (d) Y is true and Z is the correct reason for Y Solution Resistivity of conductors increases with increase in temperature because rate of collisions between free electrons and ions increase with increase of temperature. However, the resistivity of semiconductors decreases with increase in temperature, because more and more covalent bonds are broken at higher temperatures and free electrons increase with increase in temperature. Therefore, the correct option is (c). V

Example 23.15 An electric toaster uses nichrome for its heating element. When a negligibly small current passes through it. Its resistance at room temperature ( 27.0 o C ) is found to be 75 .3 Ω. When the toaster is connected to a 230 V supply, the current settles, after a few seconds, to a steady value of 2.68 A. What is the steady temperature of the nichrome element? The temperature coefficient of resistance of nichrome averaged over the temperature o range involved, is 1.70 × 10 −4 C −1 . Solution

Given, T0 = 27° C and R 0 = 75.3 Ω V At temperature T, R T = T iT =

V  R =   i

230 = 85.82 Ω 2.68

R T = R 0 [1 + α (T − T0 )]

Using the equation,

85.82 = 75.3 [1 + (1.70 × 10−4 )(T − 27)]

We have Solving this equation, we get

T ≈ 850 o C

Ans.

Thus, the steady temperature of the nichrome element is 850 o C .

INTRODUCTORY EXERCISE

23.5

1. A piece of copper and another of germanium are cooled from room temperature to 80 K. The resistance of (a) each of them increases (b) each of them decreases (c) copper increases and germanium decreases (d) copper decreases and germanium increases

(JEE 1988)

2. The resistance of a copper wire and an iron wire at 20°C are 4.1 Ω and 3.9 Ω, respectively. Neglecting any thermal expansion, find the temperature at which resistances of both are equal. α Cu = 4.0 × 10−3 K −1 and α Fe = 5.0 × 10−3 K −1.

Chapter 23

Current Electricity — 15

23.7 Ohm’s Law The equation,V = iR is not Ohm’s law. It is a mathematical relation between current passing through a resistance, value of resistance R and the potential difference V across it. V According to Ohm’s law, there are some of the materials (like metals or tan θ = slope conductors) or some circuits for which, current passing through them is =R proportional to the potential difference applied across them or i ∝ V or V ∝ i V θ ⇒ V = iR or = R = constant i i Ohmic or V - i graph for such materials and circuits is a straight line passing through Fig. 23.10 origin. Slope of this graph is called its resistance. The materials or circuits which follow this law are called ohmic. The materials or circuits which do not follow this law are called non-ohmic.V - i graph for non-ohmic V circuits is not a straight line passing through origin. or R is not constant and i is not proportional toV . i V

V2 V1

Q P i i1 i2 Non-ohmic Fig. 23.11

Note Equation V = iR is applicable for even non-ohmic circuits also. For example,

V1 = R1 = resistance at P. i1 V2 = R2 = resistance at Q , but i2 R1 ≠ R2

V

Example 23.16 The current-voltage graphs for a given metallic wire at two different temperatures T1 and T2 are shown in the figure. The temperature T2 is (JEE 1985) greater than T1 . Is this statement true or false? I T1 T2 V

Fig. 23.12

16 — Electricity and Magnetism I 1 = Slope of given graph = V R 1 R= ⇒ ( Slope ) T2 < ( Slope ) T1 Slope

Solution

or

⇒

1 1 > ( Slope ) T2 ( Slope ) T1

Resistance of a metallic wire increases with increase in temperature. or R T2 > R T1 or T2 > T1 Therefore, the statement is true.

23.8 The Battery and the Electromotive Force Before studying the electromotive force (emf) of a cell let us take an example of a pump which is more easy to understand. Suppose we want to recycle water between a overhead tank and a ground water tank. Water flows from overhead tank to ground water tank by itself (by gravity). No external agent is required for this purpose. But to raise the water from ground water tank to overhead tank a pump is required or some external work has to be done. In an electric circuit, a battery or a cell plays the same role as the pump played in the above example. Suppose a resistance (R) is connected across the terminals of a battery. A potential difference is developed across its ends. Current (or positive charge) flows from higher potential to lower potential across the resistance by itself. But inside the battery, work has to be done to bring the positive charge from lower potential to higher potential. The influence that makes current flow from lower to higher potential (inside the battery) is called electromotive force (abbreviated emf). If H W work is done by the battery in taking a charge q from negative terminal to positive terminal, then work done by the battery per unit charge is called emf (E) of the battery. W Thus, E= q

Overhead tank

Pump

Ground water tank Fig. 23.13 H

L R

L

Fig. 23.14

The name electromotive force is misleading in the sense that emf is not a force it is work done per unit charge. The SI unit of emf is J/C or V ( 1 V = 1 J /C ).

23.9 Direct Current Circuits, Kirchhoff’s Laws Single current in a simple circuit (single loop) can be found by the relation, 6V

10V

6V i

i

i

3Ω (a)

E

2Ω (b)

Fig. 23.15

R (c)

Chapter 23 i=

Current Electricity — 17

E net emf = net net resistance R net

For example : In Fig. (a) Net emf is 6 V and net resistance is 3 Ω. Therefore, 6 i = =2A 3 In Fig. (b) Net emf = (10 – 6) V = 4 V Net resistance = 2 Ω 4 Therefore, i = =2A 2 In Fig. (c) We have n cells each of emf E. Of these polarity of m cells (where n > 2m) is reversed. Then, net emf in the circuit is ( n – 2m) E and resistance of the circuit is R. Therefore, and

i=

( n – 2m) E R

Resistors in Series and in Parallel In series :

A

A i

i V1

R1

⇒

V

V

R

V2

R2

B

B

Fig. 23.16

Figure represents a circuit consisting of a source of emf and two resistors connected in series. We are interested in finding the resistance R of the network lying between A and B. That is, what single equivalent resistor R would have the same resistance as the two resistors linked together. Because there is only one path for electric current to follow, i must have the same value everywhere in the circuit. The potential difference between A and B is V. This potential difference must somehow be divided into two parts V1 and V2 as shown, ∴

V = V1 + V2 = iR1 + iR 2

or

V = i ( R1 + R 2 )

…(i)

Let R be the equivalent resistance between A and B, then V = iR From Eqs. (i) and (ii), R = R1 + R 2 for resistors in series This result can be readily extended to a network consisting of n resistors in series. ∴

R = R1 + R 2 + KK + R n

…(ii)

18 — Electricity and Magnetism In parallel :

i

i i1

V

i2 R2

R1

⇒

V

R

Fig. 23.17

In Fig. 23.17, the two resistors are connected in parallel. The voltage drop across each resistor is equal to the source voltage V. The current i, however, divides into two branches, which carry currents i1 and i2 . …(iii) i = i1 + i2 If R be the equivalent resistance, then V V V and i2 = i = , i1 = R R1 R2 Substituting in Eq. (iii), we get 1 1 1 for resistors in parallel = + R R1 R 2 This result can also be extended to a network consisting of n resistors in parallel. The result is 1 1 1 1 = + +… K + R R1 R 2 Rn V

Example 23.17 Compute the equivalent resistance of the network shown in figure and find the current i drawn from the battery. 18V i

i 6Ω

4Ω 3Ω Fig. 23.18

Solution

The 6 Ω and 3 Ω resistances are in parallel. Their equivalent resistance is 18V

4Ω

2Ω Fig. 23.19

i

Chapter 23

Current Electricity — 19

1 1 1 or R = 2 Ω = + R 6 3 Now, this 2 Ω and 4 Ω resistances are in series and their equivalent resistance is 4 + 2 = 6 Ω. Therefore, equivalent resistance of the network = 6 Ω.

Ans.

18V

i

6Ω

Fig. 23.20

Current drawn from the battery is i=

net emf 18 = net resistance 6

= 3A

Ans.

Kirchhoff’s Laws Many electric circuits cannot be reduced to simple series-parallel combinations. For example, two circuits that cannot be so broken down are shown in Fig. 23.21. R1

E1

A

C

B

A

R1

R2

R2

E2

D

C

B R3

D R4 E3

E2 E1

R3

F G

E

E

F

I R5

(a)

H (b)

Fig. 23.21

However, it is always possible to analyze such circuits by applying two rules, devised by Kirchhoff in 1845 and 1846 when he was still a student. First there are two terms that we will use often.

Junction A junction in a circuit is a point where three or more conductors meet. Junctions are also called nodes or branch points. For example, in Fig. (a) points D and C are junctions. Similarly, in Fig. (b) points B and F are junctions. Loop A loop is any closed conducting path. For example, in Fig. (a) ABCDA, DCEFD and ABEFA are loops. Similarly, in Fig. (b), CBFEC, BDGFB are loops.

20 — Electricity and Magnetism Kirchhoff’s rules consist of the following two statements :

Kirchhoff’s Junction Rule The algebraic sum of the currents into any junction is zero.

Σ

That is,

i =0

junction

This law can also be written as, “the sum of all the currents directed towards a point in a circuit is equal to the sum of all the currents directed away from that point.” Thus, in figure i1 + i2 = i3 + i4 The junction rule is based on conservation of electric charge. No charge can accumulate at a junction, so the total charge entering the junction per unit time must equal to charge leaving per unit time. Charge per unit time is current, so if we consider the currents entering to be positive and those leaving to be negative, the algebraic sum of currents into a junction must be zero.

i1

i2 i3 i4

Fig. 23.22

Kirchhoff’s Loop Rule The algebraic sum of the potential differences in any loop including those associated emf’s and those of resistive elements, must equal zero.

Σ

That is,

∆V = 0

closed loop

E E Kirchhoff’s second rule is based on the fact that the A B A B electrostatic field is conservative in nature. This result states that there is no net change in electric potential Path Path ∆V = VB – VA = –E around a closed path. Kirchhoff’s second rule applies only ∆V = VB – VA = +E for circuits in which an electric potential is defined at each Fig. 23.23 point. This criterion may not be satisfied if changing electromagnetic fields are present. In applying the loop rule, we need sign conventions. First assume a direction for the current in each branch of the circuit. Then starting at any point in the circuit, we imagine, travelling around a loop, adding emf’s and iR terms as we come to them. When we travel through a source in the direction from – to +, the emf is considered to be positive, when we travel from + to –, the emf is considered to be negative. When we travel through a resistor in the same direction as the assumed current, the iR term is negative because the current goes in the direction of decreasing potential. When we travel through a resistor in the direction opposite to the assumed current, the iR term is positive because this represents a rise of potential. R

R

i

A

B

i

A

Path

B Path

∆V = VB – VA = –iR

∆V = VB – VA = +iR

Fig. 23.24

Chapter 23

Current Electricity — 21

Note It is advised to write H (for higher potential) and L (for lower potential) across all the batteries and resistances of the loop under consideration while using the loop law. Then write – while moving from H to L and + for L to H. Across a battery write H on positive terminal and L on negative terminal. Across a resistance keep in mind the fact that current always flows from higher potential (H) to lower potential (L). For example, in the loop shown in figure we have marked H and L across all batteries and resistances. Now let us apply the second law in the loop ADCBA. E1

R1 C

B L

H

L

H

L E2

i

H i H

L A

D R2

Fig. 23.25

+ i R 2 − E2 + i R 1 + E1 = 0

The equation will be V

Example 23.18 Find currents in different branches of the electric circuit shown in figure. 4Ω

2Ω

B

A

C 4V

2V

6V

D

F

2Ω

E

4Ω

Fig. 23.26

In this problem there are three wires EFAB, BE and BCDE. Therefore, we have three unknown currents i1 , i2 and i3 . So, we require three equations. One equation will be obtained by applying Kirchhoff’s junction law (either at B or at E) and the remaining two equations, we get from the second law (loop law). We can make three loops ABEFA, ACDFA and BCDEB. But we have to choose any two of them. Initially, we can choose any arbitrary directions of i1 , i2 and i3 . Solution Applying Kirchhoff’s first law (junction law) at junction B, HOW TO PROCEED

H

A

4Ω L

H

B

2Ω L C

i3

i1 i2 H L

2V

H L

2

4V

H i1 L

F

L

1

i3

H 2Ω

E

6V

H

L

D 4Ω

Fig. 23.27

i1 = i 2 + i 3 Applying Kirchhoff’s second law in loop 1 ( ABEFA ), – 4 i1 + 4 – 2 i1 + 2 = 0

…(i) …(ii)

22 — Electricity and Magnetism Applying Kirchhoff’s second law in loop 2 ( BCDEB ), – 2 i3 – 6 – 4 i3 – 4 = 0

…(iii)

Solving Eqs. (i), (ii) and (iii), we get i1 = 1 A 8 i2 = A 3 5 i3 = – A 3 Here, negative sign of i 3 implies that current i 3 is in opposite direction of what we have assumed. V

Ans.

Example 23.19 In example 23.18, find the potential difference between points F and C. HOW TO PROCEED To find the potential difference between any two points of a circuit you have to reach from one point to the other via any path of the circuit. It is advisable to choose a path in which we come across the least number of resistors preferably a path which has no resistance. Solution

Let us reach from F to C via A and B,

VF + 2 – 4 i1 – 2 i 3 = VC ∴ VF – VC = 4 i1 + 2 i 3 – 2 5 Substituting, i1 = 1 A and i 3 = – A, we get 3 4 VF – VC = – volt 3 Here, negative sign implies that VF < VC .

Ans.

Internal Resistance (r ) and Potential Difference (V ) across the Terminals of a Battery The potential difference across a real source in a circuit is not equal to the emf of the cell. The reason is that charge moving through the electrolyte of the cell encounters resistance. We call this the internal resistance of the source, denoted by r. If this resistance behaves according to Ohm’s law r is constant and independent of the current i. As the current moves through r, it experiences an associated drop in potential equal to ir. Thus, when a current is drawn through a source, the potential difference between the terminals of the source is V = E – ir This can also be shown as below. E

r

A

i

Fig. 23.28

V A – E + ir = VB or

V A – VB = E – ir

B

Chapter 23

Current Electricity — 23

The following three special cases are possible : (i) If the current flows in opposite direction (as in case of charging of a battery), then V = E + ir (ii) V = E , if the current through the cell is zero. (iii) V = 0, if the cell is short circuited. This is because current in the circuit E Short i= circuited r or E = ir ∴

E – ir = 0

or

V =0

E

Thus, we can summarise it as follows : E

r

V = E – ir

or

V E

V =E

if

i =0

V =0

if short circuited

i E

r i E

r

i=

E

E r

r 2A r

+ 12 V

2Ω

4Ω

12 V ir = 4V 8V

r

Fig. 23.29

E = 12V iR = 8V

O Fig. 23.30 Potential rise and fall in a circuit.

24 — Electricity and Magnetism Extra Points to Remember ˜

˜

˜

A In figure (a) : There are eight wires and hence, will have eight currents or 1 eight unknowns. The eight wires are AB, BC, CE, EA, AD, BD, CD and ED. 2 4 D Number of independent loops are four. Therefore, from the second law we 3 E can make only four equations. Total number of junctions are five (A, B, C, D (a) and E). But by using the first law, we can make only four equations (one A less). So, the total number of equations are eight. 1 In figure (b) : Number of wires are six (AB, BC, CDA, BE, AE and CE). 2 3 E Number of independent loops are three so, three equations can be obtained D C from the second law. Number of junctions are four (A, B, C and E) so, we can (b) make only three (one less) equations from the first law. But total number of Fig. 23.31 equations are again six.

C B

R1

Short circuiting : Two points in an electric circuit directly connected by a conducting wire are called short circuited. Under such condition both points are at same potential. For example, resistance R1 in the adjoining circuit is short circuited, i.e. potential difference across it is zero. Hence, no current will flow through R1 and the current through R 2 is therefore, E / R 2 . Earthing : be zero.

B

R2 E Fig. 23.32 6V

If some point of a circuit is earthed, then its potential is taken to

D

E

For example, in the adjoining figure,

2Ω 3V

VA = VB = 0

B

C

VF = VC = VD = – 3 V VE = – 9 V ∴

F

Fig. 23.33

or current through 2 Ω resistance is VB – VE 2

or

A

4Ω

VB – VE = 9 V 9 A 2

(from B to E)

Similarly,

˜

VA – VF = 3 V V – VF 3 and the current through 4 Ω resistance is A or A (from A to F) 4 4 For a current flow through a resistance there must be a potential difference across it but between any two points of a circuit the potential difference may be zero.

A 1V

For example, in the circuit, net emf = 3 V and net resistance = 6 Ω 3 1 ∴ current in the circuit, i = = A 6 2 1 VA + 1 – 2 × = VB or VA – VB = 0 VA – VB 2 or by symmetry, we can say that VA = VB = VC

2Ω 1V

2Ω i B

2Ω

C

1V Fig. 23.34

So, the potential difference across any two vertices of the triangle is zero, while the current in the circuit is non-zero.

Current Electricity — 25

Chapter 23 ˜

Distribution of current in parallel connections : When more than one resistances are connected in parallel, the potential difference across them is equal and the current is distributed among them in inverse ratio of their resistance as V i = R 1 or for same value of V i ∝ R

R i1 i

i2

2R

i3

3R Fig. 23.35

e.g. in the figure, i1 : i 2 : i 3 =

1 1 1 : : = 6: 3:2 R 2 R 3R

  6 6 i1 =  i  i = 6 + 3 + 2 11  

∴

  3 3 i2 =  i  i = 11  6 + 3 + 2   2 2 i3 =  i  i = 11  6 + 3 + 2

and

Note ˜

In case of only two resistances,

i1 R2 = i2 R1

Distribution of potential in series connections : When more than one resistances are connected in series, the current through them is same and the potential is distributed in the direct ratio of i their resistance as V = iR or

R

2R

3R

V1

V2

V3

V ∝ R for same value of i.

For example in the figure, V1 : V2 : V3 = R : 2 R : 3R = 1 : 2 : 3 ∴

  1 V V1 =   V= 6  1 + 2 + 3   2 V V2 =   V= 1 + 2 + 3 3  

and

V

  3 V V3 =   V= 2  1 + 2 + 3

Example 23.20 In the circuit shown in figure, E1 r 1

E2

r2

R Fig. 23.37

E1 = 10 V , E2 = 4 V , r1 = r2 = 1 Ω and R = 2 Ω. Find the potential difference across battery 1 and battery 2.

V Fig. 23.36

26 — Electricity and Magnetism Solution

E1

Net emf of the circuit = E1 – E 2 = 6 V

Total resistance of the circuit = R + r1 + r2 = 4 Ω net emf 6 ∴ Current in the circuit, i = = = 1.5 A total resistance 4 Now,

E2

r1

r2

i V1

V2

R Fig. 23.38

V1 = E1 – ir1 = 10 – (1.5) (1) = 8.5 V

and

Ans.

V2 = E 2 + ir2 = 4 + (1.5) (1) = 5.5 V

INTRODUCTORY EXERCISE

Ans.

23.6

1. Find the current through 2 Ω and 4 Ω resistance. 10V

2Ω

4Ω

Fig. 23.39

2. In the circuit shown in figure, find the potentials of A, B, C and D and

1Ω

C

the current through 1 Ω and 2 Ω resistance.

B

5V 2V 10 V D

2Ω

A

Fig. 23.40

3. For what value of E the potential of A is equal to

15V

1Ω

E

A

2Ω

the potential of B?

B

5Ω Fig. 23.41

4. Ten cells each of emf 1 V and internal resistance 1 Ω are connected in series. In this arrangement, polarity of two cells is reversed and the system is connected to an external resistance of 2 Ω. Find the current in the circuit.

5. In the circuit shown in figure, R1 = R2 = R3 = 10 Ω. Find the

R1

currents through R1 and R2. 10 V

R2

Fig. 23.42

R3

10 V

Chapter 23

Current Electricity — 27

23.10 Heating Effects of Current An electric current through a resistor increases its thermal energy. Also, there are other situations in which an electric current can produce or absorb thermal energy.

Power Supplied or Power Absorbed by a Battery When charges are transported across a source of emf, their potential energy changes. If a net charge ∆ q moves through a potential difference E in a time ∆ t, the change in electric potential energy of the charge is E ∆ q. Thus, the source of emf does work, ∆W = E ∆ q Dividing both sides by ∆t, then taking the limit as ∆t → 0, we find dq dW =E dt dt dq dW By definition, = P , the power output of (or input to) the = i, the current through the battery and dt dt battery. Hence, P = Ei The quantity P represents the rate at which energy is transferred from a discharging battery or to a charging battery. In Fig. 23.43, energy is transferred from the source at a rate Ei E

i

Fig. 23.43

In Fig. 23.44, energy is transferred to the source at a rate Ei E

i

Fig. 23.44

Power dissipated across a resistance Now, let’s consider the power dissipated in a conducting element. Suppose it has a resistance R and the potential difference between its ends is V. In moving from higher to lower potential, a positive charge ∆ q loses energy ∆ U = V ∆ q. This electric energy is absorbed by the conductor through collisions between its atomic lattice and the charge carriers, causing its temperature to rise. This effect is commonly called Joule heating. Since, power is the rate at which energy is transferred, we have, ∆q ∆U P= =V ⋅ =V i ∆t ∆t ∴

P =V i

which with the help of equation V = iR can also be written in the forms, P = i2R

or

P=

V2 R

28 — Electricity and Magnetism Power is always dissipated in a resistance. With this rate, the heat produced in the resistor in time t is H = Pt

or

V2 H = Vit = i Rt = t R

V L

H

i

R Fig. 23.45

2

Joule heating occurs whenever a current passes through an element that has resistance. To prevent the overheating of delicate electronic components, many electric devices like video cassette recorders, televisions and computer monitors have fans in their chassis to allow some of the heat produced to escape.

Extra Points to Remember ˜

We have seen above that power may be supplied or consumed by a battery. It depends on the direction of current. i E Fig. 23.46

In the above direction of current power is supplied by the battery (= Ei ) i E Fig. 23.47

˜

In the opposite direction of current shown in Fig. 23.47, power is consumed by the battery. This normally happens during charging of a battery. A resistance always consumes power. It does not depend on the direction of current. i

or Fig. 23.48

i

In both cases shown in figure, power is only consumed and this power consumed is given by the formula. P = i 2R =

˜

V

V2 = Vi R

In the above equations V and i are the values across a resistance in which we wish to find the power consumed. In any electrical circuit, law of conservation of energy is followed. Net power supplied by all batteries of the circuit = net power consumed by all resistors in the circuit.

Example 23.21 In the circuit shown in figure, find 10V

4V

i 3Ω Fig. 23.49

(a) the power supplied by 10 V battery (b) the power consumed by 4 V battery and (c) the power dissipated in 3 Ω resistance.

Chapter 23

Current Electricity — 29

Net emf of the circuit = (10 – 4 ) V = 6 V ∴ Current in the circuit net emf 6 i= = = 2A total resistance 3

Solution

(a) Power supplied by 10 V battery = Ei = (10) ( 2) = 20 W

Ans.

(b) Power consumed by 4 V battery = Ei = ( 4 ) ( 2) = 8 W

Ans.

2

2

(c) Power consumed by 3 Ω resistance = i R = ( 2) ( 3) = 12 W

Ans.

Note Here, we can see that total power supplied by 10 V battery (i.e. 20 W) = power consumed by 4 V battery and 3 Ω resistance. Which proves that conservation of energy holds good in electric circuits also. V

Example 23.22 In the circuit shown in figure, find the heat developed across each resistance in 2 s. 6Ω 3Ω 3Ω

5Ω

20V Fig. 23.50

Solution

The 6 Ω and 3 Ω resistances are in parallel. So, their combined resistance is 1 1 1 1 = + = R 6 3 2

or R = 2Ω The equivalent simple circuit can be drawn as shown. 3Ω

2Ω V

i

5Ω

20 V Fig. 23.51

Current in the circuit, i=

i.e.

net emf 20 = = 2A total resistance 3 + 2 + 5

V = iR = ( 2) ( 2) = 4 V Potential difference across 6 Ω and 3 Ω resistances are 4 V. Now,

H 3 Ω (which is connected in series) = i 2 Rt = ( 2) 2 ( 3) ( 2) = 24 J H6Ω =

(4 )2 V2 16 t= ( 2) = J R 6 3

30 — Electricity and Magnetism H 3 Ω (which is connected in parallel) =

( 4 ) 2 ( 2) 32 V2 t= = J R 3 3

H 5 Ω = i 2 Rt = ( 2) 2 ( 5) ( 2) = 40 J

and

INTRODUCTORY EXERCISE

Ans.

23.7

1. In the circuit shown in figure, a 12 V battery with unknown internal resistance r is connected to another battery with unknown emf E and internal resistance 1 Ω and to a resistance of 3 Ω carrying a current of 2 A. The current through the rechargeable battery is 1 A in the direction shown. Find the unknown current i, internal resistance r and the emf E. 12 V

r i

E

1Ω 1A 3Ω 2A Fig. 23.52

2. In the above example, find the power delivered by the 12 V battery and the power dissipated in 3 Ω resistor.

23.11 Grouping of Cells Cells are usually grouped in the following three ways :

Series Grouping Suppose n cells each of emf E and internal resistance r are connected in series as shown in figure. E

E

r

E

r

r

i R Fig. 23.53

Net emf = nE

Then,

Total resistance = nr + R ∴

Current in the circuit, i =

net emf total resistance

or

i=

nE nr + R

Note If polarity of m cells is reversed, then equivalent emf = ( n – 2 m) E, while total resistance is still nr + R ∴

i=

( n – 2 m) E nr + R

Chapter 23

Current Electricity — 31

Parallel Grouping Here, three cases are possible. When E and r of each cell has same value and positive terminals of all cells are connected at one junction while negative at the other. r In this situation, the net emf is E. The net internal resistance is as n n resistances each of r are in parallel. Net external resistance is R. Therefore, r  total resistance is  + R  and so the current in the circuit will be, n 

E r

Case 1

i=

Net emf Total resistance

i=

or

E r E

r

i

i R Fig. 23.54

E R + r/ n

Note A comparison of series and parallel grouping reveals that to get maximum current, cells must be connected in series if effective internal resistance is lesser than external and in parallel if effective internal resistance is greater than external.

If E and r of each cell are different but still the positive terminals of all cells are connected at one junction while negative at the other. Case 2

A

E1 r1

F

i1 i2

E2 r2 E

B i3

E3 r3 i

i R

D

C Fig. 23.55

Applying Kirchhoff’s second law in loop ABCDEFA, E1 – iR – i1 r1 = 0 or i1 = – i

R E1 + r1 r1

…(i)

Similarly, we can write i2 = – i

R E2 + r2 r2

… … … … Adding all above equations, we have  1 E ( i1 + i2 +… + in ) = – iR Σ   + Σ    r r But ∴

i1 + i2 +… + in = i  1 E i = – iR Σ   + Σ    r r

…(ii)

32 — Electricity and Magnetism ∴ where,

i= E eq =

Σ ( E / r) Σ ( E / r )/ Σ (1/ r ) E eq = = 1 + R Σ (1/ r ) {1/ Σ (1/ r )} + R R eq Σ (E / r ) Σ (1/ r )

and

R eq = R +

1 Σ (1/ r )

E if n cells of same emf E and internal R + r/ n resistance r are connected in parallel. This is because, Σ ( E / r ) = nE / r and Σ (1/ r ) = n / r nE / r ∴ i= 1 + nR / r From the above expression, we can see that i =

Multiplying the numerator and denominator by r / n, we have E i= R + r/ n

Exercise In parallel grouping (Case 2) prove that, E eq = E if E1 = E 2 = K = E and r1 = r2 = K = r This is the most general case of parallel grouping in which E and r of different cells are different and the positive terminals of few cells are connected to the negative terminals of the others as shown figure. Case 3

E1 r1 i1 i2 i3

E2 r 2 E3 r3 i

i R Fig. 23.56

Kirchhoff’s second law in different loops gives the following equations : E iR E1 – iR – i1 r1 = 0 or i1 = 1 – r1 r1 – E 2 – iR – i2 r2 = 0 or i2 = – i3 =

Similarly,

E 2 iR – r2 r2

E 3 iR – r3 r3

Adding Eqs. (i), (ii) and (iii), we get i1 + i2 + i3 = ( E1 / r1 ) – ( E 2 / r2 ) + ( E 3 / r3 ) – iR (1/ r1 + 1/ r2 + 1/ r3 ) or

i [1 + R (1/ r1 + 1/ r2 + 1/ r3 )] = ( E1 / r1 ) – ( E 2 / r2 ) + ( E 3 / r3 )

…(i) …(ii) …(iii)

Chapter 23

Current Electricity — 33

∴

i=

( E1 / r1 ) – ( E 2 / r2 ) + ( E 3 / r3 ) 1 + R (1 / r1 + 1 / r2 + 1 / r3 )

or

i=

( E1 / r1 – E 2 / r2 + E 3 / r3 ) / (1/ r1 + 1/ r2 + 1/ r3 ) R + 1 / (1 / r1 + 1 / r2 + 1 / r3 )

E

r

Mixed Grouping The situation is shown in figure.

i

R

i

Fig. 23.57

There are n identical cells in a row and number of rows are m. Emf of each cell is E and internal resistance is r. Treating each row as a single cell of emf nE and internal resistance nr, we have Net emf = nE nr Total internal resistance = m Total external resistance = R ∴ Current through the external resistance R is nE i= nr R+ m This expression after some rearrangements can also be written as mnE i= ( mR – nr ) 2 + 2 mnrR If total number of cells are given then mn is fixed. E , r and R are also given, we have liberty to arrange the given number of cells in different rows. Then in the above expression the numerator nmE and in the denominator 2 mnrR all are fixed. Only the square term in the denominator is variable. Therefore, i is maximum when, nr mR = nr or R = m or total external resistance = total internal resistance Thus, we can say that the current and hence power transferred to the load is maximum when load resistance is equal to internal resistance. This is known as maximum power transfer theorem.

34 — Electricity and Magnetism Extra Points to Remember ˜

Regarding maximum current in the circuit or maximum power consumed by the external resistance R, there are three special cases. One we have discussed above, where we have to arrange the cells in such a manner that current and power in the circuit should be maximum. And this happens when we arrange the cells in such a manner that total internal resistance comes out to be equal to total external resistance. Rest two cases are discussed below.

E, r

i

R Fig. 23.58

In the figure shown,

i =

E R+ r 2

 E  PR = i 2 R =   R  R + r (i) Now if r is variable. E and R are fixed, then i and PR both are maximum when r = 0. (ii) If R is variable. E and r are fixed. Then, current in the circuit is maximum when R = 0. But PR will be maximum, when R = r or external resistance = internal resistance V

Example 23.23 Find the emf and internal resistance of a single battery which is equivalent to a combination of three batteries as shown in figure. 10V 2Ω 6V 1Ω

4V 2Ω Fig. 23.59

The given combination consists of two batteries in parallel and resultant of these two in series with the third one. For parallel combination we can apply, E1 E 2 10 4 – – r1 r2 E eq = = 2 2 = 3V 1 1 1 1 + + r1 r2 2 2 Solution

Further,

1 1 1 1 1 = + = + =1 req r1 r2 2 2

∴

req = 1 Ω

Now this is in series with the third one, i.e. 6V

3V 1Ω

1Ω

Fig. 23.60

E = 3V r = 2Ω Fig. 23.61 V

Example 23.24 In the circuit shown in figure E1 = 3V, E2 = 2V, E3 = 1 V and (JEE 1981) R = r1 = r2 = r3 = 1 Ω.

R

(a) Find the potential difference between the points A and B and the currents through each branch. (b) If r2 is short-circuited and the point A is connected to point B, find the currents through E1 , E2 , E3 and the resistor R. Solution (a) Equivalent emf of three batteries would be Σ ( E / r ) ( 3 / 1 + 2 / 1 + 1/ 1) E eq = = = 2V Σ (1/ r ) (1/ 1 + 1 / 1 + 1 / 1) Ω as all three are in parallel. A

2V

R

1Ω 3

B

Fig. 23.63

The equivalent circuit is therefore shown in the figure. Since, no current is taken from the battery. V AB = 2 V Further,

V A − VB = E1 − i1 r1

∴

i1 =

VB − V A + E1 − 2 + 3 = = 1A r1 1

Similarly,

i2 =

VB − V A + E 2 − 2 + 2 = =0 r2 1

and

i3 =

VB − V A + E 3 − 2 + 1 = = − 1A r3 1

(From V = E − i r)

36 — Electricity and Magnetism (b) r2 is short circuited means resistance of this branch becomes zero. Making a closed circuit with a battery and resistance R. Applying R =1 Ω Kirchhoff’s second law in three loops so formed. (i1+i2 +i3) …(i) 3 − i1 − ( i1 + i 2 + i 3 ) = 0 …(ii) 2 − ( i1 + i 2 + i 3 ) = 0 …(iii) 1 − i 3 − ( i1 + i 2 + i 3 ) = 0 From Eq. (ii) i1 + i 2 + i 3 = 2 A ∴ Substituting in Eq. (i), we get i1 = 1A Substituting in Eq. (iii), we get i3 = − 1A ∴ i2 = 2 A

INTRODUCTORY EXERCISE

3V

1Ω i1

i3

i2

2V

1Ω 1V

Fig. 23.64

23.8

1. Find the emf (V) and internal resistance (r) of a single battery which is equivalent to a parallel combination of two batteries of emfs V1 and V2 and internal resistances r1 and r2respectively, with polarities as shown in figure V2

r2 +

B

A r1

–

V1

Fig. 23.65

2. Find the net emf of the three batteries shown in figure. 2V 1Ω 4V

0.5 Ω 1Ω 6V Fig. 23.66

3. Find the equivalent emf and internal resistance of the arrangement shown in figure. 10V

1Ω 2Ω

4V

6V

2Ω

Fig. 23.67

Chapter 23

Current Electricity — 37

23.12 Electrical Measuring Instruments So far we have studied about current, resistance, potential difference and emf. Now, in this article we will study how these are measured. The basic measuring instrument is galvanometer, whose pointer shows a deflection when current passes through it. A galvanometer can easily be converted into an ammeter for measuring current, into a voltmeter for measuring potential difference. For accurate measurement of potential difference or emf, a potentiometer is more preferred. Resistances are accurately measured by using post office box or meter bridge which are based on the principle of “Wheatstone bridge”. All these are discussed here one by one in brief.

Galvanometer Many common devices including car instrument panels, battery chargers measure potential difference, current or resistance using d’Arsonval Galvanometer. It consists of a pivoted coil placed in the magnetic field of a permanent magnet. Attached to the coil is a spring. In the equilibrium position, with no current in the coil, the pointer is at zero and spring is relaxed. When there is a current in the coil, the magnetic field exerts a torque on the coil that is proportional to current. As the coil turns, the spring exerts a restoring torque that is proportional to the angular displacement. Thus, the angular deflection of the coil and pointer is directly proportional to the coil current and the device can be calibrated to measure current. The maximum deflection, typically 90° to 120° is called full scale deflection. The essential electrical characteristics of the galvanometer are the current ig required for full scale deflection (of the order of 10 µA to 10 mA) and the resistance G of the coil (of the order of 10 to 1000 Ω). The galvanometer deflection is proportional to the current in the coil. If the coil obeys Ohm’s law, the current is proportional to potential difference. The corresponding potential difference for full scale deflection is V = igG

Ammeter A current measuring instrument is called an ammeter. A galvanometer can be converted into an ammeter by connecting a small resistance S (called shunt) in parallel with it. Suppose we want to convert a galvanometer with full scale current ig and coil resistance G into an ammeter with full scale reading i. To determine the shunt resistance S needed, note that, at full scale deflection the total current through the parallel combination is i, the current through the galvanometer is ig and the current through the shunt is i – ig . The potential difference Vab ( = Va – Vb ) is the same for both paths, so igG = ( i – ig ) S ∴

 ig   G S =   i – ig 

G ig i

S + a

i – ig

Fig. 23.68

– b

38 — Electricity and Magnetism Voltmeter A voltage measuring device is called a voltmeter. It measures the potential difference between two points. A galvanometer can be converted into a voltmeter by connecting a high resistance (R) in series with it. The whole assembly called the voltmeter is connected in parallel between the points where potential difference has to be measured. For a voltmeter with full scale reading V, we need a series resistor R such that

G R ig ig +

i

–

Circuit element

a

b

i

V Fig. 23.69

V = ig (G + R ) R=

or

V –G ig

Extra Points to Remember ˜

Conversion of galvanometer into an ammeter. (i) A galvanometer is converted into an ammeter by connecting a low resistance (called shunt) in parallel with galvanometer. This assembly (called ammeter) is connected in series in the wire in which current is to be found. Resistance of an ideal ammeter should be zero. In parallel, current distributes in the inverse ratio of resistance. Therefore, i S = g G i − ig ∴

 i  S = shunt =  g  G − i i  g 

(ii) Resistance of an ammeter is given by 1 1 1 = + A G S

⇒ A=

GS G+S

S i – ig i ig

G

Fig. 23.70

Current Electricity — 39

Chapter 23

(iii) The reading of an ammeter is always lesser than actual current in the circuit. i¢ R

R

A

i¢

i

S

G i¢

E

E

(a)

(b) Fig. 23.71

For example, in Fig. (a), actual current through R is E i = R

…(i)

 GS  while the current after connecting an ammeter of resistance A  =  in series with R is  G + S i′ =

˜

E R+ A

…(ii)

From Eqs. (i) and (ii), we see that i ′ < i and i ′ = i when A = 0. i.e. resistance of an ideal ammeter should be zero. Conversion of a galvanometer into a voltmeter (i) A galvanometer is converted into a voltmeter by connecting a high resistance in series with galvanometer. The whole assembly called voltmeter is connected in parallel across the two points between which potential difference is to be found. Resistance of an ideal voltmeter should be infinite.

R G ig

Fig. 23.72

V = Ig (G + R ) ∴

R = high resistance required in series V = −G ig

(ii) Resistance of a voltmeter is R V = R + G (iii) The reading of a voltmeter is always lesser than the true value. For example, if a current i is passing through a resistance r, the actual value is r

r R

i

i

i

G

RV V

Fig. 23.73

V = ir

…(i)

Now, if a voltmeter of resistance R V (= G + R ) is connected across the resistance r, the new value will be V′ =

i × (rR V ) or r + RV

From Eqs. (i) and (ii), we can see that,

V′ < V

V′ =

ir 1+

and

V′ = V

Thus, resistance of an ideal voltmeter should be infinite.

…(ii)

r RV if R V = ∞

40 — Electricity and Magnetism V

Example 23.25 What shunt resistance is required to make the 1.00 mA, 20 Ω galvanometer into an ammeter with a range of 0 to 50.0 mA? Solution Here, i g = 1.00 mA = 10–3 A, G = 20 Ω, i = 50.0 × 10–3 A  ig  (10–3 ) ( 20) G= Substituting in S =   ( 50.0 × 10–3 ) – (10–3 )  i – ig  = 0.408 Ω

Ans.

Note The resistance of ammeter is given by 1 1 1 1 1 = + = + A G S 20 0.408 or A = 0.4 Ω The shunt resistance is so small in comparison to the galvanometer resistance that the ammeter resistance is very nearly equal to the shunt resistance. This shunt resistance gives us a low resistance ammeter with the desired range of 0 to 50.0 mA. At full scale deflection i = 50.0 mA , the current through the galvanometer is 1.0 mA while the current through the shunt is 49.0 mA. If the current i is less than 50.0 mA, the coil current and the deflection are proportionally less, but the ammeter resistance is still 0.4 Ω. V

Example 23.26 How can we make a galvanometer with G = 20 Ω and i g = 1.0 mA into a voltmeter with a maximum range of 10 V? Solution

We have,

Using R =

V – G, ig R=

10 – 20 10–3

= 9980 Ω

Ans.

Thus, a resistance of 9980 Ω is to be connected in series with the galvanometer to convert it into the voltmeter of desired range. Note At full scale deflection current through the galvanometer, the voltage drop across the galvanometer Vg = igG = 20 × 10 –3 volt = 0.02 volt and the voltage drop across the series resistance R is V = ig R = 9980 × 10 –3 volt = 9.98 volt or we can say that most of the voltage appears across the series resistor. V

Example 23.27 Resistance of a milliammeter is R1 of an ammeter is R 2 of a voltmeter is R3 and of a kilovoltmeter is R 4 . Find the correct order of R1 , R 2 , R3 and R 4 . To increase the range of an ammeter a low resistance has to be connected in parallel with galvanometer. Therefore, net resistance decreases. To increase the range of voltmeter, a high resistance has to be connected in series. So, net resistance further increases. Therefore, the correct order is Solution

R4 > R3 > R1 > R2

Chapter 23 V

Current Electricity — 41

Example 23.28 A microammeter has a resistance of 100 Ω and full scale range of 50 µA. It can be used as a voltmeter or as a higher range ammeter provided a resistance is added to it. Pick the correct range and resistance combination (s) (JEE 1991) (a) 50 V range with 10 kΩ resistance in series (b) 10 V range with 200 kΩ resistance in series (c) 5 mA range with 1 Ω resistance in parallel (d) 10 mA range with 1 Ω resistance in parallel Solution To increase the range of ammeter a parallel resistance (called shunt) is required which is given by  ig  G S =    i − ig  For option (c),

  50 × 10−6 S =  (100) ≈ 1 Ω −3 −6  5 × 10 − 50 × 10 

To change it in voltmeter, a high resistance R is put in series, where R is given by R = R=

For option (b),

V −G ig

10 − 100 ≈ 200 kΩ 50 × 10−6

Therefore, options (b) and (c) are correct. V

Example 23.29 A galvanometer gives full scale deflection with 0.006 A current. By connecting it to a 4990 Ω resistance, it can be converted into a 2n voltmeter of range 0-30 V. If connected to a Ω resistance, it becomes an 249 ammeter of range 0-1.5 A. The value of n is (JEE 2014) Solution

ig 4990 Ω G V Fig. 23.74

⇒ ⇒

i g (G + 4990) = V 6 (G + 4990) = 30 1000 30000 G + 4990 = = 5000 6 S d

c (1.5– ig) 1.5 A

a

ig

G

Fig. 23.75

b

42 — Electricity and Magnetism ⇒

G = 10 Ω Vab = Vcd

⇒

i gG = (1.5 − i g ) S 6 6   × 10 = 1.5 − S  1000 1000

⇒

60 2n = 1494 249 249 × 30 n= 1494 2490 = =5 498

⇒

S =

⇒

INTRODUCTORY EXERCISE

Ans.

23.9

1. The full scale deflection current of a galvanometer of resistance 1 Ω is 5 mA. How will you convert it into a voltmeter of range 5 V?

2. A micrometer has a resistance of 100 Ω and full scale deflection current of 50 µA. How can it be made to work as an ammeter of range 5 mA?

3. A voltmeter has a resistance G and range V. Calculate the resistance to be used in series with it to extend its range to nV.

Potentiometer The potentiometer is an instrument that can be used to measure the emf or the internal resistance of an unknown source. It also has a number of other useful applications.

Principle of Potentiometer The principle of potentiometer is schematically shown in figure. E1 i a

i

i b

c

i2 = 0 G E2, r

Fig. 23.76

A resistance wire ab of total resistance R ab is permanently connected to the terminals of a source of known emf E 1 .A sliding contact c is connected through the galvanometer G to a second source whose emf E 2 is to be measured. As contact c is moved along the potentiometer wire, the resistance R cb between points c and b varies. If the resistance wire is uniform R cb is proportional to the length of the wire between c and b. To determine the value of E 2 , contact c is moved until a position is found at

Current Electricity — 43

Chapter 23

which the galvanometer shows no deflection. This corresponds to zero current passing through E 2 . With i2 = 0, Kirchhoff’s second law gives E 2 = iR cb With i2 = 0, the current i produced by the emf E 1 has the same value no matter what the value of emf E 2 . A potentiometer has the following applications.

To find emf of an unknown battery E1

E1

i l1 a

l2 i

i

a

b

c

b

c

i2 = 0

i2 = 0

G

G EK

EU

Fig. 23.77

We calibrate the device by replacing E 2 by a source of known emf E K and then by unknown emf EU . Let the null points are obtained at lengths l1 and l2 . Then, E K = i (ρ l1 )

EU = i (ρl2 )

and

Here, ρ = resistance of wire ab per unit length. ∴

E K l1 = EU l2

l  EU =  2  E K  l1 

or

So, by measuring the lengths l1 and l2 , we can find the emf of an unknown battery.

To find the internal resistance of an unknown battery To find the internal resistance of an unknown battery let us derive a formula. E

r

i R

Fig. 23.78

In the circuit shown in figure, i=

E R +r

and V = potential difference across the terminals of the battery or V = E – ir = iR From Eqs. (i) and (ii), we can prove that E  r = R  – 1 V 

…(i)

…(ii)

44 — Electricity and Magnetism Thus, if a battery of emf E and internal resistance r is connected across a resistance R and the potential difference across its terminals comes out to be V then the internal resistance of the battery is given by the above formula. Now, let us apply it in a potentiometer for finding the internal resistance of the unknown battery. The circuit shown in Fig. 23.79 is similar to the previous one. E1

i

i l1

a

b c i2 = 0 G E

r

Fig. 23.79

Hence, …(i) E = iρl1 Now, a known resistance R is connected across the terminals of the unknown battery as shown in Fig. 23.80. E1

i

i

i l2 a

b c E i2 = 0

r

G i1 ≠ 0 R

Fig. 23.80

This time Vcb ≠ E , but Vcb = V where, V = potential difference across the terminals of the unknown battery. Hence, V = i ρl2 From Eqs. (i) and (ii), we get E l1 = V l2 E  Substituting in r = R  – 1 , we get V  l  r = R  1 – 1 l  2  So, by putting R, l1 and l2 we can determine the internal resistance r of unknown battery.

…(ii)

Chapter 23

Current Electricity — 45

Extra Points to Remember ˜

E1 1

A l

C

A

B

l1 2 E2 D

G r2 E 3

l2

R

S

Fig. 23.81

Under balanced condition (when IG = 0 ) loop-1 and loop-3 are independent with each other. All problems in this condition can be solved by a single equation, VAC = VDE

˜

˜

˜

V

or

i1 R AC = E2 − i 2 r2

or

i1 λ l = E2 − i 2 r2

…(i)

Here, λ is the resistance per unit length of potentiometer wire AB. Length l is called balance point length. Currents i1 and i 2 are independent with each other. Current i 2 = 0, if switch is open. Under balanced condition, a part of potential difference of E1 is balanced by the lower circuit. So, normally E2 < E1 for taking balance point length. Similarly, VAC = VDE ⇒ VA = VD and VC = VE . Therefore, positive terminals of both batteries should be on same side and negative terminals on the other side. From Eq. (i), we can see that null point length is E − i 2 r2 l= 2 i1 λ Now, suppose E1 is increased then i1 will also increase and null point length l will decrease. Similarly, we can make some other cases also. If we do not get any balanced condition(IG ≠ 0 ), then the given circuit is simply a three loops problem, which can be solved with the help of Kirchhoff's laws.

Example 23.30 A potentiometer wire of length 100 cm has a resistance of 10 Ω. It is connected in series with a resistance R and a cell of emf 2 V and of negligible internal resistance. A source of emf 10 mV is balanced against a length of 40 cm of the potentiometer wire. What is the value of R? Solution From the theory of potentiometer, Vcb = E, if no current is drawn from the battery  E1  or   R cb = E  R + R ab  Here,

 40  E 1 = 2 V, R ab = 10 Ω, R cb =   × 10 = 4 Ω  100

46 — Electricity and Magnetism E = 10 × 10–3 V

and

E1 i R a

b c E G

Fig. 23.82

Substituting in above equation, we get R = 790 Ω V

Ans.

Example 23.31 When the switch is open in lowermost loop of a potentiometer, the balance point length is 60 cm. When the switch is closed with a known resistance of R = 4 Ω, the balance point length decreases to 40 cm. Find the internal resistance of the unknown battery. Solution

Using the result, l  r = R  1 − 1 l  2   60  = 4  − 1  40  = 2Ω

V

Ans.

Example 23.32 2Ω

20 V

i1

A l C

A

B

G IG = 0 2Ω

D

8V

l2 6Ω

Fig. 23.83

In the figure shown, wire AB has a length of 100 cm and resistance 8 Ω. Find the balance point length l.

Chapter 23 Solution

Current Electricity — 47

i 1 λl = E 2 − i 2 r2

Using the equation,

We have,  20     2 + 8

 8   8   ( 2)   l = 8−   100  2 + 6

Solving this equation, we get l = 37.5 cm

INTRODUCTORY EXERCISE

Ans.

23.10

1. In a potentiometer experiment it is found that no current passes through the galvanometer when the terminals of the cell are connected across 0.52 m of the potentiometer wire. If the cell is shunted by a resistance of 5 Ω a balance is obtained when the cell is connected across 0.4 m of the wire. Find the internal resistance of the cell.

2. The potentiometer wire AB is 600 cm long. E

r

R = 15 r

A

J

B

r G E 2

Fig. 23.84

(a) At what distance from A should the jockey J touch the wire to get zero deflection in the galvanometer. (b) If the jockey touches the wire at a distance 560 cm from A, what will be the current through the galvanometer.

Principle of Wheatstone’s Bridge

VB = VD Under this condition, or or Similarly,

VA – VB = VA – VD i1 P = i2 R i1 R = i2 P VB – VC = VD − VC

B i1 P

ig = 0

The scientist Wheatstone designed a circuit to find unknown resistance. Such a circuit is popularly known as Wheatstone’s bridge. This is an arrangement of four resistances which can be used to measure one of them in terms of the rest. The figure shows the circuit designed by him. The bridge is said to be balanced when deflection in galvanometer is zero, i.e. ig = 0, and hence,

A

i1 Q

C

G R

S i2

i2 i E

D

Fig. 23.85

…(i)

48 — Electricity and Magnetism or From Eqs. (i) and (ii),

i1Q = i2 S

or

R S = P Q

or

i1 S = i2 Q

…(ii)

P R = Q S

So, this is a condition for which a Wheatstone’s bridge is balanced. To measure the resistance of an unknown resistor, it is connected as one of the four resistors in the bridge. One of the other three should be a variable resistor. Let us suppose P is the unknown resistance and Q is the variable resistance. The value of Q is so adjusted that deflection through the galvanometer is zero. In this case, the bridge is balanced and R P =   ⋅Q S Knowing R, S and Q, the value of P is calculated. Following two points are important regarding a Wheatstone’s bridge. (i) In Wheatstone’s bridge, cell and galvanometer arms are interchangeable. B

B Q

P A

C

A

C

G

Q

P

Þ R

S

R

S D

D G

Fig. 23.86

In both the cases, condition of balanced bridge is P R = Q S (ii) If bridge is not balanced current will flow from D to B in Fig. 23.85 if, PS > RQ

Exercise Try and prove the statements of both the points yourself.

Meter Bridge Experiment Meter bridge works on Wheatstone's bridge principle and is used to find the unknown resistance (X) and its specific resistance (or resistivity).

Theory As the meter bridge wire AC has uniform material density and area of cross-section, its resistance is proportional to its length. Hence, AB and BC are the ratio arms and their resistances correspond to P and Q respectively. Resistance of AB P λl l Thus, = = = Resistance of BC Q λ (100 – l) 100 – l

Current Electricity — 49

Chapter 23 Here, λ is the resistance per unit length of the bridge wire. Unknown resistance X Resistance box R D

D G Galvanometer 0

10

20

30

40

50

60

70

80

R A

A

P

B

Q

C

G

C

(100 – l )

l

X

90 100

P

Q B

E + – K

Fig. 23.87

Hence, according to Wheatstone’s bridge principle, When current through galvanometer is zero or bridge is balanced, then Q P R or X = R = Q X P  100 – l  X = R  l 

∴

…(i)

So, by knowing R and l unknown resistance X can be determined. Specific Resistance From resistance formula, L X =ρ A ρ=

or

XA L

For a wire of radius r or diameter D = 2r, A = πr 2 = or

ρ=

πD 2 4

XπD 2 4L

…(ii)

By knowing X , D and L we can find specific resistance of the given wire by Eq. (ii).

Precautions 1. The connections should be clean and tight. 2. Null point should be brought between 40 cm and 60 cm. 3. At one place, diameter of wire (D) should be measured in two mutually perpendicular directions. 4. The jockey should be moved gently over the bridge wire so that it does not rub the wire.

50 — Electricity and Magnetism End Corrections In meter bridge, some extra length (under the metallic strips) comes at points A and C. Therefore, some additional length (α and β) should be included at the ends. Here, α and β are called the end corrections. Hence, in place of l we use l + α and in place of 100 − l we use 100 − l + β. To find α and β, use known resistors R 1 and R 2 in place of R and X and suppose we get null point length equal to l1 . Then, R1 l1 + α ...(i) = R 2 100 − l1 + β Now, we interchange the positions of R 1 and R 2 and suppose the new null point length is l2 . Then, R2 l2 + α ...(ii) = R 1 100 − l2 + β Solving Eqs. (i) and (ii), we get

and V

α=

R 2 l1 − R 1 l2 R1 − R2

β=

R 1 l1 − R 2 l2 − 100 R1 − R2

Example 23.33 If resistance R1 in resistance box is 300 Ω, then the balanced length is found to be 75.0 cm from end A. The diameter of unknown wire is 1 mm and length of the unknown wire is 31.4 cm. Find the specific resistance of the unknown wire. Solution

⇒

R l = X 100 − l  100 − l X = R  l   100 − 75 =  ( 300) = 100 Ω  75 

Now,

X =

ρl ρl = A ( πd 2 / 4 )

∴

ρ=

πd 2 X 4l

=

( 22/ 7) (10−3 ) 2 (100) ( 4 )( 0.314 )

= 2.5 × 10−4 Ω -m V

Ans.

Example 23.34 In a meter bridge, null point is 20 cm, when the known resistance R is shunted by 10 Ω resistance, null point is found to be shifted by 10 cm. Find the unknown resistance X.

Chapter 23 Solution

R l = X 100 − l

∴

 100 − l X = R  l 

or

 100 − 20 X =  R = 4R  20 

Current Electricity — 51

...(i)

When known resistance R is shunted, its net resistance will decrease. Therefore, resistance parallel to this (i.e. P) should also decrease or its new null point length should also decrease. R′ l′ ∴ = X 100 − l′ = or From Eqs. (i) and (ii), we have

20 − 10 1 = 100 − ( 20 − 10) 9

X = 9 R′

...(ii)

 10 R  4R = 9R ′ = 9   10 + R  Solving this equation, we get R=

50 Ω 4

Now, from Eq. (i), the unknown resistance  50 X = 4R = 4    4 X = 50 Ω

or Note

Ans.

R′ is resultant of R and 10 Ω in parallel. ∴ or V

1 1 1 = + R′ 10 R 10 R R′ = 10 + R

Example 23.35 If we use 100 Ω and 200 Ω in place of R and X we get null point deflection, l = 33 cm. If we interchange the resistors, the null point length is found to be 67 cm. Find end corrections α and β. R l − R 1 l2 Solution α= 2 1 R1 − R2 =

( 200)( 33) − (100)( 67) 100 − 200

= 1cm

Ans.

52 — Electricity and Magnetism β= =

R 1 l1 − R 2 l2 − 100 R1 − R2 (100)( 33) − ( 200) ( 67) − 100 100 − 200

= 1cm

INTRODUCTORY EXERCISE

Ans.

23.11

1. A resistance of 2 Ω is connected across one gap of a meter bridge (the length of the wire is 100 cm) and an unknown resistance, greater than 2 Ω, is connected across the other gap. When these resistances are interchanged, the balance point shifts by 20 cm. Neglecting any corrections, the unknown resistance is (JEE 2007) (a) 3 Ω (b) 4 Ω (c) 5 Ω (d) 6 Ω

2. A meter bridge is setup as shown in figure, to determine an unknown resistance X using a standard 10 Ω resistor. The galvanometer shows null point when tapping key is at 52 cm mark. The end corrections are 1 cm and 2 cm respectively for the ends A and B. The determined value of X is

10 Ω

X

A

B

Fig. 23.88

(JEE 2011)

(a) 10.2 Ω (b) 10.6 Ω (c) 10.8 Ω (d) 11.1 Ω 3. R1, R2 , R3 are different values of R. A, B and C are the null points obtained corresponding to R1, R2 and R3 respectively. For which resistor, the value of X will be the most accurate and why? (JEE 2005) X R G A

B

Fig. 23.89

C

Current Electricity — 53

Chapter 23

Post Office Box Post office box also works on the principle of Wheatstone's bridge. P R then the bridge is balanced. So, unknown resistance In a Wheatstone's bridge circuit, if = Q X Q X = R. P P and Q are set in arms AB and BC where we can have, 10 Ω,100 Ω or1000 Ω resistances to set any Q ratio . P A

B 100

1000

B P

P

10

10

100

C 1000

X

Q

E 5000 2000

R

C

500

200

200

100

D

X D

2000 1000

R

G

A

Q

1

2

Shunt

2

5 K2

10 K1

20

20

50

G

Fig. 23.90

Q =1. The unknown P resistance ( X ) is connected between C and D and battery is connected across A and C, These arms are called ratio arm, initially we take Q =10 Ω and P =10 Ω to set

Now, adjust resistance in part A to D such that the bridge gets balanced. For this, keep on increasing the resistance with 1 Ω interval, check the deflection in galvanometer by first pressing key K 1 then galvanometer key K 2 . Suppose at R = 4 Ω, we get deflection towards left and at R = 5 Ω, we get deflection towards right. Then, we can say that for balanced condition, R should lie between 4 Ω to 5 Ω. Q 10 Now, X = R = R = R = 4 Ω to 5 Ω P 10 To get closer value of X , in the second observation, let us choose

 P = 100  Q 1 = i.e.   P 10  Q = 10 

Suppose, now at R = 42 we get deflection towards left and at R = 43 deflection is towards right. So R ∈( 42, 43). Q Q 10 1 1 Now, X = R = R = R , where R ∈( 4.2, 4.3 Ω ). Now, to get further closer value take = P 100 10 P 100 and so on.

54 — Electricity and Magnetism The observation table is shown below. Table 23.2 Resistance in the ratio arm

1

2

Unknown resistance Q X = × R (ohm) P 4 to 5

AB (P) (ohm)

BC (Q) (ohm)

10

10

4

Left

5

Right

40

Left (large)

50

Right (large)

42

Left

43

Right

420

Left

424

Left

425

No deflection

426

Right

100

3

Direction of deflection

Resistance in arm (AD (R) (ohm)

S.No

1000

10

10

(4.2 to 4.3)

4.25

So, the correct value of X is 4.25 Ω V

Example 23.36 To locate null point, deflection battery key ( K 1 ) is pressed before the galvanometer key ( K 2 ). Explain why? Solution If galvanometer key K 2 is pressed first then just after closing the battery key K 1 current suddenly increases. So, due to self-induction, a large back emf is generated in the galvanometer, which may damage the galvanometer.

V

Example 23.37 What are the maximum and minimum values of unknown resistance X, which can be determined using the post office box shown in the Fig. 23.90? Solution

∴

QR P Qmax R max = Pmin

X = X max

1000 (11110) 10 = 1111 kΩ Q R = min min Pmax

=

X min

=

Ans.

(10) (1) 1000

= 0.01Ω

Ans.

Chapter 23

INTRODUCTORY EXERCISE

Current Electricity — 55

23.12

Q 1 = . In R if 142 Ω is used then we get deflection towards right P 10 and if R = 143 Ω, then deflection is towards left. What is the range of unknown resistance?

1. In post office box experiment,

2. What is the change in post office box experiment if battery is connected between B and C and galvanometer is connected across A and C?

3. For the post office box arrangement to determine the value of unknown resistance, the unknown resistance should be connected between B

(JEE 2004)

C

D

A

B1

C1

Fig. 23.91

(a) B and C

(b) C and D

(c) A and D

(d) B1 and C1

Extra Topics For Other Examinations 23.13 Colour Codes for Resistors Resistors are of the following two major types : (i) wire bound resistors and (ii) carbon resistors First type of resistors are made by winding the wires of an alloy like nichrome, manganin or constantan etc. Materials are so chosen that their resistivities are relatively less sensitive to temperature. In carbon resistors, carbon with a suitable binding agent is molded into a cylinder. Wire leads are attached to this cylinder and the entire resistor is encased in a ceramic or plastic jacket. The two leads connect the resistor to a circuit. Carbon resistors are compact and inexpensive. Their values are given using a colour code. Table 23.3 Colour

Number

Multiplier

Black

0

1

Brown

1

101

Red

2

102

Orange

3

103

Yellow

4

104

Tolerance (%)

56 — Electricity and Magnetism Colour

Number

Multiplier

Tolerance (%)

5

Green

5

10

Blue

6

106

Violet

7

107

Gray

8

108

White

9

109

Gold

10−1

Silver

−2

10

No colour

5 10 20

The resistors have a set of four (or three) co-axial coloured rings, whose significance are listed in above table. The colours are noted from left to right. 1 2 3 4 Fig. 23.92

Colour 1 → First significant figure Colour 2 → Second significant figure Colour 3 → Decimal multiplier

Colour 4 (or no colour ) → Tolerance or possible variation in percentage.

Extra Points to Remember To remember the value of colour coding used for carbon resistor, the following sentences are found to be of great help (where bold letters stand for colours) B B ROY Great Britain Very Good Wife wearing Gold Silver necklace.

˜

OR Black Brown Rods Of Your Gate Become Very Good When Given Silver colour V

Example 23.38 The four colours on a resistor are : brown, yellow, green and gold as read from left to right. What is resistance corresponding to these colours. Solution

From the table we can see that Brown colour → 1 Yellow colour → 4 Green colour → 105 and Gold colour → 5 %

∴

R = (14 × 105 ± 5%) Ω

INTRODUCTORY EXERCISE

Ans.

23.13

1. For the given carbon resistor, let the first strip be yellow, second strip be red, third strip be orange and fourth be gold. What is its resistance?

2. The resistance of the given carbon resistor is (24 × 106 ± 5%) Ω. What is the sequence of colours on the strips provided on resistor?

Chapter 23

Current Electricity — 57

Final Touch Points 1. Mobility The physical significance of mobility means how mobile the charge carriers are for the current flow. If mobility of charge carriers is more than we can say that current flow will be more. In metals, the mobile charge carriers are electrons. In an ionised gas they are electrons and positive charged ions. In an electrolyte, these can be positive and negative ions. In semiconductors, charge carriers are electrons and holes. Later, we will see that mobility of electrons (in semiconductor) is more than the mobility of holes. Mobility (µ ) is defined as the magnitude of drift velocity per unit electric field. µ=

vd E

vd =

eEτ m

Thus,

But, ∴

µ=

eτ m

The SI units of mobility are m 2 / V - s. Therefore, practical units of mobility is cm 2 / V - s. Mobility of any charge carrier (whether it is electron, ion or hole) is always positive.

2. Deduction of Ohm’s law We know that i = neAvd

eτE where, vd = m

i =

∴

ne 2τAE m

If V is the potential difference across the conductor and l is its length, then V E = l ∴

Here, ∴ where, R = ρ

i =

ne 2τAV ml

 m  l or V =  2  i  ne τ  A

m 1 = or ρ ne 2τ σ  ρl  V =   ⋅i  A

or V = Ri

l is the resistance of the conductor. A

3. Thermistor The temperature coefficient of resistivity is negative for semiconductors. This means that the resistivity decreases as we raise the temperature of such a material. The magnitude of the temperature coefficient of resistivity is often quite large for a semiconducting material. This fact is used to construct thermometers to detect small changes in temperatures. Such a device is called a thermistor. The variation of resistivity of a semiconductor with temperature is shown in figure. A typical thermistor can easily measure a change in temperature of the order of 10–3 °C

Hence proved.

r

Semiconductor

T

58 — Electricity and Magnetism 4. Superconductors Superconductivity was first discovered in 1911 by the Dutch

r

physicist Heike Kamerlingh Onnes. There are certain materials, including several metallic alloys and oxides for which as the temperature decreases, the resistivity first decreases smoothly, like that of any metal. But then at a certain critical temperatureTc a phase transition occurs, and the resistivity suddenly drops to zero as shown in figure.

T Tc Superconductor

Once, a current has been established in a superconducting ring, it continues indefinitely without the presence of any driving field. Possible applications of superconductors are ultrafast computer switches and transmission of electric power through superconducting power lines. However, the requirement of low temperature is posing difficulty. For instance the critical temperature for mercury is 4.2 K. Scientists are putting great effort to construct compounds and alloys which would be superconducting at room temperature (300 K). Superconductivity at around 125 K has already been achieved.

5. The ρ-T equation derived in article 23.6 can be derived from the relation, dρ dρ = αρ or = α dT dT ρ ρ T dρ ∫ρ0 ρ = α ∫T 0 dT

∴

…(i)

ρ ln   = α (T − T0 )  ρ0 

∴

ρ = ρ 0 e α (T

∴ If α is

(if α = constant)

small, e α(T − T 0 )

− T0 )

can approximately be written as1 + α(T − T0 ). Hence, ρ = ρ 0[1 + α (T − T0 )]

Which is the same result as we have discussed earlier. In the above discussion, we have assumed α to be constant. If it is function of temperature it will come inside the integration in Eq. (i).

6. The principle of superposition Complex network problems can sometimes be solved easily by using the principle of superposition. This principle essentially states that when a number of emf’s act in a network, the solution is the same as the superposition of the solutions for one emf acting at a time, the others being shorted.

7. Figure shows a network with two loops. The currents in various

10.8V

branches can be calculated using Kirchhoff’s laws. We can get the same solution by considering only one battery at a time and then superposing the two solutions. If a battery has an internal resistance, it must be left in place when the emf of the battery is removed. Figure shows how the superposition principle can be applied to the present problem. 10.8V

4Ω

4Ω 12Ω 0.4A

8Ω

+

(a)

14.4V

1.2A 8Ω

0.6A 2Ω

2Ω

2Ω

12 Ω

0.8A

14.4V (b)

8Ω

12Ω

0.4A

1A 1A

4Ω

Current Electricity — 59

The current values in Fig. (a) and (b) are easily verified. For example when the 10.8 V battery alone is acting, the total resistance in the circuit is 12 × 8 4+ + 2 = 10.8 Ω 12 + 8

⇒

Chapter 23

10.8 V

4Ω

1.8 A 8Ω

12 Ω 1.8 A

10.8 V = 1 A. This current splits 10.8 Ω between 8 Ω and 12 Ω in the ratio 3 : 2. Similarly, the total resistance when only the14.4 V battery is acting is 12 × 6 8+ = 12 Ω 12 + 6

This makes the total current

Therefore, the total current is

2Ω

14.4 V (c)

14.4 V = 1.2 A. 12 Ω

The superposition principle shows that there is no current in the 12 Ω resistance. Only a current of 1.8 A flows through the outer loop. All these conclusions can be verified by analyzing the circuit using Kirchhoff’s laws.

8. The equivalent emf of a cell can also be found by the following method. 10V 2Ω A

i

B

i=0

i=0

⇒

E A

r B

4V 1Ω

Suppose we wish to find the equivalent emf of the above circuit. We apply the fact that E =V When no current is drawn from the cell. But current in the internal circuit may be non-zero. This current is, 10 + 4 14 i = = A 2 +1 3 Now, ∴ ∴ ∴

VA + 4 – 1 ×

14 = VB 3 14 2 VA – VB = –4= V 3 3 2 E = VA – VB = V = V 3 2 E = V 3

Further,VA – VB is positive, i.e.VA > VB or A is connected to the positive terminal of the battery and B to the negative. Internal resistance of the equivalent battery is found by the normal procedure. For example, here 2 Ω and 1 Ω resistances are in parallel. Hence, their combined resistance is 2 1 1 1 3 or r = Ω = + = 3 r 1 2 2

Solved Examples TYPED PROBLEMS Type 1. Based on potential difference across the terminals of a battery

Concept Potential difference V across the terminals of a battery is given by if i = 0 V =E if battery is short-circuited V =0 V = E − ir if normal current ( i ) flows through the battery and ) V = E + ir if current flows in the opposite direction ( i V

Example 1 Find the potential difference across each of the four batteries B1 , B2 , B3 and B4 as shown in the figure. 10V, 1Ω a

4V, 1Ω

d 6 V,2 Ω

B3

B2

B1

2Ω

i B4

b

c 5V, 2 Ω

Solution Across B1 This battery does not make any closed circuit. ∴ Across B2

i = 0 or V = E = 4 volt This battery is short-circuited. Therefore,

Across B3 and B4 current is

Ans.

Ans. V =0 A current in anti-clockwise direction flows in the closed loop abcda. This i=

net emf 10 − 5 = net resistance 1 + 2 + 2

=1A Now, current flows through B3 in normal direction. Hence, V = E − ir = 10 − 1 × 1 = 9 volt From B4, current flows in opposite direction. Hence, V = E + ir = 5 + 1 × 2 = 7 volt

Ans. Ans.

Chapter 23 V

Current Electricity — 61

Example 2 Draw (a) current versus load and (b) current versus potential difference (across its two terminals) graph for a cell. Solution (a) i =

E R+ r

i

V

E r

i versus R graph is shown in Fig. (a). (b) V = E − ir V versus i graph is shown in Fig. (b).

E ir

E 2r

v r=R (a)

i

R

(b)

E r

Type 2. To find values of V, i and R across all resistors of a complex circuit if values across one resistance are known

Concept (i) In series, current remains same. But the potential difference distributes in the direct ratio of resistance. (ii) In parallel, potential difference is same. But the current distributes in the inverse ratio of resistance. V

Example 3 In the circuit shown in figure potential difference across 6 Ω resistance is 4 volt. Find V and i values across each resistance. Also find emf E of the applied battery. 6Ω

12 Ω

3Ω

4Ω

8Ω

E

Solution

6Ω

12 Ω

8Ω

4V

6V

16 V

4V

6V

3Ω 2 Ω, 4 V

4Ω 3 Ω, 6 V

E

8 Ω , 2 Ω (Resultant of 6 Ω and 3 Ω) and 3 Ω (resultant of 12 Ω and 4 Ω ) are in series. Therefore, potential drop across them should be in direct ratio of resistance. So, using this concept we can find the potential difference across other resistors. For example, potential across 2 Ω was 4 V. So, potential difference across 8 Ω (which is four times of 2 Ω ) should be 16 V. Similarly, potential difference across 3 Ω (which is 1.5 times of 2 Ω ) should be 1.5 times or 6 V.

62 — Electricity and Magnetism Once V and R are known, we can find i across that resistance. For example, 6 1 i12Ω = = A 12 2 16 i 8Ω = = 2 A etc 8

V  i =   R

Type 3. To find equivalent value of temperature coefficient α if two or more than two resistors are connected in series or parallel l

This can be explained by the following example :

V

Example 4 Two resistors with temperature coefficients of resistance α1 and α2 have resistances R 01 and R 02 at 0°C. Find the temperature coefficient of the compound resistor consisting of the two resistors connected (a) in series and (b) in parallel. Solution (a) In Series At 0°C At t° C

⇒ R01

R01 (1 + α 1t )

R0 = R01 + R02

R02 R02 (1 + α2t)

R0 (1 + αt )

R01 (1 + α 1t ) + R02 (1 + α 2t ) = R0 (1 + αt ) or ∴ or (b) In Parallel

R01 (1 + α 1t ) + R02 (1 + α 2t ) = (R01 + R02) (1 + αt ) R01 + R01α 1t + R02 + R02α 2t = R01 + R02 + (R01 + R02) αt R α + R02α 2 α = 01 1 R01 + R02 R01

R=

Ans.

R01R02 R01 + R02

⇒ R02

At t° C, or

1 1 1 = + R0 (1 + αt ) R01 (1 + α 1t ) R02 (1 + α 2t ) R01 + R02 1 1 = + R01R02 (1 + αt ) R01 (1 + α 1t ) R02 (1 + α 2t )

Using the Binomial expansion, we have 1 1 1 1 (1 – αt ) + (1 – αt ) = (1 – α 1t ) + (1 – α 2t ) R02 R01 R01 R02 i.e.

 1 1  α α αt  +  = 1 t+ 2 t R02  R01 R02 R01

or

α=

α 1R02 + α 2R01 R01 + R02

Ans.

Chapter 23

Current Electricity — 63

Type 4. Based on the verification of Ohm’s law

Concept V  For verification of Ohm’s law  = constant = R , we need an ohmic resistance, which i  follows this law. A voltmeter which will measure potential difference across this resistance, an ammeter which will measure current through this resistance and a variable battery which can provide a variable current in the circuit. Now, for different values of i, we have to measure different values of V and then prove that, V = constant V ∝ i or i and this constant is called resistance of that. V

Example 5 Draw the circuit for experimental verification of Ohm’s law using a source of variable DC voltage, a main resistance of 100 Ω, two galvanometers and two resistances of values 10 6 Ω and 10 −3 Ω respectively. Clearly show the positions of the voltmeter and the ammeter. (JEE 2004) Solution

Voltmeter 106 Ω

Ammeter 10-3 Ω G2

G1

100 Ω

Variable DC voltage

Type 5. Theory of bulbs or heater etc.

Concept (i) From the rated (written) values of power ( P ) and potential difference (V ) we can determine resistance of filament of bulb.

⇒

P=

V2 R

R=

V2 P

…(i)

For example, if rated values on a bulb are 220 V and 60 W, it means this bulb will consume 60 W of power (or 60 J in 1 s) if a potential difference of 220 V is applied across it. Resistance of this bulb will be ( 220) 2 R= = 806.67 Ω 60

64 — Electricity and Magnetism (ii) Normally, rated value of V remains same in different bulbs. In India, it is 220 volt. Therefore, from Eq. (i) 1 or R60 watt > R100 watt R∝ P (iii) Actual value of potential difference may be different from the rated value. Therefore, actual power consumption may also be different. (iv) After finding resistance of the bulb using Eq. (i), we can apply normal Kirchhoff's laws for finding current passing through the bulb or actual power consumed by the bulb. V

Example 6 Prove that 60 W bulb glows more brightly than 100 W bulb if by mistake they are connected in series. Solution In series, we can use the formula P = i 2R

for the power consumption

⇒ we have seen above that,

P∝R

(as i is same in series)

R60 W > R100 W P60 W > P100 W

∴

Hence Proved.

Note In parallel 100 W bulb glows more brightly than 60 W bulb. Think why? V

Example 7 The rated values of two bulbs are ( P1 , V ) and ( P2 , V ). Find actual power consumed by both of them if they are connected in (a) series (b) parallel and V potential difference is applied across both of them. Solution (a) Q R1 =

V2 P1

and R2 =

V2 P2

PP V2 V2 V2 = = 2 = 1 2 Rnet R1 + R2 V V 2 P1 + P2 + P1 P2

In series,

P=

or

1 1 1 = + P P1 P2

Ans.

(b) In parallel, P= or V

 1  1 V2 2 1 + =V2    =V  Rnet  R1 R2  Rnet 

P  P P = V 2 12 + 22 V V 

or

P = P1 + P2

Ans.

Example 8 Heater-1, takes 3 minutes to boil a given amount of water. Heater-2 takes 6-minutes. Find the time taken if, (a) they are connected in series (b) they are connected in parallel. Potential difference V in all cases is same.

Chapter 23

Current Electricity — 65

Solution Here, the heat required (say H) to boil water is same. Let P1 and P2 are the powers of the heaters. Then, H = P1t1 = P2t2 H H P1 = = t1 3

∴

(a) In series,

P2 =

H H = t2 6

P=

P1P2 (H / 3)(H / 6) = P1 + P2 (H / 3) + (H / 6)

= t=

Now,

H 9

H H = = 9 min P (H / 9)

Ans.

P = P1 + P2 H H H = + = 3 6 2 H H t= = = 2 min P H /2

(b) In parallel,

∴ V

(Refer Example 7)

Ans.

Example 9 A 100 W bulb B1 , and two 60 W bulbs B2 and B3 , are connected to a 250 V source as shown in the figure. Now W1 , W2 and W3 are the output powers of the bulbs B1 , B2 and B3 respectively. Then, (JEE 2002) B1

B2

B3 250 V

(a) W1 > W 2 = W3 Solution ∴ Now,

V2 P= R

(b) W1 > W 2 > W3 so,

(c) W1 < W 2 = W3

V2 R= P V2 V2 and R2 = R3 = 100 60 (250)2 W1 = ⋅ R1 (R1 + R2)2 R1 =

W2 =

(250)2 (250)2 ⋅ R2 and W3 = 2 R3 (R1 + R2)

W1 : W 2 : W3 = 15 : 25 : 64 ∴ The correct option is (d).

Note We have used W = i 2 R for W1 and W2 and W =

or

V2 for W3 . R

W1 < W 2 < W3

(d) W1 < W 2 < W3

66 — Electricity and Magnetism V

Example 10 An electric bulb rated for 500 W at 100 V is used in a circuit having a 200 V supply. The resistance R that must be put in series with the bulb, so that the bulb delivers 500 W is ……Ω. (JEE 1987) Solution Resistance of the given bulb Rb

R

100 V

100 V 200 V

V 2 (100)2 = = 20 Ω P 500 To get 100 V out of 200 V across the bulb, Rb =

R = Rb = 20 Ω V

Ans.

Example 11 A heater is designed to operate with a power of 1000 W in a 100 V line. It is connected in combination with a resistance of 10 Ω and a resistance R, to a 100 V mains as shown in the figure. What will be the value of R so that the heater operates with a power of 62.5 W? (JEE 1978) 10 Ω

B

Heater

C

R 100 V

Solution From P =

V2 , R

Resistance of heater,

R=

V 2 (100)2 = = 10 Ω P 1000

P = i 2R Current required across heater for power of 62.5 W, P 62.5 i= = = 2.5 A R 10 100 100 (10 + R) 10 (10 + R) Main current in the circuit, I = = = 10R 100 + 20R 10 + 2R 10 + 10 + R From

This current will distribute in inverse ratio of resistance between heater and R.  R  ∴ i=  I  10 + R or

 R  10 (10 + R)  10R 2.5 =    =  10 + R  10 + 2R  10 + 2R

Solving this equation, we get R=5 Ω

Ans.

Chapter 23

Current Electricity — 67

Type 6. To find current through a single external resistance in a complex circuit of batteries using the concepts of equivalent value of emf of battery

Concept 10 V

6V

2Ω

2V

2Ω

2Ω

5Ω

i

In the above circuit, if we have to find only i then two parallel batteries may be converted into a single battery. Then, this battery is in series with the third battery of emf 10 volt. Now, we can find current i from the equation, Net emf i= Total resistance V

Example 12 Find the value of i in the circuit shown above. Solution Equivalent emf of the parallel combination is E= =

E1 /r1 + E 2 /r2 1 /r1 + 1 /r2 6 /2 + 2 /2 = 4 volt 1 /2 + 1 /2

Equivalent internal resistance of the parallel combination is rr (2) (2) r= 12 = =1 Ω r1 + r2 2 + 2 Now, the equivalent simple circuit is as shown below 10 V 2 Ω

4V

1Ω

5Ω

i

Net emf Total resistance 10 + 4 = 2+1+5

i=

= 1.75 A

Ans.

Note By this concept, we can find only i. To find other currents (across 6V battery or 2V battery) we will have to apply Kirchhoff's laws.

Miscellaneous Examples V

Example 13 Two sources of current of equal emf are connected in series and have different internal resistances r1 and r2 ( r2 > r1 ). Find the external resistance R at which the potential difference across the terminals of one of the sources becomes equal to zero. Solution V = E – ir E and i for both the sources are equal. Therefore, potential difference (V) will be zero for a source having greater internal resistance, i.e. r2. ∴

0 = E – ir2   2E E = ir2 =   ⋅ r2  R + r1 + r2

or ∴ or V

2r2 = R + r1 + r2 R = r2 – r1

Example 14

5A

2Ω

D

1Ω

C

Ans. 3V

2A

E 4Ω

12 V 3Ω

B

6Ω

6A

Figure shows the part of a circuit. Calculate the power dissipated in 3 Ω resistance. What is the potential difference VC – V B ? Solution Applying Kirchhoff’s junction law at E current in wire DE is 8 A from D to E. Now further applying junction law at D, the current in 3 Ω resistance will be 3 A towards D. Power dissipated in 3 Ω resistance = i 2R = (3)2 (3) = 27 W 5A C

1Ω

2Ω

D 12V 3A

8A

2A

E 3V

3Ω

VC – VB

Ans.

6A

4Ω

B

6Ω

VC – 5 × 1 + 12 – 8 × 2 – 3 – 4 × 2 = VB

∴

VC – VB = 5 – 12 + 16 + 3 + 8

or

VC – VB = 20 V

Ans.

Chapter 23 V

Current Electricity — 69

Example 15 The emf of a storage battery is 90 V before charging and 100 V after charging. When charging began the current was 10 A. What is the current at the end of charging if the internal resistance of the storage battery during the whole process of charging may be taken as constant and equal to 2 Ω? Solution The voltage supplied by the charging plant is here constant which is equal to, V = Ei + ii ⋅ r = (90) + (10) (2) = 110 V Let i f be the current at the end of charging. Then, V = E f + if r or

V – Ef r 110 – 100 = 2

if =

=5 A V

Ans.

Example 16 A battery has an open circuit potential difference of 6 V between its terminals. When a load resistance of 60 Ω is connected across the battery, the total power supplied by the battery is 0.4 W. What should be the load resistance R, so that maximum power will be dissipated in R. Calculate this power. What is the total power supplied by the battery when such a load is connected? Solution When the circuit is open, V = E ∴ E =6 V Let r be the internal resistance of the battery. Power supplied by the battery in this case is E2 P= R+ r Substituting the values, we have

0.4 =

E

R

r

(6)2 60 + r

Solving this, we get r = 30 Ω Maximum power is dissipated in the circuit when net external resistance is equal to net internal resistance or R=r ∴

R = 30 Ω

Ans.

Further, total power supplied by the battery under this condition is PTotal =

E2 (6)2 = R + r 30 + 30

= 0.6 W

Ans.

Of this 0.6 W half of the power is dissipated in R and half in r. Therefore, maximum power dissipated in R would be 0.6 Ans. = 0.3 W 2

70 — Electricity and Magnetism V

Example 17 In which branch of the circuit shown in figure a 11 V battery be inserted so that it dissipates minimum power. What will be the current through the 2 Ω resistance for this position of the battery?

2Ω

4Ω

6Ω

Solution Suppose, we insert the battery with 2 Ω resistance. Then, we can take 2 Ω as the internal resistance (r) of the battery and combined resistance of the other two as the external resistance (R). The circuit in that case is shown in figure,

R

E r

Now power, P =

E2 R+ r

This power will be minimum where R + r is maximum and we can see that (R + r ) will be maximum when the battery is inserted with 6 Ω resistance as shown in figure. i1

i i2

2Ω

4Ω

6Ω 11V

Net resistance in this case is 6+ ∴

2 × 4 22 = Ω 2+4 3 i=

11 = 1.5 A 22 /3

This current will be distributed in 2 Ω and 4 Ω in the inverse ratio of their resistances. i1 4 ∴ = =2 i2 2 ∴

 2  i1 =   (1.5) = 1.0 A  2 + 1

Ans.

Chapter 23 V

Current Electricity — 71

Example 18 An ammeter and a voltmeter are connected in series to a battery of emf E = 6.0 V . When a certain resistance is connected in parallel with the voltmeter, the reading of the voltmeter decreases two times, whereas the reading of the ammeter increases the same number of times. Find the voltmeter reading after the connection of the resistance. Solution Let R = resistance of ammeter A

V

i

6V

Potential difference across voltmeter = 6 − potential difference across ammeter In first case, V = 6 − iR V In second case, = 6 − (2i ) R 2 Solving these two equations, we get

K(ii)

V = 4 volt V /2 = 2 volt

∴ V

K(i)

Ans.

Example 19 A voltmeter of resistance R1 and an ammeter of resistance R2 are connected in series across a battery of negligible internal resistance. When a resistance R is connected in parallel to voltmeter, reading of ammeter increases three times while that of voltmeter reduces to one third. Find R1 and R2 in terms of R. Solution Let E be the emf of the battery.

E

In the second case main current increases three times while current through voltmeter will reduce to i /3. Hence, the remaining 3i – i /3 = 8i /3 passes through R as shown in figure.  i  8i  VC – VD =   R1 =   R  3  3 R1 = 8R

or

Ans. E

A

G

3i

B

A R2

C

i 3

D

V R1

8i 3 R

F

i R2 A

R1 V

72 — Electricity and Magnetism In the second case, main current becomes three times. Therefore, total resistance becomes

1 3

times or R2 +

RR1 1 = (R1 + R2) R + R1 3

Substituting R1 = 8R, we get R2 = V

8R 3

Ans.

Example 20 Find the current in each branches of the circuit. 5Ω

4Ω

A

21V

6Ω 5V 8Ω

E

B

C

1Ω

D 2V

16Ω

Solution It is possible to use Kirchhoff’s laws in a slightly different form, which may simplify the solution of certain problems. This method of applying Kirchhoff’s laws is called the loop current method. In this method, we assign a current to every closed loop in a network. 5Ω

4Ω

A

21V

6Ω i2

i1

5V

8Ω E 1Ω

B

C i3

D 2V

16Ω

Suppose currents i1 , i 2 and i3 are flowing in the three loops. The clockwise or anti-clockwise sense given to these currents is arbitrary. Applying Kirchhoff’s second law to the three loops, we get 21 – 5i1 – 6 (i1 + i 2) – i1 = 0 5 – 4i 2 – 6 (i1 + i 2) – 8 (i 2 + i3 ) = 0 and 2 – 8 (i 2 + i3 ) – 16 i3 = 0 Solving these three equations, we get 1 1 and i2 = – A i3 = A i1 = 2 A , 2 4

…(i) …(ii) …(iii)

Chapter 23

Current Electricity — 73

Therefore, current in different branches are as shown in figure given below. 5Ω

4Ω

A

0.5A

21V

6Ω

2A

5V

1.5A 8Ω

2A

E

C

1Ω

0.25 A

B

0.25A D 2 V 16Ω

Note In wire AC, current is i1 + i2 and in CB it is i2 + i3 . V

Example 21 What amount of heat will be generated in a coil of resistance R due to a charge q passing through it if the current in the coil (a) decreases down to zero uniformly during a time interval t0? (b) decreases down to zero halving its value every t0 seconds?

Heat generated in a resistance is given by H = i 2 Rt We can directly use this formula provided i is constant. Here, i is varying. So, first we will calculate i at any time t, then find a small heat dH in a short interval of time dt. Then by integrating it with proper limits we can obtain the total heat produced.

HOW TO PROCEED

Solution (a) The corresponding i-t graph will be a straight line with i decreasing from a peak value (say i 0) to zero in time t0. i-t equation will be as i  i = i0 –  0  t  t0 

(y = – mx + c)

Here, i 0 is unknown, which can be obtained by using the fact that area the flow of charge. Hence, i i0

t0

t

1 (t0 ) (i 0 ) 2 2q i0 = t0 q=

∴ Substituting in Eq. (i), we get

i=

2q  t 1 –  t0  t0 

…(i) under i-t graph gives

74 — Electricity and Magnetism  2q 2qt  i= – 2 t0   t0

or

Now, at time t, heat produced in a short interval dt is dH = i 2R dt 2

 2q 2qt  = – 2  Rdt t0   t0 ∴

Total heat produced = ∫

or

H =∫ =

t0 0 t0

0

dH 2

 2q 2qt  – 2  R dt  t0   t0

4 q2R 3 t0

Ans.

(b) Here, current decreases from some peak value (say i 0) to zero exponentially with half life t0. i i0

t0

t

i-t equation in this case will be i = i 0 e– λt ln (2) λ= t0

Here,

∞ ∞ i  q = ∫ i dt = ∫ i 0 e– λt dt =  0  0 0  λ

Now, ∴ ∴ ∴

i0 = λ q i = (λq) e– λ t dH = i 2R dt = λ2q2e–2λt R dt ∞

Substituting λ =

∞

H = ∫ dH = λ2q2R ∫ e–2λt dt =

or

0

0

ln (2) , we have t0

H =

2

q R ln (2) ⋅ 2t0

Note In radioactivity, half-life is given by ln 2 λ ln 2 λ= t1

t1 / 2 = ∴

q2λR 2

2

Ans.

Exercises LEVEL 1 Assertion and Reason Directions : (a) (b) (c) (d)

Choose the correct option.

If both Assertion and Reason are true and the Reason is correct explanation of the Assertion. If both Assertion and Reason are true but Reason is not the correct explanation of Assertion. If Assertion is true, but the Reason is false. If Assertion is false but the Reason is true.

1. Assertion : If potential difference across two points is zero, current between these two points should be zero. Reason : Current passing from a resistor I=

V R 3R

2. Assertion : In the part of the circuit shown in figure, maximum power is produced across R.

I

Power P =

Reason :

2R

2

V R

R

3. Assertion : Current I is flowing through a cylindrical wire of non-uniform cross-section as shown. Section of wire near A will be more heated compared to the section near B. B A I

Reason :

Current density near A is more.

4. Assertion : In the circuit shown in figure after closing the switch S reading of ammeter will increase while that of voltmeter will decrease. S V A

Reason :

Net resistance decreases as parallel combination of resistors is increase`d.

76 — Electricity and Magnetism 5. Assertion : In the circuit shown in figure ammeter and voltmeter are non-ideal. When positions of ammeter and voltmeter are changed, reading of ammeter will increase while that of voltmeter will decrease. V A

Reason :

Resistance of an ideal ammeter is zero while that of an ideal voltmeter is infinite.

6. Assertion : In the part of a circuit shown in figure, given that V b > V a . The current should flow from b to a. a

Reason : terminal.

b

Direction of current inside a battery is always from negative terminal to positive

7. Assertion : In the circuit shown in figure R is variable. Value of current I is maximum when R = r.

E, r I

R

Reason :

At R = r, maximum power is produced across R.

8. Assertion : If variation in resistance due to temperature is taken into consideration, then current in the circuit I and power produced across the resistance P both will decrease with time. R I

Reason : V = IR is Ohm’s law.

9. Assertion : When a potential difference is applied across a conductor, free electrons start travelling with a constant speed called drift speed. Reason : Due to potential difference an electric field is produced inside the conductor, in which electrons experience a force.

10. Assertion : When temperature of a conductor is increased, its resistance increases. Reason :

Free electrons collide more frequently.

11. Assertion : Two non-ideal batteries are connected in parallel with same polarities on same side. The equivalent emf is smaller than either of the two emfs. Reason : Two non-ideal batteries are connected in parallel, the equivalent internal resistance is smaller than either of the two internal resistances.

Chapter 23

Current Electricity — 77

Objective Questions 1. An ammeter should have very low resistance (a) (b) (c) (d)

to show large deflection to generate less heat to prevent the galvanometer so that it may not change the value of the actual current in the circuit

2. A steady current flows in a metallic conductor of non-uniform cross-section. The quantity/ quantities which remain constant along the length of the conductor is/are (a) current, electric field and drift speed (c) current and drift speed

(b) drift speed only (d) current only

3. If M = mass, L = length, T = time and I = electric current, then the dimensional formula of resistance R will be given by (a) [R] = [ML2 T−3 I−2] (c) [R] = [ML2 T3 I−2]

(b) [R] = [ML2 T−3 I2] (d) [R] = [ML2 T3 I2]

4. The unit of electrical conductivity is (a) ohm-m−2 (c) ohm−1 -m−1

(b) ohm × m (d) None of these

5. Through an electrolyte an electrical current is due to drift of (a) free electrons (c) free electrons and holes

(b) positive and negative ions (d) protons

6. The current in a circuit with an external resistance of 3.75 Ω is 0.5 A. When a resistance of 1 Ω is introduced into the circuit, the current becomes 0.4 A. The emf of the power source is (a) 1 V (c) 3 V

(b) 2 V (d) 4 V

7. The deflection in a galvanometer falls from 50 divisions to 20 divisions, when a 12 Ω shunt is applied. The galvanometer resistance is (a) 18 Ω (c) 30 Ω

(b) 24 Ω (d) 36 Ω

8. If 2% of the main current is to be passed through the galvanometer of resistance G, the resistance of shunt required is G 49 (c) 49 G

(a)

G 50 (d) 50 G

(b)

9. If the length of the filament of a heater is reduced by 10%, the power of the heater will (a) (b) (c) (d)

increase by about 9% increase by about 11% increase by about 19% decrease by about 10%

10. N identical current sources each of emf E and internal resistance r are connected

B

to form a closed loop as shown in figure. The potential difference between points A and B which divides the circuit into n and ( N − n ) units is (a) NE (c) nE

(b) (N − n )E (d) zero

A

78 — Electricity and Magnetism 11. A 2.0 V potentiometer is used to determine the internal resistance of a 1.5 V cell. The balance

point of the cell in the open circuit is 75 cm. When a resistor of 10 Ω is connected across the cell, the balance point shifts to 60 cm. The internal resistance of the cell is (b) 2.5 Ω (d) 4.5 Ω

(a) 1.5 Ω (c) 3.5 Ω

12. Three resistances are joined together to form a letter Y , as shown in figure. If the potentials of the terminals A, B and C are 6 V, 3 V and 2 V respectively, then the potential of the point O will be A +6V +3V B 3Ω

6Ω O

2Ω +2 V C

(a) 4 V (c) 2.5 V

(b) 3 V (d) 0 V

13. The drift velocity of free electrons in a conductor is v, when a current i is flowing in it. If both the radius and current are doubled, then the drift velocity will be (a) v (c) v /4

(b) v /2 (d) v /8

14. A galvanometer is to be converted into an ammeter or voltmeter. In which of the following cases the resistance of the device is largest? (a) (b) (c) (d)

an ammeter of range 10 A a voltmeter of range 5 V an ammeter of range 5 A a voltmeter of range 10 V

15. In the given circuit the current flowing through the resistance 20 Ω is 0.3 A, while the ammeter reads 0.8 A. What is the value of R1? R1

20 Ω A 15 Ω

(a) 30 Ω (c) 50 Ω

(b) 40 Ω (d) 60 Ω

16. An ammeter and a voltmeter are joined in series to a cell. Their readings are A and V respectively. If a resistance is now joined in parallel with the voltmeter, then (a) (b) (c) (d)

both A and V will increase both A and V will decrease A will decrease, V will increase A will increase, V will decrease

Chapter 23

Current Electricity — 79

17. A resistor R has power of dissipation P with cell voltage E. The resistor is cut in n equal parts and all parts are connected in parallel with same cell. The new power dissipation is (b) nP 2 (d) n /P

(a) nP (c) n 2P

18. In the circuit diagram shown in figure, a fuse bulb can cause all other bulbs to go out. Identify the bulb A

B

C

+ D

E

(a) B (c) A

(b) C (d) D or E

19. Two batteries one of the emf 3 V, internal resistance 1 Ω and the other of emf 15V, internal resistance 2 Ω are connected in series with a resistance R as shown. If the potential difference between points a and b is zero, the resistance R in Ω is b

a 3 V, 1Ω

15 V, 2Ω

R

(b) 7 (d) 1

(a) 5 (c) 3

20. A part of a circuit is shown in figure. Here reading of ammeter is 5 A and voltmeter is 100 V. If voltmeter resistance is 2500 ohm, then the resistance R is approximately R A

V

(a) 20 Ω (c) 100 Ω

(b) 10 Ω (d) 200 Ω

21. A copper wire of resistance R is cut into ten parts of equal length. Two pieces each are joined in series and then five such combinations are joined in parallel. The new combination will have a resistance (a) R

(b)

R 4

(c)

R 5

(d)

R 25

22. Two resistances are connected in two gaps of a meter bridge. The balance point is 20 cm from the zero end. A resistance of 15 Ω is connected in series with the smaller of the two. The null point shifts to 40 cm. The value of the smaller resistance in Ω is

(a) 3 (c) 9

(b) 6 (d) 12

80 — Electricity and Magnetism 23. In the given circuit, the voltmeter records 5 volt. The resistance of the voltmeter in Ω is V

100 Ω

50 Ω

10 V

(b) 100 (d) 50

(a) 200 (c) 10

24. The wire of potentiometer has resistance 4 Ω and length 1 m. It is connected to a cell of emf 2 volt and internal resistance 1 Ω. If a cell of emf 1.2 volt is balanced by it, the balancing length will be

(a) 90 cm (c) 50 cm

(b) 60 cm (d) 75 cm

25. The potential difference between points A and B, in a section of a circuit shown, is 1A

2A 2Ω

2Ω 2Ω 3A

(a) 5 volt (c) 10 volt

2Ω A

1Ω 3V

1Ω 2V

B

(b) 1 volt (d) 17 volt

26. Two identical batteries, each of emf 2 V and internal resistance r = 1 Ω

1Ω

are connected as shown. The maximum power that can be developed across R using these batteries is

1Ω

(a) (b) (c) (d)

3.2 W 8.2 W 2W 4W

2V

R

2V

27. For a cell, the terminal potential difference is 2.2 V, when circuit is open and reduces to 1.8 V. When cell is connected to a resistance R = 5 Ω, the internal resistance of cell (r ) is 10 Ω 9 11 (c) Ω 9

(a)

9 Ω 10 5 (d) Ω 9 (b)

28. The potential difference between points A and B in the circuit shown

25 Ω

in figure, will be (a) (b) (c) (d)

1V 2V –3V None of the above

15Ω

A

10V, 2.5Ω

B 5V, 2.5Ω

Chapter 23

Current Electricity — 81

29. Potentiometer wire of length 1 m is connected in series with 490 Ω resistance and 2 Vbattery. If 0.2 mV/ cm is the potential gradient, then resistance of the potentiometer wire is approximately (a) 4.9 Ω

(b) 7.9 Ω

(c) 5.9 Ω

(d) 6.9 Ω

30. Find the ratio of currents as measured by ammeter in two cases when the key is open and when the key is closed 2R

R

k 2R

R

A

(a) 9/8

(b) 10/11

(c) 8/9

(d) None of these

31. A galvanometer has a resistance of 3663 Ω. A shunt S is connected across it such that (1/ 34) of the total current passes through the galvanometer. Then, the value of the shunt is (a) 222 Ω

(b) 111 Ω

(c) 11 Ω

(d) 22 Ω

Note Attempt the following questions after reading the chapter of capacitors. r

32. The network shown in figure is an arrangement of nine identical resistors. A and B is 1.5 Ω. The

The resistance of the network between points resistance r is (a) (b) (c) (d)

r

r

r r

1.1 Ω 3.3 Ω 1.8 Ω 1.6 Ω

B

A r r

r

r

33. The equivalent resistance of the hexagonal network as shown in figure between points A and B is r

r r r

r A

B r

r r

r

(a) r

(b) 0.5 r

r

(c) 2 r

(d) 3 r

34. A uniform wire of resistance 18 Ω is bent in the form of a circle. The a

effective resistance across the points a and b is (a) (b) (c) (d)

3Ω 2Ω 2.5 Ω 6Ω

60° b

82 — Electricity and Magnetism 35. Each resistor shown in figure is an infinite network of resistance 1 Ω. The effective resistance between points A and B is 1Ω

1Ω

1Ω

1Ω

1Ω

1Ω

1Ω

1Ω

A

1Ω

(a) less than 1 Ω (c) more than 1 Ω but less than 3 Ω

1Ω

1Ω

B

1Ω

1Ω

(b) 1 Ω (d) 3 Ω

36. In the circuit shown in figure, the total resistance between points A and B is R0. The value of resistance R is R

R

A R

R0

B

(a) R0

(b)

(c)

3 R0

R0 2

(d)

R0 3

37. In the circuit shown in the figure, R = 55 Ω, the equivalent resistance between the points P and Q is P R

R

R

R

R R

Q

(a) 30 Ω (c) 55 Ω

(b) 35 Ω (d) 25 Ω

38. The resistance of all the wires between any two adjacent dots is R. Then,

A

equivalent resistance between A and B as shown in the figure is (a) (7 /3) R (b) (7 /6) R (c) (14 /8) R

B

(d) None of the above

39. A uniform wire of resistance 4 Ω is bent into a circle of radius r. A specimen of the same wire is connected along the diameter of the circle. What is the equivalent resistance across the ends of this wire? 4 Ω (4 + π ) 2 (c) Ω (2 + π )

(a)

3 Ω (3 + π ) 1 (d) Ω (1 + π ) (b)

Chapter 23

Current Electricity — 83

40. In the network shown in figure, each resistance is R. The equivalent resistance between points A and B is 20 R 11 19 (b) R 20 8 (c) R 15 R (d) 2 (a)

A

41. The equivalent resistance between the points A and B is (R is the

B

A R

resistance of each side of smaller square) 3R 2 R (d) 2

(a) R

(b)

(c) 2R

R

B

Subjective Questions

1. When a steady current passes through a cylindrical conductor, is there an electric field inside the conductor?

2. Electrons in a conductor have no motion in the absence of a potential difference across it. Is this statement true or false? 3. In the Bohr model of hydrogen atom, the electron is pictured to rotate in a circular orbit of radius 5 × 10−11 m, at a speed 2.2 × 106 m/ s. What is the current associated with electron motion?

4. A 120 V house circuit has the following light bulbs switched on : 40 W, 60 W and 75 W. Find the equivalent resistance of these bulbs.

5. Assume that the batteries in figure have negligible internal resistance. Find R1 = 4.0 Ω +

R2 = 8.0 Ω

–

–

+

E2 = 6 V

E1 = 12 V

(a) the current in the circuit, (b) the power dissipated in each resistor and (c) the power of each battery, stating whether energy is supplied by or absorbed by it.

6. The potentiometer wire AB shown in figure is 40 cm long. Where the free end of the galvanometer should be connected on AB so that the galvanometer may show zero deflection? 8Ω

12 Ω

G A

B

84 — Electricity and Magnetism 7. An ideal voltmeter V is connected to a 2.0 Ω resistor and a battery with emf 5.0 V and internal resistance 0.5 Ω as shown in figure :

(a) What is the current in the 2.0 Ω resistor? (b) What is the terminal voltage of the battery? (c) What is the reading of the voltmeter?

8. In figure, E1 = 12 V and E2 = 8 V.

2.0 Ω

A

– +

0.5 Ω

5.0 V

V

B

E1

E2

– +

(a) What is the direction of the current in the resistor? (b) Which battery is doing positive work? (c) Which point, A or B, is at the higher potential?

9. In figure, if the potential at point P is 100 V, what is the potential at point Q? 3.0 Ω

Q

–

150 V

50 V

+

– +

2.0 Ω P

10. Copper has one conduction electron per atom. Its density is 8.89 g/ cm3 and its atomic mass is 63.54 g/ mol. If a copper wire of diameter 1.0 mm carries a current of 2.0 A, what is the drift speed of the electrons in the wire?

11. An aluminium wire carrying a current has diameter 0.84 mm. The electric field in the wire is 0.49 V/m. What is (a) the current carried by the wire? (b) the potential difference between two points in the wire 12.0 m apart? (c) the resistance of a 12.0 m length of this wire? Specific resistance of aluminium is 2.75 × 10−8 Ω - m.

12. A conductor of length l has a non-uniform cross-section. The radius of cross-section varies linearly from a to b. The resistivity of the material is ρ. Find the resistance of the conductor across its ends. b

a

l

13. If a battery of emf E and internal resistance r is connected across a load of resistance R. Show

that the rate at which energy is dissipated in R is maximum when R = r and this maximum power is P = E 2/4r.

14. Two identical batteries each of emf E = 2 volt and internal resistance r = 1 ohm are available to produce heat in an external resistance by passing a current through it. What is the maximum power that can be developed across an external resistance R using these batteries?

Chapter 23

Current Electricity — 85

15. Two coils connected in series have resistance of 600 Ω and 300 Ω at 20 ° C and temperature coefficient of 0.001 and 0.004 ( ° C )−1 respectively. Find resistance of the combination at a temperature of 50° C. What is the effective temperature coefficient of combination?

16. An aluminium wire 7.5 m long is connected in parallel with a copper wire 6 m long. When a current of 5 A is passed through the combination, it is found that the current in the aluminium wire is 3 A. The diameter of the aluminium wire is 1 mm. Determine the diameter of the copper wire. Resistivity of copper is 0.017 µΩ -m and that of the aluminium is 0.028 µ Ω -m.

17. The potential difference between two points in a wire 75.0 cm apart is 0.938 V, when the current density is 4.40 × 107 A/ m 2. What is

(a) the magnitude of E in the wire? (b) the resistivity of the material of which the wire is made?

18. A rectangular block of metal of resistivity ρ has dimensions d × 2d × 3d. A potential difference V is applied between two opposite faces of the block. (a) To which two faces of the block should the potential difference V be applied to give the maximum current density? What is the maximum current density? (b) To which two faces of the block should the potential difference V be applied to give the maximum current? What is this maximum current?

19. An electrical conductor designed to carry large currents has a circular cross-section 2.50 mm in diameter and is 14.0 m long. The resistance between its ends is 0.104 Ω.

(a) What is the resistivity of the material? (b) If the electric field magnitude in the conductor is 1.28 V/m, what is the total current? (c) If the material has 8.5 × 1028 free electrons per cubic metre, find the average drift speed under the conditions of part (b).

20. It is desired to make a 20.0 Ω coil of wire which has a zero thermal coefficient of resistance. To do this, a carbon resistor of resistance R1 is placed in series with an iron resistor of resistance R2. The proportions of iron and carbon are so chosen that R1 + R2 = 20.00 Ω for all temperatures near 20° C. How large are R1 and R2? Given, αC = − 0.5 × 10−3 K −1 and α Fe = 5.0 × 10−3 K −1.

21. Find the current supplied by the battery in the circuit shown in figure. 8Ω

24 V

12 Ω

22. Calculate battery current and equivalent resistance of the network shown in figure. 8Ω

24 V

6Ω

4Ω

12 Ω

86 — Electricity and Magnetism 23. Compute total circuit resistance and battery current as shown in figure. 8Ω

24 V

6Ω

12 Ω

24. Compute the value of battery current i shown in figure. All resistances are in ohm. 6

i

4

12 V

6

12 3 2

25. Calculate the potentials of points A, B, C and D as shown in Fig. (a). What would be the new potential values if connections of 6 V battery are reversed as shown in Fig. (b)? All resistances are in ohm. A 12 V

A 12 V

1

1

B G

0V

B

2

G

0V

2

C 6V

C 6V

3

3

D

D

(a)

(b)

26. Give the magnitude and polarity of the following voltages in the circuit of figure : 1 5Ω 2

200 V 10 Ω

25 Ω 3

(i) V1 (v) V1 − 2

(ii) V 2 (vi) V1 − 3

(iii) V3

(iv) V3 − 2

Chapter 23

Current Electricity — 87

27. The emf E and the internal resistance r of the battery shown in figure are 4.3 V and 1.0 Ω

respectively. The external resistance R is 50 Ω. The resistances of the ammeter and voltmeter are 2.0 Ω and 200 Ω, respectively. r

E

R A S V

(a) Find the readings of the two meters. (b) The switch is thrown to the other side. What will be the readings of the two meters now?

28. Find the current in each branch of the circuit shown in figure. 42 V 5 Ω A

4Ω 10 V

6Ω 8Ω E

1Ω

B

C 16 Ω

D

4V

29. An electrical circuit is shown in figure. Calculate the potential difference across the resistor of

400 Ω as will be measured by the voltmeter V of resistance 400 Ω either by applying Kirchhoff’s rules or otherwise. (JEE 1996) V 400 Ω 100 Ω i2

100 Ω

200 Ω

100 Ω

i1 i

10 V

30. In the circuit shown in figure V1 and V 2 are two voltmeters of resistances

E

3000 Ω and 2000 Ω, respectively. In addition R1 = 2000 Ω , R2 = 3000 Ω and E = 200 V, then

(a) Find the reading of voltmeters V1 and V 2 when (i) switch S is open (ii) switch S is closed (b) Current through S, when it is closed (Disregard the resistance of battery)

V2

V1 R1

S

R2

88 — Electricity and Magnetism 31. In figure, circuit section AB absorbs energy at the rate of 5.0 W when a current i = 1.0 A passes through it in the indicated direction. i A

X E

R = 2.0 Ω

B

(a) What is the potential difference between points A and B? (b) Emf device X does not have internal resistance. What is its emf? (c) What is its polarity (the orientation of its positive and negative terminals)?

32. The potential difference across the terminals of a battery is 8.4 V when there is a current of 1.50 A in the battery from the negative to the positive terminal. When the current is 3.50 A in the reverse direction, the potential difference becomes 9.4 V. (a) What is the internal resistance of the battery? (b) What is the emf of the battery?

33. A battery of emf 2.0 V and internal resistance 0.10 Ω is being charged with a current of 5.0 A. Find the potential difference between the terminals of the battery?

34. Find the currents in different resistors shown in figure. 8Ω

2Ω

2Ω

4Ω

8Ω

2V 2V

2V

35. A resistance box, a battery and a galvanometer of resistance G ohm are connected in series. If the galvanometer is shunted by resistance of S ohm, find the change in resistance in the box required to maintain the current from the battery unchanged.

36. Determine the resistance r if an ammeter shows a current of I = 5 A and a voltmeter 100 V. The internal resistance of the voltmeter is R = 2,500 Ω. A

A

r

B

V

37. In the circuit, a voltmeter reads 30 V when it is connected across 400 Ω resistance. Calculate what the same voltmeter will read when it is connected across the 300 Ω resistance? 60 V

300 Ω

400 Ω

V

Chapter 23

Current Electricity — 89

38. Resistances R1 and R2 , each 60 Ω , are connected in series. The potential difference between points A and B is 120 V. Find the reading of voltmeter connected between points C and D if its resistance r = 120 Ω. A

B

R1

C

R2 D

V

39. A moving coil galvanometer of resistance 20 Ω gives a full scale deflection when a current of 1 mA is passed through it. It is to be converted into an ammeter reading 20 A on full scale. But the shunt of 0.005 Ω only is available. What resistance should be connected in series with the galvanometer coil?

40. A cell of emf 3.4 V and internal resistance 3 Ω is connected to an ammeter having resistance

2 Ω and to an external resistance of 100 Ω. When a voltmeter is connected across the 100 Ω resistance, the ammeter reading is 0.04 A. Find the voltage reading by the voltmeter and its resistance. Had the voltmeter been an ideal one what would have been its reading?

41. (a) A voltmeter with resistance RV is connected across the terminals of a battery of emf E and internal resistance r. Find the potential difference measured by the voltmeter. (b) If E = 7.50 V and r = 0.45 Ω , find the minimum value of the voltmeter resistance RV so that the voltmeter reading is within 1.0% of the emf of the battery. (c) Explain why your answer in part (b) represents a minimum value.

42. (a) An ammeter with resistance RA is connected in series with a resistor R, a battery of emf ε and internal resistance r. The current measured by the ammeter is IA. Find the current through the circuit if the ammeter is removed so that the battery and the resistor form a complete circuit. Express your answer in terms of IA , r , RA and R. Show that more “ideal” the ammeter, the smaller the difference between this current and the current IA. (b) If R = 3.80 Ω , ε = 7.50 V and r = 0.45 Ω , find the maximum value of the ammeter resistance RA so that IA is within 99% of the current in the circuit when the ammeter is absent. (c) Explain why your answer in part (b) represents a maximum value.

43. Each of three resistors in figure has a resistance of 2.4 Ω and can dissipate a maximum of 36 W without becoming excessively heated. What is the maximum power the circuit can dissipate?

44. A storage battery with emf 2.6 V loaded with external resistance produces a current 1 A. In this case, the potential difference between the terminals of the storage battery equals 2 V. Find the thermal power generated in the battery and the net power supplied by the battery for external circuit.

90 — Electricity and Magnetism 45. In the circuit shown in figure E1 = 7 V , E2 = 1 V , R1 = 2 Ω, R2 = 2 Ω and R3 = 3 Ω respectively. Find the power supplied by the two batteries. R1

R2

E1

E2

R3

+

46. In the circuit shown in figure, find a

1.0 Ω

12.0 V

5.0 Ω

b

–

d

c

(a) the rate of conversion of internal (chemical) energy to electrical energy within the battery (b) the rate of dissipation of electrical energy in the battery (c) the rate of dissipation of electrical energy in the external resistor.

47. Three resistors having resistances of 1.60 Ω, 2.40 Ω and 4.80 Ω are connected in parallel to a 28.0 V battery that has negligible internal resistance. Find (a) (b) (c) (d) (e) (f)

the equivalent resistance of the combination. the current in each resistor. the total current through the battery. the voltage across each resistor. the power dissipated in each resistor. which resistor dissipates the maximum power the one with the greatest resistance or the least resistance? Explain why this should be.

48. (a) The power of resistor is the maximum power the resistor can safely dissipate without too rise in temperature. The power rating of a 15 kΩ resistor is 5.0 W. What is the maximum allowable potential difference across the terminals of the resistor? (b) A 9.0 kΩ resistor is to be connected across a 120 V potential difference. What power rating is required?

Note

Attempt the following questions after reading the chapter of capacitors.

49. Find the equivalent resistance between points A and B in the following circuits : 4Ω

8Ω

4Ω

2Ω

6Ω

R

R

R R

R R

A

4Ω

(a)

A B

(b)

B

Chapter 23

Current Electricity — 91

4Ω

10 Ω 5Ω

2Ω

3Ω

10 Ω

2Ω

3Ω

5Ω 5Ω

5Ω

10 Ω

10 Ω A

1Ω

1Ω

A

B

B (d)

(c) 2Ω

R

8Ω

R

R B

R

R 10 Ω

R

R

R 4Ω

R R

R

2Ω

A

R

A

B

(e)

(f)

B 2Ω

1Ω

2Ω

1Ω

2Ω

2Ω 2Ω

1Ω

A

2Ω (g)

50. What will be the change in the resistance of a circuit between A and F consisting of five identical conductors, if two similar conductors are added as shown by the dashed line in figure? B

A

D

C

F

E

51. Find RAB in the circuit, shown in figure. B

A 2Ω

15 Ω

30 Ω

40 Ω

8Ω

20 Ω

15 Ω

6Ω

92 — Electricity and Magnetism 52. Find the equivalent resistance of the networks shown in figure between the points a and b. r

b

r

r a

r

r

r

r

r

r b a

r r a

r

b

r

r

r

(b)

(a)

(c)

b r

r r

r

a

r

r

r

a r

r

(d)

b

(e)

53. Find the equivalent resistance of the circuits shown in figure between the points a and b. Each resistor has a resistance r.

a a

b

b

(a)

(b)

LEVEL 2 Single Correct Option 1. Two cells A and B of emf 1.3 V and 1.5 V respectively are arranged as shown in figure. The voltmeter reads 1.45 V. The voltmeter is assumed to be ideal. Then A E1

r1

V E2 B

(a) r1 = 2r2

(b) r1 = 3r2

r2

(c) r2 = 2r1

(d) r2 = 3r1

Chapter 23

Current Electricity — 93

2. A voltmeter connected in series with a resistance R1 to a circuit indicates a voltage V1 = 198 V.

When a series resistor R2 = 2R1 is used, the voltmeter indicates a voltage V 2 = 180 V. If the resistance of the voltmeter is RV = 900 Ω , then the applied voltage across A and B is A

A R 2 = 2 R1

R1 V 198 V B

V 180 V B

(b) 200 V (d) 240 V

(a) 210 V (c) 220 V

3. All bulbs in the circuit shown in figure are identical. Which bulb glows most brightly?

A

B

C

D

(a) B (c) D

(b) A (d) C

4. A student connects an ammeter A and a voltmeter V to measure a resistance R as shown in figure. If the voltmeter reads 20 V and the ammeter reads 4 A, then R is V

4A

R

A

(a) (b) (c) (d)

equal to 5 Ω greater than 5 Ω less than 5 Ω greater or less than 5 Ω depending upon the direction of current

5. The given figure represents an arrangement of potentiometer for the calculation of internal resistance (r ) of the unknown battery ( E ). The balance length is 70.0 cm with the key opened and 60.0 cm with the key closed. R is 132.40 Ω. The internal resistance (r ) of the unknown cell will be E0

A

B E

r

R

(a) 22.1 Ω (c) 154.5 Ω

G

K

(b) 113.5 Ω (d) 10 Ω

94 — Electricity and Magnetism 6. Switch S is closed at time t = 0. Which one of the following statements is correct? r1

E1

r2

E2 S R

(a) (b) (c) (d)

Current in the resistance R increases if E1r2 < E 2(R + r1 ) Current in the resistance R increases if E1r2 > E 2(R + r1 ) Current in the resistance R decreases if E1r2 > E 2(R + r1 ) Current in the resistance R decreases if E1r2 = E 2(R + r1 )

7. A, B and C are voltmeters of resistances R , 1.5R and 3R respectively. When some potential difference is applied between x and y, the voltmeter readings are V A , V B and VC , then B A

x

y C

(a) V A = VB = VC

(b) V A ≠ VB = VC

(c) V A = VB ≠ VC

(d) V A + VB = VC

8. In the circuit shown, the voltage drop across the 15 Ω resistor is 30 V having the polarity as indicated. The ratio of potential difference across 5 Ω resistor and resistance R is 2A G 5A B

A

5Ω C 3A

– 15Ω +

R H

100 V

(a) 2/7

F

(b) 0.4

(c) 5/7

(d) 1

9. In an experiment on the measurement of internal resistance of a cell by using a potentiometer, when the key K is kept open then balancing length is obtained at y metre. When the key K is closed and some resistance R is inserted in the resistance box, then the balancing length is found to be x metre. Then, the internal resistance is +

Rb

– EP

A

B E, r

J K

G

RB R

(a)

(x − y) R y

(b)

( y − x) R x

(c)

( y − x) R y

(d)

(x − y) R x

Chapter 23

Current Electricity — 95

10. A source of emf E = 10 V and having negligible internal resistance is connected to a variable resistance. The resistance varies as shown in figure. The total charge that has passed through the resistor R during the time interval from t1 to t2 is R 40 Ω 20 Ω

t1 = 10 s

t2 = 30 s

t

(b) 30 log e 3 (d) 10 log e 2

(a) 40 log e 4 (c) 20 log e 2

11. In order to increase the resistance of a given wire of uniform cross-section to four times its value, a fraction of its length is stretched uniformly till the full length of the wire becomes

3 2

times the original length. What is the value of this fraction? 1 4 1 (c) 16

1 8 1 (d) 6 (b)

(a)

12. The figure shows a meter bridge circuit with AB = 100 cm, X = 12 Ω and R = 18 Ω and the jockey

J in the position of balance. If R is now made 8 Ω, through what distance will J have to be moved to obtain balance? +

–

R

X J

A

B

(b) 20 cm (d) 40 cm

(a) 10 cm (c) 30 cm

13. A milliammeter of range 10 mA and resistance 9 Ω is joined in a circuit as shown. The meter gives full scale deflection for current I when A and B are used as its terminals, i.e. current enters at A and leaves at B (C is left isolated). The value of I is 9 Ω, 10mA

0.9 Ω

0.1Ω

A

(a) 100 mA (c) 1 A

B

C

(b) 900 mA (d) 1.1 A

96 — Electricity and Magnetism 14. A battery of emf E0 = 12 V is connected across a 4 m long uniform

wire having resistance 4 Ω /m. The cell of small emfs ε1 = 2 V and ε 2 = 4 V having internal resistance 2 Ω and 6 Ω respectively are connected as shown in the figure. If galvanometer shows no A deflection at the point N , the distance of point N from the point A is equal to

5 m 3 3 (c) m 2

(b)

(a)

E0

R = 4Ω

N ε1

B

r1

4 m 3

G

(d) None of these

ε2

r2

15. In the circuit shown, when keys K 1 and K 2 both are closed, the ammeter reads I 0. But when K 1 is open and K 2 is closed, the ammeter reads I 0 2. Assuming that ammeter resistance is much less than R2, the values of r and R1 in Ω are K1

100 Ω

K2

R2 = 100 Ω

R1 A E, r

(b) 25, 100 (d) 0, 50

(a) 25, 50 (c) 0, 100

16. In the circuit shown in figure, V must be

+

4Ω

20 Ω

100 Ω

25 Ω

V 6Ω –

4A

(b) 80 V (d) 1290 V

(a) 50 V (c) 100 V

17. In the circuit shown in figure ammeter and voltmeter are ideal. If E = 4 V, R = 9 Ω and r = 1 Ω, then readings of ammeter and voltmeter are V R

R R

E, r A

(a) 1 A, 3 V (c) 3 A, 4 V

(b) 2 A, 3 V (d) 4 A, 4 V

Chapter 23

Current Electricity — 97

18. A moving coil galvanometer is converted into an ammeter reading up to 0.03 A by connecting a r . What is the maximum current which can be sent through this 4 galvanometer, if no shunt is used. (Here, r = resistance of galvanometer) shunt of resistance

(b) 0.005 A (d) 0.008 A

(a) 0.004 A (c) 0.006 A

19. The potential difference between points A and B is 8Ω

B

6Ω

4Ω

A

3Ω

10 V

20 V 7 10 V (c) 7

(a)

(b)

40 V 7

(d) zero

20. Two wires A and B made of same material and having their lengths in the ratio 6 : 1 are connected in series. The potential difference across the wires are 3 V and 2 V respectively. If rA r and rB are the radii of A and B respectively, then B is rA 1 4 (c) 1

1 2 (d) 2 (b)

(a)

21. A galvanometer of resistance 50 Ω is connected to a battery of 3 V along with resistance of

2950 Ω in series. A full scale deflection of 30 divisions is obtained in the galvanometer. In order to reduce this deflection to 20 divisions, the above series resistance should be

(a) 4450 Ω (c) 5550 Ω

(b) 5050 Ω (d) 6050 Ω

22. Figure shows a potentiometer arrangement with RAB = 10 Ω and rheostat of variable

resistance x. For x = 0 null deflection point is found at 20 cm from A. For unknown value of x null deflection point was at 30 cm from A, then the value of x is

E

C

A E0

(a) 10 Ω (c) 2 Ω

1Ω

G

(b) 5 Ω (d) 1 Ω

B

98 — Electricity and Magnetism 23. In the given potentiometer arrangement, the null point (a) (b) (c) (d)

V

can be obtained for any value of V can be obtained only if V < V 0 can be obtained only if V > V 0 can never be obtained

r G

A

24. In the given figure the current through 4 Ω resistor is 20Ω

B

V0

R0

4Ω

1.4 A

15Ω 50Ω

10Ω

(b) 0.4 A (d) 0.7 A

(a) 1.4 A (c) 1.0 A

25. All resistances shown in circuit are 2 Ω each. The current in the resistance between D and E is A

B D

C

E

10 V F

H

G

(b) 2.5 A (d) 7.5 A

(a) 5 A (c) 1 A

26. In the circuit shown in figure, the resistance of voltmeter is 6 kΩ. The voltmeter reading will be 10 V

2 kΩ

3 kΩ

V

(a) 6 V (c) 4 V

(b) 5 V (d) 3 V

27. For what ratio of R1 , R2 and R3 power developed across each resistor is equal? R2 R1 i

i R3

(a) 1 : 1 : 1 (c) 4 : 1 : 1

(b) 4 : 4 : 1 (d) 1 : 4 : 4

Chapter 23

Current Electricity — 99

More than One Correct Options 1. Two heaters designed for the same voltage V have different power ratings. When connected individually across a source of voltage V , they produce H amount of heat each in time t1 and t2 respectively. When used together across the same source, they produce H amount of heat in time t (a) If they are in series, t = t1 + t2 t1t2 (c) If they are in parallel, t = (t1 + t2)

(b) If they are in series, t = 2 (t1 + t2) t1t2 (d) If they are in parallel, t = 2(t1 + t2)

2. Two cells of emf E1 = 6 V and E2 = 5 V are joined in parallel with same polarity on same side,

without any external load. If their internal resistances are r1 = 2 Ω and r2 = 3 Ω respectively, then

(a) (b) (c) (d)

terminal potential difference across any cell is less than 5 V terminal potential difference across any cell is 5.6 V current through the cells is 0.2 A current through the cells is zero if E1 = E 2

3. Three ammeters A, B and C of resistances RA , RB and RC respectively are joined as shown. When some potential difference is applied across the terminals T1 and T2 , their readings are I A , I B and IC respectively. Then, A

B T2

T1

C

(a) I A = IB I R (c) A = C IC RA

(b) I ARA + IB RB = IC RC I RC (d) B = IC RA + RB

4. Three voltmeters all having different resistances, are joined as shown. When some potential difference is applied across A and B, their readings are V1 , V 2 and V3 . Then, V1

V2

A

B

V3

(a) V1 = V 2 (c) V1 + V 2 = V3

(b) V1 ≠ V 2 (d) V1 + V 2 > V3

5. Two conductors made of the same material have lengths L and 2L but have equal resistances. The two are connected in series in a circuit in which current is flowing. Which of the following is/are correct? (a) (b) (c) (d)

The potential difference across the two conductors is the same The drift speed is larger in the conductor of length L The electric field in the first conductor is twice that in the second The electric field in the second conductor is twice that in the first

100 — Electricity and Magnetism 6. In the figure shown, (a) (b) (c) (d)

A

current will flow from A to B current may flow A to B current may flow from B to A the direction of current will depend on E

r

E

20 V

B 2V

7. In the potentiometer experiment shown in figure, the null point length is l. Choose the correct options given below. E1

J

l E2 G S

(a) (b) (c) (d)

If jockey J is shifted towards right, l will increase If value of E1 is increased, l is decreased If value of E 2 is increased, l is increased If switch S is closed, l will decrease

8. In the circuit shown in figure, reading of ammeter will S1

R A R E

E

r

S2

r z

(a) increase if S1 is closed (c) increase if S 2 is closed

(b) decrease if S1 is closed (d) decrease if S 2 is closed\

9. In the circuit shown in figure it is given that V b − V a = 2 volt. Choose the correct options. (a) (b) (c) (d)

2Ω

10 V c

a

Current in the wire is 6 A Direction of current is from a to b V a − V c = 12 volt V c − V a = 12 volt

b

a

10. Each resistance of the network shown in figure is r. Net resistance between 7 r 3 (b) a and c is r (c) b and d is r r (d) b and d is 2 (a) a and b is

b

d

c

Current Electricity — 101

Chapter 23

Comprehension Based Questions Passage (Q. No. 1 and 2) The length of a potentiometer wire is 600 cm and it carries a current of 40 mA. For a cell of emf 2V and internal resistance 10 Ω, the null point is found to be at 500 cm. On connecting a voltmeter across the cell, the balancing length is decreased by 10 cm.

1. The voltmeter reading will be (a) 1.96 V

(b) 1.8 V

(c) 1.64 V

(d) 0.96 V

(c) 490 Ω

(d) 20 Ω

2. The resistance of the voltmeter is (a) 500 Ω

(b) 290 Ω

Match the Columns 1. For the circuit shown in figure, match the two columns. b

+4 V

2Ω 1Ω

a

e

c +6 V

1Ω

+2 V 2Ω +4 V

Column I (a) (b) (c) (d)

d

Column II

current in wire ae current is wire be current in wire ce current in wire de

(p) (q) (r) (s)

1A 2A 0.5 A None of these

2. Current i is flowing through a wire of non-uniform cross-section as shown. Match the following two columns. 1

i

Column I

2

i

Column II

(a) Current density

(p) is more at 1

(b) Electric field

(q) is more at 2

(c) Resistance per unit length

(r) is same at both sections 1 and 2

(d) Potential difference per unit length

(s) data is insufficient

102 — Electricity and Magnetism 3. In the circuit shown in figure, after closing the switch S, match the following two columns. R3

S

R1

R2

Column I

Column II

(a) current through R1

(p) will increase

(b) current through R2

(q) will decrease

(c) potential difference across R1 (r) will remain same (d) potential difference across R2 (s) data insufficient

4. Match the following two columns. Column I

Column II

(a) Electrical resistance

(p) [MLT−2A 2]

(b) Electric potential

(q) [ML2 T−3 A −2]

(c)

(r) [ML2 T−3 A −1 ]

Specific resistance

(d) Specific conductance

(s) None of these

5. In the circuit shown in figure, match the following two columns : A

B

4V, 1Ω

1V, 1Ω

1Ω

Column I

Column II (In SI units)

(a) potential difference across battery A

(p) zero

(b) potential difference across battery B

(q) 1

(c) net power supplied/ consumed by A

(r) 2

(d) net power supplied/ consumed by B

(s) 3

Chapter 23

Current Electricity — 103

Subjective Questions 1. Find the equivalent resistance of the triangular bipyramid between the points. D

A

C B

E

(a) A and C

(b) D and E

Assume the resistance of each branch to be R.

2. Nine wires each of resistance r are connected to make a prism as shown in figure. Find the equivalent resistance of the arrangement across E

A D

B

F C

(a) AD

(b) AB

3. The figure shows part of certain circuit, find : 2Ω

1Ω C

5A

4Ω

2A B

12 V 5Ω

3V 6Ω 4A

(a) Power dissipated in 5 Ω resistance. (b) Potential difference VC − VB . (c) Which battery is being charged?

4. A 6 V battery of negligible internal resistance is connected across a uniform wire AB of length 100 cm. The positive terminal of another battery of emf 4 V and internal resistance 1 Ω is joined to the point A A as shown in figure. Take the potential at B to be zero.

6V

D

(a) What are the potentials at the points A and C? (b) At which point D of the wire AB, the potential is equal to the potential C at C? G (c) If the points C and D are connected by a wire, what will be the current 4V 1Ω through it? (d) If the 4V battery is replaced by 7.5 V battery, what would be the answers of parts (a) and (b)?

B

104 — Electricity and Magnetism 5. A thin uniform wire AB of length 1 m, an unknown resistance X and

a resistance of 12 Ω are connected by thick conducting strips, as shown in the figure. A battery and a galvanometer (with a sliding jockey connected to it) are also available. Connections are to be made to measure the unknown resistance X. Using the principle of A Wheatstone bridge answer the following questions :

12 Ω

X B

C

D

(a) Are there positive and negative terminals on the galvanometer? (b) Copy the figure in your answer book and show the battery and the galvanometer (with jockey) connected at appropriate points. (c) After appropriate connections are made, it is found that no deflection takes place in the galvanometer when the sliding jockey touches the wire at a distance of 60 cm from A. Obtain the value of the resistance X.

6. A galvanometer (coil resistance 99 Ω) is converted into an ammeter using a shunt of 1 Ω and connected as shown in figure (a). The ammeter reads 3 A. The same galvanometer is converted into a voltmeter by connecting a resistance of 101 Ω in series. This voltmeter is connected as shown in figure (b). Its reading is found to be 4/5 of the full scale reading. Find : 12 V

12 V

r

r 2Ω

A V 2Ω

(a)

(b)

(a) internal resistance r of the cell (b) range of the ammeter and voltmeter (c) full scale deflection current of the galvanometer.

7. In a circuit shown in figure if the internal resistances of the sources are negligible then at what value of resistance R will the thermal power generated in it will be the maximum. What is the value of maximum power? 10 V

R 6V 3Ω 6Ω

8. In the circuit shown in figure, find : 2.00 A

E1

+

4.00 Ω 3.00 A

R E2 +

3.00 Ω

6.00 Ω 5.00 A

(a) the current in the 3.00 Ω resistor, (b) the unknown emfs E1 and E 2 and (c) the resistance R.

Chapter 23

Current Electricity — 105

9. In the circuit shown, all the ammeters are ideal. 4Ω

20 V A6

2Ω

A1

15 V

A2

A5

6V

A4

2Ω

S

1Ω

A3

10 V

(a) If the switch S is open, find the reading of all ammeters and the potential difference across the switch. (b) If the switch S is closed, find the current through all ammeters and the switch also.

10. An accumulator of emf 2 V and negligible internal resistance is connected across a uniform wire of length 10 m and resistance 30 Ω. The appropriate terminals of a cell of emf 1.5 V and internal resistance 1 Ω is connected to one end of the wire and the other terminal of the cell is connected through a sensitive galvanometer to a slider on the wire. What is the length of the wire that will be required to produce zero deflection of the galvanometer? How will the balancing length change? (a) When a coil of resistance 5 Ω is placed in series with the accumulator. (b) The cell of 1.5 V is shunted with 5 Ω resistor?

11. A circuit shown in the figure has resistances 20 Ω and 30 Ω. At what value of resistance Rx will the thermal power generated in it be practically independent of small variations of that resistance? The voltage between points A and B is supposed to be constant in this case. 20Ω A 30Ω

Rx

B

12. In the circuit shown in figure, the emfs of batteries are E1 and E2 which have internal resistances R1 and R2. At what value of the resistance R will the thermal power generated in it be the highest? What it is?

E1

E2 R

R1

R2

13. A conductor has a temperature independent resistance R and a total heat capacity C. At the moment t = 0 it is connected to a DC voltage V. Find the time dependence of the conductor's temperature T assuming the thermal power dissipated into surrounding space to vary as q = k (T – T0 ), where k is a constant, T0 is the surrounding temperature (equal to conductor’s temperature at the initial moment).

Answers Introductory Exercise 23.1 1. 4.375 × 1018 2. 38880 C 5. 300 C

3. (a) 337.5 C

(b) 2.1 × 10 21

4. 6.6 × 1015 rps, 1.06 mA

6. Yes, from left to right

Introductory Exercise 23.2 1. False

Introductory Exercise 23.3 1. 6.0 × 10 −4 m/s

2. 0.735 µ m/s, 431.4 yr.

Introductory Exercise 23.4 1. 0.18 Ω

2. True

3. 15 g

4. (c)

Introductory Exercise 23.5 1. (d)

2. 85° C

Introductory Exercise 23.6 1. 5 A, 2.5 A

2. 0, 2 V, 5 V, 15 V, 3 A from C to B, 7.5 A from D to A. 1 4. 5. Zero, 1 A A 2

3. 5 V

Introductory Exercise 23.7 1. 3 A, 2 Ω, – 5 V

2. 36 W, 12 W

Introductory Exercise 23.8 1. V =

V1r2 − V2 r1 r1 + r2

,r =

r1r2 r1 + r2

2. 2 V

3. 7.5 V, 0.5 Ω

Introductory Exercise 23.9 1. By connecting a resistance of 999 Ω in series with galvanometer 2. By connecting 1 Ω resistance in parallel with it 3. (n – 1) G

Introductory Exercise 23.10 1. 1.5 Ω

2. (a) 320 cm

(b)

3E 22r

Introductory Exercise 23.11 1. (a)

2. (b)

3. B is most accurate

Introductory Exercise 23.12 1. 14.2 Ω to 14.3 Ω

2. See the hints

3. (c)

Introductory Exercise 23.13 1. (42 × 103 ± 5%) Ω

2. Red, Yellow, Blue, Gold

Exercises LEVEL 1 Assertion and Reason 1. (d)

2. (a,b)

3. (b)

4. (d)

5. (b)

6. (c)

7. (d)

8. (c)

9. (d)

10. (a)

11. (d)

Objective Questions 1.(d)

2.(d)

3.(a)

4.(c)

5.(b)

6.(b)

7.(a)

8.(a)

9.(b)

10.(d)

11.(b)

12.(b)

13.(b)

14.(d)

15.(d)

16.(d)

17.(c)

18.(c)

19.(c)

20.(a)

21.(d)

22.(c)

23.(b)

24.(d)

25.(d)

26.(c)

27.(a)

28.(d)

29.(a)

30.(c)

31.(b)

32.(b)

33.(b)

34.(c)

35.(c)

36.(d)

37.(d)

38.(b)

39.(a)

40.(d)

41.(b)

Subjective Questions 1. Yes

2. False

1 5. (a) A (b) 1 W, 2 W 2 7. (a) zero (b) 5.0 V 9. −10 V

6. 16 cm from A

(c) 6 W (supplied), 3 W (absorbed)

8. (a) Anti-clockwise

(c) 5.0 V

10. 1.9 × 10

ρl 12. πab

4. 82 Ω

3. 1.12 mA

–4

11. (a) 9.9 A

m/ s

19. (a) 3.65 × 10 –8 Ω -m (b) 172.3 A 20. R1 = 18.18 Ω , R 2 = 1.82 Ω

16. 0.569 mm

18. (a) 2d × 3d ,

17. (a) 1.25 V/m (b) 2.84 × 10 –8 Ω -m

(c) 0.60 Ω

(b) 5.88 V

15. 954 Ω , 0.002 /° C

14. 2 W

(b) E1 (c) Point B

V ρd

(b) 2d × 3d ,

6vd ρ

(c) 2.58 × 10 –3 m /s

21. 5 A

8 Ω 5

22. 15 A,

23.

8 Ω, 9 A 3

24.

13 A 3

25. VA = 12 V, VB = 9 V, VC = 3 V, VD = − 6 V, VA′ = 12 V, VB′ = 11 V, VC′ = 9 V, VD′ = 6 V 26. − 75 V, −50 V, 125 V,175 V, −25 V, −200 V 28.

29.

27. (a) 0.1 A, 4.0 V

(b) 0.08 A, 4.2 V

Resistance

5Ω

8Ω

6Ω

16 Ω

4Ω

Current

4A

0.5 A

3.0 A

0.5 A

1.0 A

4A

Towards

A

C

C

C

B

E

20 V 3

31. (a) 5 V 33. 2.5 V 38. 48 V

30. (a) (i) 120 V, 80 V

(ii) 100 V, 100 V

(b)

1 A 60 32. (a) 0.20 Ω

(b) 3 V (c) positive terminal on left side 34. current in all resistors is zero 39. 80 Ω

G2 35. G + S

40. 400 Ω , 3.2 V, 3.238 V

 RA  (b) 0.0045 Ω 42. (a) IA 1 + R + r  

1Ω

36. 20.16 Ω 41. (a)

43. 54 W

ER v Rv + r

44. 0.6 W, 2 W

(b) 8.7 V 37. 22.5 V (b) 4.5 × 10 –3 Ω

108 — Electricity and Magnetism 45. +14 W, −1 W 47. (a) 0.80 Ω

46. (a) 24 W

(b) 4 W

(c) 20 W

(b) 1.60 Ω resistor 17.5 A, 2.40 Ω resistor 11.7 A, 4.80 Ω resistor 5.8 A

(c) 35.0 A

(d) 28.0 V for each

(e) 1.60 Ω resistor 490 W, 2.40 Ω resistor 327 W, 4.80 Ω resistor 163 W 48. (a) 273.8 V (b) 1.6 W 42 R 32 (c) 49. (a) Ω (b) Ω 31 2 21

(d)

25 Ω 6

(e) 6.194 Ω

(f)

5R 4

(g)

(f) least resistance

5 Ω 7

50. The new equivalent resistance will become 0.6 times

51. 23.32 Ω

5 52. (a) r 8

53. (a) r /2 (b) 4r /5

4 (b) r 3

(c) r

r (d) 4

(e) r

LEVEL 2 Single Correct Option 1.(b)

2.(c)

3.(b)

4.(c)

5.(a)

6.(b)

7.(a)

8.(d)

9.(b)

10.(d)

11.(b)

12.(b)

13.(c)

14.(d)

15.(d)

16.(b)

17.(a)

18.(c)

19.(d)

20.(b)

21.(a)

22.(b)

23.(d)

24.(c)

25.(b)

26.(b)

27.(d)

More than One Correct Options 1. (a,c)

2. (b,c,d)

3. (a,b,d)

4. (b,c)

5. (a,b,c)

6. (b,c,d) 7. (a,b,c,d)

8. (a,c)

9. (a,d) 10. (b,d)

Comprehension Based Questions 1. (a)

2. (c)

Match the Columns 1. (a) → q

(b) → s

(c) → q

(d) → s

2. (a) → p

(b) → p

(c) → p

(d) → p

3. (a) → q

(b) → p

(c) → q

(d) → p

4. (a) → q

(b) → r

(c) → s

(d) → s

5. (a) → s

(b) → r

(c) → s

(d) → r

Subjective Questions 1. (a)

2 R 5

(b)

4. (a) 6 V, 2V

2 R 3

2. (a)

8 r 15

(b)

3 r 5

3. (a) 605 W

(b) 6 V

(b) AD = 66.7 cm (c) zero (d) 6 V, –1.5 V, no such point D exists.

5. (a) No (c) 8 Ω

6. (a) 1.01 Ω

7. 2 Ω , 4.5 W

8. (a) 8 A

9. (a) 9.5 A, 9.5 A, 2 A, 5 A, 5 A, 2 A, 12 V 10. 7.5 m (a) 8.75 m (b) 6.25 m 13. T = T 0 + (1 – e – kt /C )

V2 kR

(b) 5 A, 9.95 V

(b) 36 V, 54 V

(c) 0.05 A

(c) 9 Ω

(b) 12.5 A, 2.5 A, 10 A, 7 A, 8 A, 5 A, 15 A R1R 2 (E1R 2 + E 2R1)2 11. 12 Ω 12. R = , Pmax = R1 + R 2 4R1R 2 (R1 + R 2 )

(c) both

Electrostatics Chapter Contents 24.1

Introduction

24.2

Electric charge

24.3

Conductor and Insulators

24.4

Charging of a body

24.5

Coulomb's law

24.6

Electric field

24.7 Electric potential energy 24.8 Electric potential 24.9

Relation between electric field and potential

24.10 Equipotential surfaces 24.11 Electric dipole 24.12 Gauss's law 24.13 Properties of a conductor 24.14 Electric field and potential due to charged spherical shell or solid conducting sphere 24.15 Electric field and potential due to a solid sphere of charge

110 — Electricity and Magnetism

24.1 Introduction When we comb our hair on a dry day and bring the comb close to tiny pieces of paper, we note that they are swiftly attracted by the comb. Similar phenomena occur if we rub a glass rod or an amber rod with a cloth or with a piece of fur. Why does this happens? What really happens in an electric circuit? How do electric motors and generators work? The answers to all these questions come from a branch of physics known as electromagnetism, the study of electric and magnetic interactions. These interactions involve particles that have a property called electric charge, an inherent property of matter that is as fundamental as mass. We begin our study of electromagnetism in this chapter by the electric charge. We will see that it is quantized and obeys a conservation principle. Then we will study the interactions of electric charges that are at rest, called electrostatic interactions. These interactions are governed by a simple relationship known as Coulomb’s law. This law is more conveniently described by using the concept of electric field.

24.2 Electric Charge The electrical nature of matter is inherent in atomic structure. An atom consists of a small, relatively massive nucleus that contains particles called protons and neutrons. A proton has a mass 1.673 × 10 −27 kg, while a neutron has a slightly greater mass1.675 × 10 –27 kg. Surrounding the nucleus is a diffuse cloud of orbiting particles called electrons. An electron has a mass of 9.11 × 10 −31 kg. Like mass, electric charge is an intrinsic property of protons and electrons, and only two types of charge have been discovered positive and negative. A proton has a positive charge, and an electron has a negative charge. A neutron has no net electric charge. The magnitude of the charge on the proton exactly equals the magnitude of the charge on the electron. The proton carries a charge +e and the electron carries a charge −e. The SI unit of charge is coulomb (C) and e has the value e = 1.6 × 10 −19 C Regarding charge the following points are worth noting: 1. Like charges repel each other and unlike charges attract each other. 2. Charge is a scalar and can be of two types positive or negative. 3. Charge is quantized. The quantum of charge is e. The charge on any body will be some integral

multiple of e, i.e. q = ± ne

where,

n = 1, 2, 3…

1  Charge on any body can never be  e , 1.5e, etc. 3  Note

(i) Apart from charge, energy, angular momentum and mass are also quantized. The quantum of energy is h . Quantum of mass is not known till date. hν and that of angular momentum is 2π (ii) The protons and neutrons are combination of other entities called quarks, which have charges ± 1 e and 3

2 ± e. However, isolated quarks have not been observed. So, quantum of charge is still e. 3

Chapter 24

Electrostatics — 111

4. During any process, the net electric charge of an isolated system remains constant or we can say

that charge is conserved. Pair production and pair annihilation are two examples of conservation of charge. 5. A charged particle at rest produces electric field. A charged particle in an unaccelerated motion produces both electric and magnetic fields but does not radiate energy. But an accelerated charged particle not only produces an electric and magnetic fields but also radiates energy in the form of electromagnetic waves. V

Example 24.1 How many electrons are there in one coulomb of negative charge? The negative charge is due to the presence of excess electrons, since they carry negative charge. Because an electron has a charge whose magnitude is e = 1.6 × 10−19 C, the number of electrons is equal to the charge q divided by the charge e on each electron. Therefore, the number n of electrons is q 1.0 Ans. = 6.25 × 1018 n= = − 19 e 1.6 × 10 Solution

24.3 Conductors and Insulators For the purpose of electrostatic theory, all substances can be divided into two main groups, conductors and insulators. In conductors, electric charges are free to move from one place to another, whereas in insulators they are tightly bound to their respective atoms. In an uncharged body, there are equal number of positive and negative charges. The examples of conductors of electricity are the metals, human body and the earth and that of insulators are glass, hard rubber and plastics. In metals, the free charges are free electrons known as conduction electrons. Semiconductors are a third class of materials and their electrical properties are somewhere between those of insulators and conductors. Silicon and germanium are well known examples of semiconductors.

24.4 Charging of a Body Mainly there are the following three methods of charging a body :

Charging by Rubbing The simplest way to experience electric charges is to rub certain bodies against each other. When a glass rod is rubbed with a silk cloth, the glass rod acquires some positive charge and the silk cloth acquires negative charge by the same amount. The explanation of appearance of electric charge on rubbing is simple. All material bodies contain large number of electrons and equal number of protons in their normal state. When rubbed against each other, some electrons from one body pass onto the other body. The body that donates the electrons becomes positively charged while that which receives the electrons becomes negatively charged. For example, when glass rod is rubbed with silk cloth, glass rod becomes positively charged because it donates the electrons while the silk cloth

112 — Electricity and Magnetism becomes negatively charged because it receives electrons. Electricity so obtained by rubbing two objects is also known as frictional electricity. The other places where the frictional electricity can be observed are when amber is rubbed with wool or a comb is passed through a dry hair. Clouds also become charged by friction.

Charging by Contact When a negatively charged ebonite rod is rubbed on a metal object, such as a sphere, some of the excess electrons from the rod are transferred to the sphere. Once the electrons are on the metal sphere, where they can move readily, they repel one another and spread out over the sphere’s surface. The insulated stand prevents them from flowing to the earth. When the rod is removed, the sphere is left with a negative charge distributed over its surface. In a similar manner, the sphere will be left with a positive charge after being rubbed with a positively charged rod. In this case, electrons from the sphere would be transferred to the rod. The process of giving one object a net electric charge by placing it in contact with another object that is already charged is known as charging by contact. ––– – Ebonite – –––– –––––– –

rod

– –– –––– Metal sphere

– – – – –

–– – – –

–– ––

– – – – –

Insulated stand

Fig. 24.1

Charging by Induction It is also possible to charge a conductor in a way that does not involve contact. ––– –Ebonite rod – –––– +++ –– –––––– – +

+ + + +

– –

++

–

–

––– – – –––– –––––– –

Metal – – sphere –

+ + + + +

+++ –– – – –

++

Insulated stand

–

–

– –

+ ++ ++ + + Grounding + + + + wire + + + + ++ ++ Earth

(a)

(b)

(c)

Fig. 24.2

In Fig. (a), a negatively charged rod brought close to (but does not touch) a metal sphere. In the sphere, the free electrons close to the rod move to the other side (by repulsion). As a result, the part of the sphere nearer to the rod becomes positively charged and the part farthest from the rod negatively charged. This phenomenon is called induction. Now, if the rod is removed, the free electrons return to their original places and the charged regions disappear. Under most conditions the earth is a good electric conductor. So, when a metal wire is attached between the sphere and the ground as in figure (b) some of the free electrons leave the sphere and distribute themselves on the much larger earth. If

Chapter 24

Electrostatics — 113

the grounding wire is then removed, followed by the ebonite rod, the sphere is left with a net positive charge. The process of giving one object a net electric charge without touching the object to a second charged object is called charging by induction. The process could also be used to give the sphere a net negative charge, if a positively charged rod were used. Then, electrons would be drawn up from the ground through the grounding wire and onto the sphere. If the sphere were made from an insulating material like plastic, instead of metal, the method of producing a net charge by induction would not work, because very little charge would flow through the insulating material and down the grounding wire. However, the electric force of the charged rod would have some effect as shown in figure. The electric force would cause the positive and negative charges in the molecules of the insulating material to separate slightly, with the negative charges being pushed away from the negative rod. The surface of the plastic sphere does acquire a slight induced positive charge, although no net charge is created. Ebonite rod – – – – – –

+ – + – + – Plastic + – + –

Fig. 24.3 V

Example 24.2 If we comb our hair on a dry day and bring the comb near small pieces of paper, the comb attracts the pieces, why? This is an example of frictional electricity and induction. When we comb our hair, it gets positively charged by rubbing. When the comb is brought near the pieces of paper some of the electrons accumulate at the edge of the paper piece which is closer to the comb. At the farther end of the piece there is deficiency of electrons and hence, positive charge appears there. Such a redistribution of charge in a material, due to presence of a nearby charged body is called inducion. The comb exerts larger attraction on the negative charges of the paper piece as compared to the repulsion on the positive charge. This is because the negative charges are closer to the comb. Hence, there is a net attraction between the comb and the paper piece. Solution

V

Example 24.3 Does the attraction between the comb and the piece of papers last for longer period of time? No, because the comb loses its net charge after some time. The excess charge of the comb transfers to earth through our body after some time. Solution

V

Example 24.4

Can two similarly charged bodies attract each other?

Yes, when the charge on one body ( q1 ) is much greater than that on the other ( q 2 ) and they are close enough to each other so that force of attraction between q1 and induced charge on the other exceeds the force of repulsion between q1 and q 2 . However, two similar point charges can never attract each other because no induction will take place here. Solution

V

Example 24.5 Solution

a mass.

Does in charging the mass of a body change?

Yes, as charging a body means addition or removal of electrons and electron has

114 — Electricity and Magnetism V

Example 24.6

Why a third hole in a socket provided for grounding?

All electric appliances may end with some charge due to faulty connections. In such a situation charge will be accumulated on the appliance. When the user touches the appliance, he may get a shock. By providing the third hole for grounding all accumulated charge is discharged to the ground and the appliance is safe. Solution

INTRODUCTORY EXERCISE 1. 2. 3. 4.

24.1

Is attraction a true test of electrification? Is repulsion a true test of electrification? Why does a phonograph record attract dust particles just after it is cleaned? What is the total charge, in coulombs, of all the electrons in three gram mole of hydrogen atom?

24.5 Coulomb’s Law The law that describes how charges interact with one another was discovered by Charles Augustin de Coulomb in 1785. With a sensitive torsion balance, Coulomb measured the electric force between charged spheres. In Coulomb’s experiment, the charged spheres were much smaller than the distance between them so that the charges could be treated as point charges. The results of the experiments of Coulomb and others are summarized in Coulomb’s law. The electric force Fe exerted by one point charge on another acts along the line between the charges. It varies inversely as the square of the distance separating the charges and is proportional to the product of charges. The force is repulsive if the charges have the same sign and attractive if the charges have opposite signs. The magnitude of the electric force exerted by a charge q1 on another charge q 2 a distance r away is thus, given by k | q1 q 2 | …(i) Fe = r2 The value of the proportionality constant k in Coulomb’s law depends on the system of units used. In SI units the constant k is k = 8.987551787 × 10 9 ≈ 8.988 × 10 9

N-m 2 C2

N-m 2 C2

The value of k is known to such a large number of significant digits because this value is closely related to the speed of light in vacuum. This speed is defined to be exactly c = 2.99792458 × 10 8 m /s. The numerical value of k is defined in terms of c to be precisely.  N- s 2  2  c k = 10 −7  C2 

Chapter 24

Electrostatics — 115

1 , where ε 0 (“epsilon-nought”) is another constant. This 4πε 0 appears to complicate matters, but it actually simplifies many formulae that we will encounter in later chapters. Thus, Eq. (i) can be written as 1 | q1 q 2 | …(ii) Fe = 4πε 0 r 2 This constant k is often written as

 N-s 2  1 = 10 −7 2  c 2 4πε 0  C 

Here,

Substituting value of c = 2.99792458 × 10 8 m /s, we get 1 = 8.99 × 10 9 N-m /C 2 4π ε 0 In examples and problems, we will often use the approximate value 1 = 9.0 × 10 9 N-m 2 /C 2 4πε 0 Here, the quantity ε 0 is called the permittivity of free space. It has the value, ε 0 = 8.854 × 10 –12 C 2 / N-m 2 Regarding Coulomb’s law, the following points are worth noting: 1. Coulomb’s law stated above describes the interaction of two point charges. When two charges exert forces simultaneously on a third charge, the total force acting on that charge is the vector sum of the forces that the two charges would exert individually. This important property, called the principle of superposition of forces, holds for any number of charges. Thus, F net = F1 + F2 +… + Fn 2. The electric force is an action reaction pair, i.e. the two charges exert equal and opposite forces on each other. 3. The electric force is conservative in nature. q2 q1 Fe 4. Coulomb’s law as we have stated above can be used for point Fe r charges in vacuum. If some dielectric is present in the space In vacuum between the charges, the net force acting on each charge is Fig. 24.4 altered because charges are induced in the molecules of the intervening medium. We will describe this effect later. Here at this moment it is enough to say that the force decreases K times if the medium extends till infinity. Here, K is a dimensionless constant which depends on the medium and called dielectric constant of the medium. Thus, q q 1 (in vacuum) ⋅ 122 Fe = 4πε 0 r Fe q q 1 1 q1 q 2 = ⋅ 1 2 = ⋅ K 4πε 0 K r 2 4πε r 2 Here, ε = ε 0 K is called permittivity of the medium. Fe′ =

(in medium)

116 — Electricity and Magnetism Extra Points to Remember ˜

˜

F2 In few problems of electrostatics Lami’s theorem is very useful. According to this theorem, “if three concurrent forces F1, F2 and F3 as shown in Fig. 24.5 are in equilibrium or if F1 + F2 + F3 = 0, then α F1 F F = 2 = 3 sinα sinβ sin γ

Suppose the position vectors of two chargesq1 and q 2 arer1 and r2 , then electric force on charge q1 due to charge q 2 is, q1q 2 1 F1 = ⋅ (r1 – r2 ) 4 πε0 |r1 – r2|3

γ

F1 β

F3

Fig. 24.5

Similarly, electric force on q 2 due to charge q1 is q1q 2 1 F2 = ⋅ (r2 – r1 ) 4 πε0 |r2 – r1|3 Here, q1 and q 2 are to be substituted with sign. r1 = x1 $i + y1 $j + z1k$ and r2 = x2 $i + y2 $j + z2 k$ where ( x1, y1, z1 ) and ( x2 , y2 , z2 ) are the coordinates of charges q1 and q 2 . V

Example 24.7 What is the smallest electric force between two charges placed at a distance of 1.0 m? Solution

Fe =

1 q1 q 2 ⋅ 4πε 0 r 2

…(i)

For Fe to be minimum q1 q 2 should be minimum. We know that ( q1 ) min = ( q 2 ) min = e = 1.6 × 10−19 C Substituting in Eq. (i), we have ( Fe ) min =

( 9.0 × 109 ) (1.6 × 10−19 ) (1.6 × 10−19 ) (1.0) 2

= 2.304 × 10−28 N V

Ans.

Example 24.8 Three charges q1 = 1 µC, q2 = – 2 µC and q3 = 3 µC are placed on the vertices of an equilateral triangle of side 1.0 m. Find the net electric force acting on charge q1 . q3

q1

q2

Fig. 24.6

Charge q2 will attract charge q1 (along the line joining them) and charge q3 will repel charge q1 . Therefore, two forces will act on q1 , one due to q2 and another due to q3 . Since, the force is a vector quantity both of these forces (say F1 and F2 ) will be added by vector method. The following are two methods of their addition. HOW TO PROCEED

Electrostatics — 117

Chapter 24

q3

Solution Method 1. In the figure,

| F1 | = F1 =

1 q1 q 2 ⋅ 4πε 0 r 2

= magnitude of force between q1 and q 2 =

q1

(9.0 × 109 ) (1.0 × 10−6 ) ( 2.0 × 10−6 )

q2

120°

F2

(1.0) 2

F net

Fig. 24.7

= 1.8 × 10−2 N | F2 | = F2 =

Similarly,

F1 α

1 q1 q 3 ⋅ 4πε 0 r 2

= magnitude of force between q1 and q 3 =

(9.0 × 109 ) (1.0 × 10−6 ) ( 3.0 × 10−6 ) (1.0) 2

= 2.7 × 10−2 N Now,

| Fnet | = F12 + F22 + 2F1 F2 cos 120°   =  (1.8) 2 + (2.7) 2 + 2 (1.8) (2.7)  –  

1  −2   × 10 N  2 

= 2.38 × 10−2 N and

tan α = =

or

F2 sin 120° F1 + F2 cos 120° ( 2.7 × 10−2 ) ( 0.87)  1 (1.8 × 10−2 ) + ( 2.7 × 10−2 )  −   2

α = 79.2°

Thus, the net force on charge q1 is 2.38 × 10−2 N at an angle α = 79.2° with a line joining q1 and Ans. q 2 as shown in the figure. Method 2. In this method let us assume a coordinate axes with q1 at origin as shown in figure. The coordinates of q1 , q 2 and q 3 in this coordinate system are (0, 0, 0), ( 1 m, 0, 0) and (0.5 m, 0.87 m, 0) respectively. Now, y

q3

q1

q2

Fig. 24.8

x

118 — Electricity and Magnetism F1 = force on q1 due to charge q 2 q1 q 2 1 = ⋅ ( r1 – r 2 ) 4πε 0 | r1 – r 2 | 3 =

( 9.0 × 109 ) (1.0 × 10–6 ) (–2.0 × 10–6 ) (1.0) 3

[( 0 – 1) $i + ( 0 – 0) $j + ( 0 – 0) k$ ]

= (1.8 × 10−2 $i ) N F2 = force on q1 due to charge q 3 q1 q 3 1 ( r1 – r 3 ) = ⋅ 4πε 0 | r1 – r 3 |3

and

=

(9.0 × 109 ) (1.0 × 10–6 ) (3.0 × 10–6 ) (1.0) 3

[( 0 – 0.5) $i + ( 0 – 0.87) $j + ( 0 – 0) k$ ]

= ( – 1.35 i$ – 2.349 $j ) × 10−2 N F = F1 + F2 = ( 0.45 $i – 2.349 $j ) × 10–2 N

Therefore, net force on q1 is

Ans.

$ there is no need of writing the magnitude and direction Note Once you write a vector in terms of $i , $j and k, of vector separately. V

Example 24.9 Two identical balls each having a density ρ are suspended from a common point by two insulating strings of equal length. Both the balls have equal mass and charge. In equilibrium each string makes an angle θ with vertical. Now, both the balls are immersed in a liquid. As a result the angle θ does not change. The density of the liquid is σ. Find the dielectric constant of the liquid. Each ball is in equilibrium under the following three forces : (iii) weight (i) tension, (ii) electric force and So, Lami’s theorem can be applied. Solution

θ

θ

θ T

θ T′ Fe′

Fe

W′ In liquid

W In vacuum

Fig. 24.9

In the liquid, where, K = dielectric constant of liquid

Fe′ =

Fe K

and W ′ = W − upthrust

Chapter 24

Electrostatics — 119

Applying Lami’s theorem in vacuum Fe W = sin ( 90° + θ ) sin (180° − θ ) or

F W = e cos θ sin θ

…(i)

Similarly in liquid,

F′ W′ = e cos θ sin θ

…(ii)

Dividing Eq. (i) by Eq. (ii), we get F W = e W′ F ′ e K=

or

= K=

or

W W – upthrust Vρg Vρg – Vσg

 F  as e = K     Fe′  (V = volume of ball)

ρ ρ−σ

Ans.

Note In the liquid Fe and W have changed. Therefore, T will also change.

INTRODUCTORY EXERCISE 1. The mass of an electron is 9.11 × 10

−31

24.2

kg, that of a proton is 1.67 × 10−27 kg. Find the ratio Fe /Fg

of the electric force and the gravitational force exerted by the proton on the electron.

2. Find the dimensions and units of ε 0. 3. Three point charges q are placed at three vertices of an equilateral triangle of side a. Find magnitude of electric force on any charge due to the other two.

4. Three point charges each of value + q are placed on three vertices of a square of side a metre. What is the magnitude of the force on a point charge of value − q coulomb placed at the centre of the square?

5. Coulomb’s law states that the electric force becomes weaker with increasing distance. Suppose that instead, the electric force between two charged particles were independent of distance. In this case, would a neutral insulator still be attracted towards the comb.

6. A metal sphere is suspended from a nylon thread. Initially, the metal sphere is uncharged. When a positively charged glass rod is brought close to the metal sphere, the sphere is drawn towards the rod. But if the sphere touches the rod, it suddenly flies away from the rod. Explain, why the sphere is first attracted then repelled?

7. Is there any lower limit to the electric force between two particles placed at a certain distance? 8. Does the force on a charge due to another charge depend on the charges present nearby? 9. The electric force on a charge q 1 due to q 2 is (4 $i − 3 $j) N. What is the force on q 2 due to q 1?

120 — Electricity and Magnetism

24.6 Electric Field A charged particle cannot directly interact with another particle kept at a distance. A charge produces something called an electric field in the space around it and this electric field exerts a force on any other charge (except the source charge itself) placed in it. Thus, the region surrounding a charge or distribution of charge in which its electrical effects can be observed is called the electric field of the charge or distribution of charge. Electric field at a point can be defined in terms of either a vector function E called ‘electric field strength’ or a scalar function V called ‘electric potential’. The electric field can also be visualised graphically in terms of ‘lines of force’. Note that all these are functions of position r ( x, y, z ). The field propagates through space with the speed of light, c. Thus, if a charge is suddenly moved, the force it exerts on another charge a distance r away does not change until a time r /c later. In our forgoing discussion, we will see that electric field strength E and electric potential V are interrelated. It is similar to a case where the acceleration, velocity and displacement of a particle are related to each other.

Electric Field Strength (E ) Like its gravitational counterpart, the electric field strength (often called electric field) at a point in an electric field is defined as the electrostatic force Fe per unit positive charge. Thus, if the electrostatic force experienced by a small test charge q 0 is Fe , then field strength at that point is defined as Fe q0 → 0 q 0

E = lim

The electric field is a vector quantity and its direction is the same as the direction of the force Fe on a positive test charge. The SI unit of electric field is N/C. Here, it should be noted that the test charge q 0 should be infinitesimally small so that it does not disturb other charges which produces E. With the concept of electric field, our description of electric interactions has two parts. First, a given charge distribution acts as a source of electric field. Second, the electric field exerts a force on any charge that is present in this field.

An Electric Field Leads to a Force Suppose there is an electric field strength E at some point in an electric field, then the electrostatic force acting on a charge +q is qE in the direction of E, while on the charge – q it is qE in the opposite direction of E. V

Example 24.10 An electric field of 105 N/C points due west at a certain spot. What are the magnitude and direction of the force that acts on a charge of + 2 µC and − 5 µC at this spot? Solution

Force on + 2µC = qE = ( 2 × 10–6 ) (105 ) = 0.2 N

(due west)

Ans.

(due east)

Ans.

Force on – 5 µ C = (5 × 10–6 ) (105 ) = 0.5 N

Chapter 24

Electrostatics — 121

Electric Field Due to a Point Charge The electric field produced by a point charge q can be obtained in general terms from Coulomb’s law. First note that the magnitude of the force exerted by the charge q on a test charge q 0 is q +

q0

r

Fe

q +

E E

q –

Fig. 24.10

Fe =

qq 1 ⋅ 20 4π ε 0 r

then divide this value by q 0 to obtain the magnitude of the field. q 1 E= ⋅ 2 4πε 0 r If q is positive, E is directed away from q. On the other hand, if q is negative, then E is directed towards q. The electric field at a point is a vector quantity. Suppose E1 is the field at a point due to a charge q1 and E 2 in the field at the same point due to a charge q 2 . The resultant field when both the charges are present is E = E1 + E 2 If the given charge distribution is continuous, we can use the technique of integration to find the resultant electric field at a point. V

Example 24.11 Two positive point charges q1 = 16 µC and q2 = 4 µC, are separated in vacuum by a distance of 3.0 m. Find the point on the line between the charges where the net electric field is zero. Between the charges the two field contributions have opposite directions, and the net electric field is zero at a point (say P) where the magnitudes of E1 and E2 are equal. However, since q 2 < q1 , point P must be closer to q 2 , in order that the field of the smaller charge can balance the field of the larger charge. Solution

q1 +

E2

E1

+ q2

P r1

r2

Fig. 24.11

At P, E1 = E 2 or

1 q1 1 q2 = ⋅ 4πε 0 r12 4πε 0 r22

122 — Electricity and Magnetism r1 = r2

∴

q1 16 = =2 q2 4

…(i)

r1 + r2 = 3.0 m

Also, Solving these equations, we get

r1 = 2 m and

…(ii)

r2 = 1 m

Thus, the point P is at a distance of 2 m from q1 and 1 m from q 2 .

Ans.

Electric Field of a Ring of Charge A conducting ring of radius R has a total charge q uniformly distributed over its circumference. We are interested in finding the electric field at point P that lies on the axis of the ring at a distance x from its centre. y dl, dq r

r= x2+R2

R 90° O

θ x

P dEy

q

θ

dEx

x

dE

Fig. 24.12

We divide the ring into infinitesimal segments of length dl. Each segment has a charge dq and acts as a point charge source of electric field. Let dE be the electric field from one such segment; the net electric field at P is then the sum of all contributions dE from all the segments that make up the ring. If we consider two ring segments at the top and bottom of the ring, we see that the contributions dEto the field at P from these segments have the same x-component but opposite y-components. Hence, the total y-component of field due to this pair of segments is zero. When we add up the contributions from all such pairs of segments, the total field E will have only a component along the ring’s symmetry axis (the x-axis) with no component perpendicular to that axis (i.e. no y or z-component). So, the field at P is described completely by its x-component E x .

Calculation of E x

∴

 q  dq =   ⋅ dl  2πR  dq 1 dE = ⋅ 2 4πε 0 r  1   dq   dE x = dE cos θ =    4π ε 0   x 2 + R 2  =

( dq ) x 1 ⋅ 4π ε 0 (x 2 + R 2 ) 3/2

 x   x2 + R 2 

   

Chapter 24 x

∴

E x = ∫ dE x =

or

 1  qx Ex =   2  4πε 0  ( x + R 2 ) 3/2

4πε 0 (x + R 2 ) 3/2 2

Electrostatics — 123

∫ dq

From the above expression, we can see that (i) E x = 0 at x = 0, i.e. field is zero at the centre of the ring. We should expect this, charges on opposite sides of the ring would push in opposite directions on a test charge at the centre, and the forces would add to zero. q 1 (ii) E x = for x >> R , i.e. when the point P is much farther from the ring, its field is the ⋅ 4πε 0 x 2 same as that of a point charge. To an observer far from the ring, the ring would appear like a point, and the electric field reflects this. dE x (iii) E x will be maximum where = 0. Differentiating E x w. r. t. x and putting it equal to zero we dx q  R 2  1 get x = and E max comes out to be, 3  ⋅ 2. 2 3  4πε 0 R  Ex

Emax

x R 2

Fig. 24.13

Electric Field of a Line Charge Positive charge q is distributed uniformly along a line with length 2a, lying along the y-axis between y = – a and y = + a. We are here interested in finding the electric field at point P on the x-axis. y r = x2 + y 2 dy r

y O

x

θ

P dEy

Fig. 24.14

dEx θ

x

124 — Electricity and Magnetism λ = charge per unit length =

q 2a

q dy 2a q dy dq 1 dE = ⋅ 2 = 4πε 0 2a ( x 2 + y 2 ) 4πε 0 r dq = λ dy =

dE x = dE cos θ =

q x dy ⋅ 4πε 0 2a ( x 2 + y 2 ) 3/ 2

dE y = – dE sin θ = –

q y dy ⋅ 2 4πε 0 2a (x + y 2 ) 3/ 2

qx a dy q 1 1 ⋅ ∫ = ⋅ / 2 2 3 2 4π ε 0 2a – a ( x + y ) 4πε 0 x x 2 + a 2

∴

Ex =

and

Ey = –

q a y dy 1 ⋅ ∫ =0 4πε 0 2a – a ( x 2 + y 2 ) 3/ 2

Thus, electric field is along x-axis only and which has a magnitude, q Ex = 4πε 0 x x 2 + a 2

…(i)

From the above expression, we can see that q 1 (i) if x >> a, E x = ⋅ 2 , i.e. if point P is very far from the line charge, the field at P is the same 4πε 0 x as that of a point charge. (ii) if we make the line of charge longer and longer, adding charge in proportion to the total length so that λ, the charge per unit length remains constant. In this case, Eq. (i) can be written as Ex = = Now, x 2 / a 2 → 0 as a >> x, E x =

1 2πε 0

1  q ⋅  ⋅  2a  x x 2 /a 2 + 1 λ

2πε 0 x x 2 /a 2 + 1 λ 2πε 0 x

Thus, the magnitude of electric field depends only on the distance of point P from the line of charge, so we can say that at any point P at a perpendicular distance r from the line in any direction, the field has magnitude λ (due to infinite line of charge) E= 2πε 0 r

Chapter 24 E∝

or

Electrostatics — 125

1 r

Thus, E-r graph is as shown in Fig. 24.15. E

1 E∝ r

r

Fig. 24.15

The direction of E is radially outward from the line. Note Suppose a charge q is placed at a point whose position vector is rq and we want to find the electric field at a point P whose position vector is rP. Then, in vector form the electric field is given by 1 q E= ⋅ (rP – rq ) 4 πε 0 | rP – rq |3 rP = xP$i + y P r = x $i + y

Here, and

q

$j + z k$ P $ $ q j + zq k

q

In this equation, q is to be substituted with sign. V

Example 24.12 A charge q = 1 µC is placed at point (1 m, 2 m, 4 m). Find the electric field at point P (0, – 4 m, 3 m). Solution

r q = $i + 2$j + 4k$

Here,

r P = – 4$j + 3k$

and ∴

r P − r q = – i$ – 6$j – k$

or

| r P – r q | = (–1) 2 + (–6) 2 + (–1) 2 = 38 m =

Now,

q 1 ⋅ ( rP – rq ) 4πε 0 | r P – r q |3

Substituting the values, we have E=

( 9.0 × 109 ) (1.0 × 10–6 ) ( 38) 3/ 2

(– $i – 6$j – k$ )

= (–38.42 $i – 230.52 $j – 38.42 k$ ) N/ C

Ans.

Electric Field Lines As we have seen, electric charges create an electric field in the space surrounding them. It is useful to have a kind of “map” that gives the direction and indicates the strength of the field at various places. Field lines, a concept introduced by Michael Faraday, provide us with an easy way to visualize the electric field.

126 — Electricity and Magnetism “An electric field line is an imaginary line or curve drawn through a region of space so that its tangent at any point is in the direction of the electric field vector at that point. The relative closeness of the lines at some place give an idea about the intensity of electric field at that point.” EQ Q EP

B A

P

|EA | > |EB |

Fig. 24.16

The electric field lines have the following properties : 1. The tangent to a line at any point gives the direction of E at that point. This is also the path on which a positive test charge will tend to move if free to do so. 2. Electric field lines always begin on a positive charge and end on a negative charge and do not start or stop in mid-space. 3. The number of lines leaving a positive charge or entering a negative charge is proportional to the magnitude of the charge. This means, for example that if 100 lines are drawn leaving a + 4 µC charge then 75 lines would have to end on a –3 µC charge. 4. Two lines can never intersect. If it happens then two tangents can be drawn at their point of intersection, i.e. intensity at that point will have two directions which is absurd. 5. In a uniform field, the field lines are straight parallel and uniformly spaced.

q

–q

(a)

(b)

+ q

q

(d)

+

q

–

+

q

(c)

– –q

q

(e)

–

+

2q

q

–

(f)

Fig. 24.17

6. The electric field lines can never form closed loops as a line can never start and end on the

same charge.

Chapter 24

Electrostatics — 127

7. Electric field lines also give us an indication of the equipotential surface (surface which has the

same potential) 8. Electric field lines always flow from higher potential to lower potential. 9. In a region where there is no electric field, lines are absent. This is why inside a conductor (where

electric field is zero) there, cannot be any electric field line. 10. Electric lines of force ends or starts normally from the surface of a conductor.

INTRODUCTORY EXERCISE

24.3

1. The electric field of a point charge is uniform. Is it true or false? 2. Electric field lines are shown in Fig. 24.18. State whether the electric potential is greater at A or B.

B

A

Fig. 24.18

3. A charged particle always move in the direction of electric field. Is this statement true or false? 4. The trajectory of a charged particle is the same as a field line. Is this statement true or false? 5. Figure shows some of the electric field lines due to three point charges q 1, q 2 and q 3 of equal magnitude. What are the signs of each of the three charges?

q1

q2

q3

Fig. 24.19

6. Four particles each having a charge q, are placed on the four vertices of a regular pentagon. The distance of each corner from the centre is a. Find the electric field at the centre of the pentagon.

7. A charge q = − 2.0 µC is placed at origin. Find the electric field at (3 m, 4 m, 0).

24.7 Electric Potential Energy The electric force between two charges is directed along the line of the charges and depends on the inverse square of their separation, the same as the gravitational force between two masses. Like the gravitational force, the electric force is conservative, so there is a potential energy function U associated with it. When a charged particle moves in an electric field, the field exerts a force that can do work on the particle. This work can always be expressed in terms of electric potential energy. Just as gravitational potential energy depends on the height of a mass above the earth’s surface, electric potential energy depends on the position of the charged particle in the electric field, when a force F acts on a particle that moves from point a to point b, the work Wa → b done by the force is given by

128 — Electricity and Magnetism b

Wa → b = ∫ F ⋅ ds = ∫ a

b

a

F cos θ ds

where, ds is an infinitesimal displacement along the particle’s path and θ is the angle between F and ds at each point along the path. Second, if the force F is conservative, the work done by F can always be expressed in terms of a potential energy U. When the particle moves from a point where the potential energy isU a to a point where it isU b , the change in potential energy is, ∆U = U b – U a . This is related by the work Wa → b as …(i) Wa→ b = U a – U b = – (U b – U a ) = – ∆U Here, Wa→ b is the work done in displacing the particle from a to b by the conservative force (here electrostatic) not by us. Moreover we can see from Eq. (i) that if Wa → b is positive, ∆U is negative and the potential energy decreases. So, whenever the work done by a conservative force is positive, the potential energy of the system decreases and vice-versa. That’s what happens when a particle is thrown upwards, the work done by gravity is negative, and the potential energy increases. V

Example 24.13 A uniform electric field E 0 is directed along positive y-direction. Find the change in electric potential energy of a positive test charge q 0 when it is displaced in this field from yi = a to y f = 2a along the y-axis. Solution

∴ or

E0

Electrostatic force on the test charge, (along positive y-direction) Fe = q 0 E 0

q 0E 0

W i − f = – ∆U ∆U = – Wi − f = – [ q 0 E 0 ( 2a – a )]

+ q0

= – q0 E0 a

Ans. Fig. 24.20

Note Here, work done by electrostatic force is positive. Hence, the potential energy is decreasing.

Electric Potential Energy of Two Charges The idea of electric potential energy is not restricted to the special case of a uniform electric field as in example 24.13. Let us now calculate the work done on a test charge q 0 moving in a non-uniform electric field caused by a single, stationary point charge q. a q

b q0

r ra rb

Fig. 24.21

The Coulomb’s force on q 0 at a distance r from a fixed charge q is qq 1 ⋅ 20 F= 4πε 0 r If the two charges have same signs, the force is repulsive and if the two charges have opposite signs, the force is attractive. The force is not constant during the displacement, so we have to integrate to calculate the work Wa→ b done on q 0 by this force as q 0 moves from a to b.

Chapter 24 ∴

rb

rb

ra

ra

Wa→ b = ∫ F dr = ∫

Electrostatics — 129

qq 0  1 qq 1 1 ⋅ 0 dr =  –  4πε 0  ra rb  4πε 0 r 2

Being a conservative force this work is path independent. From the definition of potential energy, qq 0  1 1 U b – U a = − Wa – b =  –  4πε 0  rb ra  We choose the potential energy of the two charge system to be zero when they have infinite separation. This means U ∞ = 0. The potential energy when the separation is r is U r qq 0  1 1  Ur – U∞ = ∴  –  4πε 0  r ∞  or

Ur =

qq 0 1 4πε 0 r

This is the expression for electric potential energy of two point charges kept at a separation r. In this expression both the charges q and q 0 are to be substituted with sign. The potential energy is positive if the charges q and q 0 have the same sign and negative if they have opposite signs. Note that the above equation is derived by assuming that one of the charges is fixed and the other is displaced. However, the potential energy depends essentially on the separation between the charges and is independent of the spatial location of the charged particles. We emphasize that the potential energy U given by the above equation is a shared property of two charges q and q 0 , it is a consequence of the interaction between these two charges. If the distance between the two charges is changed from ra to rb , the change in the potential energy is the same whether q is held fixed and q 0 is moved or q 0 is held fixed and q is moved. For this reason we will never use the phrase ‘the electric potential energy of a point charge’.

Electric Potential Energy of a System of Charges The electric potential energy of a system of charges is given by qi q j 1 U = ∑ 4πε 0 i < j rij This sum extends over all pairs of charges. We don’t let i = j, because that would be an interaction of a charge with itself, and we include only terms with i < j to make sure that we count each pair only once. q2 Thus, to account for the interaction between q 5 and q 4 , we include a term with i = 4 and j = 5 but not a term with i = 5 and j = 4. q3 For example, electric potential energy of four point charges q1 , q 2 , q 3 and q 4 would q 1 q4 be given by Fig. 24.22 1  q 4 q 3 q 4 q 2 q 4 q1 q 3 q 2 q 3 q1 q 2 q1  …(ii) U = + + + + +   r21  4πε 0  r43 r42 r41 r32 r31 Here, all the charges are to be substituted with sign. Note Total number of pairs formed by n point charges are

n (n – 1) . 2

130 — Electricity and Magnetism V

Example 24.14 Four charges q1 = 1 µC, q2 = 2 µC, q3 = – 3 µC and q4 = 4 µC are kept on the vertices of a square of side 1m. Find the electric potential energy of this system of charges. q4

q3

1m

1m q1

q2

Fig. 24.23

Solution

In this problem, r41 = r43 = r32 = r21 = 1 m r42 = r31 = (1) 2 + (1) 2 = 2 m

and

Substituting the proper values with sign in Eq. (ii), we get

 ( 4)(–3) ( 4)(2) ( 4)(1) (–3)(2) (–3)(1) (2)(1)  + + U = (9.0 × 10 9 )(10 –6 )(10 –6 )  + + + 1 1  1 2 2  1 5  = ( 9.0 × 10–3 ) –12 +  2  = – 7.62 × 10–2 J

Ans.

Note Here, negative sign of U implies that positive work has been done by electrostatic forces in assembling these charges at respective distances from infinity. V

Example 24.15 Two point charges are located on the x-axis, q1 = – 1 µC at x = 0 and q2 = + 1 µC at x = 1 m. (a) Find the work that must be done by an external force to bring a third point charge q3 = + 1 µC from infinity to x = 2 m. (b) Find the total potential energy of the system of three charges. Solution (a) The work that must be done on q 3 by an external force is equal to the difference of potential energy U when the charge is at x = 2m and the potential energy when it is at infinity. ∴ W = U f – Ui

=

1 4πε 0

 q3 q2 q q q q + 3 1 + 2 1  ( r31 ) f ( r21 ) f  ( r32 ) f

Here,

( r21 ) i = ( r21 ) f

and

( r32 ) i = ( r31 ) i = ∞

∴

W=

1 4πε 0

 q q q q  1  q3 q2 + 3 1 + 2 1 –   4πε 0  ( r32 ) i ( r31 ) i ( r21 ) i 

 q3 q2 q q + 3 1  ( r31 ) f  ( r32 ) f

  

Chapter 24

Electrostatics — 131

Substituting the values, we have  (1) (1) (1) (–1) W = ( 9.0 × 109 ) (10–12 )  + ( 2.0)   (1.0) = 4.5 × 10–3 J

Ans.

(b) The total potential energy of the three charges is given by, U=

1  q 3 q 2 q 3 q1 q 2 q1  + +   r31 r21  4πε 0  r32

 (1) (1) (1) (–1) (1) (–1) (10–12 ) = ( 9.0 × 109 )  + + (2.0) (1.0)   (1.0) = – 4.5 × 10–3 J V

Ans.

Example 24.16 Two point charges q1 = q2 = 2 µC are fixed at x1 = + 3 m and x2 = – 3 m as shown in figure. A third particle of mass 1 g and charge q3 = – 4 µC are released from rest at y = 4.0 m . Find the speed of the particle as it reaches the origin. y y = 4m

q3

q2

q1 x

O

x 2 = –3 m

x1 = 3m

Fig. 24.24

Here, the charge q3 is attracted towards q1 and q2 both. So, the net force on q3 is towards origin.

HOW TO PROCEED

y q3

Fnet q2

q1 x O Fig. 24.25

By this force, charge is accelerated towards origin, but this acceleration is not constant. So, to obtain the speed of particle at origin by kinematics we will have to first find the acceleration at some intermediate position and then will have to integrate it with proper limits. On the other hand, it is easy to use energy conservation principle, as the only forces are conservative.

132 — Electricity and Magnetism Solution

Let v be the speed of particle at origin. From conservation of mechanical energy, Ui + K i = U f + K f

q q q q  1  q3 q2 1 + 3 1 + 2 1  + 0=  4πε 0  ( r32 ) i ( r31 ) i ( r21 ) i  4πε 0

or

 q3 q2 q q q q + 3 1 + 2 1  ( r31 ) f ( r21 ) f  ( r32 ) f

 1 2  + mv  2

( r21 ) i = ( r21 ) f

Here,

Substituting the proper values, we have  (– 4 ) ( 2) (– 4 ) ( 2)  (– 4 ) ( 2) (– 4 ) ( 2) –12 ( 9.0 × 109 )  + × 10–12 = ( 9.0 × 109 )  +   × 10 ( 5.0 ) ( 5.0 ) ( 3.0 ) ( 3.0 )     1 + × 10–3 × v 2 2  16  16 1 ( 9 × 10–3 )  –  = ( 9 × 10–3 )  –  + × 10–3 × v 2  5  3 2

∴

 2 1 ( 9 × 10–3 ) (16)   = × 10–3 × v 2  15 2 ∴

v = 6.2 m/s

INTRODUCTORY EXERCISE

Ans.

24.4

1. A point charge q 1 = 1.0 µC is held fixed at origin. A second point charge q 2 = − 2.0 µC and a

mass 10−4 kg is placed on the x-axis, 1.0 m from the origin. The second point charge is released from rest. What is its speed when it is 0.5 m from the origin?

2. A point charge q 1 = − 1.0 µC is held stationary at the origin. A second point charge q 2 = + 2.0 µC moves from the point (1.0 m, 0, 0) to (2.0 m, 0, 0). How much work is done by the electric force on q 2? 3. A point charge q 1 is held stationary at the origin. A second charge q 2 is placed at a point a, and the electric potential energy of the pair of charges is –6.4 × 10−8 J. When the second charge is moved to point b, the electric force on the charge does 4.2 × 10−8 J of work. What is the electric potential energy of the pair of charges when the second charge is at point b?

4. Is it possible to have an arrangement of two point charges separated by finite distances such that the electric potential energy of the arrangement is the same as if the two charges were infinitely far apart? What if there are three charges?

24.8 Electric Potential As we have discussed in Article 24.6 that an electric field at any point can be defined in two different ways: (i) by the field strength E, and (ii) by the electric potential V at the point under consideration. Both E and V are functions of position and there is a fixed relationship between these two. Of these, the field strength Eis a vector quantity while the electric potential V is a scalar quantity. In this article,

Chapter 24

Electrostatics — 133

we will discuss about the electric potential and in the next, the relationship between E and V. “Potential is the potential energy per unit charge.” Electric potential at any point in an electric field is defined as the potential energy per unit charge, same as the field strength is defined as the force per unit charge. Thus, U or U = q 0V V= q0 The SI unit of potential is volt (V) which is equal to joule per coulomb. So, 1 V = 1 J/C The work done by the electrostatic force in displacing a test charge q 0 from a to b in an electric field is defined as the negative of change in potential energy between them, or ∆U = – Wa – b ∴ We divide this equation by q 0

U b – U a = – Wa – b Wa – b Ub Ua − =– q0 q0 q0

or

Va – Vb =

as

V=

Wa – b q0 U q0

Thus, the work done per unit charge by the electric force when a charged body moves from a to b is equal to the potential at a minus the potential at b. We sometimes abbreviate this difference as Vab = Va – Vb . Another way to interpret the potential difference Vab is that the potential at a minus potential at b, equals the work that must be done to move a unit positive charge slowly from b to a against the electric force. (Wb – a ) external force Va – Vb = q0

Absolute Potential at Some Point Suppose we take the point b at infinity and as a reference point assign the value Vb = 0, the above equations can be written as (Wa – b ) electric force (Wb – a ) external force Va – Vb = = q0 q0 or

Va =

(Wa – ∞ ) electric force q0

=

(W∞ – a ) external

force

q0

Thus, the absolute electric potential at point a in an electric field can be defined as the work done in displacing a unit positive test charge from infinity to a by the external force or the work done per unit positive charge in displacing it from a to infinity.

134 — Electricity and Magnetism Note The following three formulae are very useful in the problems related to work done in electric field. (Wa – b ) electric force = q0 (Va – Vb ) (Wa – b ) external force = q0 (Vb – Va ) = – (Wa – b ) electric force (W∞ – a ) external force = q0Va Here, q0 , Va and Vb are to be substituted with sign. V

Example 24.17 The electric potential at point A is 20 V and at B is – 40 V. Find the work done by an external force and electrostatic force in moving an electron slowly from B to A. Solution

Here, the test charge is an electron, i.e. q 0 = – 1.6 × 10–19 C VA = 20 V VB = – 40 V

and Work done by external force

(WB – A ) external

force

= q 0 (VA – VB ) = (– 1.6 × 10 –19 ) [(20) – (– 40)] = – 9.6 × 10 –18 J

Ans.

Work done by electric force

(WB – A ) electric force = – (WB – A ) external force = – (– 9.6 × 10 –18 J ) = 9.6 × 10 –18 J

Ans.

Note Here, we can see that the electron (a negative charge) moves from B (lower potential) to A (higher potential) and the work done by electric force is positive. Therefore, we may conclude that whenever a negative charge moves from a lower potential to higher potential work done by the electric force is positive or when a positive charge moves from lower potential to higher potential the work done by the electric force is negative. V

Example 24.18 Find the work done by some external force in moving a charge q = 2 µC from infinity to a point where electric potential is 104 V . Solution

Using the relation, (W∞– a ) external force = qVa

We have,

(W∞– a ) external

force

= ( 2 × 10–6 ) (104 ) = 2 × 10–2 J

Electric Potential Due to a Point Charge q From the definition of potential,

q q0 1 ⋅ 4πε 0 r U V= = q0 q0

Ans.

Chapter 24 V=

or

Electrostatics — 135

q 1 ⋅ 4πε 0 r

Here, r is the distance from the point charge q to the point at which the potential is evaluated. If q is positive, the potential that it produces is positive at all points; if q is negative, it produces a potential that is negative everywhere. In either case, V is equal to zero at r = ∞.

Electric Potential Due to a System of Charges Just as the electric field due to a collection of point charges is the vector sum of the fields produced by each charge, the electric potential due to a collection of point charges is the scalar sum of the potentials due to each charge. qi 1 V= ∑ 4πε 0 i ri In this expression, ri is the distance from the i th charge, q i , to the point at which V is evaluated. For a continuous distribution of charge along a line, over a surface or through a volume, we divide the charge into elements dq and the sum in the above equation becomes an integral, dq 1 V= ∫ 4πε 0 r qi 1 dq 1 , if the whole charge is at equal distance r0 from the or V = Σ ∫ 4 πε 0 r 4 πε 0 i ri point where V is to be evaluated, then we can write, 1 q V= ⋅ net 4 πε 0 r0

Note In the equation V =

where, qnet is the algebraic sum of all the charges of which the system is made.

Here there are few examples :

Example (i) Four charges are placed on the vertices of a square as shown in figure. The electric potential at centre of the square is zero as all the charges are at same distance from the centre and

+ 4 µC

– 2 µC

– 4 µC

+2 µC

qnet = 4 µC – 2 µC + 2 µC – 4 µC = 0

Fig. 24.26

Example (ii) A charge q is uniformly distributed over the circumference of a ring in Fig. (a) and is non-uniformly distributed in Fig. (b). +

+

+ + +

+ +

+q

+

+ +

R

q + + + +

+

+

+ ++ +

+ + + +

R

+ +

+

+ + + ++

+

(a)

(b)

Fig. 24.27

+

136 — Electricity and Magnetism The electric potential at the centre of the ring in both the cases is 1 q V = ⋅ 4 πε 0 R

R 2+ r 2

(where, R = radius of ring)

r P

C

and at a distance r from the centre of ring on its axis would be V= V

q 1 ⋅ 4πε 0 R 2 + r2

Fig. 24.28

Example 24.19 Three point charges q1 = 1 µC , q2 = – 2 µC and q3 = 3 µC are placed at (1 m, 0, 0), (0, 2 m, 0) and (0, 0, 3 m) respectively. Find the electric potential at origin. Solution

The net electric potential at origin is 1  q1 q 2 q 3  V= +   + 4πε 0  r1 r2 r3 

Substituting the values, we have 2 3  1 –6 V = ( 9.0 × 109 )  – +  × 10  1.0 2.0 3.0 = 9.0 × 103 V V

Ans.

Example 24.20 A charge q = 10 µC is distributed uniformly over the circumference of a ring of radius 3 m placed on x-y plane with its centre at origin. Find the electric potential at a point P (0, 0, 4 m). Solution

The electric potential at point P would be z P 4m

r0 +

+

+

y

+

+

+

3m

+

+ +

+ + +

+

+

+

x

q +

Fig. 24.29

v

q 1 V= ⋅ 4πε 0 r0 r0 = distance of point P from the circumference of ring

Here,

= ( 3) 2 + ( 4 ) 2 = 5 m q = 10 µC = 10–5 C

and Substituting the values, we have V=

( 9.0 × 109 ) (10–5 ) = 1.8 × 104 V ( 5.0)

Ans.

Chapter 24

Electrostatics — 137

Variation of Electric Potential on the Axis of a Charged Ring We have discussed earlier that the electric potential at the centre of a charged ring (whether charged q 1 uniformly or non-uniformly) is ⋅ and at a distance r from the centre on the axis of the ring is 4πε 0 R q 1 .From these expressions, we can see that electric potential is maximum at the centre ⋅ 2 4πε 0 R + r2 and decreases as we move away from the centre on the axis. Thus, potential varies with distance r as shown in figure. V V0

r r=0

Fig. 24.30

V0 =

In the figure,

q 1 ⋅ 4πε 0 R

Electric Potential on the Axis of a Uniformly Charged Disc Let us find the electric potential at any point P, a distance x on the axis of a uniformly charged circular disc, having surface charge density σ. Let us divide the disc into a large number of thin circular strips and consider a strip of radius r and width dr. Each point of this strip can be assumed to be at equal distance r 2 + x 2 from point P. Potential at P due to this circular strip is dr x

P

r

Fig. 24.31

dV = Here, ∴

dq 1 ⋅ 4πε 0 r2 + x2

dq = σ ( area of strip) or σ (2πrdr ) 1 dV = ⋅ 4πε 0 r2 + x2

Thus, the potential due to the whole disc is R σ R rdr V = ∫ dV = ∫ 0 0 2ε 0 r2 + x2

or

V=

dq = σ (2πrdr )

σ [ R 2 + x 2 – x] 2ε 0

138 — Electricity and Magnetism (i) At the centre of the disc, x = 0 ∴

V (centre ) =

σR 2ε 0

…(i)

(ii) For x >> R , using the Binomial expansion for  R2  R + x = x 1 + 2   x  2

1/ 2

≈x+

2

R2 2x

∴

V=

 σR 2 R2 σ  π R 2σ x + – x = = 2ε0  2x  4 ε 0 x 4π ε 0 x

or

V=

q 4πε 0 x

as πR 2σ = q, the total charge on the disc. This is the relation as obtained due to a point charge. Thus, at far away points, the distribution of charge becomes insignificant. It is difficult to calculate the potential at the points other than on the axis. However, potential on the edge of the disc can be calculated as under.

Potential on the Edge of the Disc dr

To calculate the potential at point P, let us divide the disc in large number of rings with P as centre. The potential due to one segment between r and r + dr is given as dq 1 dV = ⋅ 4πε 0 r Here, ∴

Further, ∴ Hence, ∴ Solving, we get

P

dq = σ (Area of ring) = σ (2r θ ) dr σ (2r θ ) dr 1 dV = ⋅ 4πε 0 r =

r θ R

C

Fig. 24.32

σ ⋅ θ dr 2πε 0

r = 2R cos θ dr = – 2R sin θ dθ σ dV = – 2Rθ sin θ dθ 2πε 0 V =∫

0

π/2

dV =

σR πε 0

V=

σR πε 0

π/2

∫0

θ sin θ dθ …(ii)

Comparing Eqs. (i) and (ii), we see that potential at the centre of the disc is greater than the potential at the edge.

Chapter 24 V

Electrostatics — 139

Example 24.21 Find out the points on the line joining two charges + q and – 3q (kept at a distance of 1.0 m) where electric potential is zero. Let P be the point on the axis either to the left or to the right of charge + q at a distance r where potential is zero. Hence, Solution

+q

P

–3q

1.0 m

+q

–3q

P

or r

r

1.0 – r

Fig. 24.33

VP =

q 3q – =0 4πε 0 r 4πε 0 (1 + r )

VP =

q 3q =0 – 4πε 0 r 4πε 0 (1 – r )

Solving this, we get r = 0.5 m Further,

which gives r = 0.25 m Thus, the potential will be zero at point P on the axis which is either 0.5 m to the left or 0.25 m to the right of charge + q. Ans.

INTRODUCTORY EXERCISE

24.5

1. FindVba if 12 J of work has to be done against an electric field to take a charge of 10–2 C from a to b.

2. A rod of length L lies along the x-axis with its left end at the origin. It has a non-uniform charge density λ = αx, where α is a positive constant. (a) What are the units of α? (b) Calculate the electric potential at point A where x = – d .

3. A charge q is uniformly distributed along an insulating straight wire of length 2l as shown in Fig. 24.34. Find an expression for the electric potential at a point located a distance d from the distribution along its perpendicular bisector. P

d

2l

Fig. 24.34

4. A cone made of insulating material has a total charge Q spread uniformly over its sloping surface. Calculate the work done in bringing a small test charge q from infinity to the apex of the cone. The cone has a slope length L.

140 — Electricity and Magnetism

24.9 Relation Between Electric Field and Potential As we have discussed above, an invisible space is produced across a charge or system of charges in which any other test charge experiences an electrical force. The vector quantity related to this force is known as electric field. Further, a work is done by this electrostatic force when this test charge is moved from one point to another point. The scalar quantity related to this work done is called potential. Electric field ( E) and potential (V ) are different at different positions. So, they are functions of position. In a cartesian coordinate system, position of a particle can be represented by three variable coordinates x, y and z. Therefore, E and V are functions of three variables x, y and z. In physics, we normally keep least number of variables. So, sometimes E andV are the functions of a single variable x or r. Here, x is the x-coordinate along x-axis and r normally a distance from a point charge or from the centre of a charged sphere or charged spherical shell. From the x-coordinate, we can cover only x -axis. But, from the variable r, we can cover the whole space. Now, E and V functions are related to each other either by differentiation or integration. As far as differentiation is concerned, if there are more than one variables then partial differentiation is done and in case of single variable direct differentiation is required. In case of integration, some limit is required. Limit means value of the function which we get after integration should be known to us at some position. For example, after integrating E, we get V . So, value of V should be known at some given position. Without knowing some limit, an unknown in the form of constant of integration remains in the equation. One known limit of V is : potential is zero at infinity.

Conversion of V function into E function This requires differentiation. Case 1

When variables are more than one In this case, E = E x $i + E y $j + E z k$ Here,

∂V = – (partial derivative of V w. r. t. x ) ∂x ∂V Ey = – = – (partial derivative of V w. r. t. y) ∂y ∂V Ez = – = – (partial derivative of V w. r. t. z ) ∂z Ex = –

 ∂V $ ∂V $ ∂V $  E=− i+ j+ k ∂y ∂z   ∂x

∴ This is also sometimes written as

E = – gradient V = – grad V = – ∇ V V

Example 24.22 The electric potential in a region is represented as V = 2x + 3 y – z obtain expression for electric field strength.

Electrostatics — 141

Chapter 24  ∂V $ ∂V $ ∂V $  E= –  i+ j+ k ∂y ∂z   ∂x ∂V ∂ = ( 2x + 3 y – z ) = 2 ∂x ∂x ∂V ∂ = ( 2x + 3 y – z ) = 3 ∂y ∂y

Solution

Here,

∂V ∂ = ( 2x + 3 y – z ) = – 1 ∂z ∂z E = −2$i − 3$j + k$

∴

Ans.

When variable is only one In this case, electric potential is function of only one variable (say r ) and we can write the expression like : dV E =− dr

Case 2

E = − slope of V - r graph

or

Example Electric potential due to a point charge q at distance r is given as q q dV 1 1 ⇒ =− ⋅ 2 V= ⋅ dr 4π ε 0 r 4πε 0 r E=−

∴

1 dV q = ⋅ 2 dr 4 πε 0 r

and we know that this is the expression of electric field due to a point charge. Note E is a vector quantity. In the above method, if single variable is x and E comes out to be positive, then direction of E is towards positive x-axis. Negative value of E means direction is towards negative x- axis. If variable is r, then positive value of E means away from the point charge or away from the centre of charged spherical body and negative value of E means towards the charge or towards the centre of charged spherical body.

Let us take an another example : We wish to find E - r graph corresponding to V - r graph shown in Fig. 24.35. Electric field E = – 5 V/m for 0 ≤ r ≤ 2 m as slope of V-r graph is 5 V/m. E = 0 for 2 m ≤ r ≤ 4 m as slope of V-r graph in this region is zero. Similarly, E = 5 V/m for 4 m ≤ r ≤ 6 m as slope in this region is – 5 V/m. So, the corresponding E - r graph is as shown in Fig. 24.36. E(V/m) +5 2 4 –5

Fig. 24.36

6

r (m)

V (volt) 10

0

2

4 6 Fig. 24.35

r (m)

142 — Electricity and Magnetism V

Example 24.23 The electric potential V at any point x, y, z (all in metre) in space is given by V = 4x 2 volt. The electric field at the point (1m, 0, 2 m) is ………V/m. (JEE 1992)  ∂ V $ ∂ V $ ∂V $  E=− i+ j+ k  ⇒ V = 4x 2 ∂y ∂z  ∂x ∂V ∂V ∂V Therefore, = 8x and = 0= ∂x ∂y ∂z

Solution

E = − 8x$i or E at (1 m, 0, 2 m) is −8 $i V/m.

Conversion of E into V We have learnt, how to find electric field E from the electrostatic potential V. Let us now discuss how to calculate potential difference or absolute potential if electric field E is known. For this, use the relation dV = – E ⋅ d r B

B

or

∫A dV = – ∫A E ⋅ d r

or

VB – V A = – ∫ E ⋅ d r

B

A

dr = dx $i + dy $j + dz k$

Here,

When E is Uniform Let us take this case with the help of an example. V

$) Example 24.24 Find V ab in an electric field E = (2 $i + 3 $j + 4 k where Solution

or

N , C

ra = ( $i – 2 $j + k$ ) m and rb = (2 $i + $j – 2 k$ ) m Here, the given field is uniform (constant). So using, dV = – E⋅ d r Vab = Va – Vb = – =−∫

(1 ,–2 , 1)

=–∫

(1 ,–2 , 1)

(2 , 1 ,–2)

(2 , 1 ,–2)

a

∫b

E⋅ d r

( 2 $i + 3 $j + 4 k$ ) ⋅ ( dx $i + dy $j + dz k$ ) ( 2 dx + 3 dy + 4 dz ) ( 1, – 2, 1)

= − [ 2x + 3 y + 4 z ] (2, 1, – 2) = – 1V

Ans.

V = Ed Here, V is the potential difference between any two points, E is the magnitude of uniform electric field and d is the projection of the distance between two points along the electric field.

Note In uniform electric field, we can also apply

Chapter 24

Electrostatics — 143

For example, in the figure for finding the potential difference between points A and B we will have to keep two points in mind, B

A

E

C d

Fig. 24.37

(i) V A > VB as electric lines always flow from higher potential to lower potential. (ii) d ≠ AB but d = AC Hence, in the above figure, V A – VB = Ed V

Example 24.25 In uniform electric field E = 10 N /C , find A

E

2m B

2m C

2m

Fig. 24.38

(b) V B – VC

(a) V A – V B Solution

(a) VB > VA , So, VA – VB will be negative.

Further d AB = 2 cos 60° = 1 m ∴

VA – VB = – Ed AB = (–10) (1) = – 10 volt

Ans.

(b) VB > VC , so VB – VC will be positive. Further, d BC = 2.0 m ∴ V

VB – VC = (10) ( 2) = 20 volt

Ans.

Example 24.26 A uniform electric field of 100 V/m is directed at 30° with the positive x-axis as shown in figure. Find the potential difference V BA if OA = 2 m and OB = 4 m. y B O A

Fig. 24.39

30°

x

144 — Electricity and Magnetism Solution

This problem can be solved by both the methods discussed above.

Method 1.

Electric field in vector form can be written as E = (100 cos 30° $i + 100 sin 30° $j ) V/m = ( 50 3 i$ + 50$j ) V/m

and ∴

A ≡ (–2m, 0, 0) B ≡ ( 0, 4m, 0) VBA = VB – VA = – =–

B

∫A

E⋅ d r

(0 , 4 m , 0)

∫(−2 m, 0, 0) ( 50

3 i$ + 50 $j ) ⋅ ( dx $i + dy $j + dz k$ )

= – [ 50 3 x + 50 y] ((–02, 4mm, 0, ,0)0) = – 100 ( 2 + 3 ) V Method 2.

Ans.

V = Ed

We can also use,

With the view that VA > VB or VB – VA will be negative. Here, d AB = OA cos 30° + OB sin 30° = 2× ∴ V

3 1 + 4 × = ( 3 + 2) 2 2

VB – VA = – Ed AB = – 100 ( 2 + 3 )

Ans.

Example 24.27 A uniform electric field pointing in positive x-direction exists in a region. Let A be the origin, B be the point on the x-axis at x = + 1 cm and C be the point on the y-axis at y = + 1 cm. Then, the potentials at the points A, B and C satisfy (JEE 2001) (a) V A < V B (c) V A < VC

(b) V A > V B (d) V A > VC Solution Potential decreases in the direction of electric field. Dotted lines are equipotential lines. y

C A

X

B E

Fig. 24.40

∴ V A = VC Hence, the correct option is (b).

and

V A > VB

Chapter 24 V

Electrostatics — 145

Example 24.28 A non-conducting ring of radius 0.5 m carries a total charge of 1.11 × 10 −10 C distributed non-uniformly on its circumference producing an electric field E everywhere in space. The value of the integral (l = 0 being centre of the ring) in volt is (b) − 1 (c) − 2

(JEE 1997)

(a) + 2

Solution

−∫

l=0 l=∞

E⋅ dl = ∫

l=0 l=∞

l=0

∫ l = ∞ − E ⋅ dl

(d) zero

dV = V (centre) − V (infinity)

V (infinity) = 0

but ∴ −∫

l=0 l=∞

E⋅ dl corresponds to potential at centre of ring. V (centre) =

and

1 q ( 9 × 109 ) (1.11 × 10−10 ) ⋅ = ≈ 2V 0.5 4 πε 0 R

Therefore, the correct answer is (a).

INTRODUCTORY EXERCISE

24.6

1. Determine the electric field strength vector if the potential of this field depends on x, y coordinates as (a) V = a ( x 2 – y 2 )

(b) V = axy

where, a is a constant.

2. The electrical potential function for an electrical field directed parallel to the x-axis is shown in the given graph. V (volt) 20 10

–2

0

2

4

8

x (m)

Fig. 24.41

Draw the graph of electric field strength.

3. The electric potential decreases uniformly from 100 V to 50 V as one moves along the x-axis from x = 0 to x = 5 m. The electric field at x = 2 m must be equal to 10 V/m. Is this statement true or false.

4. In the uniform electric field shown in figure, find : (a) (b) (c) (d)

VA – VD VA – VC VB – VD VC − VD

A

B 1m

D

1m

C

Fig. 24.42

E = 20 V/m

146 — Electricity and Magnetism

24.10 Equipotential Surfaces The equipotential surfaces in an electric field have the same basic idea as topographic maps used by civil engineers or mountain climbers. On a topographic map, contour lines are drawn passing through the points having the same elevation. The potential energy of a mass m does not change along a contour line as the elevation is same everywhere. By analogy to contour lines on a topographic map, an equipotential surface is a three-dimensional surface on which the electric potential V is the same at every point on it. An equipotential surface has the following characteristics. 1. Potential difference between any two points in an equipotential surface is zero. 2. If a test charge q 0 is moved from one point to the other on such a surface, the electric potential energy q 0V remains constant. 3. No work is done by the electric force when the test charge is moved along this surface. 4. Two equipotential surfaces can never intersect each other because otherwise the point of intersection will have two potentials which is of course not possible. 5. As the work done by electric force is zero when a test charge is moved along the equipotential surface, it follows that E must be perpendicular to the surface at every point so that the electric force q 0 E will always be perpendicular to the displacement of a charge moving on the surface. Thus, field lines and equipotential surfaces are always mutually perpendicular. Some equipotential surfaces are shown in Fig. 24.43. 10V 20V 30V 40V

+

40V 30V 20V 10V

E

–

40 V

30 V

20 V

Fig. 24.43

The equipotential surfaces are a family of concentric spheres for a point charge or a sphere of charge and are a family of concentric cylinders for a line of charge or cylinder of charge. For a special case of a uniform field, where the field lines are straight, parallel and equally spaced the equipotential surfaces are parallel planes perpendicular to the field lines. Note While drawing the equipotential surfaces we should keep in mind the two main points. (i) These are perpendicular to field lines at all places. (ii) Field lines always flow from higher potential to lower potential. V

Example 24.29 Equipotential spheres are drawn round a point charge. As we move away from the charge, will the spacing between two spheres having a constant potential difference decrease, increase or remain constant.

Chapter 24

Electrostatics — 147

Solution V1 > V2

V1 =

q 1 ⋅ 4πε 0 r1

and V2 =

q 1 ⋅ 4πε 0 r2

Now,

q  1 1 q  r2 – r1  V1 – V2 =  – =   4πε 0  r1 r2  4πε 0  r1 r2 

∴

( r2 – r1 ) =

r2

( 4πε 0 ) (V1 – V2 ) ( r1 r2 ) q

q + r1

V1 V2

Fig. 24.44

For a constant potential difference (V1 – V2 ) , r2 – r1 ∝ r1 r2 i.e. the spacing between two spheres ( r2 – r1 ) increases as we move away from the charge, because the product r1 r2 will increase.

24.11 Electric Dipole A pair of equal and opposite point charges ±q, that are separated by a fixed distance is known as electric dipole. Electric dipole occurs in nature in a variety of situations. The hydrogen fluoride molecule (HF) is typical. When a hydrogen atom combines with a fluorine atom, the single electron of the former is strongly attracted to the later and spends most of its time near the fluorine atom. As a result, the molecule consists of a strongly negative fluorine ion some (small) distance away from a strongly positive ion, though the molecule is electrically neutral overall. p +q Every electric dipole is characterized by its electric dipole moment which is –q – + 2a a vector p directed from the negative to the positive charge. Fig. 24.45 The magnitude of dipole moment is p = (2a ) q Here, 2a is the distance between the two charges.

Electric Potential and Field Due to an Electric Dipole Consider an electric dipole lying along positive y-direction with its centre at origin. p = 2aq $j y x 2 + z 2 + (y – a)2 +q

+

A (x, y, z)

x 2 + z 2 + (y + a)2

a

x

a –q – z

Fig. 24.46

The electric potential due to this dipole at point A ( x, y, z ) as shown is simply the sum of the potentials due to the two charges. Thus,

148 — Electricity and Magnetism V=

q 1   – 2 4πε 0  x + ( y – a ) 2 + z 2 

  2 2 2  x + ( y + a) + z  q

By differentiating this function, we obtain the electric field of the dipole.  q  ∂V x x Ex = – = – 2  2  2 2 2 3 / 2 2 2 3 / ∂x 4πε 0 [x + ( y – a ) + z ] [x + ( y + a ) + z ]  Ey = –

 q  y– a y+a ∂V = – 2  2  2 2 3 / 2 2 2 3 / 2 ∂y 4πε 0 [ x + ( y – a ) + z ] [x + ( y + a ) + z ] 

Ez = –

 q  ∂V z z = –  2  ∂z 4πε 0 [ x + ( y – a ) 2 + z 2 ]3/ 2 [ x 2 + ( y + a ) 2 + z 2 ]3/ 2 

Special Cases 1. On the axis of the dipole (say, along y-axis) x = 0, z = 0 ∴

V=

or

V=

2aq q  1 1  = –   4πε 0  y – a y + a  4πε 0 ( y 2 – a 2 ) p

(as 2aq = p)

4πε 0 ( y 2 – a 2 )

i.e. at a distance r from the centre of the dipole ( y = r ) V=

p 4πε 0 ( r – a ) 2

2

Vaxis ≈

or

p 4πε 0 r 2

(for r >> a)

V is positive when the point under consideration is towards positive charge and negative if it is towards negative charge. Moreover the components of electric field are as under ( as x = 0, z = 0)

E x = 0, E z = 0 and

Ey = =

q 4πε 0

  1 1 –  2 2 ( y + a)   ( y – a) 4ayq

4πε 0 ( y – a ) 2

2 2

or

Ey =

2 py 1 2 4πε 0 ( y – a 2 ) 2

Note that E y is along positive y-direction or parallel to p. Further, at a distance r from the centre of the dipole ( y = r ). Ey =

2 pr 1 4πε 0 ( r 2 – a 2 ) 2

or

E axis ≈

2p 1 ⋅ 4πε 0 r 3

(for r >> a)

Chapter 24

Electrostatics — 149

2. On the perpendicular bisector of dipole

Say along x-axis (it may be along z-axis also). y = 0, z = 0 q 1   – ∴ V= 4πε 0  x 2 + a 2 

  =0 x 2 + a 2  q

or V⊥ bisector = 0 Moreover the components of electric field are as under, E x = 0, E z = 0 Ey =

and

=

 q  –a a – 2  2  / / 2 3 2 2 3 2 4πε 0  ( x + a ) (x + a )  – 2aq 4πε 0 ( x 2 + a 2 ) 3/ 2 p 1 ⋅ 2 4πε 0 ( x + a 2 ) 3/ 2

Ey = –

or

Here, negative sign implies that the electric field is along negative y-direction or antiparallel to p. Further, at a distance r from the centre of dipole ( x = r ), the magnitude of electric field is E=

p 1 2 4πε 0 ( r + a 2 ) 3/ 2

or

E⊥

bisector

≈

p 1 ⋅ 3 4πε 0 r

(for r >> a)

Electric Dipole in Uniform Electric Field As we have said earlier also, uniform electric field means, at every point the direction and magnitude of electric field is constant. A uniform electric field is shown by parallel equidistant lines. The field due to a point charge or due to an electric dipole is non-uniform in nature. Uniform electric field is found between the plates of a parallel plate capacitor. Now, let us discuss the behaviour of a dipole in uniform electric field.

Force on Dipole Suppose an electric dipole of dipole moment | p| = 2aq is placed in a uniform electric field E at an angle θ. Here, θ is the angle between p and E. A force F1 = qE will act on positive charge and F2 = – q E on negative charge. Since, F1 and F2 are equal in magnitude but opposite in direction. E +q F1 a O a F2

p

A θ

–q B

Fig. 24.47

E

150 — Electricity and Magnetism Hence, F1 + F2 = 0 or F net = 0 Thus, net force on a dipole in uniform electric field is zero. While in a non-uniform electric field it may or may not be zero.

Torque on Dipole τ 1 = OA × F1 = q (OA × E) τ 2 = OB × F2 = – q (OB × E) = q( BO × E) The net torque acting on the dipole is τ = τ 1 + τ 2 = q (OA × E) + q ( BO × E) = q (OA + BO) × E The torque of F1 about O, and torque of F2 about O is,

= q ( BA × E) τ = p× E

or

Thus, the magnitude of torque is τ = pE sin θ. The direction of torque is perpendicular to the plane of paper inwards. Further this torque is zero at θ = 0° or θ = 180°, i.e. when the dipole is parallel or antiparallel to E and maximum at θ = 90°.

Potential Energy of Dipole When an electric dipole is placed in an electric field E, a torque τ = p × E acts on it. If we rotate the dipole through a small angle dθ, the work done by the torque is dW = τ dθ dW = – pE sin θ dθ The work is negative as the rotation dθ is opposite to the torque. The change in electric potential energy of the dipole is therefore dU = – dW = pE sin θ dθ Now, at angle θ = 90°, the electric potential energy of the dipole may be assumed to be zero as net work done by the electric forces in bringing the dipole from infinity to this position will be zero. +q 90°

–q

Fig. 24.48

dU = pE sin θ dθ

Integrating, From 90° to θ, we have or ∴

θ

θ

∫90° dU = ∫90° pE sin θ dθ θ

U (θ ) – U (90° ) = pE [– cos θ ]90° U (θ ) = – pE cos θ = – p ⋅ E

Chapter 24

Electrostatics — 151

If the dipole is rotated from an angle θ 1 to θ 2 , then Work done by external forces = U (θ 2) – U (θ 1) or

Wext. forces = – pE cos θ 2 – (– pE cos θ 1 )

or

Wext. forces = pE (cos θ 1 – cos θ 2)

and work done by electric forces, Welectric force = – Wext. force = pE (cos θ 2 – cos θ 1)

Equilibrium of Dipole When an electric dipole is placed in a uniform electric field net force on it is zero for any position of the dipole in the electric field. But torque acting on it is zero only at θ = 0° and 180°. Thus, we can say that at these two positions of the dipole, net force and torque on it is zero or the dipole is in equilibrium E

E +q

F1 –q

⇒

+q p

Restoring torque

–q F2

θ = 0° U = minimum = − pE Fnet = 0, τ = 0

When displaced from equilibrium position a restoring torque acts on the dipole E

E –q

F1 +q

–q

⇒

Torque in opposite direction

p +q θ = 180° U = maximum = + pE Fnet = 0, τ = 0

F2

When displaced from equilibrium position, torque acts in opposite direction

Fig. 24.49

Of this, θ = 0° is the stable equilibrium position of the dipole because potential energy in this position is minimum (U = – pE cos 0° = – pE ) and when displaced from this position a torque starts acting on it which is restoring in nature and which has a tendency to bring the dipole back in its equilibrium position. On the other hand, at θ = 180°, the potential energy of the dipole is maximum (U = – pE cos 180° = + pE ) and when it is displaced from this position, the torque has a tendency to rotate it in other direction. This torque is not restoring in nature. So, this equilibrium is known as unstable equilibrium position.

Important Formulae 1. As there are too many formulae in electric dipole, we have summarised them as under : |p| = (2a) q Direction of p is from −q to + q.

2. If a dipole is placed along y-axis with its centre at origin, then 1  q –  4 πε0  x2 + ( y – a)2 + z2  ∂V ∂V , Ey = – Ex = – ∂x ∂y ∂V Ez = – ∂z

V( x, y, z) =

and

  x + ( y + a) + z  q

2

2

2

3. On the axis of dipole x = 0, z = 0 1 4 πε0 1 = 4 πε0 1 ≈ 4 πε0

V=

(i)

or (ii)

or

Vaxis

p ( y2 – a 2 ) p ⋅ 2 r – a2 p r2

E x = 0 = Ez and 1 2 py E = Ey = ⋅ 4 πε0 ( y2 – a2 )2 1 2 pr = 4 πε0 (r 2 – a2 )2 1 2p ⋅ Eaxis ≈ 4 πε0 r 3

if y = r if r >> a

(along p ) if y = r for r >> a

4. On the perpendicular bisector of dipole Along x-axis, y = 0, z = 0 V⊥ bisector = 0 E x = 0,

(i) (ii)

Ez = 0 and 1 p ⋅ 4 πε0 ( x2 + a2 )3 / 2 1 p E= 4 πε0 (r 2 + a2 )3 / 2 1 p ≈ ⋅ 4 πε0 r 3

Ey = – or

5. Dipole in uniform electric field (i) Fnet = 0 (ii) τ = p × E and |τ| = pE sin θ (iii) U (θ) = – p ⋅ E = – pE cos θ with U (90° ) = 0 (iv) (Wθ1 → θ 2 )ext. force = pE (cos θ1 – cos θ2 )

(v) (Wθ1 → θ 2 )electric force = pE (cos θ2 – cos θ1 ) = – (Wθ1 → θ 2 )ext. force

(vi) At θ = 0° ,Fnet = 0, τ net = 0, U = minimum (stable equilibrium position) (vii) At θ = 180°, Fnet = 0, τnet = 0, U = maximum (unstable equilibrium position)

(opposite to p) for r >> a

Chapter 24 V

Electrostatics — 153

Example 24.30 Draw electric lines of forces due to an electric dipole. Solution

Electric lines of forces due to an electric dipole are as shown in figure.

– q

+ q

Fig. 24.50 V

Example 24.31 Along the axis of a dipole, direction of electric field is always in the direction of electric dipole moment p. Is this statement true or false? False. In the above figure, we can see that direction of electric field is in the opposite direction of p between the two charges. Solution

V

Example 24.32 At a far away distance r along the axis from an electric dipole electric field is E. Find the electric field at distance 2r along the perpendicular bisector. Solution

Along the axis of dipole, E=

1 2p 4πε 0 r 3

…(i)

This electric field is in the direction of p. Along the perpendicular bisector at a distance 2r, p 1 …(ii) E′ = 4πε 0 ( 2r ) 3 From Eqs. (i) and (ii), we can see that E′ =

E 16

Moreover, E ′ is in the opposite direction of p. Hence, E E′ = − 16

Ans.

24.12 Gauss’s Law Gauss’s law is a tool of simplifying electric field calculations where there is symmetrical distribution of charge. Many physical systems have symmetry, for example a cylindrical body doesn’t look any different if we rotate it around its axis. Before studying the detailed discussion of Gauss's law let us understand electric flux.

Electric Flux (φ) (i) Electric flux is a measure of the field lines crossing a surface. N (ii) It is a scalar quantity with SI units - m 2 or V- m. C

154 — Electricity and Magnetism (iii) Electric flux passing through a small surface dS is given by E

dS θ

Fig. 24.51

…(i)

dφ = E ⋅ dS = E dS cos θ

Here, dS is an area vector, whose magnitude is equal to dS and whose direction is perpendicular to the surface. Note If the surface is open, then dS can be taken in either of the two directions perpendicular to the surface, but it should not change even if we rotate the surface. If the surface is closed then by convention, dS is normally taken in outward direction.

(iv) From Eq. (i), we can see that maximum value of dφ is E dS, if θ = 90° or electric lines are perpendicular to the surface. Electric flux is zero, if θ = 90° or electric lines are tangential to the surface. E E

d φ = E dS

dφ = 0

Fig. 24.52

(v) Electric flux passing through a large surface is given by φ = ∫ dφ = ∫ E ⋅ dS = ∫ E dS cos θ

…(ii)

This is basically surface integral of electric flux over the given surface. But normally we do not study surface integral in detail in physics. Here, are two special cases for calculating the electric flux passing through a surface S of finite size (whether closed or open) φ = ES

Case 1

F

E 90°

90°

E

E

90° Closed surface

S 90° E

E 90°

90°

E E

Fig. 24.53

If at every point on the surface, the magnitude of electric field is constant and perpendicular (to the surface).

Chapter 24

Electrostatics — 155

φ =0

Case 2

E E

S

Closed surface

Fig. 24.54

If at all points on the surface the electric field is tangential to the surface.

Gauss’s Law This law gives a relation between the net electric flux through a closed surface and the charge enclosed by the surface. According to this law, “the net electric flux through any closed surface is equal to the net charge inside the surface divided by ε 0 .” In symbols, it can be written as q …(i) φe = ∫ E ⋅ dS = in ε0 S where, q in represents the net charge inside the closed surface and E represents the electric field at any point on the surface. In principle, Gauss’s law is valid for the electric field of any system of charges or continuous distribution of charge. In practice however, the technique is useful for calculating the electric field only in situations where the degree of symmetry is high. Gauss’s law can be used to evaluate the electric field for charge distributions that have spherical, cylindrical or plane symmetry.

Simplified Form of Gauss's Theorem Gauss’s law in simplified form can be written as under q q ES = in or E = in ε0 Sε 0

…(ii)

but this form of Gauss’s law is applicable only under the following two conditions : (i) The electric field at every point on the surface is either perpendicular or tangential. (ii) Magnitude of electric field at every point where it is perpendicular to the surface has a constant value (say E). Here, S is the area where electric field is perpendicular to the surface.

Applications of Gauss’s Law As Gauss’s law does not provide expression for electric field but provides only for its flux through a closed surface. To calculate E we choose an imaginary closed surface (called Gaussian surface) in which Eq. (ii) can be applied easily. Let us discuss few simple cases.

156 — Electricity and Magnetism Electric field due to a point charge The electric field due to a point charge is everywhere radial. We wish to find the electric field at a distance r from the charge q. We select Gaussian surface, a sphere at distance r from the charge. At every point of this sphere the electric field has the same magnitude E and it is perpendicular to the surface itself. Hence, we can apply the simplified form of Gauss’s law, q ES = in ε0 Here,

∴

q E r

Fig. 24.55

S = area of sphere = 4πr 2 and q in = net charge enclosing the Gaussian surface = q q E ( 4π r 2 ) = ε0

∴

E=

q 1 ⋅ 2 4πε 0 r

It is nothing but Coulomb’s law.

Electric field due to a linear charge distribution Consider a long line charge with a linear charge density (charge per unit length) λ. We have to calculate the electric field at a point, a distance r from the line charge. We construct a Gaussian surface, a cylinder of any arbitrary length l of radius r and its axis coinciding with the axis of the line charge. This cylinder have three surfaces. One is curved surface and the two plane parallel surfaces. Field lines at plane parallel surfaces are tangential (so flux passing through these surfaces is zero). The magnitude of electric field is having the same magnitude (say E) at curved surface and simultaneously the electric field is perpendicular at every point of this surface. Hence, we can apply the Gauss’s law as q ES = in ε0 Here,

S = area of curved surface = (2πrl)

E E

Curved surface

Plane surface

Fig. 24.57

+ + + +

r l

E

E

+ + +

Fig. 24.56

Chapter 24 and ∴

q in = net charge enclosing this cylinder = λl λl E (2πrl) = ε0

Electrostatics — 157

E

∴

E=

λ 2πε 0 r

i.e.

E∝

1 r

r

Fig. 24.58

or E-r graph is a rectangular hyperbola as shown in Fig. 24.58.

Electric field due to a plane sheet of charge Figure shows a portion of a flat thin sheet, infinite in size with constant surface charge density σ (charge per unit area). By symmetry, since the sheet is infinite, the field must have the same magnitude and the opposite directions at two points equidistant from the sheet on opposite sides. Let us draw a Gaussian surface (a cylinder) with one end on one side and other end on the other side and of cross-sectional area S 0 . Field lines will be tangential to the curved surface, so flux passing through this surface is zero. At plane surfaces electric field has same magnitude and perpendicular to surface.

E

+ + + + + + + + + +

+ + + + + + + + + +

+ + + + + + + + + +

+ + + + + + + + + +

+ + + + + + + + + +

+ + + + + + + + + +

+ + + + + + + + + +

+ + + + + + + + + +

S0 E

Fig. 24.59

Hence, using ∴ ∴

ES = E (2S 0 ) = E=

q in ε0 (σ ) ( S 0 ) ε0

σ 2ε 0

Thus, we see that the magnitude of the field is independent of the distance from the sheet. Practically, an infinite sheet of charge does not exist. This result is correct for real charge sheets if points under consideration are not near the edges and the distances from the sheet are small compared to the dimensions of sheet.

Electric field near a charged conducting surface When a charge is given to a conducting plate, it distributes itself over the entire outer surface of the plate. The surface density σ is uniform and is the same on both surfaces if plate is of uniform thickness and of infinite size.

158 — Electricity and Magnetism This is similar to the previous one the only difference is that this time charges are on both sides.

E

++ + ++ + + + + ++ +++++ ++ ++++ + ++ ++++++ ++ +++ ++ ++ ++++++ ++ + ++++ ++ ++++++ ++ ++++++ ++ ++++++ ++ ++++++ ++ ++++++ ++ ++++ ++ ++

+ + + + + + + + + + + +

+ + + + + + + + + + +

S0 E

Fig. 24.60

ES =

Hence, applying S = 2S 0

q in ε0

q in = (σ ) (2S 0 ) (σ ) (2S 0 ) E (2S 0 ) = ε0

Here, ∴ ∴

and

E=

σ ε0

Thus, field due to a charged conducting plate is twice the field due to plane sheet of charge. It also has same limitations. Later, we will see that the electric field near a charged conducting surface of any shape is σ/ε 0 and it is normal to the surface. Note In case of closed symmetrical body with charge q at its centre, the electric flux linked with each half will φ q . If the symmetrical closed body has n identical faces with point charge at its centre, flux = 2 2ε0 φ q linked with each face will be = . n n ε0

be

Extra Points to Remember ˜

V

Net electric flux passing through a closed surface in uniform electric field is zero.

Example 24.33 An electric dipole is placed at the centre of a sphere. Find the electric flux passing through the sphere. Net charge inside the sphere q in = 0. Therefore, according to Gauss’s law net flux passing through the sphere is zero. Ans.

Solution

– –q

+ +q

Fig. 24.61

Chapter 24 V

Electrostatics — 159

Example 24.34 A point charge q is placed at the centre of a cube. What is the flux linked (a) with all the faces of the cube? (b) with each face of the cube? (c) if charge is not at the centre, then what will be the answers of parts (a) and (b)? Solution

(a)

According to Gauss’s law, φtotal =

q in q = ε0 ε0

Ans.

(b) The cube is a symmetrical body with 6 faces and the point charge is at its centre, so electric flux linked with each face will be φ q Ans. φeach face = total = 6 6ε 0 (c) If charge is not at the centre, the answer of part (a) will remain same while that of part (b) will change.

INTRODUCTORY EXERCISE

24.7

1. In figure (a), a charge q is placed just outside the centre of a closed hemisphere. In figure (b), the same charge q is placed just inside the centre of the closed hemisphere and in figure (c), the charge is placed at the centre of hemisphere open from the base. Find the electric flux passing through the hemisphere in all the three cases. (a)

(b)

(c) q q

q

Fig. 24.62

2. Net charge within an imaginary cube drawn in a uniform electric field is always zero. Is this statement true or false?

3. A hemispherical body of radius R is placed in a uniform electric field E. What is the flux linked with the curved surface if, the field is (a) parallel to the base, (b) perpendicular to the base.

4. A cube has sides of length L = 0.2 m. It is placed with one corner at the origin as shown in figure. The electric field is uniform and given by E = (2.5 N/C) $i − (4.2 N/C) $j. Find the electric flux through the entire cube. z

y x

Fig. 24.63

v

160 — Electricity and Magnetism

24.13 Properties of a Conductor Conductors (such as metals) possess free electrons. If a resultant electric field exists in the conductor these free charges will experience a force which will set a current flow. When no current flows, the resultant force and the electric field must be zero. Thus, under electrostatic conditions the value of E at all points within a conductor is zero. This idea, together with the Gauss’s law can be used to prove several interesting facts regarding a conductor.

Excess Charge on a Conductor Resides on its Outer Surface Consider a charged conductor carrying a charge q and no currents are flowing in it. Now, consider a Gaussian surface inside the conductor everywhere on which E = 0. + +

+ + +

+ + +

++ +

+

+

+

Gaussian (E = 0) surface +

+ + +q + +

Conductor +

+

+ +

+ + +

+ ++ + +

Fig. 24.64

Thus, from Gauss’s law, q in

∫ E ⋅ dS = ε 0 S

We get,

q in = 0, as E = 0

Thus, the sum of all charges inside the Gaussian surface is zero. This surface can be taken just inside the surface of the conductor, hence, any charge on the conductor must be on the surface of the conductor. In other words, “Under electrostatic conditions, the excess charge on a conductor resides on its outer surface.”

Electric Field at Any Point Close to the Charged Conductor is

σ ε0

Consider a charged conductor of irregular shape. In general, surface charge density will vary from point to point. At a small surface ∆S, let us assume it to be a constant σ. Let us construct a Gaussian surface in the form of a cylinder of cross-section ∆S. One plane face of the cylinder is inside the conductor and other outside the conductor close to it. The surface inside the conductor does not contribute to the flux as E is zero everywhere inside the conductor. The curved surface outside the conductor also does not contribute to flux as Eis always normal to the charged conductor and hence parallel to the curved surface. Thus, the only contribution to the flux is through the plane face outside the conductor. Thus, from Gauss’s law,

E

∆S E=0

Fig. 24.65

Chapter 24

Electrostatics — 161

q in

∫ E ⋅ dS = ε 0 S

or

E∆ S =

or

E=

Note

(σ ) ( ∆ S ) ε0 σ ε0

(i) Electric field changes discontinuously at the surface of a conductor. Just inside the conductor it is zero σ σ to zero in a small and just outside the conductor it is . In fact, the field gradually decreases from ε0 ε0 thickness of about 4 to 5 atomic layers at the surface. (ii) For a non-uniform conductor the surface charge density (σ) varies inversely as the radius of curvature (ρ) of that part of the conductor, i.e. σ∝

E2

1 Radius of curvature ( ρ )

++ + + + ++ ++ + 2 ++ + + + 1 + + + + + ++ + + ++ + + + ++

E1

Fig. 24.66

For example in the figure,

ρ1 < ρ2

∴

or

E1 > E2

as

σ1 > σ2 σ E= ε0

Electric Field and Field Lines are Normal to the Surface of a Conductor Net field inside a conductor is zero. It implies that no field lines enter a conductor. On the surface of a conductor, electric field and hence field lines are normal to the surface of the conductor.

– – – – – –

E=0

+ + + + + + + +

+ + + + + +

+

90° +

+ +

+ +

+

+ +

+ +

+ +

Fig. 24.67

If a conducting box is immersed in a uniform electric field, the field lines near the box are somewhat distorted. Similarly, if a conductor is positively charged, the field lines originate from the surface and are normal at every point and if it is negatively charged the field lines terminate on the surface normally at every point.

162 — Electricity and Magnetism Cavity Inside a Conductor Consider a charge + q suspended in a cavity in a conductor. Consider a Gaussian surface just outside the cavity and inside the conductor. E = 0 on this Gaussian surface as it is inside the conductor. Hence, from Gauss’s law, q in ∫ E ⋅ dS = ε 0 gives q in = 0 +

+

+

+ + + + + +

+

+

+

+

– +

–

+

+ +

+

+

+

– – – – – +q – – –q – – – ––

+

+

Gaussian surface

+

–

+

+q

+

–

+

+

+ +

+

(a)

+

–

+

+

+

(b)

Fig. 24.68

This concludes that a charge of – q must reside on the metal surface of the cavity so that the sum of this induced charge – q and the original charge + q within the Gaussian surface is zero. In other words, a charge q suspended inside a cavity in a conductor induces an equal and opposite charge – q on the surface of the cavity. Further as the conductor is electrically neutral a charge + q is induced on the outer surface of the conductor. As field inside the conductor is zero, the field lines coming from q cannot penetrate into the conductor. The field lines will be as shown in Fig. (b). The same line of approach can be used to show that the field inside the cavity of a conductor is zero when no charge is suspended in it.

Electrostatic shielding Suppose we have a very sensitive electronic instrument that we want to protect from external electric fields that might cause wrong measurements. We surround the instrument with a conducting box or we keep the instrument inside the cavity of a conductor. By doing this charge in the conductor is so distributed that the net electric field inside the cavity becomes zero and the instrument is protected from the external fields. This is called electronic shielding.

The Potential of a Charged Conductor Throughout its Volume is Same In any region in which E = 0 at all points, such as the region very far from all charges or the interior of a charged conductor, the line integral of Eis zero along any path. It means that the potential difference between any two points in the conductor are at the same potential or the interior of a charged conductor is an equipotential region.

Chapter 24

Electrostatics — 163

24.14 Electric Field and Potential Due to Charged Spherical Shell

or Solid Conducting Sphere Electric Field At all points inside the charged spherical conductor or hollow spherical shell, electric field E = 0, as there is no charge inside such a sphere. In an isolated charged spherical conductor any excess charge on it is distributed uniformly over its outer surface same as that of charged spherical shell or hollow sphere. The field at external points has the same symmetry as that of a point charge. We can construct a Gaussian surface (a sphere) of radius r > R . At all points of this sphere the magnitude of electric field is the same and its direction is perpendicular to the surface. q + + +

+ + + + +

+ + + + +

r R

E

+ + + + +

Gaussian surface

Fig. 24.69

Thus, we can apply ES =

q in ε0

or

E ( 4π r 2 ) =

q ε0

q 1 ⋅ 2 4πε 0 r Hence, the electric field at any external point is the same as if the total charge is concentrated at centre. At the surface of sphere r = R , q 1 ∴ E= ⋅ 2 4πε 0 R ∴

Thus, we can write

E=

E inside = 0 E surface =

q 1 4πε 0 R 2

E outside =

q 1 ⋅ 4πε 0 r 2

The variation of electric field (E) with the distance from the centre ( r ) is as shown in Fig. 24.70. Note

(i) At the surface graph is discontinuous σ 1 q q/ 4 πR 2 (ii) Esurface = ⋅ 2 = = ε0 ε0 4 πε 0 R

E σ 1 q = 4πε0 R 2 ε0

O

E∝

R

Fig. 24.70

1 r2

r

164 — Electricity and Magnetism Potential As we have seen, ∴ ∴ ∴

q 1 ⋅ 2 4πε 0 r q  – dVoutside  1 ⋅  =   4πε 0 r 2 dr V – q r dr ∫0 dVoutside = 4πε 0 ∫∞ r 2 q 1 1 or V ∝ V= ⋅ 4πε 0 r r E outside =

dV   E = –   dr  (V∞ = 0)

Thus, at external points, the potential at any point is the same when the whole charge is assumed to be concentrated at the centre. At the surface of the sphere, r = R q 1 ∴ V= ⋅ 4πε 0 R At some internal point electric field is zero everywhere, therefore, the potential is same at all points which is equal to the potential at surface. Thus, we can write q 1 Vinside = Vsurface = ⋅ 4πε 0 R and

Voutside

V σR 1 q = ε0 4πε0 R

q 1 = ⋅ 4πε 0 r

V∝

O R Fig. 24.71

1 r

r

The potential (V ) varies with the distance from the centre ( r ) as shown in Fig. 24.71.

24.15 Electric Field and Potential Due to a Solid Sphere of Charge Electric Field Positive charge q is uniformly distributed throughout the volume of a solid sphere of radius R. For finding the electric field at a distance r ( < R ) from the centre let us choose as our Gaussian surface a sphere of radius r, concentric with the charge distribution. From symmetry, the magnitude E of electric field has the same value at every point on the Gaussian surface and the direction of E is radial at every point on the surface. So, applying Gauss’s law q …(i) ES = in ε0 4  Here, S = 4πr 2 and q in = (ρ)  πr 3  3  q Here, ρ = charge per unit volume = 4 πR 3 3

+ + + + Gaussian + + + surface + + + + + + + + + + + + + +r ++ + + + + + + + + + + + r ++ + + + + + R + + + + + + + +

Fig. 24.72

Chapter 24

Electrostatics — 165

Substituting these values in Eq. (i) We have, At the centre

E=

q 1 ⋅ 3 ⋅ r or 4πε 0 R so r = 0,

E ∝r

r = R,

E=

At surface

so

E =0 q 1 ⋅ 2 4πε 0 R

To find the electric field outside the charged sphere, we use a spherical Gaussian surface of radius r ( > R ). This surface encloses the entire charged sphere, so q in = q, and Gauss’s law gives q q 1 or E = E ( 4π r 2 ) = ⋅ 2 ε0 4πε 0 r E∝

or

1

r2 Notice that if we set r = R in either of the two expressions for E (outside and inside the sphere), we q 1 E get the same result, E= ⋅ 2 4πε 0 R

E surface =

q 1 ⋅ 2 4πε 0 R

E outside =

q 1 ⋅ 2 4πε 0 r

E∝

∝

r

1 q 4πε0 R 2

E

this is because E is continuous function of r in this case. By contrast, for the charged conducting sphere the magnitude of electric field is discontinuous at r = R (it jumps from E = 0 to E = σ/ε 0 ). Thus, for a uniformly charged solid sphere we have the following formulae for magnitude of electric field : q 1 E inside = ⋅ ⋅r 4πε 0 R 3

O

R

1 r2

r

Fig. 24.73

The variation of electric field (E) with the distance from the centre of the sphere (r) is shown in Fig. 24.73.

Potential The field intensity outside the sphere is E outside =

∴

q 1 ⋅ 2 4πε 0 r

dVoutside = – E outside dr dVoutside = – E outside dr

166 — Electricity and Magnetism V

∴

r

1

q

∫∞ dVoutside = – ∫∞ 4πε 0 ⋅ r 2 dr

or V=

q 1 ⋅ 4πε 0 r

At r = R ,

as V∞ = 0 or V ∝ V=

1 r

q 1 ⋅ 4πε 0 R

i.e. at the surface of the sphere potential is VS =

q 1 ⋅ 4πε 0 R

The electric intensity inside the sphere, E inside =

q 1 ⋅ 3 ⋅r 4πε 0 R

dVinside = – E inside dr dVinside = – E inside dr V q 1 ∫VS dVinside = – 4πε 0 ⋅ R 3

∴ ∴

r

∫R r dr r

q r2  1 V – VS = – ⋅ 3  4πε 0 R  2  R

∴ Substituting VS =

q 1 ⋅ , we get 4πε 0 R

q 1 (1.5 R 2 − 0.5 r 2 ) 3 4πε 0 R q 3 3 1 At the centre r = 0 andVc =  ⋅  = Vs , i.e. potential at 2  4πε 0 R  2 the centre is 1.5 times the potential at surface. Thus, for a uniformly charged solid sphere we have the following formulae for potential : V=

and

V 3 2

Voutside =

q 1 ⋅ 4πε 0 r

Vsurface =

q 1 ⋅ 4πε 0 R

Vinside =

q 3 1 r 2  1 ⋅  –  4πε 0 R  2 2 R 2 

1 q 4πε0 R 1 q 4πε0 R

O

R Fig. 24.74

r

The variation of potential (V) with distance from the centre (r) is as shown in Fig. 24.74. For inside points variation is parabolic.

Electrostatics — 167

Chapter 24

 1  List of formulae for field strength E and potential V  k =  4 πε 0   Table 24.1

S.No. 1.

E

Charge Distribution

V

Formula

Point charge

E=

kq r2

Graph

Formula V=

E

kq r

Graph V

r

2.

Uniformly charged spherical shell

Ei = 0

Vi = Vs =

E

Es = k ⋅ Eo =

q σ = 2 ε0 R

σR ε0 kq Vo = r

4.

Solid sphere charge

On the uniformly ring

of

Es

kq r2

axis of charged

Ei =

kqr

Es =

kq R2

Eo =

Kq

E=

R3

kqx

Infinitely long line charge

r

λ 2 π ε0 r

(15 . R 2 − 0.5 r 2 )

V 1.5Vs

E

PD =

r

r

R

kq

V

R + x2 2

At centre x=0 kq ∴ V= R

x

r

Vs

V=

E

At centre x=0 ∴ E=0

E=

kq

R3 kq Vs = R kq Vo = r

Es

R

Vs

R

Vi =

R 2

5.

V

r

E

r2

(R 2 + x2 )3 / 2

kq R

=

R

3.

r

r  λ ln  2  2 π ε0  r1 

x

Not required

168 — Electricity and Magnetism

Final Touch Points 1. Permittivity Permittivity or absolute permittivity is a measure of resistance that is encountered when forming an electric field in a medium. Thus, permittivity relates to a material's ability to resist an electric field (while unfortunately, the word “permit” suggests the inverse quantity). The permittivity of a medium describes how much electric field (more correctly, flux) is generated per unit charge in that medium. More electric flux (per unit charge) exists in a medium with a low permittivity. Vacuum has the lowest permittivity (therefore maximum electric flux per unit charge). Any other dielectric medium has K -times (K = dielectric constant) the permittivity of vacuum. This is because, due to polarization effects electric flux per unit charge deceases K - times (K > 1).

2. Dielectric constant ( K ) Also known as relative permittivity of a given material is the ratio of permittivity of the material to the permittivity of vacuum. This is the factor by which the electric force between the two charges is decreased relative to vacuum. Similarly, in the chapter of capacitors we will see that it is the ratio of capacitance of a capacitor using that material as a dielectric compared to a similar capacitor that has vacuum as its dielectric.

3. Electric field and potential due to a dipole at polar coordinates ( r , θ ) E Eθ

φ

Er

A r

p θ O

V =

or

p cos θ 4πε 0 r 2

The electric field E can be resolved into two components E r and E θ , where or

Er =

1 2p cos θ ⋅ 4πε 0 r3

and

Eθ =

1 p sin θ 4πε 0 r 3

The magnitude of resultant electric field E = E r2 + E θ2 or

E =

p 1 + 3 cos 2 θ 4πε 0 r 3

Its inclination φ to OA is given by

or

tan φ =

Eθ p sin θ/4πε 0r 3 = Er 2p cos θ/4πε 0r 3

tan φ =

tan θ 2

Chapter 24

Electrostatics — 169

4. Force between two dipoles The force between two dipoles varies inversely with the fourth power of the distance between their centres or F ∝

1 r4

p1

p2 r

In the, figure, a dipole on left with dipole moment p1 interacts with the dipole on the right with dipole moment p 2. We assume that the distance between them is quite large. The electric field of the dipole on the left hand side exerts a net force on the dipole on the right hand side. Let us now calculate the net force on the dipole on right hand side. The electric field at the centre of this dipole E = ∴

1 2p ⋅ 1 4πε 0 r 3

dE = –

1 6p ⋅ 41 dr 4πε 0 r

Now, the electric field at the point where – q charge of the dipole lies is given by E 1 = E + | dE | force on – q is qE 1

and

(towards left)

Similarly, electric field at the point where + q charge of the dipole lies is E 2 = E – | dE | force on + q is qE 2

and

p2

–q – dr

(towards right) +q +

dr

r E

∴

Net force on the dipole is F = q E1 – q E 2

(towards left)

= 2q | dE | 6 ( 2qdr ) p1 = 4πε 0 r 4 or

F =

6p1 p2 4πε 0 r 4

[as 2q (dr ) = p2]

Thus, if p1 | | p 2, the two dipoles attract each other with a force given by the above relation.

5. Earthing a conductor Potential of earth is often taken to be zero. If a conductor is connected to the earth, the potential of the conductor becomes equal to that of the earth, i.e. zero. If the conductor was at some other potential, charges will flow from it to the earth or from the earth to it to bring its potential to zero.

Solved Examples TYPED PROBLEMS Type 1. To find electric potential due to charged spherical shells

Concept To find the electric potential due to a conducting sphere (or shell) we should keep in mind the following two points (i) Electric potential on the surface and at any point inside the sphere is 1 q V = ⋅ (R = radius of sphere) 4πε 0 R (ii) Electric potential at any point outside the sphere is 1 q V = ⋅ (r = distance of the point from the centre) 4πε 0 r For example, in the figure shown, potential at A is q q  1 qA VA = + B + C  4πε 0  rA rB rC  q q  1 qA Similarly, potential at B is V B = + B + C  4πε 0  rB rB rC  and potential at C is, V

VC =

C

qC B

qB qA

A

q q  1 qA + B + C  4πε 0  rC rC rC 

Example 1 Three conducting spherical shells have charges q, − 2q and 3q as shown in figure. Find electric potential at point P as shown in figure. 3q –2q q r

R

P 2R

3R

Solution Potential at P, VP = V q + V −2q + V3 q kq k (2q) k (3q) = − + r r 3R

Chapter 24  1 1 = kq  −   R r 1 k= 4πε 0

Here,

V

Electrostatics — 171 Ans.

Example 2 Figure shows two conducting thin concentric shells of radii r and 3r. The outer shell carries a charge q. Inner shell is neutral. Find the charge that will flow from inner shell to earth after the switch S is closed. q

r S 3r

Solution Let q′ be the charge on inner shell when it is earthed. Potential of inner shell is zero. ∴

1  q′ q + =0 4πε 0  r 3r 

∴

q′ = –

i.e. +

q 3

q charge will flow from inner shell to earth. 3

Ans.

Type 2. Based on the principle of generator

Concept A generator is an instrument for producing high voltages in the million volt region. Its design is based on the principle that if a charged conductor (say A) is brought into contact with a hollow conductor (say B), all of its charge transfers to the hollow conductor no matter how high the potential of the later may be. This can be shown as under: B A rA

qA

qB

rB

In the figure,

VA =

q  1 qA – B  4πε 0  rA rB 

and

VB =

q  1 qA – B  4πε 0  rB rB 

172 — Electricity and Magnetism ∴

VA – VB =

qA  1 1 –   4πε 0 rA rB 

From this expression the following conclusions can be drawn : qB qA B A

(i) The potential difference (PD) depends on q A only. It does not depend on q B . (ii) If q A is positive, then V A – V B is positive (as rA < rB ), i.e. V A > V B . So if the two spheres are connected by a conducting wire charge flows from inner sphere to outer sphere (positive charge flows from higher potential to lower potential) till V A = V B or V A – V B = 0. But potential difference will become zero only when q A = 0, i.e. all charge q A flows from inner sphere to outer sphere. (iii) If q A is negative, V A – V B is negative, i.e. V A < V B . Hence, when the two spheres are connected by a thin wire all charge q A will flow from inner sphere to the outer sphere. Because negative charge flows from lower potential to higher potential. Thus, we see that the whole charge q A flows from inner sphere to the outer sphere, no matter how high q B is. Charge always flows from A to B, whether q A > q B or q B > q A , V A > V B or V B > V A . V

Example 3 Initially the spheres A and B are at potentials V A and VB . Find the potential of A when sphere B is earthed. B A S

Solution As we have studied above that the potential difference between these two spheres depends on the charge on the inner sphere only. Hence, the PD will remain unchanged because by earthing the sphere B charge on A remains constant. Let V′A be the new potential at A. Then, VA – VB = V ′A – VB′ but VB′ = 0 as it is earthed. Hence, V ′A = V A – VB

Ans.

Type 3. Based on the charges appearing on different surfaces of concentric spherical shells

Concept Figure shows three concentric thin spherical shells A, B and C of radii a, b and c. The shells A and C are given charges q1 and q 2 and the shell B is earthed. We are interested in finding

Chapter 24

Electrostatics — 173

the charges on inner and outer surfaces of A, B and C. To solve such type of problems we should keep the following points in mind : q6 q5 q4 q3 q1

C B A q1

a b

q2

c

(i) The whole charge q1 will come on the outer surface of A unless some charge is kept inside A. To understand it let us consider a Gaussian surface (a sphere) through the material of A. As the electric field in a conducting material is zero. The flux through this Gaussian surface is zero. Using Gauss’s law, the total charge enclosed must be zero. q1

Gaussian surface A

(ii) Similarly, if we draw a Gaussian surface through the material of B we can see that q3 + q1 = 0 or q3 = – q1 and if we draw a Gaussian surface through the material of C, then q5 + q 4 + q3 + q1 = 0 or q5 = – q 4 (iii) q5 + q 6 = q 2 . As q 2 charge was given to shell C. (iv) Potential of B should be zero, as it is earthed. Thus, VB = 0 1  q1 q3 + q 4 q5 + q 6  or + + =0 4πε 0  b b c  So, using the above conditions we can find charges on different surfaces. We can summarise the above points as under 1. Net charge inside a closed Gaussian surface drawn in any shell is zero. (provided the shell is conducting). 2. Potential of the conductor which is earthed is zero. 3. If two conductors are connected, they are at same potential. 4. Charge remains constant in all conductors except those which are earthed. 5. Charge on the inner surface of the innermost shell is zero provided no charge is kept inside it. In all other shells charge resides on both the surfaces. 6. Equal and opposite charges appear on opposite faces.

174 — Electricity and Magnetism V

Example 4 A charge q is distributed uniformly on the surface of a solid sphere of radius R. It is covered by a concentric hollow conducting sphere of radius 2R. Find the charges on inner and outer surfaces of hollow sphere if it is earthed.

q

R 2R

Solution The charge on the inner surface of the hollow sphere should be – q, because if we draw a closed Gaussian surface through the material of the hollow sphere the total charge enclosed by this Gaussian surface should be zero. Let q′ be the charge on the outer surface of the hollow sphere. Since, the hollow sphere is earthed, its potential should be zero. The potential on it is due to the charges q, – q and q′, Hence, 1  q q q′  – =0 V = + 4πε 0 2R 2R 2R  ∴ q′ = 0 Therefore, there will be no charge on the outer surface of the hollow sphere. V

q′ –q q

Ans.

Example 5 Solve the above problem if thickness of the hollow sphere is considerable. q′ q

–q

R r R2

P

R3

Solution In this case, we can set V = 0 at any point on the hollow sphere. Let us select a point P a distance r from the centre, were R2 < r < R3 . So,

∴

VP = 0  1 q q q′   – +  =0 4πε 0  r r R3 

∴ q′ = 0 i.e. in this case also there will be no charge on the outer surface of the hollow sphere. V

Ans.

Example 6 Figure shows three concentric thin spherical shells A, B and C of radii R, 2R and 3R. The shell B is earthed and A and C are given charges q and 2q, respectively. Find the charges appearing on all the surfaces of A, B and C.

A B C

Chapter 24

Electrostatics — 175

Solution Since, there is no charge inside A. The whole charge q given to the shell A will appear on its outer surface. Charge on its inner surface will be zero. Moreover if a Gaussian surface is drawn on the material of shell B, net charge enclosed by it should be zero. Therefore, charge on its inner surface will be – q. Now let q′ be the charge on its outer surface, then charge on the inner surface of C will be – q′ and on its outer surface will be, 2q – (– q′ ) = 2q + q′ as total charge on C is 2q. Shell B is earthed. Hence, its potential should be zero. ∴

2q + q′ q′

–q′ q –q

R 2R 3R

VB = 0 1  q q q′ q′ 2q + q′  =0 – + – +  4πε 0 2R 2R 2R 3R 3R 

Solving this equation, we get 4 q 3 4 2 ∴ 2q + q′ = 2q – q = q 3 3 Therefore, charges on different surfaces in tabular form are given below : q′ = –

Table 24.2 A

B

C

Inner surface

0

–q

4 q 3

Outer surface

q

4 – q 3

2 q 3

Type 4. Based on finding electric field due to spherical charge distribution

Concept According to Gauss’s theorem, at a distance r from centre of sphere, kq E = 2in r

 1  k =  4 πε 0  

Here, q in is the net charge inside the sphere of radius r . If volume charge density (say ρ) is constant, then 4 q in = ( volume of sphere of radius r )(ρ) = πr 3 ρ 3 If ρ is variable, then q in can be obtained by integration. ρ (r) Passage (Ex. 7 to Ex. 9) The nuclear charge ( Ze) is non-uniformly distributed within a nucleus of radius R. The charge density ρ(r ) (charge per unit volume) is dependent only on the radial distance r from the centre of the nucleus as shown in figure. The electric field is only along the radial direction.

d

a

r R

176 — Electricity and Magnetism V

Example 7 The electric field at r = R is

(JEE 2008)

(a) independent of a (b) directly proportional to a (c) directly proportional to a 2 (d) inversely proportional to a Solution At r = R, from Gauss’s law E (4πR2) =

qin Ze or = ε0 ε0

E=

1 Ze ⋅ 4 π ε 0 R2

E is independent of a. ∴ The correct option is (a). V

Example 8 For a = 0, the value of d (maximum value of ρ as shown in the figure) is (JEE 2008) ρ (r) d

a

(a)

3Ze

(b)

4πR Solution For a = 0, 3

3Ze

r R

(c)

πR

3

4Ze 3πR

3

(d)

Ze 3πR3 ρ

 d  ρ(r ) =  − ⋅ r + d  R  R

∫0 (4πr

Now, Solving this equation, we get

2

d

d   )  d − r dr = net charge = Ze  R  3Ze d= πR3

r R

∴ The correct option is (b). V

Example 9 The electric field within the nucleus is generally observed to be linearly dependent on r. This implies (JEE 2008) (a) a = 0

(b) a =

R 2

(c) a = R

(d) a =

2R 3

Solution In case of solid sphere of charge of uniform volume ρ(r) density

E=

1 q ⋅ ⋅ r or 4πε 0 R3

E ∝r

Thus, for E to be linearly dependent on r, volume charge density should be constant. or a=R ∴ The correct option is (c).

r R

Chapter 24

Electrostatics — 177

Type 5. Based on calculation of electric flux

Concept (i) To find electric flux from any closed surface, direct result of Gauss's theorem can be used, q φ = in ε0 (ii) To find electric flux from an open surface, result of Gauss's theorem and concept of symmetry can be used. (iii) To find electric flux from a plane surface in uniform electric field, φ = E ⋅ S or ES cos θ can be used. (iv) Net electric flux from a closed surface in uniform electric field is always zero. V

Example 10 The electric field in a region is given by E = a$i + b$j . Here, a and b are constants. Find the net flux passing through a square area of side l parallel to y-z plane. Solution A square area of side l parallel to y-z plane in vector form can be written as, S = l2 $i E = a i$ + b $j

Given,

∴ Electric flux passing through the given area will be, φ = E⋅ S = (a $i + b $j) ⋅ (l 2 $i ) = al2 V

Ans.

Example 11 Figure shows an imaginary cube of side a. A uniformly charged rod of length a moves towards right at a constant speed v. At t = 0, the right end of the rod just touches the left face of the cube. Plot a graph between electric flux passing through the cube versus time.

λ +++++ v a

Solution The electric flux passing through a closed surface depends on the net charge inside the surface. Net charge in this case first increases, reaches a maximum value and finally decreases to zero. The same is the case with the electric flux. The electric flux φ versus time graph is as shown in figure below. φ λa ε0

a v

2a v

t

178 — Electricity and Magnetism V

Example 12 The electric field in a region is given by E = αx $i. Here, α is a constant of proper dimensions. Find (a) the total flux passing through a cube bounded by the surfaces, x = l , x = 2l , y = 0, y = l, z = 0 , z = l. (b) the charge contained inside the above cube. y Solution (a) Electric field is along positive x-direction. B

Therefore, field lines are perpendicular to faces ABCD and EFGH. At all other four faces field lines are tangential. So, net flux passing through these four faces will be zero. Flux entering at face ABCD At this face x = l z

F E

A

x C

D

G H

B E = αl

A

C D

∴

E = α l$i

∴ Flux entering the cube from this face, φ1 = ES = (αl) (l2) = αl3 Flux leaving the face EFGH At this face x = 2l ∴ E = 2α li$ ∴ Flux coming out of this face

F E = 2αl

E

φ2 = ES = (2 αl) (l ) = 2αl3 ∴ Net flux passing through the cube, φnet = φ2 – φ1 = 2 αl3 – αl3 = αl3 (b) From Gauss’s law, q φnet = in ε0 2

G H

Ans.

qin = (φnet ) (ε 0 ) = α ε 0l3 V

Example 13 Consider the charge configuration and a spherical Gaussian surface as shown in the figure. When calculating the flux of the electric field over the spherical surface, the electric field will (JEE 2004) be due to (a) q2 (b) only the positive charges (c) all the charges (d) + q1 and − q1

Ans.

+ q1

– q1

q2

Chapter 24

Electrostatics — 179

Solution At any point over the spherical Gaussian surface, net electric field is the vector sum

of electric fields due to + q1 , − q1 and q2 ∴ The correct option is (c)..

Note Don't confuse with the electric flux which is zero (net) passing over the Gaussian surface as the net charge enclosing the surface is zero. V

Example 14 A point charge q is placed on the top of a cone of semi vertex angle q ( 1 – cos θ ) . θ. Show that the electric flux through the base of the cone is 2ε 0 This problem can be solved by the method of symmetry. Consider a Gaussian surface, a sphere with its centre at the top and radius the slant length of the cone. The flux through the whole sphere is q/ ε 0 . Therefore, the flux through the base of the cone can be calculated by using the following formula, S q φe =   ⋅ S0  ε0

HOW TO PROCEED

Here, S 0 = area of whole sphere and

S = area of sphere below the base of the cone.

Solution Let R = slant length of cone = radius of Gaussian sphere

c q θ θ

R B

A

∴ S 0 = area of whole sphere = (4 πR2) S = area of sphere below the base of the cone = 2πR2 (1 – cos θ ) ∴ The desired flux is,

S q φ= ⋅  S0  ε 0 (2πR2) (1 – cos θ ) q ⋅ ε0 (4πR2) q (1 – cos θ ) = 2ε 0

=

Note S = 2 πR 2 ( 1 – cos θ ) can be calculated by integration. At and

θ = 0°,

S = 2πR2 (1 – cos 0° ) = 0

θ = 90°,

S = 2πR2 (1 – cos 90° ) = 2πR2

θ = 180°,

S = 2πR2 (1 – cos 180° ) = 4πR2

Proved

180 — Electricity and Magnetism Proof dS = (2πr ) Rdα = (2πR sin α ) Rdα = (2πR2) sin α dα

as r = R sin α

θ

∴

S = ∫ (2πR2) sin α dα

∴

S = 2πR2 (1 – cos θ )

0

r

R

θq α

Rd α dα

c

Students are advised to remember this result.

Type 6. Based on E-r and V-r graphs due to two point charges

Concept E=

(i)

r

V =±

and

 1  k =  4πε 0  

kq 2

kq r

(due to a point charge)

(ii) As r → 0, E → ∝ and V → ± ∝ As r → ∝, E → 0 and V → 0 (iii) E is a vector quantity. Due to a point charge, its direction is away from the charge and due to negative charge it is towards the charge. Along one dimension if one direction is taken as positive direction then the other direction is taken as the negative direction. +ve

+ve

+q E = –ve

–q E = +ve

E = +ve

E = –ve

(iv) V is a scalar quantity. On both sides of a positive charge it is positive and it is negative due to negative charge. +q V = +ve

–q V = +ve

V = –ve

V = –ve

(v) Between zero and zero value, normally we get either a maximum or minimum value. V

Example 15 Draw E - r and V - r graphs due to two point charges + q and −2q kept at some distance along the line joining these two charges.

Electrostatics — 181

Chapter 24 Solution E - r graph

+q E r

0 0

P E r

–∝ E 0 r

–2q +∝ 0

(I)

E r

+∝ E 0 r

–∝ 0

(II)

E r

0 0

(III)

In region I E due to + q is towards left (so negative) and E due to −2q is towards right (so positive). Near + q, electric field of + q will dominate. So, net value will be negative. At some point say P both positive and negative values are equal. So, E p = 0. Beyond this point, electric field due to −2q will dominate due to its higher magnitude. So, net value will be positive. E p = 0 and E ∝ (towards left) is also zero. Between zero and zero we will get a maximum positive value. In region II E due to + q and due to −2q is towards right (so positive). Between the value + ∝ and + ∝ the graph is as shown in figure. In region III E due to + q is towards right (so positive) and E due to −2q is towards left (so negative). But electric field of −2q will dominate due to its higher magnitude and lesser distance. Hence, net electric field is always negative. V - r graph

M V r

0 ∞

+q V r

+∞ V r 0

P

–2q

+∞ 0

–∞ V 0 r

V r

–∞ 0

V r

0 ∞

The logics developed in E - r graph can also be applied here with V - r graph. At point P, positive potential due to + q is equal to negative potential due to −2q. Hence, V p = 0, so this point is near 2q. Same is the case at M.

Type 7. E - r and V - r graphs due to charged spherical shells of negligible thickness

Concept According to Gauss’s theorem, E=

kq in 2

r So, only inside charges contribute in the electric field. kq V = = constant R kq V = ≠ constant r Here, q is the charge on shell.

 1  k =  4πε 0   (inside the shell) (outside the shell)

182 — Electricity and Magnetism V

Example 16 Draw E - r and V - r graphs due to two charged spherical shells as shown in figure (along the line between C and ∝). –2q q C

∞

R

2R

Solution E - r graph –2q q C R

PM

∞

NT

2R E E0

E0 =

E0 4 O E0 – 4

Kq R2 r

⇒ E =0 kq At M E = 2 ( radially outwards, say positive) = E 0 (say) R kq kq E At N (radially outwards) E= = = 0 4 (2R)2 4R2 E From M to N Value will decrease from E 0 to 0 4 k (−2q + q) (radially inwards) At T E= (2R)2 E =− 0 4 E0 From T to ∞ Value changes from − to zero. 4

C to P

V - r graph From C to P constant.

qin = 0

Points are lying inside both the shells. Hence, potential due to both shells is

Chapter 24

Electrostatics — 183

kq k (2q) − =0 R 2R From M to N Potential of −2q will remain constant but potential of q will decrease. So, net value comes out to be negative. At N or T kq k (2q) V = − 2R 2R kq =− = − V0 (say ) 2R From T to ∞ Value will change from −V 0 to zero. The correct graph is as shown below.

∴

V =

C

V

∞

V0 =

kq 2R

O

r

–V0

Type 8. Based on motion of a charged particle in uniform electric field

Concept (i) In uniform electric field, force on the charged particle is F = qE or qE force acts in the direction of electric field if q is positive and in the opposite direction of electric field if q is negative. (ii) Acceleration of the particle is therefore, F qE a= = m m This acceleration is constant. So, path is therefore either a straight line or parabola. If initial velocity is zero or parallel to acceleration or antiparallel to acceleration, then path is straight line. Otherwise in all other cases, path is a parabola. V

Example 17 An electron with a speed of 5.00 × 10 6 m/ s enters an electric field of magnitude 103 N /C, travelling along the field lines in the direction that retards its motion. (a) How far will the electron travel in the field before stopping momentarily?

184 — Electricity and Magnetism (b) How much time will have elapsed? (c) If the region with the electric field is only 8.00 mm long (too short from the electron to stop with in it), what fraction of the electron’s initial kinetic energy will be lost in that region? Solution (a) s =

u2 u2 mu 2 = = 2a 2 (qE /m) 2qE =

(9.1 × 10− 31 ) (5 × 106 )2 2 × 1.6 × 10− 19 × 103

= 7.1 × 10−2 m = 7.1 cm u u mu t= = = a qE /m qE

(b)

=

Ans.

(9.1 × 10− 31 ) (5 × 106 ) (1.6 × 10− 19 ) (103 )

= 2.84 × 10−8 s

Ans.

(c) Loss of energy (in fraction) 1 1 mu 2 − mv2 u2 2 2 = =1 − 2 1 u mu 2 2 u 2 − 2as 2as 2qEs =1 − = 2 = u2 u mu 2 2 × 1.6 × 10− 19 × 103 × 8 × 10− 3 = 9.1 × 10−31 × (5 × 106 )2 = 0.11 V

Ans.

Example 18 A charged particle of mass m = 1 kg and charge q = 2 µC is thrown from a horizontal ground at an angle θ = 45° with speed 20 m/s. In space a horizontal electric field E = 2 × 10 7 V /m exist. Find the range on horizontal ground of the projectile thrown. Solution The path of the particle will be a parabola, but along x-axis also motion of the particle will be accelerated. Time of flight of the projectile is 2uy 2uy 2 × 20 cos 45° T= = = =2 2 s ay g 10

y

E u θ

x

Horizontal range of the particle will be 1 a xT 2 2 qE (2 × 10–6 ) (2 × 107 ) ax = = = 40 m/s 2 m 1 1 R = (20 cos 45° ) (2 2 ) + (40) (2 2 )2 2 R = uxT +

Here, ∴

= 40 + 160 = 200 m

Ans.

Chapter 24

Electrostatics — 185

Type 9. To find potential difference between two points when electric field is known

Concept In Article 24.9, we have already read the relation between Eand V. There we have taken a simple case when electric field was uniform. Here, two more cases are possible depending on the nature of E. When E has a Function Like f 1 (x)i$ + f 2 ( y)$j + f 3 (z )k$ In this case also, we will use the same approach. Let us take an example. V

Example 19 Find the potential difference VAB between A (2m, 1m, 0) and B ( 0, 2m , 4m ) in an electric field, E = ( x $i – 2 y $j + z k$ ) V / m dV = – E ⋅ dr

Solution A

∫B

dV = –

( 2, 1 , 0 )

∫( 0, 2, 4) (xi$ – 2 y$j + z k$ ) ⋅ (dx$i + dy$j + dz k$ )

V A − VB = − ∫

∴

( 2, 1 , 0 )

( 0 , 2, 4 )

(xdx − 2 ydy + zdz ) ( 2, 1 , 0 )

 x2 z2  V AB = –  – y2 +  2 ( 0, 2, 4) 2

or

Ans. = 3 volt When E ⋅ dr becomes a Perfect Differential. Same method is used when E ⋅ dr becomes a perfect differential. The following example will illustrate the theory. V

Example 20 Find potential difference V AB between A (0, 0, 0) and B (1m, 1m, 1m) in an electric field (a) E = y$i + x $j (b) E = 3x 2 y$i + x 3 $j dV = – E ⋅ dr

Solution (a) A

∫B

( 0, 0, 0)

( yi$ + x$j) ⋅ (dx$i + dy$j + dz k$ )

( 0, 0, 0)

( y dx + x dy)

dV = –

∫(1, 1, 1)

or

V A – VB = –

∫(1, 1, 1)

or

V AB = –

∫(1, 1, 1)

∴

V AB = – [xy]((01 ,, 01 ,, 01)) = 1 V

∴

( 0, 0, 0)

d (xy)

[as y dx + x dy = d (xy)] Ans.

dV = – E ⋅ dr

(b) ∴ or

∴

A

∫B

( 0, 0, 0)

(3x2yi$ + x3 $j) • (dx$i + dy$j + dz k$ )

( 0, 0, 0)

(3x2ydx + x3 dy)

( 0, 0, 0)

d (x3 y)

dV = –

∫(1, 1, 1)

V A – VB = –

∫(1, 1, 1)

=–

∫(1, 1, 1)

V AB = – [x3 y]((01 ,, 10 ,, 10)) = 1 V

Ans.

186 — Electricity and Magnetism Type 10. Based on oscillations of a dipole

Concept In uniform electric field, net force on a dipole is zero at all angles. But net torque is zero for θ = 0° or 180°. Here, θ = 0° is the stable equilibrium position and θ = 180° is unstable equilibrium position. If the dipole is released from any angle other than 0° or180°, it rotates towards 0°. In this process electrostatic potential energy of the dipole decreases. But rotational kinetic energy increases. At two angles θ1 and θ 2 , we can apply the equation U θ1 + K θ1 = U θ2 + K θ2 1 1 or − pE cos θ1 + Iω 12 = − pE cos θ 2 + Iω 22 2 2 Moreover, if the dipole is displaced from stable equilibrium position (θ = 0° ), then it starts rotational oscillations. For small value of θ, these oscillations are simple harmonic in nature. V

Example 21 An electric dipole of dipole moment p is placed in a uniform electric field E in stable equilibrium position. Its moment of inertia about the centroidal axis is I. If it is displaced slightly from its mean position, find the period of small oscillations. Solution When displaced at an angle θ from its mean position, the magnitude of restoring torque is τ = – pE sin θ For small angular displacement sin θ ≈ θ The angular acceleration is

τ = – pE θ τ  pE  2 α= =–  θ=–ω θ  I  I E

–q

+q τ

⇒

+q p

E

θ

–q

where,

ω2 =

pE I

⇒

T=

2π I = 2π ω pE

Ans.

Type 11. Based on the work done (by external forces) in moving a charge from one point to another point

Concept If kinetic energy of the particle is not changed, then W = ∆U = U f − U i = q(V f − V i ) or q( ∆U ) Here, q is the charge to be displaced and V i and V f are the initial and final potentials.

Chapter 24 V

Electrostatics — 187

Example 22 Two identical thin rings, each of radius R, are coaxially placed a distance R apart. If Q1 and Q2 are respectively the charges uniformly spread on the two rings, the work done in moving a charge q from the centre of one ring to that of the other is (JEE 1992) (a) zero (c)

(b)

q 2(Q1 + Q2 ) ( 4πε 0R )

q(Q1 − Q2 ) ( 2 − 1) 2 ( 4πε 0 R )

(d) q(Q1 / Q2 ) ( 2 + 1) 2 ( 4πε 0R )

Solution VC1 = VQ1 + VQ2 =

Q2 1 Q1 1 1 + = 4 πε 0 R 4π ε 0 R 2 4π ε 0R

Q2   Q1 +   2

Q2

Q1

R√2

R

C1

R C2

1 Q1    Q2 +   4π ε 0R 2

Similarly,

VC2 =

∴

∆V = VC1 − VC2 1  1  = (Q1 − Q2) − (Q1 − Q2) 4πε 0R  2  Q1 − Q2 = ( 2 − 1) 2 (4π ε 0R) W = q∆V = q (Q1 − Q2) ( 2 − 1) / 2 (4π ε 0R)

∴ The correct option is (b).

Miscellaneous Examples V

Example 23 Five point charges each of value + q are placed on five vertices of a regular hexagon of side ‘a’ metre. What is the magnitude of the force on a point charge of value – q coulomb placed at the centre of the hexagon? Solution

q1

q2

r

–q

q3 r

q5

a

60° q4

a /2

188 — Electricity and Magnetism a /2 1 = cos 60° = r 2 ∴

a=r

q1 = q2 = … = q5 = q Net force on – q is only due to q3 because forces due to q1 and due to q4 are equal and opposite so cancel each other. Similarly, forces due to q2 and q5 also cancel each other. Hence, the net force on – q is 1 (q) (q) (towards q3 ) ⋅ F = 4π ε 0 r2 F =

or

V

1 q2 ⋅ 2 4π ε 0 r

Ans.

Example 24 A point charge q1 = 9.1 µC is held fixed at origin. A second point charge q2 = – 0.42 µC and a mass 3.2 × 10 −4 kg is placed on the x-axis, 0.96 m from the origin. The second point charge is released at rest. What is its speed when it is 0.24 m from the origin? Solution From conservation of mechanical energy, we have Decrease in electrostatic potential energy = Increase in kinetic energy 1 q q 1 1 or mv2 = Ui – U f = 1 2  –  2 4π ε 0  ri rf  = ∴

v= =

q1q2  rf – ri    4π ε 0  rr i f  q1q2 2π ε 0m

 rf – ri     rr i f 

(9.1 × 10–6 ) (– 0.42 × 10–6 ) × 2 × 9 × 109  0.24 – 0.96     (0.24) (0.96) 3.2 × 10–4

= 26 m/s V

Ans.

Example 25 A point charge q1 = – 5.8 µC is held stationary at the origin. A second point charge q2 = + 4.3 µC moves from the point ( 0.26 m, 0, 0) to ( 0.38 m, 0, 0). How much work is done by the electric force on q2 ? Solution Work done by the electrostatic forces = Ui – U f 1 1  –   ri rf 

=

q1q2 4π ε 0

=

q1q2  rf – ri    4π ε 0  ri rf 

=

(– 5.8 × 10–6 ) (4.3 × 10–6 ) (9 × 109 ) (0.38 – 0.26) (0.38) (0.26)

= – 0.272 J

Ans.

Chapter 24 V

Electrostatics — 189

Example 26 A uniformly charged thin ring has radius 10.0 cm and total charge + 12.0 µC. An electron is placed on the ring’s axis a distance 25.0 cm from the centre of the ring and is constrained to stay on the axis of the ring. The electron is then released from rest. (a) describe the subsequent motion of the electron. (b) find the speed of the electron when it reaches the centre of the ring. Solution (a) The electron will be attracted towards the centre C of the ring. At C net force is zero, but on reaching C, electron has some kinetic energy and due to inertia it crosses C, but on the other side it is further attracted towards C. Hence, motion of electron is oscillatory about point C. +

+

+ + +

+ +

C

e– P

+

+ + +

R

+

+

r

+ +

+

+

(b) As the electron approaches C, its speed (hence, kinetic energy) increases due to force of attraction towards the centre C. This increase in kinetic energy is at the cost of electrostatic potential energy. Thus, 1 mv2 = U i – U f 2 …(i) = U P – U C = (– e) [VP – VC ] Here, V is the potential due to ring. 1 q (q = charge on ring) VP = ⋅ 4π ε 0 r = VC = =

(9 × 109 ) (12 × 10–9 ) ( (10)2 + (25)2 ) × 10–2

= 401 V

1 q ⋅ 4π ε 0 R (9 × 109 ) (12 × 10–9 ) = 1080 V 10 × 10–2

Substituting the proper values in Eq. (i), we have 1 × 9.1 × 10–31 × v2 = (– 1.6 × 10–19 ) (401 – 1080) 2 ∴ V

v = 15.45 × 106 m/s

Ans.

Example 27 Two points A and B are 2 cm apart and a uniform electric field E acts along the straight line AB directed from A to B with E = 200 N / C. A particle of charge + 10 –6 C is taken from A to B along AB. Calculate (a) the force on the charge (b) the potential difference VA – VB and (c) the work done on the charge by E

190 — Electricity and Magnetism Solution (a) Electrostatic force on the charge, F = qE = (10–6 ) (200) = 2 × 10–4 N (b) In uniform electric field, PD, or

V = E⋅d VA – VB = 200 × 2 × 10–2 =4 V

(c) W = (2 × 10 ) (2 × 10 ) cos 0° = 4 × 10–6 J –4

V

Ans.

Ans.

–2

Ans.

Example 28 An alpha particle with kinetic energy 10 MeV is heading towards a stationary tin nucleus of atomic number 50. Calculate the distance of closest approach. Initially they were far apart. Solution Due to repulsion by the tin nucleus, the kinetic energy of the α-particle gradually decreases at the expense of electrostatic potential energy. 2e +

v

v=0 +

+50e r

∴ or or

Decrease in kinetic energy = increase in potential energy 1 mv2 = U f – Ui 2 1 1 qq mv2 = ⋅ 1 2–0 2 4πε 0 r r=

∴

1 (2e) (50e) ⋅ 4πε 0 (KE)

Substituting the values, r=

(9 × 109 ) (2 × 1.6 × 10–19 ) (1.6 × 10–19 × 50) 10 × 106 × 1.6 × 10–19

= 14.4 × 10–15 m V

Ans.

Example 29 Three point charges of 1 C , 2 C and 3 C are placed at the corners of an equilateral triangle of side 1 m. Calculate the work required to move these charges to the corners of a smaller equilateral triangle of side 0.5 m. Solution

Work done = U f – U i 1  1 1  –  [q3 q2 + q3 q1 + q2q1 ] = 4π ε 0  rf ri  1  1 = 9 × 109  –  [(3)(2) + (3)(1) + (2)(1)]  0.5 1 = 99 × 109 J

Ans.

Note Work done by electrostatic forces is Ui – Uf but work done by external forces is Uf – Ui . Sometimes in a simple way it is asked, find the work done. It means Uf – Ui .

Electrostatics — 191

Chapter 24 V

Example 30 Consider a spherical surface of radius 4 m centred at the origin. Point charges + q and – 2q are fixed at points A ( 2 m, 0, 0) and B (8 m, 0, 0) respectively. Show that every point on the spherical surface is at zero potential. Solution Let P (x, y, z ) be any point on the sphere. From the property of the sphere, x2 + y2 + z 2 = (42) = 16

…(i)

Further,

PA = (x – 2) + y + z

2

…(ii)

and

PB = (x – 8)2 + y2 + z 2

…(iii)

2

2

1  q 2q  – VP =  4πε 0  PA PB    (x – 8)2 + y2 + z 2 

=

1  q  – 2 4πε 0  (x – 2) + y2 + z 2 

=

1  q  – 2 2 4πε 0  x + y + z 2 + 4 – 4x 

=

q 1  –  4πε 0  16 + 4 – 4x

=

1  q –  4πε 0  20 – 4x

2q

  x2 + y2 + z 2 + 64 – 16x  2q

 2q  16 + 64 – 16x 

 q  20 – 4x 

=0 V

Proved

Example 31 The intensity of an electric field depends only on the coordinates x and y as follows E=

a ( x $i + y $j) x2 + y2

where, a is a constant and $i and $j are the unit vectors of the x and y-axes. Find the charge within a sphere of radius R with the centre at the origin. Solution At any point P (x, y, z ) on the sphere a unit vector perpendicular to the sphere radially outwards is n$ = =

x x + y +z 2

2

2

$i +

x $ y $ z $ i+ j+ k R R R

y x + y +z 2

2

2

$j +

y

z x + y +z 2

2

2

x O

as x2 + y2 + z 2 = R2

Let us find the electric flux passing through a small area dS at point P on the sphere,   ax2 ay2 dφ = E ⋅ n$ dS =  dS + 2 2 2 2  R x y R x y ( + ) ( + )    a =   dS  R

dS

k$

∧ n P (x,y,z)

z

192 — Electricity and Magnetism Here, we note that dφ is independent of the coordinates x, y and z. Therefore, total flux passing through the sphere a  a φ = ∫ dφ = ∫ dS =   (4πR2)  R R = 4πaR From Gauss’s law, φ= ∴ V

qin ε0

or

(4πaR) =

qin ε0

qin = 4πε 0aR

Ans.

Example 32 Find the electric field caused by a disc of radius a with a uniform surface charge density σ (charge per unit area), at a point along the axis of the disc a distance x from its centre. Solution We can assume this charge distribution as a collection of concentric rings of charge. dA = (2πr ) dr dq = σ dA = (2πσr ) dr 1 (dq)x dE x = ⋅ 4πε 0 (x2 + r 2)3/ 2

dr x

 1  (2πσr dr ) x =   4πε 0  (x2 + r 2)3/ 2

dEx

a

E x = ∫ dE x

∴

0

=∫ = Ex =

or

a 0 4π

σx 2ε 0

(2πσr dr ) x ε 0 (x2 + r 2)3/ 2 a

r dr

∫0 (x2 + r 2)3/ 2

σ  1 – 2ε 0  

  a /x + 1  1

2 2

If the charge distribution gets very large, i.e. a >> x, the term small, and we get E =

P

r

σ . 2ε 0

1 a /x + 1 2 2

becomes negligibly

Thus, we can say that electric field produced by an infinite plane sheet of charge is independent of the distance from the sheet. Thus, the field is uniform, its direction is everywhere perpendicular to the sheet. V

Example 33 A non-conducting disc of radius a and uniform positive surface charge density σ is placed on the ground with its axis vertical. A particle of mass m and positive charge q is dropped, along the axis of the disc from a height H with zero initial velocity. The particle has q/m = 4 ε 0 g / σ. (a) Find the value of H if the particle just reaches the disc. (b) Sketch the potential energy of the particle as a function of its height and find its equilibrium position.

Chapter 24

Electrostatics — 193

Solution Potential at a height H on the axis of the disc V(P). The charge dq contained in the ring shown in figure, dq = (2πrdr )σ Potential of P due to this ring 1 dq dV = ⋅ , where x = H 2 + r 2 4πε 0 x dV =

1 (2πrdr ) σ σ ⋅ = 4πε 0 H 2 + r 2 2ε 0

VP =

r=a r=0

x

H

rdr H 2 + r2

O dr

r

∴ Potential due to the complete disc, VP = ∫

P (q, m)

dV =

σ 2ε 0

r=a

a

rdr

∫r = 0

H 2 + r2

σ [ a2 + H 2 – H ] 2ε 0

Potential at centre, O will be VO =

σa 2ε 0

(H = 0)

(a) Particle is released from P and it just reaches point O. Therefore, from conservation of mechanical energy decrease in gravitational potential energy = increase in electrostatic potential energy (∆ KE = 0 because K i = K f = 0) ∴ or

mgH = q [VO – VP ]  q  σ  gH =     [a –  m  2ε 0  q 4ε 0 g = m σ

⇒

a2 + H 2 + H ]

qσ = 2g 2ε 0m

…(i)

P q,m

Substituting in Eq. (i), we get gH = 2 g [a + H – or or or or

H = (a + H ) – 2 H a2 + H 2 = a + 2 a2 + H 2 = a2 + 3 2 H = aH 4

a2 + H 2]

H

a2 + H 2 a

O

H2 + aH 4 or

H =

4 a and H = 0 3

Ans. ∴ H = (4 /3) a (b) Potential energy of the particle at height H = Electrostatic potential energy + gravitational potential energy ∴

U = qV + mgH

Here, V = Potential at height H ∴

U =

σq [ a 2 + H 2 – H ] + mgH 2ε 0

…(ii)

194 — Electricity and Magnetism – dU =0 dH

F =

At equilibrium position, Differentiating Eq. (ii) w.r.t. H, mg +

or

σq 2ε 0

∴ 1+

or

  1   (2H )  2

 – 1 = 0  a2 + H 2

 σq  = 2mg   2 ε  0 

1

  H mg + 2mg  – 1 = 0  a 2 + H 2  2H 2H – 2 =0 ⇒ =1 a2 + H 2 a2 + H 2 H2 1 = a + H2 4

or

2

H=

or

or 3H 2 = a 2

a 3

Ans.

From Eq. (ii), we can see that U = 2 mga at H = 0 and U = U min = 3 mga at H =

U

a 3

2mga

Therefore, U-H graph will be as shown. a Note that at H = , U is minimum. 3 a is stable equilibrium position. Therefore, H = 3 V

3mga

O

a/ 3

H

Example 34 Four point charges + 8 µC, – 1 µC, – 1 µC and + 8 µC are fixed at the points – 27/2 m, – 3 /2 m, + 3 /2 m and + 27/2 m respectively on the Y-axis. A particle of mass 6 × 10 –4 kg and charge + 0.1 µC moves along the –X direction. Its speed at x = + ∞ is v0 . Find the least value of v0 for which the particle will cross the origin. Find also the kinetic energy of the particle at the origin. Assume that space is gravity free. Solution In the figure,

y

q = 1 µC = 10– 6 C q0 = + 0.1 µC = 10–7 C m = 6 × 10–4 kg and Q = 8 µC = 8 × 10–6 C Let P be any point at a distance x from origin O. Then, 3 AP = CP = + x2 2 BP = DP =

27 + x2 2

27/2 m B

+Q

3/2 m A

–q v0

x O P – 3/2 m C

–q

– 27/2 m D

+Q

m q0

x

Chapter 24

Electrostatics — 195

Electric potential at point P will be 2kQ 2kq – BP AP 1 k= = 9 × 109 Nm2/ C2 4πε 0

V = where,

 8 × 10 V = 2 × 9 × 109  – 27  + x2  2 –6

∴

∴ Electric field at P is E=–

8 V = 1.8 × 104  –  27 + x2   2

10–6   3 + x2  2   3 + x2  2 1

…(i)

– 3/ 2 – 3/ 2   1  27 dV  1  3   = − 1.8 × 104 (8)  –   + x2 – (1)  –   + x2  (2x)  2  2   dx   2  2 

E = 0 on x-axis where 8  27  + x2  2 

3/ 2

=

3/ 2

=

(4)3/ 2

⇒

 27  + x2  2 

1 3 2  +x 2 

3/ 2

1 3 2  +x 2 

3/ 2

 27  3  + x2 = 4  + x2  2  2 

⇒

x=±

This equation gives

5 m 2

The least value of kinetic energy of the particle at infinity should be enough to take the particle 5 upto x = + m because 2 at x = + for x >

5 m, E = 0 ⇒ Electrostatic force on charge q0 is zero or Fe = 0 2 5 m, E is repulsive (towards positive x-axis) 2

and for x
− a

(b) x < − a (d) 0 < x < a

y

–q

+ 2q x

15. Five point charges (+ q each) are placed at the five vertices of a regular hexagon of side 2a. What is the magnitude of the net electric field at the centre of the hexagon? 1 q 4πε 0 a 2 2q (c) 4πε 0a 2

(a)

q 16πε 0a 2 5q (d) 16πε 0a 2

(b)

16. Two identical small conducting spheres having unequal positive charges q1 and q2 are separated by a distance r. If they are now made to touch each other and then separated again to the same distance, the electrostatic force between them in this case will be (a) less than before (c) more than before

(b) same as before (d) zero

17. Three concentric conducting spherical shells carry charges + 4Q on the inner shell − 2Q on the middle shell and + 6Q on the outer shell. The charge on the inner surface of the outer shell is

(a) 0 (c) − Q

(b) 4 Q (d) − 2 Q

18. 1000 drops of same size are charged to a potential of 1 V each. If they coalesce to form a single drop, its potential would be (a) V (c) 100 V

(b) 10 V (d) 1000 V

19. Two concentric conducting spheres of radii R and 2R are carrying

charges Q and − 2Q, respectively. If the charge on inner sphere is doubled, the potential difference between the two spheres will

(a) (b) (c) (d)

become two times become four times be halved remain same

Q

–2Q

20. Charges Q , 2 Q and − Q are given to three concentric conducting spherical shells A, B and C respectively as shown in figure. The ratio of charges on the inner and outer surfaces of shell C will be (a) + (c)

3 2

3 4

−3 4 −3 (d) 2 (b)

A B C

Chapter 24

Electrostatics — 201

21. The electric field in a region of space is given by E = 5$i + 2$j N/ C . The flux of E due to this field through an area 1 m2 lying in the y -z plane, in SI units, is

(a) 5 (c) 2

(b) 10 (d) 5 29

22. A charge Q is placed at each of the two opposite corners of a square. A charge q is placed at each of the other two corners. If the resultant force on each charge q is zero, then (a) q = 2 Q

(b) q = − 2 Q

(c) q = 2 2 Q

(d) q = − 2 2 Q

23. A and B are two concentric spherical shells. If A is given a charge + q while

B

B is earthed as shown in figure, then (a) (b) (c) (d)

charge on the outer surface of shell B is zero the charge on B is equal and opposite to that of A the field inside A and outside B is zero All of the above

A

24. A solid sphere of radius R has charge ‘q’ uniformly distributed over its volume. The distance from its surface at which the electrostatic potential is equal to half of the potential at the centre is (a) R R (c) 3

(b) 2R R (d) 2

25. Four dipoles each of magnitudes of charges ± e are placed inside a sphere. The total flux of E coming out of the sphere is (a) zero (c)

8e ε0

(b)

4e ε0

(d) None of these

26. A pendulum bob of mass m carrying a charge q is at rest with its string making an angle θ with the vertical in a uniform horizontal electric field E. The tension in the string is mg sin θ qE (c) sin θ (a)

(b) mg (d)

qE cos θ

27. Two isolated charged conducting spheres of radii a and b produce the same electric field near their surfaces. The ratio of electric potentials on their surfaces is a b a2 (c) 2 b (a)

b a b2 (d) 2 a

(b)

28. Two point charges + q and − q are held fixed at ( − a , 0) and ( a , 0) respectively of a x - y coordinate system, then (a) the electric field E at all points on the x-axis has the same direction (b) E at all points on the y-axis is along $i (c) positive work is done in bringing a test charge from infinity to the origin (d) All of the above

202 — Electricity and Magnetism 29. A conducting shell S1 having a charge Q is surrounded by an uncharged concentric conducting spherical shell S 2. Let the potential difference between S1 and that S 2 be V . If the shell S 2 is now given a charge − 3Q, the new potential difference between the same two shells is (b) 2 V (d) − 2 V

(a) V (c) 4 V

30. At a certain distance from a point charge, the field intensity is 500 V/m and the potential is − 3000 V. The distance to the charge and the magnitude of the charge respectively are (a) 6 m and 6 µC (c) 6 m and 4 µC

(b) 4 m and 2 µC (d) 6 m and 2 µC

31. Two point charges q1 and q2 are placed at a distance of 50 m from each other in air, and interact with a certain force. The same charges are now put in oil whose relative permittivity is 5. If the interacting force between them is still the same, their separation now is (b) 22.3 m (d) 25.0 cm

(a) 16.6 m (c) 28.4 m

32. An infinite line of charge λ per unit length is placed along the y-axis. The work done in moving a charge q from A ( a , 0) to B ( 2a , 0) is qλ  1 ln    2 2π ε 0 qλ (d) ln 2 4π ε 0

qλ ln 2 2π ε 0 qλ (c) ln 2 4π ε 0

(a)

(b)

+ + + + (a) The dipole is attracted towards the line charge + + (b) The dipole is repelled away from the line charge + (c) The dipole does not experience a force + + (d) The dipole experiences a force as well as a torque + An electrical charge 2 × 10−8 Cis placed at the point (1, 2, 4) m. At the point (4, 2, 0) m, ++

33. An electric dipole is placed perpendicular to an infinite line of charge at some distance as shown in figure. Identify the correct statement.

34.

p

the electric (a) (b) (c) (d)

potential will be 36 V field will be along y-axis field will increase if the space between the points is filled with a dielectric All of the above

35. If the potential at the centre of a uniformly charged hollow sphere of radius R is V , then electric field at a distance r from the centre of sphere will be (r > R ) r

R

(a)

VR r2

(b)

Vr R2

(c)

VR r

(d)

VR R2 + r 2

Chapter 24

Electrostatics — 203

36. There is an electric field E in x-direction. If the work done on moving a charge of 0.2 C through a distance of 2 m along a line making an angle 60° with x-axis is 4 J, then what is the value of E? (a) 3 N/C (c) 5 N/C

(b) 4 N/C (d) 20 N/C

37. Two thin wire rings each having radius R are placed at a distance d apart with their axes

coinciding. The charges on the two rings are + Q and − Q. The potential difference between the centres of the two rings is

(a) zero

(c)

Q 4πε 0d 2

(b)

Q 1  − 4πε 0  R 

  R +d 

(d)

Q 2πε 0

1  −  R

  R2 + d 2 

1

2

2

1

38. The electric field at a distance 2 cm from the centre of a hollow spherical conducting shell of radius 4 cm having a charge of 2 × 10−3 C on its surface is

(a) 1.1 × 1010 V /m

(b) 4.5 × 10−10 V /m

(c) 4.5 × 1010 V /m

(d) zero

39. Charge Q is given a displacement r = a$i + b$j in an electric field E = E1$i + E2$j. The work done is (a) Q (E1a + E 2b)

(b) Q (E1a )2 + (E 2b)2

(c) Q (E1 + E 2) a 2 + b2

(d) Q E12 + E 22

a 2 + b2

Subjective Questions Note You can take approximations in the answers.

1. A certain charge Q is divided into two parts q and Q − q , which are then separated by a certain distance. What must q be in terms of Q to maximize the electrostatic repulsion between the two charges?

2. An α-particle is the nucleus of a helium atom. It has a mass m = 6.64 × 10−27 kg and a charge

q = + 2e = 3.2 × 10−19 C. Compare the force of the electric repulsion between two α-particles with the force of gravitational attraction between them.

3. What is the charge per unit area in C/ m 2 of an infinite plane sheet of charge if the electric field produced by the sheet of charge has magnitude 3.0 N/C?

4. A circular wire loop of radius R carries a total charge q distributed uniformly over its length. A small length x (> a ) along the y-axis is, 1 3qa2 $ E= – j 4πε 0 y 4

P y a q

a –2q

q

x

Note This charge distribution which is essentially that of two electric dipoles is called an electric quadrupole. Note that E varies as r – 4 for a quadrupole compared with variations of r – 3 for the dipole and r – 2 for a monopole (a single charge).

8. A charge q is placed at point D of the cube. Find the electric flux passing through the face EFGH and face AEHD. B

F E

A

G

C D

H

9. Point charges q1 and q2 lie on the x-axis at points x = − a and x = + a respectively.

(a) How must q1 and q2 be related for the net electrostatic force on point charge + Q, placed at x = + a / 2, to be zero? (b) With the same point charge +Q now placed at x = + 3a /2 .

10. Two particles (free to move) with charges +q and +4q are a distance L apart. A third charge is placed so that the entire system is in equilibrium. (a) Find the location, magnitude and sign of the third charge. (b) Show that the equilibrium is unstable.

11. Two identical beads each have a mass m and charge q. When placed in a hemispherical bowl of radius R with frictionless, non-conducting walls, the beads move, and at equilibrium they are a distance R apart (figure). Determine the charge on each bead. R

R

m

m R

12. Three identical small balls, each of mass 0.1 g, are suspended at one point on silk thread having a length of l = 20cm . What charges should be imparted to the balls for each thread to form an angle of α = 30° with the vertical?

13. Three charges, each equal to q, are placed at the three corners of a square of side a. Find the electric field at fourth corner.

14. A point charge q = − 8.0 nC is located at the origin. Find the electric field vector at the point x = 1.2 m , y = − 1.6 m.

15. Find the electric field at the centre of a uniformly charged semicircular ring of radius R. Linear charge density is λ.

16. Find the electric field at a point P on the perpendicular bisector of a uniformly charged rod. The length of the rod is L, the charge on it is Q and the distance of P from the centre of the rod is a.

Chapter 24

Electrostatics — 205

17. Find the direction of electric field at point P for the charge distribution as shown in figure. y

y

y

–Q –Q

+Q P

P

x

P

x +Q

+Q

+Q

(a)

x

(c)

(b)

18. A clock face has charges − q , − 2q , − 3q ,…−12q fixed at the position of the corresponding numerals on the dial. The clock hands do not disturb the net field due to point charges. At what time does the hour hand point in the direction of the electric field at the centre of the dial.

19. A charged particle of mass m = 1 kg and charge q = 2 µC is thrown from a horizontal ground at

an angle θ = 45° with the speed 25 m/ s. In space, a horizontal electric field E = 2 × 107 V/ m exists in the direction of motion. Find the range on horizontal ground of the projectile thrown. Take g = 10 m/s 2.

20. Protons are projected with an initial speed vi = 9.55 × 103 m/ s into a region where a uniform electric field E = (–720 $j) N/ C is present, as shown in figure. The protons are to hit a target that lies at a horizontal distance of 1.27 mm from the point where the protons are launched. Find E = (–720 ^j ) N/C Vi

Proton beam

θ

Target

1.27 mm

(a) the two projection angles θ that result in a hit and (b) the total time of flight for each trajectory.

21. At some instant the velocity components of an electron moving between two charged parallel plates are vx = 1.5 × 105 m/ s and v y = 3.0 × 106 m/ s. Suppose that the electric field between the plates is given by E = (120 N/ C) $j . (a) What is the acceleration of the electron? (b) What will be the velocity of the electron after its x-coordinate has changed by 2.0 cm?

22. A point charge q1 = + 2 µC is placed at the origin of coordinates. A second charge, q2 = − 3 µC, is placed on the x-axis at x = 100 cm. At what point (or points) on the x-axis will the absolute potential be zero?

100 cm x q1

q2

Q on the sides of an 3a equilateral triangle of perimeter 3a. Calculate the potential at the centroid C of the triangle.

23. A charge Q is spread uniformly in the form of a line charge density λ =

206 — Electricity and Magnetism 24. A uniform electric field of magnitude 250 V/ m is directed in the positive x-direction. A + 12µC charge moves from the origin to the point ( x , y ) = ( 20.0 cm, 5.0 cm ).

(a) What was the change in the potential energy of this charge? (b) Through what potential difference did the charge move?

25. A small particle has charge −5.00 µC and mass 2.00 × 10−4 kg. It moves from point A, where the electric potential is V A = + 200 V, to point B, where the electric potential is V B = + 800 V. The electric force is the only force acting on the particle. The particle has speed 5.00 m/ s at point A. What is its speed at point B? Is it moving faster or slower at B than at A? Explain.

26. A plastic rod has been formed into a circle of radius R. It has a positive charge

P

+Q uniformly distributed along one-quarter of its circumference and a negative charge of −6Q uniformly distributed along the rest of the circumference –6Q (figure). With V = 0 at infinity, what is the electric potential

(a) at the centre C of the circle and (b) at point P, which is on the central axis of the circle at distance z from the centre?

z

R

C +Q

27. A point charge q1 = + 2.40 µC is held stationary at the origin. A second point charge

q2 = − 4.30 µC moves from the point x = 0.150 m, y = 0 to the point x = 0.250 m , y = 0.250 m. How much work is done by the electric force on q2?

28. A point charge q1 = 4.00 nC is placed at the origin, and a second point charge q2 = − 3.00 nC is

placed on the x-axis at x = + 20.0 cm. A third point charge q3 = 2.00 nC is placed on the x-axis between q1 and q2. (Take as zero the potential energy of the three charges when they are infinitely far apart).

(a) What is the potential energy of the system of the three charges if q3 is placed at x = + 10.0 cm? (b) Where should q3 be placed to make the potential energy of the system equal to zero?

29. Three point charges, which initially are infinitely far apart, are placed at the corners of an equilateral triangle with sides d. Two of the point charges are identical and have charge q. If zero net work is required to place the three charges at the corners of the triangles, what must the value of the third charge be? 30. The electric field in a certain region is given by E = ( 5 i$ − 3 $j) kV/ m . Find the difference in potential VB − VA . If A is at the origin and point B is at (a) (0, 0, 5) m, (b) (4, 0, 3) m.

31. In a certain region of space, the electric field is along + y-direction and has a magnitude of 400 V/ m . What is the potential difference from the coordinate origin to the following points? (a) x = 0, y = 20 cm, z = 0 (c) x = 0, y = 0, z = 15 cm

(b) x = 0, y = −30 cm, z = 0

32. An electric field of 20 N/C exists along the x-axis in space. Calculate the potential difference V B − V A where the points A and B are given by (a) A = (0, 0), B = (4 m, 2 m)

(b) A = (4 m, 2 m), B = ( 6 m, 5 m)

33. The electric potential existing in space is V ( x , y , z ) = A ( xy + yz + zx ). (a) Write the dimensional formula of A. (b) Find the expression for the electric field. (c) If A is 10 SI units, find the magnitude of the electric field at (1 m, 1 m, 1 m)

34. An electric field E = ( 20 $i + 30 $j) N/C exists in the space. If the potential at the origin is taken to be zero, find the potential at ( 2 m , 2 m ).

Chapter 24

Electrostatics — 207

35. In a certain region of space, the electric potential is V ( x , y , z ) = Axy − Bx 2 + Cy , where A, B and C are positive constants. (a) Calculate the x, y and z - components of the electric field. (b) At which points is the electric field equal to zero?

36. A sphere centered at the origin has radius 0.200 m. A −500 µC point charge is on the x-axis at x = 0.300 m. The net flux through the sphere is 360 N-m 2 / C. What is the total charge inside the sphere?

37. (a) A closed surface encloses a net charge of −3.60 µC. What is the net electric flux through the surface? (b) The electric flux through a closed surface is found to be 780 N-m2 /C. What quantity of charge is enclosed by the surface? (c) The closed surface in part (b) is a cube with sides of length 2.50 cm. From the information given in part (b), is it possible to tell where within the cube the charge is located? Explain.

3 $ 4 $ E0 i + E0 j with E0 = 2.0 × 103 N/C. Find the flux 5 5 of this field through a rectangular surface of area 0.2 m 2 parallel to the y-z plane.

38. The electric field in a region is given by E =

E0x $ i. Find the charge contained inside a cubical l volume bounded by the surfaces x = 0, x = a , y = 0, y = a , z = 0 and z = a. Take E0 = 5 × 103 N/ C, l = 2 cm and a = 1 cm .

39. The electric field in a region is given by E =

40. A point charge Q is located on the axis of a disc of radius R at a distance b from the plane of the disc (figure). Show that if one-fourth of the electric flux from the charge passes through the disc, then R = 3 b.

R b Q 41. A cube has sides of length L. It is placed with one corner at the origin as shown in figure. The electric field is uniform and given by E = − B $i + C $j − D k$ , where B, C and D are positive constants. z S2 (top) S6 (back) S3 (right side)

S1 (left side) L

y L x

L S5 (front)

S4 (bottom)

208 — Electricity and Magnetism (a) Find the electric flux through each of the six cube faces S1 , S 2, S3 , S 4 , S5 and S 6. (b) Find the electric flux through the entire cube.

42. Two point charges q and −q are separated by a distance 2l. Find the flux of electric field strength vector across the circle of radius R placed with its centre coinciding with the mid-point of line joining the two charges in the perpendicular plane.

43. A point charge q is placed at the origin. Calculate the electric flux through the open hemispherical surface : ( x − a )2 + y 2 + z 2 = a 2 , x ≥ a

44. A charge Q is distributed over two concentric hollow spheres of radii r and R (> r ) such that the surface charge densities are equal. Find the potential at the common centre.

45. A charge q0 is distributed uniformly on a ring of radius R. A sphere of equal radius R is constructed with its centre on the circumference of the ring. Find the electric flux through the surface of the sphere.

46. Two concentric conducting shells A and B are of radii R and 2R. A charge + q is placed at the centre of the shells. Shell B is earthed and a charge q is given to shell A. Find the charge on outer surface of A and B.

47. Three concentric metallic shells A, B and C of radii a , b and c ( a < b < c) have surface charge densities, σ , − σ and σ respectively.

σ –σ σ

A B C

(a) Find the potentials of three shells A , B and C. (b) It is found that no work is required to bring a charge q from shell A to shell C, then obtain the relation between the radii a , b and c.

48. A charge Q is placed at the centre of an uncharged, hollow metallic sphere of radius a, (a) Find the surface charge density on the inner surface and on the outer surface. (b) If a charge q is put on the sphere, what would be the surface charge densities on the inner and the outer surfaces? (c) Find the electric field inside the sphere at a distance x from the centre in the situations (a) and (b).

49. Figure shows three concentric thin spherical shells A, B and C of radii a , b and c respectively.

The shells A and C are given charges q and −q respectively and the shell B is earthed. Find the charges appearing on the surfaces of B and C.

b a q A c

B C

–q

Chapter 24

Electrostatics — 209

50. Three spherical shells have radii R , 2R and 3R respectively. Total charge on A and C is 3q. Find the charges on different surfaces of A, B and C . The connecting wire does not touch the shell B.

C

B

A R 2R 3R

51. In the above problem, the charges on different surfaces if a charge q is placed at the centre of the shell with all other conditions remaining the same.

52. A solid sphere of radius R has a charge +2Q. A hollow spherical shell of radius 3R placed concentric with the first sphere that has net charge −Q. +2Q

R 3R –Q

(a) Find the electric field between the spheres at a distance r from the centre of the inner sphere. [R < r < 3R] (b) Calculate the potential difference between the spheres. (c) What would be the final distribution of charges, if a conducting wire joins the spheres? (d) Instead of (c), if the inner sphere is earthed, what is the charge on it?

53. Three concentric conducting spherical shells of radii R , 2R and 3R carry charges Q , − 2Q and 3Q, respectively. 3Q –2Q Q 2R

R

3R

(a) Find the electric potential at r = R and r = 3R, where r is the radial distance from the centre. 5 (b) Compute the electric field at r = R 2 (c) Compute the total electrostatic energy stored in the system. The inner shell is now connected to the external one by a conducting wire, passing through a very small hole in the middle shell. (d) Compute the charges on the spheres of radii R and 3R. 5 (e) Compute the electric field at r = R. 2

LEVEL 2 Single Correct Option 1. In the diagram shown, the charge + Q is fixed. Another charge + 2q and mass M is projected from a distance R from the fixed charge. Minimum separation between the two charges if the 1 times of the projected velocity, at this moment is (Assume gravity to be velocity becomes 3 absent) V 30° +2q

+Q R

(a)

3 R 2

(b)

1 R 3

(c)

1 R 2

(d) None of these

2. A uniform electric field of strength E exists in a region. An electron enters a point A with velocity v as shown. It moves through the electric field and reaches at point B. Velocity of particle at B is 2 v at 30° with x-axis. Then, y

2v 30° v

(0, 0)

(a) electric field E = −

B (2a, d)

A(a, 0)

x

3mv2 $ i 2ea

(b) rate of doing work done by electric field at B is

3mv3 2ea

(c) Both (a) and (b) are correct (d) Both (a) and (b) are wrong

3. Two point charges a and b whose magnitudes are same, positioned at a certain distance along the positive x-axis from each other. a is at origin. Graph is drawn between electrical field strength and distance x from a. E is taken positive if it is along the line joining from a to b. From the graph it can be decided that E x

(a) a is positive, b is negative (c) a and b both are negative

(b) a and b both are positive (d) a is negative, b is positive

Chapter 24

Electrostatics — 211

Note Graph is drawn only between a and b.

4. Six charges are placed at the vertices of a rectangular hexagon as shown in the figure. The electric field on the line passing through point O and perpendicular to the plane of the figure as a function of distance x from point O is ( x > > a ) +Q

–Q

–Q

+Q O –Q

+Q

(a) 0

(b)

a

Qa π ε 0x3

2Qa π ε 0x3

(c)

(d)

3Qa π ε 0x3

5. If the electric potential of the inner shell is 10 V and that of the outer shell is 5 V, then the potential at the centre will be

a 2a

(a) 10 V

(b) 5 V

(c) 15 V

(d) zero

6. A solid conducting sphere of radius a having a charge q is surrounded by a concentric conducting spherical shell of inner radius 2a and outer radius 3a as shown in figure. Find the  1  amount of heat produced when switch is closed  k =  4πε 0  

a S 3a

(a)

kq2 2a

(b)

kq2 3a

2a

(c)

kq2 4a

(d)

kq2 6a

7. There are four concentric shells A, B, C and D of radii a , 2a , 3a and 4a respectively. Shells

B and D are given charges + q and − q respectively. Shell C is now earthed. The potential  1  difference V A − VC is  k =  4πε 0  

(a)

kq 2a

(b)

kq 3a

(c)

kq 4a

(d)

kq 6a

212 — Electricity and Magnetism 8. Potential difference between centre and surface of the sphere of radius R and uniform volume charge density ρ within it will be

(a)

ρ R2 6 ε0

(b)

ρ R2 4 ε0

(c)

ρ R2 3 ε0

(d)

ρ R2 2 ε0

9. A positively charged disc is placed on a horizontal plane. A charged particle is released from a certain height on its axis. The particle just reaches the centre of the disc. Select the correct alternative. (a) Particle has negative charge on it (b) Total potential energy (gravitational + electrostatic) of the particle first increases, then decreases (c) Total potential energy of the particle first decreases, then increases (d) Total potential energy of the particle continuously decreases

10. The curve represents the distribution of potential along the straight line joining the two charges Q1 and Q2 (separated by a distance r) then which of the following statements are correct? y Q2 Q1

1. |Q1| > |Q2| 2. Q1 is positive in nature 3. A and B are equilibrium points 4. C is a point of unstable equilibrium (a) 1 and 2 (c) 1, 2 and 4

A

x B

C

(b) 1, 2 and 3 (d) 1, 2, 3 and 4

11. A point charge q1 = q is placed at point P. Another point charge q2 = − q is placed at point Q. At some point R( R ≠ P , R ≠ Q ), electric potential due to q1 is V1 and electric potential due to q2 is V 2. Which of the following is correct? (a) (b) (c) (d)

Only for some points V1 > V 2 Only for some points V 2 > V1 For all points V1 > V 2 For all points V 2 > V1

12. The variation of electric field between two charges q1 and q2 along the line joining the charges is plotted against distance from q1 (taking rightward direction of electric field as positive) as shown in the figure. Then, the correct statement is (a) (b) (c) (d)

q1 and q2 are positive and q1 < q2 q1 and q2 are positive and q1 > q2 q1 is positive and q2 is negative q1 E3 > E 4 (d) E1 > E 2 = E3 < E 4

32. An isolated conducting sphere whose radius R = 1 m has a charge q =

1 nC. The energy density 9

at the surface of the sphere is (a)

ε0 J/m3 2

(b) ε 0 J/m3

(c) 2 ε 0 J/m3

(d)

ε0 J/m3 3

33. Two conducting concentric, hollow spheres A and B have radii a and b respectively, with A inside B. Their common potentials is V . A is now given some charge such that its potential becomes zero. The potential of B will now be (b) V (1 − a /b) (d) Vb /a

(a) 0 (c) Va /b

34. In a uniform electric field, the potential is 10 V at the origin of coordinates and 8 V at each of the points (1, 0, 0), (0, 1, 0) and ( 0, 0, 1). The potential at the point (1, 1, 1) will be (a) 0

(b) 4 V

(c) 8 V

(d) 10 V

35. There are two uncharged identical metallic spheres 1 and 2 of radius r separated by a distance

d ( d >> r ). A charged metallic sphere of same radius having charge q is touched with one of the sphere. After some time it is moved away from the system. Now, the uncharged sphere is earthed. Charge on earthed sphere is

q 2 qr (c) − 2d

q 2 qd (d) − 2r

(a) +

(b) −

36. Figure shows a closed dotted surface which intersects a conducting uncharged sphere. If a positive charge is placed at the point P, the flux of the electric field through the closed surface

P

(a) will remain zero (c) will become negative

(b) will become positive (d) data insufficient

218 — Electricity and Magnetism 37. Two concentric conducting thin spherical shells A and B having radii rA and rB (rB > rA ) are charged to QA and − QB (|QB|>|QA|). The electrical field along a line passing through the centre is

E

(a)

(b)

E

rB O

rA

O

x

rA

rB

x

E

(d) None of these

(c) rB O

rA

x

38. The electric potential at a point ( x , y ) in the x-y plane is given by V = − kxy. The field intensity at a distance r in this plane, from the origin is proportional to (a) r 2 (c) 1/r

(b) r (d) 1 /r 2

More than One Correct Options 1. Two concentric shells have radii R and 2 R charges qA and qB and potentials 2 V and ( 3/ 2) V respectively. Now, shell B is earthed and let charges on them become qA ′ and qB ′. Then, B A

(a) (b) (c) (d)

qA /qB = 1 /2 qA′ /qB ′ = 1 potential of A after earthing becomes (3 /2) V potential difference between A and B after earthing becomes V /2

2. A particle of mass 2 kg and charge 1 mC is projected vertically with a velocity 10 ms −1. There is a uniform horizontal electric field of 104 N/C, then (a) the horizontal range of the particle is 10 m (c) the maximum height reached is 5 m

(b) the time of flight of the particle is 2 s (d) the horizontal range of the particle is 5 m

3. At a distance of 5 cm and 10 cm from surface of a uniformly charged solid sphere, the potentials are 100 V and 75 V respectively. Then, 50 × 10−10 C 3 (c) the electric field on the surface is 1500 V/m (d) the electric potential at its centre is 25 V (a) potential at its surface is 150 V

(b) the charge on the sphere is

Chapter 24

Electrostatics — 219

4. Three charged particles are in equilibrium under their electrostatic forces only. Then, (a) (b) (c) (d)

the particles must be collinear all the charges cannot have the same magnitude all the charges cannot have the same sign the equilibrium is unstable

5. Charges Q1 and Q2 lie inside and outside respectively of a closed surface S. Let E be the field at any point on S and φ be the flux of E over S.

(a) (b) (c) (d)

If Q1 changes, both E and φ will change If Q2 changes, E will change but φ will not change If Q1 = 0 and Q2 ≠ 0, then E ≠ 0 but φ = 0 If Q1 ≠ 0 and Q2 = 0, then E = 0 but φ ≠ 0

6. An electric dipole is placed at the centre of a sphere. Mark the correct options. (a) (b) (c) (d)

The flux of the electric field through the sphere is zero The electric field is zero at every point of the sphere The electric field is not zero at anywhere on the sphere The electric field is zero on a circle on the sphere

7. Mark the correct options. (a) Gauss’s law is valid only for uniform charge distributions (b) Gauss’s law is valid only for charges placed in vacuum (c) The electric field calculated by Gauss’s law is the field due to all the charges (d) The flux of the electric field through a closed surface due to all the charges is equal to the flux due to the charges enclosed by the surface

8. Two concentric spherical shells have charges + q and − q as shown in figure. Choose the correct options. –q +q A

(a) (b) (c) (d)

B

C

At A electric field is zero, but electric potential is non-zero At B electric field and electric potential both are non-zero At C electric field is zero but electric potential is non-zero At C electric field and electric potential both are zero

9. A rod is hinged (free to rotate) at its centre O as shown in figure.

Two point charges + q and + q are kept at its two ends. Rod is placed in uniform electric field E as shown. Space is gravity free. Choose the correct options.

(a) (b) (c) (d)

Net force from the hinge on the rod is zero Net force from the hinge on the rod is leftwards Equilibrium of rod is neutral Equilibrium of rod is stable

+q

O

+q

E

220 — Electricity and Magnetism 10. Two charges + Q each are fixed at points C and D. Line AB is the

A

bisector line of CD. A third charge + q is moved from A to B, then from B to C.

(a) (b) (c) (d)

From A to B electrostatic potential energy will decrease From A to B electrostatic potential energy will increase From B to C electrostatic potential energy will increase From B to C electrostatic potential energy will decrease

C

B

D

Comprehension Based Questions Passage I (Q. No. 1 to 3) There are two concentric spherical shell of radii r and 2r. Initially, a charge Q is given to the inner shell and both the switches are open. 2r

S2 S1 r

1. If switch S1 is closed and then opened, charge on the outer shell will be (a) Q (c) − Q

(b) Q/2 (d) − Q/2

2. Now, S 2 is closed and opened. The charge flowing through the switch S 2 in the process is (a) Q (c) Q/2

(b) Q/4 (d) 2Q/3

3. The two steps of the above two problems are repeated n times, the potential difference between the shells will be  Q    2  4 πε 0 r  1  Q  (c) n   2 2πε 0r  (a)

1

n+1

 Q    4πε 0r  1  Q  (d) n − 1   2  2 πε 0 r  (b)

1 2n

Passage II (Q. No. 4 to 7) A sphere of charge of radius R carries a positive charge whose volume charge density depends r  only on the distance r from the ball’s centre as ρ = ρ0 1 −  , where ρ0 is a constant. Assume ε as  R the permittivity of space.

4. The magnitude of electric field as a function of the distance r inside the sphere is given by  r r2   −  3 4R  ρ r r2  (c) E = 0  +  ε 3 4R  (a) E =

ρ0 ε

 r r2   −  4 3R  ρ r r2  (d) E = 0  +  ε 4 3R  (b) E =

ρ0 ε

Chapter 24

Electrostatics — 221

5. The magnitude of the electric field as a function of the distance r outside the ball is given by ρ 0R3 8εr 2 ρ R2 (c) E = 0 3 8εr (a) E =

ρ 0R3 12εr 2 ρ R2 (d) E = 0 3 12εr (b) E =

6. The value of distance rm at which electric field intensity is maximum is given by R 3 2R (c) rm = 3

(a) rm =

3R 2 4R (d) rm = 3

(b) rm =

7. The maximum electric field intensity is (a) Em =

ρ0R 9ε

(b) Em =

ρ0ε 9R

(c) Em =

ρ0R 3ε

(d) Em =

ρ0R 6ε

Passage III (Q. No. 8 to 10) A solid metallic sphere of radius a is surrounded by a conducting spherical shell of radius b ( b > a ). The solid sphere is given a charge Q. A student measures the potential at the surface of the solid sphere as V a and the potential at the surface of spherical shell as V b. After taking these readings, he decides to put charge of − 4Q on the shell. He then noted the readings of the potential of solid sphere and the shell and found that the potential difference is ∆V . He then connected the outer spherical shell to the earth by a conducting wire and found that the charge on the outer surface of the shell as q1. He then decides to remove the earthing connection from the shell and earthed the inner solid sphere. Connecting the inner sphere with the earth he observes the charge on the solid sphere as q2. He then wanted to check what happens if the two are connected by the conducting wire. So he removed the earthing connection and connected a conducting wire between the solid sphere and the spherical shell. After the connections were made he found the charge on the outer shell as q3 . Answer the following questions based on the readings taken by the student at various stages.

8. Potential difference ( ∆V ) measured by the student between the inner solid sphere and outer shell after putting a charge − 4Q is

(a) V a − 3 Vb

(b) 3(V a − Vb )

(c) V a

(d) V a − Vb

9. q2 is (a) Q (c) − 4Q

 a (b) Q    b (d) zero

10. q3 is Q (a + b) a−b Q (a − b) (c) b

(a)

Qa 2 b Qb (d) − a (b)

222 — Electricity and Magnetism Match the Columns 1. Five identical charges are kept at five vertices of a regular hexagon. Match the following two columns at centre of the hexagon. If in the given situation electric field at centre is E. Then, Column I

Column II

(a) If charge at B is removed, then electric field will become (b) If charge at C is removed, then electric field will become (c) If charge at D is removed then electric field will become (d) If charges at B and C both are removed, then electric field will become

q

(p) 2E

C

A

(q) E

D F

(r) zero q

(s)

q B

q

E q

3E

Note Only magnitudes of electric field are given. 2. In an electric field E = ( 2i$ + 4$j) N/C, electric potential at origin is 0 V. Match the following two columns. Column I

Column II

(a) Potential at (4 m, 0)

(p) 8 V

(b) Potential at (−4 m, 0)

(q) – 8 V

(c) Potential at (0, 4 m)

(r) 16 V

(d) Potential at (0, – 4 m)

(s) – 16 V

3. Electric potential on the surface of a solid sphere of charge is V . Radius of the sphere is 1m. Match the following two columns. Column I

Column II

(a) Electric potential at r =

R 2

(b) Electric potential at r = 2R (c) Electric field at r =

R 2

(d) Electric field at r = 2R

(p) V 4 (q) V 2 (r) 3V 4 (s) None of these

4. Match the following two columns. Column I

Column II

(a) Electric potential

(p) [ MLT–3 A −1 ]

(b) Electric field

(q) [ML3 T−3 A −1 ]

(c) Electric flux

(r) [ML2 T−3 A −1 ]

(d) Permittivity of free space

(s) None of these

Electrostatics — 223

Chapter 24 5. Match the following two columns. Column I

Column II

(a) Electric field due to (p) charged spherical shell r

(b) Electric potential due to (q) charged spherical shell

r

(c) Electric field due charged solid sphere

to (r)

r

(d) Electric potential due to charged solid sphere

(s) None of these

Subjective Questions

1. A 4.00 kg block carrying a charge Q = 50.0 µC is connected to a spring for which k = 100 N/ m. The block lies on a frictionless horizontal track, and the system is immersed in a uniform electric field of magnitude E = 5.00 × 105 V/ m, directed as shown in figure. If the block is released from rest when the spring is unstretched (at x = 0 ).

m, Q k

E

x=0

(a) (b) (c) (d)

By what maximum amount does the spring expand? What is the equilibrium position of the block? Show that the block’s motion is simple harmonic and determine its period. Repeat part (a) if the coefficient of kinetic friction between block and surface is 0.2.

2. A particle of mass m and charge −Q is constrained to move along the axis of a ring of radius a.

The ring carries a uniform charge density +λ along its length. Initially, the particle is in the centre of the ring where the force on it is zero. Show that the period of oscillation of the particle when it is displaced slightly from its equilibrium position is given by T = 2π

2ε 0ma 2 λQ

224 — Electricity and Magnetism 3. Three identical conducting plane parallel plates, each of area A are held with equal separation d between successive surfaces. Charges Q , 2Q , and 3Q are placed on them. Neglecting edge effects, find the distribution of charges on the six surfaces.

4. A long non-conducting, massless rod of length L pivoted at its centre and balanced with a weight w at a distance x from the left end. At the left and right ends of the rod are attached small conducting spheres with positive charges q and 2q respectively. A distance h directly beneath each of these spheres is a fixed sphere with positive charge Q. (a) Find the distance x where the rod is horizontal and balanced. (b) What value should h have so that the rod exerts no vertical force on the bearing when the rod is horizontal and balanced?

Note Ignore the force between Q (beneath q) and 2q and the force between Q (beneath 2q) and q. Also the force between Q and Q.

5. The electric potential varies in space according to the relation V = 3x + 4 y. A particle of mass 10 kg starts from rest from point (2, 3.2) m under the influence of this field. Find the velocity of the particle when it crosses the x-axis. The charge on the particle is +1 µC. Assume V ( x , y ) are in SI units.

6. A simple pendulum with a bob of mass m = 1 kg, charge q = 5 µC and string length l = 1 m is

given a horizontal velocity u in a uniform electric field E = 2 × 106 V/ m at its bottommost point A, as shown in figure. It is given that the speed u is such that the particle leaves the circle at point C. Find the speed u (Take g = 10 m/ s2) C 60° B E u

A

7. Eight point charges of magnitude Q are arranged to form the corners of a cube of side L. The arrangement is made in manner such that the nearest neighbour of any charge has the opposite sign. Initially, the charges are held at rest. If the system is let free to move, what happens to the arrangement? Does the cube-shape shrink or expand? Calculate the velocity of each charge when the side-length of the cube formation changes from L to nL. Assume that the mass of each point charge is m.

8. There are two concentric spherical shells of radii r and 2r. Initially, a charge Q is given to the inner shell. Now, switch S1 is closed and opened then S 2 is closed and opened and the process is repeated n times for both the keys alternatively. Find the final potential difference between the shells. 2r S2 S1 r

Chapter 24

Electrostatics — 225

9. Two point charges Q1 and Q2 are positioned at points 1 and 2. The field intensity to the right of the charge Q2 on the line that passes through the two charges varies according to a law that is represented schematically in the figure. The field intensity is assumed to be positive if its direction coincides with the positive direction on the x-axis. The distance between the charges is l. E

a

2

1

l

x

b

(a) Find the sign of each charge. (b) Find the ratio of the absolute values of the charges

Q1 Q2

(c) Find the value of b where the field intensity is maximum.

10. A conducting sphere S1 of radius r is attached to an insulating handle. Another conducting

sphere S 2 of radius R (> r ) is mounted on an insulating stand, S 2 is initially uncharged. S1 is given a charge Q. Brought into contact with S 2 and removed. S1 is recharged such that the charge on it is again Q and it is again brought into contact with S 2 and removed. This procedure is repeated n times.

(a) Find the electrostatic energy of S 2 after n such contacts with S1. (b) What is the limiting value of this energy as n → ∞?

11. A proton of mass m and accelerated by a potential difference V gets into a uniform electric field of a parallel plate capacitor parallel to plates of length l at mid-point of its separation between plates. The field strength in it varies with time as E = at , where a is a positive constant. Find the angle of deviation of the proton as it comes out of the capacitor. (Assume that it does not collide with any of the plates.)

12. Two fixed, equal, positive charges, each of magnitude 5 × 10−5 C are located at points A and B separated by a distance of 6 m. An equal and opposite charge moves towards them along the line COD, the perpendicular bisector of the line AB. The moving charge when it reaches the point C at a distance of 4 m from O, has a kinetic energy of 4 J. Calculate the distance of the farthest point D which the negative charge will reach before returning towards C. A

+q 3m

D

O 4m

C

3m B

+q

13. Positive charge Q is uniformly distributed throughout the volume of a sphere of radius R. A point mass having charge +q and mass m is fired towards the centre of the sphere with velocity

226 — Electricity and Magnetism v from a point A at distance r(r > R ) from the centre of the sphere. Find the minimum velocity v so that it can penetrate R 2 distance of the sphere. Neglect any resistance other than electric interaction. Charge on the small mass remains constant throughout the motion.

14. Two concentric rings placed in a gravity free region in yz-plane one of radius R carries a charge

+ Q and second of radius 4R and charge −8Q distributed uniformly over it. Find the minimum velocity with which a point charge of mass m and charge +q should be projected from a point at a distance 3R from the centre of rings on its axis so that it will reach to the centre of the rings.

15. An electric dipole is placed at a distance x from centre O on the axis of a charged ring of radius R and charge Q uniformly distributed over it. R

–q

+q

O x

2a

(a) Find the net force acting on the dipole. (b) What is the work done in rotating the dipole through 180°?

16. A point charge −q revolves around a fixed charge +Q in elliptical orbit. The minimum and maximum distance of q from Q are r1 and r2 , respectively. The mass of revolving particle is m. Q > q and assume no gravitational effects. Find the velocity of q at positions when it is at r1 and r2 distance from Q.

17. Three concentric, thin, spherical, metallic shells have radii 1, 2, and 4 cm and they are held at potentials 10, 0 and 40 V respectively. Taking the origin at the common centre, calculate the following: (a) Potential at r = 1.25 cm (b) Potential at r = 2.5 cm (c) Electric field at r = 1.25 cm

18. A thin insulating wire is stretched along the diameter of an insulated circular hoop of radius R.

A small bead of mass m and charge −q is threaded onto the wire. Two small identical charges are tied to the hoop at points opposite to each other, so that the diameter passing through them is perpendicular to the thread (see figure). The bead is released at a point which is a distance x0 from the centre of the hoop. Assume that x0 U f , i.e. there is a decrease in energy. Hence, energy is always lost in redistribution of charge. Further, ∆U = 0

if

V1 = V2

this is because no flow of charge takes place when both the conductors are at same potential. V

Example 25.3 Two isolated spherical conductors have radii 5 cm and 10 cm, respectively. They have charges of 12 µC and – 3 µC. Find the charges after they are connected by a conducting wire. Also find the common potential after redistribution.

Chapter 25 Solution +

– + +

+ + +

– –

+ +

R1 + + +

12 µC

–

– –

+ –

R2

–

⇒

V

–

– – –3 µC

+

+ +

+ + +

Capacitors — 239

+ +

+

+

+

+ + q1′

+ + + +

V +

+ + + q 2′

Fig. 25.5

Net charge = (12 – 3) µC = 9 µC Charge is distributed in the ratio of their capacities (or radii in case of spherical conductors), i.e. q1′ R 5 1 = 1 = = q 2′ R 2 10 2 ∴

 1  q1′ =   ( 9) = 3 µC  1 + 2

and

 2  q 2′ =   ( 9) = 6 µC  1 + 2 q1 + q 2 ( 9 × 10–6 ) = C1 + C 2 4πε 0 ( R1 + R 2 )

Common potential, V = =

( 9 × 10–6 ) ( 9 × 109 ) (15 × 10–2 )

= 5.4 × 105 V V

Ans.

Example 25.4 An insulated conductor initially free from charge is charged by repeated contacts with a plate which after each contact is replenished to a charge Q. If q is the charge on the conductor after first operation prove that the Qq . maximum charge which can be given to the conductor in this way is Q–q Solution Let C1 be the capacity of plate and C 2 that of the conductor. After first contact charge on conductor is q. Therefore, charge on plate will remain Q – q. As the charge redistributes in the ratio of capacities. Q – q C1 …(i) = q C2

Let q m be the maximum charge which can be given to the conductor. Then, flow of charge from the plate to the conductor will stop when, Vconductor = V plate qm Q C  ∴ = ⇒ qm =  2  Q C 2 C1  C1  Substituting

C2 from Eq. (i), we get C1 qm =

Qq Q–q

Hence Proved.

240 — Electricity and Magnetism V

Example 25.5 A conducting sphere S1 of radius r is attached to an insulating handle. Another conducting sphere S2 of radius R is mounted on an insulating stand. S2 is initially uncharged. S1 is given a charge Q, brought into contact with S2 and removed. S1 is recharged such that the charge on it is again Q and it is again brought into contact with S2 and removed. This procedure is (JEE 1998) repeated n times. (a) Find the electrostatic energy of S 2 after n such contacts with S1 . (b) What is the limiting value of this energy as n → ∞ ? Solution Capacities of conducting spheres are in the ratio of their radii. Let C1 and C 2 be the capacities of S 1 and S 2 , then C2 R = C1 r (a) Charges are distributed in the ratio of their capacities. Let in the first contact, charge acquired by S 2 is q1 . Therefore, charge on S 1 will be Q − q1 . Say it is q ′1 q1 q1 C R = = 2 = ∴ q1′ Q − q1 C1 r  R  q1 = Q    R + r

∴

…(i)

In the second contact, S 1 again acquires the same charge Q. Therefore, total charge in S 1 and S 2 will be R   Q + q1 = Q 1 +   R + r This charge is again distributed in the same ratio. Therefore, charge on S 2 in second contact, R  R   q 2 = Q 1 +    R + r  R + r 2  R  R   =Q +    R + r  R + r 

Similarly,

2 3  R  R   R   q3 = Q  +  +    R + r   R + r  R + r 

and

2 n  R  R    R  qn = Q  +    + ...+   R + r   R + r  R + r 

or

qn = Q

n R  R   1 −    r   R + r   

 a (1 − r n ) S =  n  (1− r )  

…(ii)

Chapter 25

Capacitors — 241

Therefore, electrostatic energy of S 2 after n such contacts =

q n2 2 ( 4πε 0 R )

or U n =

q n2 8πε 0 R

where, q n can be written from Eq. (ii). (b) As n → ∞ q∞ = Q ∴

U∞ = U∞ =

or

INTRODUCTORY EXERCISE

R r

q ∞2 Q 2 R 2 / r2 = 2C 8πε 0 R Q2R 8 πε 0 r 2

25.1

1. Find the dimensions of capacitance. 2. No charge will flow when two conductors having the same charge are connected to each other. Is this statement true or false?

3. Two conductors of capacitance 1µF and 2 µF are charged to +10 V and −20 V. They are now connected by a conducting wire. Find (a) their common potential (b) the final charges on them (c) the loss of energy during redistribution of charges.

25.3 Capacitors Any two conductors separated by an insulator (or a vacuum) form a a b capacitor. +q –q In most practical applications, each conductor initially has zero net charge, and electrons are transferred from one conductor to the other. This is called charging of the conductor. Then, the two conductors have Fig. 25.6 charges with equal magnitude and opposite sign, and the net charge on the capacitor as a whole remains zero. When we say that a capacitor has charge q we mean that the conductor at higher potential has charge + q and the conductor at lower potential has charge – q. In circuit diagram, a capacitor is represented by two parallel Fig. 25.7 lines as shown in Fig. 25.7. One common way to charge a capacitor is to connect the two conductors to opposite terminals of a battery. This gives a fixed potential differenceVab between the conductors, which is just equal to the q voltage of the battery. The ratio is called the capacitance of the capacitor. Hence, Vab C=

q Vab

(capacitance of a capacitor)

242 — Electricity and Magnetism Calculation of Capacitance Give a charge + q to one plate and − q to the other plate. Then, find potential differenceV between the plates. Now, q C= V

Parallel Plate Capacitor Two metallic parallel plates of any shape but of same size and separated by a small distance constitute parallel plate capacitor. Suppose the area of each plate is A and the separation between the two plates is d. Also assume that the space between the plates contains vacuum. +q + + + + + + + + + +

–q – – – – – – – – – –

q + + + + + + + + + +

or

(a)

–q – – – – – – – – – – (b)

Fig. 25.8

We put a charge q on one plate and a charge – q on the other. This can be done either by connecting one plate with the positive terminal and the other with negative plate of a battery [as shown in Fig. (a)] or by connecting one plate to the earth and by giving a charge + q to the other plate only. This charge will induce a charge – q on the earthed plate. The charges will appear on the facing surfaces. The charge density on each of these surfaces has a magnitude σ = q/ A. If the plates are large as compared to the separation between them, then the electric field between the plates (at point B) is uniform and perpendicular to the plates except for a small region near the edge. The magnitude of this uniform field E may be calculated by using the fact that both positive and negative plates produce the electric field in the same direction (from positive plate towards negative plate) of magnitude σ/2ε 0 and therefore, the net electric field between the plates will be σ σ σ E= + = 2ε 0 2ε 0 ε 0 +σ

A E=0

+ + + + + B + + σ σ +E = + 2ε0 2ε0 + + = σ + ε0 P + d Fig. 25.9

–σ – – – – – – – – – – – – Q

C E=0

Chapter 25

Capacitors — 243

Outside the plates (at points A and C) the field due to positive sheet of charge and negative sheet of charge are in opposite directions. Therefore, net field at these points is zero. The potential difference between the plates is σ qd V = E ⋅d =   d = ∴ Aε 0 ε0  ∴ The capacitance of the parallel plate capacitor is q Aε 0 C= = V d ε A or C= 0 d Note

(i) Instead of two plates if there are n similar plates at equal distances from each other and the alternate plates are connected together, the capacitance of the arrangement is given by (n – 1) ε 0A C= d (ii) From the above relation, it is clear that the capacitance depends only on geometrical factors (A and d).

Effect of Dielectrics Most capacitors have a dielectric between their conducting plates. Placing a solid dielectric between the plates of a capacitor serves the following three functions : (i) It solves the problem of maintaining two large metal sheets at a very small separation without actual contact. σ

–σ +

–

+

–

+

–

+

–

+

–

+

–

+

– –

+ E0

Fig. 25.10

(ii) It increases the maximum possible potential difference which can be applied between the plates of the capacitor without the dielectric breakdown. Many dielectric materials can tolerate stronger electric fields without breakdown than can air. (iii) It increases the capacitance of the capacitor. When a dielectric material is inserted between the plates (keeping the charge to be constant) the electric field and hence the potential difference decreases by a factor K (the dielectric constant of the dielectric). ∴

E=

E0 K

and V =

V0 K

(When q is constant)

244 — Electricity and Magnetism Electric field is decreased because an induced charge of the opposite sign appears on each surface of the dielectric. This induced charge produces an electric field inside the dielectric in opposite directions and as a result net electric field is decreased. The induced charge in the dielectric can be calculated as under E0 = E0 – Ei E = E 0 – E i or K ∴

1  E i = E 0 1 –   K

Therefore,

σi σ  1 = 1 –  ε0 ε0  K 

or

1  σi = σ 1 −   K

or

1  q i = q 1 –   K E =0

and

and otherwise

qi < q

Thus,

q i ≤ q,

σi ≤ σ

Dielectric σ – – + + – – + + – – + + – – + E0 + E=0 E= – K+ – + – – + + – – + + – – + + – – + + E E 0 –σ σi σ 0 –σi d

d

d

Conductor –σ – – – – – – – – – E0 –σ

+ + + + + + + + + σ

d

d

Fig. 25.12

Hence, we can conclude the above discussion as under: q σ (i) E vacuum = E 0 = = ε 0 Aε 0 (ii) E dielectric =

E0 K

(iii) E conductor = 0

(here, K = dielectric constant) ( as K = ∞ )

E0

Ei

E

σi – σ + – + – + – + – + – + – + – + – + –

Fig. 25.11

For a conductor K = ∞. Hence, q i = q , σi = σ

σ –σi + – + – + – + – + – + – + – + – + –

Chapter 25 If we plot a graph between potential and distance from positive plate, it will be as shown in Fig. 25.13: Modulus of,

Capacitors — 245

Potential V+ A

Slope of AB = slope of CD = slope of EF = E 0 V– E0 Slope of BC = K and slope of DE = 0 Further, the potential difference between positive O and negative plate is E V+ – V– = E 0 d + 0 ⋅ d + E 0 d + 0 + E 0 d K E = 3E 0 d + 0 ⋅ d K

B

C D

E F

2d

d

Here we have used PD = Ed.

3d 4d Fig. 25.13

5d

Distance

(in uniform electric field)

Capacitance of a Capacitor Partially Filled with Dielectric Suppose, a dielectric is partially filled with a dielectric (dielectric constant = K) as shown in figure. If a charge q is given to the capacitor, an induced charge q i is developed on the dielectric. q + + + + + + + + +

–qi – – – – – – – – –

K

t

qi + + + + + + + + +

–q – – – – – – – – –

E0

⇒

E

t

d

d–t

Fig. 25.14

1  q i = q 1 –   K

where,

Moreover, if E 0 is the electric field in the region where dielectric is absent, then electric field inside the dielectric will be E = E 0 /K . The potential difference between the plates of the capacitor is V = V+ – V– = Et + E 0 ( d – t ) E t  = 0 t + E 0 (d – t ) = E 0  d – t +   K K =

q  σ  t t d – t +  = d – t +  ε0  K  Aε 0  K

246 — Electricity and Magnetism Now, as per the definition of capacitance, ε0A q C= = t V d–t+ K

ε0A

C=

or

d–t+

t K

Different Cases (i) If more than one dielectric slabs are placed between the capacitor, then ε0A C=  t t t  ( d – t1 – t 2 – … – t n ) +  1 + 2 +… + n  K K K  1 n  2 (ii) If the slab completely filles the space between the plates, then t = d and therefore, ε A Kε 0 A C= 0 = d/K d (iii) If a conducting slab ( K = ∞ ) is placed between the plates, then ε0A ε A C= = 0 t d–t d–t+ ∞ This can be explained from the following figure: q + + + + + + +

–q – – – – – – –

–qi qi – + – + – + K =∞ – + – + – + – +

q

Fig. 25.15

–q – – – – – – –

+ + + K =∞ + + + +

⇒

K

d–t

t qi = q

Fig. 25.16

(iv) If the space between the plates is completely filled with a conductor, then t = d and K = ∞.

q Conductor

Fig. 25.17

Then,

C=

ε0A d–d+

d ∞

=∞

Chapter 25

Capacitors — 247

The significance of infinite capacitance can be explained as under: If one of the plates of a capacitor is earthed and the second one is given a charge q, then the whole charge transfers to earth and as the capacity of earth is very large compared to the capacitor we can say that the capacitance has become infinite. q

q

q q

(a)

(b)

Fig. 25.18

Alternatively, if the plates of the capacitor are connected to a battery, the current starts flowing in the circuit. Thus, is as much charge enters the positive plate of the capacitor, the same charge leaves the negative plate. So, we can say, the positive plate can accept infinite amount of charge or its capacitance has become infinite.

Energy Stored in Charged Capacitor A charged capacitor stores an electric potential energy in it, which is equal to the work required to charge it. This energy can be recovered if the capacitor is allowed to discharge. If the charging is done by a battery, electrical energy is stored at the expense of chemical energy of battery. Suppose at time t, a charge q is present on the capacitor and V is the potential of the capacitor. If dq amount of charge is brought against the forces of the field due to the charge already present on the capacitor, the additional work needed will be q dW = ( dq ) V =   ⋅ dq ( as V = q / C ) C  ∴ Total work to charge a capacitor to a charge q 0 , q0  q  q2 W = ∫ dW = ∫   ⋅ dq = 0 0 C  2C

∴ Energy stored by a charged capacitor, q 02 1 1 = CV 2 = q V 2C 2 0 2 0 0 Thus, if a capacitor is given a charge q, the potential energy stored in it is U =W =

1 1 q2 1 U = CV 2 = = qV 2 2 C 2 The above relation shows that the charged capacitor is the electrical analog of a stretched spring 1 1 whose elastic potential energy is Kx 2 . The charge q is analogous to the elongation x and , i.e. the 2 C reciprocal of capacitance to the force constant k.

248 — Electricity and Magnetism Extra Points to Remember ˜

Capacitance of a spherical conductor enclosed by an earthed concentric spherical shell If a charge q is given to the inner spherical conductor, it spreads over the outer surface of it and a charge – q appears on the inner surface of the shell. The electric field is produced only between the two. From the principle of generator, the potential difference between the two will depend on the inner charge q only and is given by –q q  1 1 V=  –  +q 4 πε0  a b  Hence, the capacitance of this system C=

b

 ab  1 1 C = 4 πε0  −  = 4 πε0    a b  b – a

or

˜

a

q V

Fig. 25.19

From this expression we see that if b = ∞, C = 4 πε0 a, which corresponds to that of an isolated sphere, i.e. the charged sphere may be regarded as a capacitor in which the outer surface has been removed to infinity. Capacitance of a cylindrical capacitor When a metallic cylinder of radius a is placed coaxially inside an earthed hollow metallic cylinder of radius b (> a) we get cylindrical capacitor. If a charge q is given to the inner cylinder, induced charge – q will reach to the inner surface of the outer cylinder. Assume that the capacitor is of very large length (l >> b ) so that the lines of force are radial. Using Gauss’s law, we can prove that

– – – – – – – –

+ + + + + + +

– – – – – – – –

+ + + + + + +

Fig. 25.20

E (r ) =

λ 2 πε0 r

for

a≤ r≤ b

Here, λ = charge per unit length Therefore, the potential difference between the cylinders a

V = – ∫ E ⋅ dr = – b

∴ Hence,

a

λ

λ

 b

∫b 2 πε0 r dr = 2 πε0 ln  a 

2 πε0 charge /length λ capacitance = = = V potential difference length ln (b/a) capacitance per unit length =

2 πε0 ln (b/a)

Chapter 25 V

Capacitors — 249

Example 25.6 A parallel-plate capacitor has capacitance of 1.0 F. If the plates are 1.0 mm apart, what is the area of the plates? ε A Solution Q C= 0 d (1) (10− 3 ) Cd ∴ A= = ε 0 8.86 × 10− 12 = 1.1 × 108 m 2

V

Example 25.7 Two parallel plate vacuum capacitors have areas A1 and A2 and equal plate spacing d. Show that when the capacitors are connected in parallel, the equivalent capacitance is the same as for a single capacitor with plate area A1 + A2 and spacing d. Note : In parallel C = C1 + C2 . Solution C = C1 + C 2 (in parallel)

ε 0 A ε 0 A1 ε 0 A 2 = + d d d A = A1 + A 2

∴ or V

Example 25.8 (a) Two spheres have radii a and b and their centres are at a distance d apart. Show that the capacitance of this system is 4πε 0 C= 1 1 2 + ± a b d provided that d is large compared with a and b. (b) Show that as d approaches infinity the above result reduces to that of two isolated 1 1 1 spheres in series. Note : In series, = + . C C1 C2 Solution

(a)

PD, V = V1 − V2 +q

–q 2

1 a

d

b

Fig. 25.21

= ∴

C=

1 4πε 0

 q q   − q q   a − d  −  b + d  

4πε 0 q = V 1 1 2 + − a b d

If − q is given to first sphere and + q to second sphere, then 4πε 0 C= 1 1 2 + + a b d

250 — Electricity and Magnetism 4πε 0 ( 4πε 0 ) ( ab ) = 1 1 a+b + a b C1 C 2 ( 4π ε 0 a ) ( 4π ε 0 b ) ( 4πε 0 ) ab C net = = = C1 + C 2 ( 4π ε 0 a ) + ( 4πε 0 b ) a+b

(b) If d → ∞, then C =

In series,

INTRODUCTORY EXERCISE

Hence Proved.

25.2

1. A capacitor has a capacitance of 7.28 µF. What amount of charge must be placed on each of its plates to make the potential difference between its plates equal to 25.0 V?

2. A parallel plate air capacitor of capacitance 245 µFhas a charge of magnitude 0.148 µC on each plate. The plates are 0.328 mm apart. (a) What is the potential difference between the plates? (b) What is the area of each plate? (c) What is the surface charge density on each plate?

3. Two parallel plates have equal and opposite charges. When the space between the plates is evacuated, the electric field is E0 = 3.20 × 105 V/m. When the space is filled with dielectric, the electric field is E = 2.50 × 105 V/ m. (a) What is the dielectric constant? (b) What is the charge density on each surface of the dielectric?

25.4 Mechanical Force on a Charged Conductor E1 We know that similar charges repel each other, hence the charge on any part of E2 P surface of the conductor is repelled by the charge on its remaining part. The Q ∆S surface of the conductor thus experiences a mechanical force. E1 The electric field at any point P near the conductor’s surface can be assumed E2 as due to a small part of the surface of area say ∆S immediately in the neighbourhood of the point under consideration and due to the rest of the surface. Let E1 and E 2 be the field intensities due to these parts respectively. Fig. 25.22 Then, total electric field, E = E1 + E 2 E has a magnitude σ/ε 0 at any point P just outside the conductor and is zero at point Q just inside the conductor. Thus,

E1 + E 2 = σ/ε 0 at P and ∴

E1 – E 2 = 0 at Q E1 = E 2 =

σ 2ε 0

Hence, the force experienced by small surface of area ∆S due to the charge on the rest of the surface is F = qE 2 = (σ∆S ) ( E 2 ) =

(σ 2 ) ( ∆S ) 2ε0

Chapter 25 ∴

Force F σ2 1 = = = ε E2 Area ∆S 2 ε 0 2 0

∴

Force 1 = ε E2 Area 2 0

Capacitors — 251  σ  as E =  ε  0

Force between the Plates of a Capacitor Consider a parallel plate capacitor with plate area A. Suppose a positive charge q is given to one plate and a negative charge – q to the other plate. The electric field on the negative plate due to positive charge is q –q q σ E= = + – 2ε 0 2 Aε 0 The magnitude of force on the charge in negative plate is F = qE =

2

q 2 Aε 0

This is the force with which both the plates attract each other. Thus,

–

+

–

+

–

+

–

+

–

+

–

Fig. 25.23

q2 F= 2 Aε 0 V

+

Example 25.9 A capacitor is given a charge q. The distance between the plates of the capacitor is d. One of the plates is fixed and the other plate is moved away from the other till the distance between them becomes 2d. Find the work done by the external force. Solution

When one plate is fixed, the other is attracted towards the first with a force F=

q2 = constant 2 Aε 0

Hence, an external force of same magnitude will have to be applied in opposite direction to increase the separation between the plates. q2d 2 Aε 0

∴

W = F ( 2d – d ) =

Alternate solution

W = ∆U = U f – U i =

Here,

Cf =

ε0 A 2d

and

q2 q2 – 2C f 2C i

Ci =

Ans. …(i)

ε0 A d

Substituting in Eq. (i), we have W=

q2 q2 q2d – =  ε A  2ε 0 A  ε A 2 0  2 0   d   2d 

Ans.

252 — Electricity and Magnetism

25.5 Capacitors in Series and Parallel In Series

C1 –

+

+

C2

q –

+

q

q

V1

V2

+

– C

⇒ –

V

+

V

–

Fig. 25.24

In a series connection, the magnitude of charge on all plates is same. The potential is distributed in the inverse ratio of the capacity ( as V = q/C or V ∝1 / C ). Thus, in the figure, if a potential difference V is applied across the two capacitors C1 and C 2 , then V1 C 2 = V2 C1  C2  V1 =  V  C1 + C 2 

or Further, in the figure, or

V = V1 + V2

 C1  and V2 =  V  C1 + C 2  or

q q q = + C C1 C 2

1 1 1 + = C C1 C 2

Here, C is the equivalent capacitance. The equivalent capacitance of the series combination is defined as the capacitance of a single capacitor for which the charge q is the same as for the combination, when the same potential difference V is applied across it. In other words, the combination can be replaced by an equivalent capacitor of capacitance C. We can extend this analysis to any number of capacitors in series. We find the following result for the equivalent capacitance. 1 1 1 1 = + + +… C C1 C 2 C 3 Following points are important in case of series combination of capacitors. (i) In a series connection, the equivalent capacitance is always less than any individual capacitance. (ii) For the equivalent capacitance of two capacitors it is better to remember the following form CC C= 1 2 C1 + C 2 For example, equivalent capacitance of two capacitors C1 = 6 µF and C 2 = 3 µF is CC  6 × 3 C = 1 2 =  µF = 2 µF C1 + C 2  6 + 3  (iii) If n capacitors of equal capacity C are connected in series, then their equivalent capacitance is

C . n

Chapter 25 V

Capacitors — 253

Example 25.10 In the circuit shown in figure, find 2 µF

3 µF

100 V Fig. 25.25

(a) the equivalent capacitance, (b) the charge stored in each capacitor and (c) the potential difference across each capacitor. Solution (a) The equivalent capacitance C1 C 2 C= C1 + C 2 C=

or

( 2) ( 3) = 1.2 µF 2+ 3

2 µF +

– V1

V2

Ans. 100 V

Fig. 25.26

= 120 µC 1 C

or

Ans. V1 C 2 = V 2 C1

∴

 C2   3  V1 =  V =  (100) = 60 V  2 + 3  C1 + C 2 

and

V2 = V – V1 = 100 – 60 = 40 V

In Parallel

– q

q = CV = (1.2 × 10–6 ) (100) C

V∝

+

q

(b) The charge q stored in each capacitor is

(c) In series combination,

3 µF

Ans.

C1 +

– q

q1

+

– C

⇒

C2 +

+

–

V

– q2

+

V

–

Fig. 25.27

The arrangement shown in figure is called a parallel connection. In a parallel combination, the potential difference for all individual capacitors is the same and the total charge q is distributed in the ratio of their capacities. (as q = CV or q ∝ C for same potential difference). Thus, q1 C1 = q2 C2

254 — Electricity and Magnetism  C1  q1 =   q and  C1 + C 2 

or

 C2  q2 =  q  C1 + C 2 

The parallel combination is equivalent to a single capacitor with the same total charge q = q1 + q 2 and potential difference V. Thus, q = q1 + q 2 or

CV = C1V + C 2V

or

C = C1 + C 2

In the same way, we can show that for any number of capacitors in parallel, C = C1 + C 2 + C 3 +… In a parallel combination, the equivalent capacitance is always greater than any individual capacitance. V

Example 25.11 In the circuit shown in figure, find 1 µF 2 µF 3 µF

100 V

Fig. 25.28

(a) the equivalent capacitance and (b) the charge stored in each capacitor. Solution

(a) The capacitors are in parallel. Hence, the equivalent capacitance is C = C1 + C 2 + C 3

or

C = (1 + 2 + 3) = 6µF

Ans.

(b) Total charge drawn from the battery, q = CV = 6 × 100µC = 600µC This charge will be distributed in the ratio of their capacities. Hence, ∴

q1 : q 2 : q 3 = C1 : C 2 : C 3 = 1: 2 : 3  1  q1 =   × 600 = 100 µC  1 + 2 + 3  2  q2 =   × 600 = 200 µC  1 + 2 + 3

and

 3  q3 =   × 600 = 300 µC.  1 + 2 + 3

Ans.

Chapter 25

Capacitors — 255

Alternate solution Since the capacitors are in parallel, the PD across each of them is 100 V. Therefore, from q = CV , the charge stored in 1µF capacitor is 100µC, in 2µF capacitor is 200µC and that in 3µF capacitor is 300µC.

INTRODUCTORY EXERCISE

25.3

1. Find charges on different capacitors. 4 µF 3 µF 2 µF

15 V Fig. 25.29

2. Find charges on different capacitors. 9 µF 4 µF 3 µF

40 V Fig. 25.30

25.6 Two Laws in Capacitors Like an electric circuit having resistances and batteries in a complex circuit containing capacitors and the batteries charges on different capacitors can be obtained with the help of Kirchhoff ’s laws.

First Law This law is basically law of conservation of charge which is normally applied across a battery or in an isolated system. (i) In case of a battery, both terminals of the battery supply equal amount of charge. (ii) In an isolated system (not connected to any source of charge like terminal of a battery or earth) net charge remains constant. For example, in the Fig. 25.31, the positive terminal of the battery supplies a positive charge q1 + q 2 . Similarly, the negative terminal supplies a negative charge of magnitude q 3 + q 4 . Hence, q1 + q 2 = q 3 + q 4 Further, the plates enclosed by the dotted lines form an isolated system, as they are neither connected to a battery terminal nor to the earth. Initially, no charge was present in these plates. Hence, after charging net charge on these plates should also be zero. Or, q 3 + q 5 – q1 = 0 and q 4 – q 2 – q 5 = 0

256 — Electricity and Magnetism So, these are the three equations which can be obtained from the first law. G

C1 + –

– C3

E

–

C2 I

H

C5 + q + 5

q1 B

q3 M

+

+ – q2

D

–

J

q C4 4 F

A V

Fig. 25.31

Second Law In a capacitor, potential drops by q/C when one moves from positive plate to the negative plate and in a battery it drops by an amount equal to the emf of the battery. Applying second law in loop ABGHEFA, we have q q – 1 – 3 +V = 0 C1 C 3 Similarly, the second law in loop GMDIG gives the equation, q q q – 1 – 5 + 2 =0 C1 C 5 C 2 V

Example 25.12 Find the charges on the three capacitors shown in figure. 2 µF

4 µF

6 µF

10 V

20 V

Fig. 25.32

Let the charges in three capacitors be as shown in Fig. 25.33. Charge supplied by 10 V battery is q1 and that from 20 V battery is q 2 . Thus,

2 µF

Solution

q1 + q 2 = q 3

B

+

–

–

q1

–

+

D

q2

+ 6 µF

q3

…(i) A

This relation can also be obtained by a different method. 10 V The charges on the three plates which are in contact add to zero. Because these plates taken together form an isolated system which can’t receive charges from the batteries. Thus, or

4 µF M

q 3 – q1 – q 2 = 0 q 3 = q1 + q 2

E F

Fig. 25.33

20 V

Chapter 25

Capacitors — 257

Applying second law in loops BMFAB and MDEFM, we have q q – 1 – 3 + 10 = 0 2 6 q 3 + 3q1 = 60 q2 q – 20 + 3 = 0 4 6

or and

or 3q 2 + 2q 3 = 240 Solving the above three equations, we have 10 q1 = µC 3 140 q2 = µC 3 and q 3 = 50 µC Thus, charges on different capacitors are as shown in Fig. 25.34.

…(ii)

…(iii) 140 µC 3 – +

10 µC 3 – + + –

50 µC

10 V

20 V

Fig. 25.34

Note In the problem q1 , q2 and q3 are already in microcoulombs.

25.7 Energy Density (u) The potential energy of a charged conductor or a capacitor is stored in the electric field. The energy per unit volume is called the energy density ( u). Energy density in a dielectric medium is given by 1 u = ε 0 KE 2 2 This relation shows that the energy stored per unit volume depends on E 2 . If E is the electric field in a space of volume dV , then the total stored energy in an electrostatic field is given by 1 U = ε 0 K ∫ E 2 dV 2 and if E is uniform throughout the volume (electric field between the plates of a capacitor is almost uniform), then the total stored energy can be given by U = u (Total volume ) = V

1 Kε E 2V 2 0

Example 25.13 Using the concept of energy density, find the total energy stored in a (a) parallel plate capacitor (b) charged spherical conductor. Solution

this field is

(a) Electric field is uniform between the plates of the capacitor. The magnitude of

258 — Electricity and Magnetism q σ = ε 0 Aε 0

E=

Therefore, the energy density ( u ) should also be constant.

E=0

q2 1 u = ε0E 2 = 2 2A 2 ε 0 ∴ Total stored energy, U = ( u ) (total volume)

– – – – – – – – – E=

 q2  q2 =  2  ( A ⋅ d ) =  Aε   2A ε 0  2 0  d 

∴

+ + + + + + + + +

E=0

σ ε0

Fig. 25.35

=

q2 2C

Aε 0   as C =   d 

U=

q2 2C

Ans.

(b) In case of a spherical conductor (of radius R) the excess charge resides on the outer surface of the conductor. The field inside the conductor is zero. It extends from surface to infinity. And since the potential energy is stored in the field only, it will be stored in the region extending from surface to infinity. But as the field is non-uniform, the energy density u is also non-uniform. So, the total energy will be calculated by integration. Electric field at a distance r from the centre is q 1 E= ⋅ 2 + + + 4πε 0 r + + + + +

1 u( r ) = ε 0 E 2 2

∴

=

q 1  1 ε0  ⋅ 2 2  4πε 0 r 

2

R ++ + +

+ + +q +

Fig. 25.36

Energy stored in a volum dV = ( 4πr ) dr is 2

dU = u dV ∴ Total energy stored, U = ∫

r=∞

r=R

dU

r dr

Substituting the values, we get

or

U=

q2 2 ( 4πε 0 R )

U=

q2 2C

(as C = 4πε 0 R )

Ans.

Fig. 25.37

Chapter 25

Capacitors — 259

25.8 C - R Circuits Charging of a Capacitor in C-R Circuit To understand the charging of a capacitor in C-R circuit, let us first consider the charging of a capacitor without resistance. q 0 = CV + –

C

⇒ S

V

V

Fig. 25.38

Consider a capacitor connected to a battery of emf V through a switch S. When we close the switch the capacitor gets charged immediately. Charging takes no time. A charge q 0 = CV appears in the capacitor as soon as switch is closed and the q-t graph in this case is a straight line parallel to t-axis as shown in Fig. 25.39 q q0

t

Fig. 25.39

If there is some resistance in the circuit charging takes some time. Because resistance opposes the charging (or current flow in the circuit). Final charge (called steady state charge) is still q 0 but it is acquired after a long period of time. The q-t equation in this case is q = q 0 (1 – e

– t/τC

R

C

S

V

Fig. 25.40

)

Here, q 0 = CV and τ C = CR = time constant. q q0 0.632 q0

t = τC

t

Fig. 25.41

q-t graph is an exponentially increasing graph. The charge q increases exponentially from 0 to q 0 . From the graph and equation, we see that at t = 0, q = 0 and at t = ∞ , q = q 0

260 — Electricity and Magnetism Definition of τC At t = τ C , q = q 0 (1 – e –1 ) ≈ 0.632 q 0 Hence, τ C can be defined as the time in which 63.2% charging is over. Note that τ C is the time. Hence, [τ C ] = [time ] or

[CR ] = [M 0 L0 T]

Proof : Now, let us derive the q-t relation discussed above. Suppose the switch is closed at time t = 0. At some instant of time, let charge in the capacitor is q ( < q 0 ) and it is still increasing and hence current is flowing in the circuit. Applying loop law in ABEDA, we get q q R – – iR + V = 0 + – B C E C dq i Here, i= dt D A q  dq  V ∴ – –  R +V = 0 C  dt  Fig. 25.42 ∴

dq

=

q C q dq t dt ∫0 q = ∫0 R V– C V–

or

dt R

q = CV (1 – e

This gives

–

t CR

)

Substituting CV = q 0 and CR = τ C , we have q = q 0 (1 – e – t/τC ).

Charging Current Current flows in a C-R circuit during charging of a capacitor. Once charging is over or the steady state condition is reached the current becomes zero. The current at any time t can be calculated by differentiating q with respect to t. Hence, dq d i= = {q 0 (1 – e – t/τC )} dt dt q or i = 0 e – t/τC τC Substituting q 0 = CV and τ C = CR , we have i=

V – t/τC e R

Chapter 25 By letting,

V =i R 0

Capacitors — 261

i = i0 e – t/τC

i.e. current decreases exponentially with time. The i-t graph is as shown in Fig. 25.43. V Here, i0 = is the current at time t = 0. This is the current which would had R been in the absence of capacitor in the circuit.

i i0

t

Fig. 25.43

Discharging of a Capacitor in C-R Circuit To understand discharging through a C-R circuit again we first consider the discharging without resistance. q0 q=0 Suppose a capacitor has a charge q 0 . The positive plate + – has a charge + q 0 and negative plate – q 0 . It implies that the positive plate has deficiency of electrons and negative ⇒ plate has excess of electrons. When the switch is closed, the extra electrons on negative plate immediately rush to S Fig. 25.44 the positive plate and net charge on both plates becomes zero. So, we can say that discharging takes place immediately. In case of a C-R circuit, discharging also takes time. Final charge on the capacitor is still zero but after sufficiently long period of time. The q-t equation in this case is q0 +

–

R

S Fig. 25.45 v

q = q 0 e – t/τC

q q0

Thus, q decreases exponentially from q 0 to zero, as shown in Fig. 25.46. From the graph and the equation, we see that 0.368 q0

At t = 0, q = q 0

t = τC

At t = ∞, q = 0.

Fig. 25.46

Definition of Time Constant ( τC )

In case of discharging, definition of τ C is changed. At time t = τ C , q = q 0 e –1 = 0.368 q 0 Hence, in this case τ C can be defined as the time when charge reduces to 36.8% of its maximum value q 0 .

t

262 — Electricity and Magnetism Discharging Current During discharging, current flows in the circuit till q becomes zero. This current can be found by differentiating q with respect to t but with negative sign because charge is decreasing with time. So, d  dq  i i = – ( q e – t/τC ) =–  dt  i0 dt 0 =

q 0 – t/τC e τC q0 = i0 τC

By letting,

t

Fig. 25.47

i = i0 e – t/τC

We have,

This is an exponentially decreasing equation. Thus, i-t graph decreases exponentially with time from i0 to 0. The i-t graph is as shown in Fig. 25.47.

25.9 Methods of Finding Equivalent Resistance and Capacitance We know that in series, R eq = R1 + R 2 +… + R n 1 1 1 1 = + +… + C eq C1 C 2 Cn

and

1 1 1 1 = + +… + R eq R1 R 2 Rn

and in parallel,

and C eq = C1 + C 2 +… + C n Sometimes there are circuits in which resistances/capacitors are in mixed grouping. To find R eq or C eq for such circuits few methods are suggested here which will help you in finding R eq or C eq .

Method of Same Potential Give any arbitrary potentials (V1 , V2 , … etc.) to all terminals of capacitors/resistors. But notice that the points connected directly by a conducting wire will have at the same potential. The capacitors/resistors having the same PD are in parallel. Make a table corresponding to the figure. Now, corresponding to this table a simplified figure can be formed and from this figure C eq and R eq can be calculated. V

Example 25.14 Find equivalent capacitance between points A and B as shown in figure.

A

B C

C

C

Fig. 25.48

C

C

C

Chapter 25

Capacitors — 263

Three capacitors have PD, V1 – V2 . So, they are in parallel. Their equivalent capacitance is 3C.

Solution

A V1

V1

V2

V2 V1

V1 V2

V2 V3

V3 V2

B V2 V4 V4

Fig. 25.49

Two capacitors have PD, V2 – V3 . So, their equivalent capacitance is 2C and lastly there is one capacitor across which PD isV2 – V4 . So, let us make a table corresponding to this information. Table 25.1 PD

Capacitance

V1 – V2

3C

V2 – V3

2C

V2 – V4

C

Now, corresponding to this table, we make a simple figure as shown in Fig. 25.50. C B V2

V4

2C

3C A V1

V2

V2

V3

Fig. 25.50

As we have to find the equivalent capacitance between points A and B, across which PD is V1 – V4 . From the simplified figure, we can see that the capacitor of capacitance 2C is out of the circuit and points A and B are as shown. Now, 3C and C are in series and their equivalent capacitance is

Ceq =

(3 C ) (C ) 3 = C 3C + C 4

Ans.

Find equivalent capacitance between points A and B.

EXERCISE

A

B C

C

C

C

C

Fig. 25.51

In case PD across any capacitor comes out to be zero (i.e. the plates are short circuited), then this capacitor will not store charge. So ignore this capacitor. HINT

Ans.

3 C 4

264 — Electricity and Magnetism Identical metal plates are located in air at equal distance d from one another as shown in figure. The area of each plate is A. Find the capacitance of the system between points P and Q if plates are interconnected as shown. EXERCISE

P

P Q

Q (a)

(b)

P

Q

P

Q

(c)

(d)

Fig. 25.52

2ε 0 A 3ε 0 A 2 ε0A 3 ε0A (b) (c) (d) 3 d 2 d d d EXERCISE Find equivalent resistance between A and B. Ans.

(a)

A

B 2Ω

6Ω

3Ω

Fig. 25.53 Ans.

1Ω

EXERCISE

Find equivalent capacitance between points A and B. C

C

A

B C C

Fig. 25.54 Ans.

5 C. 3

Infinite Series Problems This circuit consists of an infinite series of identical loops. To find C eq or R eq of such a series first we consider by ourself a value (say x) of C eq or R eq . Then, we break the chain in such a manner that only one loop is left with us and in place of the remaining portion we connect a capacitor or resistor x. Then, we find the C eq or R eq and put it equal to x. With this we get a quadratic equation in x. By solving this equation, we can find the desired value of x.

Chapter 25 V

Capacitors — 265

Example 25.15 An infinite ladder network is constructed with 1 Ω and 2 Ω resistors as shown. Find the equivalent resistance between points A and B. 1Ω

1Ω

1Ω

A 2Ω

2Ω

∞

2Ω

B

Fig. 25.55 1Ω

Let the equivalent resistance between A and B is x. We may consider the given circuit as shown in Fig. 25.56. In this diagram, 2x 2x (as R AB = x) RAB = + 1 or x = +1 2+ x 2+ x

Solution

or

x ( 2 + x ) = 2x + 2 + x or

A 2Ω B

Fig. 25.56

x2 – x – 2 = 0 x=

1± 1+ 8

Ignoring the negative value, we have

2

x

= – 1 Ω and 2 Ω

R AB = x = 2 Ω

Ans.

Note Care should be taken while breaking the chain. It should be broken from those points from where the broken chain resembles with the original chain. R1

R1

R2

R2

∞

R2

R1

R2

R1

R1

⇒

R2

R1

R1

R2

∞

R2

⇒

R2

Fig. 25.57

EXERCISE

Find equivalent resistance between A and B. R

k 2R

kR

A R

kR

k 2R

B

Fig. 25.58 HINT Ans.

Let RAB = x, then the resistance of the broken chain will be kx. R [(2k – 1) + 4k 2 + 1] / 2k

x

∞

x

266 — Electricity and Magnetism Method of Symmetry Symmetry of a circuit can be checked in the following four manners : 1. Points which are symmetrically located about the starting and last points are at same potentials. So, the resistances/capacitors between these points can be ignored. The following example will illustrate the theory. V

Example 25.16 Twelve resistors each of resistance r are connected together so that each lies along the edge of the cube as shown in figure. Find the equivalent resistance between 6

7 3

2

8

5 1

4 Fig. 25.59

(a) 1 and 4 (b) 1 and 3 Solution (a) Between 1 and 4 : Points 2 and 5 are symmetrically located w.r.t. points 1 and 4. So, they are at same potentials. Similarly, points 3 and 8 are also symmetrically located w.r.t. points 1 and 4. So, they are again at same potential. Now, we have 12 resistors each of resistance r connected across 1 and 2, 2 and 3,…, etc. So, redrawing them with the assumption that 2 and 5 are at same potential and 3 and 8 are at same potential. The new figure is as shown in Fig.25.60. Now, we had to find the equivalent resistance between 1 and 4. We can now simplify the circuit as r

1

r/ 2

2,5

4 r/ 2

4

2,5

3,8

7

6

Fig. 25.60

r

1

3,8

1

4

⇒

r/ 2

2 r 5

r 2

r 2

2r r

1

4

⇒

1

4 7 r 12

7 r 5

Fig. 25.61

Thus, the equivalent resistance between points 1 and 4 is 7 r. 12

Ans.

Capacitors — 267

Chapter 25

(b) Between 1 and 3 : Points 6 and 8 are symmetrically located w.r.t. points 1 and 3. Similarly, points 2 and 4 are located symmetrically w.r.t. points 1 and 3. So, points 6 and 8 are at same potential. Similarly, 2 and 4 are at same potentials. Redrawing the simple circuit, we have Fig. 25.62. 1 r

5

1

3r 2

r

r r

r 6,8 r

2,4

r r

r

r/2

6,8

2,4 r

r /2

r

7

3r 2

r 3

r /2

3 Fig. 25.63

Fig. 25.62

Between 1 and 3, a balanced Wheatstone bridge is formed as shown in Fig. 25.63. So, the resistance between 2 and 6 and between 4 and 8 can be removed. 1 1

3r 2

1

r /2

⇒ 3r 2

r

3r

3r 4

⇒

r /2 3

3 3

Fig. 25.64

Thus, the equivalent resistance between 1 and 3 is

3 r. 4

EXERCISE Fourteen identical resistors each of resistance r are connected as shown. Calculate equivalent resistance between A and B. Ans. 1.2r

Ans.

A

B

Fig. 25.65

EXERCISE

Eight identical resistances r each are connected along edges of a pyramid having square base ABCD as shown. Calculate equivalent resistance between A and O. 7r Ans. 15

O

B

C

A

D

Fig. 25.66

268 — Electricity and Magnetism 2. If points A and B are connected to a battery and AB is a line of symmetry, then all points lying on

perpendiculars drawn to AB are at the same potential. For example, 3 r

r 1

6

r r

r

r

A

B 4

r

r

r

r 7

2 r

r 5 V

Fig. 25.67

In Fig. 25.67, points (1, 2), (3, 4, 5) and (6, 7) are at same potential. So, we can join these points and draw a simple circuit as shown in Fig. 25.68. r

r

1, 2

r

3, 4, 5

r

r

6, 7

r

A

B r

r

r

r

r

r

Fig. 25.68

Now, the equivalent of this series combination is r r r r 3r R eq = + + + = 2 4 4 2 2 EXERCISE Solve the same problem by connection removal method (will be discussed later). 3. Even if AB is not a line of symmetry but its perpendicular bisector is, then all the points on this

perpendicular bisector are at the same potential. For example, r

r

1

r

r

r

r r

r

r

A r

3

r

⇒

2 r

r

r

r

r

B

A

r

r r r B r

(a)

r (b)

Fig. 25.69

In Fig. (a), AB is not a line of symmetry but, 1, 2 and 3 are line of symmetry. Hence, they are at same potential (if A and B are connected to a battery). This makes the resistors between 1 & 2 and 2 and 3 redundant because no current flows through them. So, the resistance between them can be

Chapter 25

Capacitors — 269

removed [as shown in Fig. (b)]. The equivalent resistance between A and B can now be easily 5r determined as ⋅ 4 d, f 4. Each wire in the cube has a resistance r. We e d are interested in calculating the equivalent c e resistance between A and B. This is a three-dimensional case and in c f ⇒ place of a line of symmetry involving points B g A and B we locate a plane of symmetry A B involving A and B. A h Such a plane is the plane ABce and for this g, h plane points d and f and g and h have the (a) (b) same potential. Fig. 25.70 The equivalent resistance between A and B can now be easily worked out (Using Wheatstone’s bridge principle) as 3r R eq = 4

Connection Removal Method This method is useful when the circuit diagram is symmetric except for the fact that the input and output are reversed. That is the flow of current is a mirror image between input and output above a particular axis. In such cases, some junctions are unnecessarily made. Even if we remove that junction there is no difference in the remaining circuit or current distribution. But after removing the junction, the problem becomes very simple. The following example illustrates the theory. V

Example 25.17 Find the equivalent resistance between points A and B. r r

r r

r r

r A

B

Fig. 25.71

Solution

A

B

Fig. 25.72

Input and output circuits are mirror images of each other about the dotted line as shown. So, if a current i enters from A and leaves from B, it will distribute as shown below.

270 — Electricity and Magnetism i3

i4

i1

i4

i1 A

i

i2

B

i2

Fig. 25.73

Now, we can see in figure that the junction where i 2 and i 4 are meeting can be removed easily and then the circuit becomes simple. r

r

r A

⇒

8r 3

r

r

r

⇒ B

r

A

A

2r

B

B 8 r 7

Fig. 25.74

Hence, the equivalent resistance between A and B is

8 r. 7

Ans.

EXERCISE

Eight identical resistances r each are connected as shown. Find equivalent resistance between A and D. O B

C

A

D

Fig. 25.75 Ans.

8r 15

EXERCISE

Twelve resistors each of resistance r are connected as shown. Find equivalent resistance between A and B.

A

B

Fig. 25.76 Ans.

( 4 / 5) r

Chapter 25 EXERCISE

Capacitors — 271

Find equivalent resistance between A and B. 10 Ω

10 Ω

6Ω

10 Ω

10 Ω

A

B 6Ω 10 Ω

10 Ω

Fig. 25.77 Ans.

20 Ω. 3

Wheatstone Bridge Circuits Wheatstone bridge in case of resistors has already been discussed in the chapter of current electricity. For capacitor, theory is same. C C If 1 = 3 , bridge is said to be balanced and in that case C2 C4 VE = VD or VE – VD or VED = 0 i.e. no charge is stored in C 5 . Hence, it can be removed from the circuit. R C EXERCISE In the circuit shown in figure, prove that VAB = 0 if 1 = 2 . R 2 C1

E C2

C1 A

B C5 C4

C3 D

Fig. 25.78

A C2

C1 R R1

R2

B

E

Fig. 25.79

By Distributing Current/Charge Sometimes none of the above five methods is applicable. So, this one is the last and final method which can be applied everywhere. Of course this method is a little bit lengthy but is applicable everywhere, under all conditions. In this method, we assume a main current/charge, i or q. Distribute it in different resistors/capacitors as i1 , i2 … (or q1 , q 2 , … , etc.). Using Kirchhoff’s laws, we find i1 , i2 , … etc., (or q1 , q 2 , … , etc.) in terms of i (or q). Then, find the potential difference between starting and end points through any path and equate it with iR net or q/C net . By doing so, we can calculate R net or C net .

272 — Electricity and Magnetism The following example is in support of the theory. V

Example 25.18 Find the equivalent capacitance between A and B. 2C

C

2C

A

B

2C

C

Fig. 25.80

The given circuit forms a Wheatstone bridge. But the bridge is not balanced. Let us suppose point A is connected to the positive terminal of a battery and B to the negative terminal of the same battery; so that a total charge q is stored in the capacitors. Just by seeing input and output symmetry, we can say that charges will be distributed as shown below. Solution

q2

C + – q1

+ 2C

A

– 2C

+

B

– q3 +

2C –

C +

q2

– q1

Fig. 25.81

q1 + q 2 = q

…(i)

Applying second law, we have –

q1 q q – 3 + 2 =0 C 2C 2C

or

q 2 – q 3 – 2q1 = 0 Plates inside the dotted line form an isolated system. Hence,

…(ii)

q 2 + q 3 – q1 = 0 Solving these three equations, we have q 2 3 q1 = q, q 2 = q and q 3 = – 5 5 5

…(iii)

Now, let C eq be the equivalent capacitance between A and B. Then, q q q V A – VB = = 1 + 2 C eq C 2C ∴ ∴

q 2q 3q 7q = + = C eq 5C 10C 10C C eq =

10 C 7

Ans.

Chapter 25

Capacitors — 273

Final Touch Points 1. Now, onwards we will come across the following integration very frequently. So, remember the result as such.

∫0 a – bx = ∫0 c dt ,

then

x =

a (1 – e –bct ) b

t dx = c dt , a – bx ∫0

then

x =

a a  –  – x 0 e –bct  b b

x

If

x

∫x

and if

0

dx

t

Here, a, b and c are constants.

2. Sometimes a physical quantity x decreases from x1 to x 2, exponentially, then x-t equation is like x x1

x1

x

x1 – x 2 x2

x2 t

t

x = x 2 + ( x1 – x 2 ) e –Kt Here, K is a constant. Similarly, if x increases from x 2 to x1 exponentially, then x-t equation is x = x 2 + ( x1 – x 2 ) (1 – e – Kt )

3. Leakage Current Through a Capacitor The space between the capacitor’s plates is filled with a dielectric and we assume that no current flows through it when the capacitor is connected to a battery as in figure (a) or if the capacitor is charged, the charge on its plates remains forever. But every insulator has some conductivity. On account of which some current flows through the capacitor if connected to a battery. This small current is known as the leakage current. Similarly, when it is charged, the charge does not remain as it is for a long period of time. But it starts discharging. Or we can say it becomes a case of discharging of a capacitor in C-R circuit.

+

–

or q0 (a)

(b)

In both the cases, we will first find the resistance of the dielectric. l R = σA Here, ∴

( σ = specific conductance)

l = d (the distance between the plates of the capacitor) R =

d σA

Thus, the leakage current in the circuit shown in the figure is V i = R

R=

i

d σA i

V

274 — Electricity and Magnetism Similarly, if the capacitor is given a charge q 0 at time t = 0, then after time t, q charge will remain on it, where q = q 0 e – t /τC +

Here,

(Discharging of a capacitor)

⇒

–

+

–

q0

q

At t = 0

At t = t

 Kε A   d  τC = CR =  0     d   σA 

τC =

or

Kε 0 σ

4. If capacitors are in series, then charges on them are equal, provided they are initially uncharged. This can be proved by the following illustration : Let us suppose that charges on two capacitors are q1 and q 2. The two plates encircled by dotted lines form an isolated system. So, net charge on them will remain constant.

+– q1

+– q2

Σqf = Σqi If initially they are uncharged, then Σqi = 0 Σqf is also zero −q1 + q 2 = 0 or q1 = q 2

or

So, this proves that charges are equal if initially they are uncharged.

5. Independent parallel circuit C1

C2

R2

R3 a

d

b

C V

Three circuits shown in figure are independently connected in parallel with the battery. Potential difference across each of the circuit is V . By this potential difference, capacitor C1 is immediately charged. Capacitor C2 is exponentially charged and current grows immediately in R 3. Thus, qC1 = C1V qC 2 = C2V (1 − e IR 3 =

V R3

(immediately) t − C 2R 2

) (immediately)

Note If any resistance or capacitance is connected between abcd, then it no longer remains an independent parallel circuit.

Solved Examples TYPED PROBLEMS Type 1. In a complex capacitor circuit method of finding values of q and V across different capacitors if values across one capacitor are known

Concept In series, q is same and V distributes in inverse ratio of capacity. q 1 As, V = ⇒ V ∝ C C If capacitance is double, then V will be half. In parallel, V is same and q distributes in direct ratio of capacity. As, q = CV ⇒ q ∝C If C is double, then q is also double. V

Example 1

9 µF

5 µF

3 µF

1 µF

(q is same)

(V is same)

4 µF

E

In the circuit shown in figure potential difference across 3 µF is 10 V . Find potential difference and charge stored in different capacitors. Also find emf of the battery E. Solution The given circuit can be simplified as under 4 µF

12 µF

6 µF

V1

10 V

V2

E

1 rd of 12 µF. Therefore, V1 is thrice of 10 V or 30 V. 3 V2: 6 µF is half of 12 µF. Therefore, V 2 is twice of 10 V or 20 V. E = V1 + 10 + V 2 = 30 + 10 + 20 = 60 V

V1 : 4 µF is

Ans.

276 — Electricity and Magnetism Potential difference and charge on different capacitors in tabular form are given below.

Table 25.2 Capacitance

Potential difference

Charge q = CV

4 µF

30 V

120 µC

9 µF

10 V

90 µC

3 µF

10 V

30 µC

5 µF

20 V

100 µC

1µF

20 V

20 µC

Type 2. Configuration of capacitor is changed and change in five quantities q, C, V, U and E is asked

Concept Some problems are asked when a capacitor is charged through a battery and then the configuration of capacitor is changed : (i) either by inserting a dielectric slab or removing the slab (if it already exists) or (ii) by changing the distance between the plates of capacitor or (iii) by both. The questions will be based on the change in electric field, potential, etc. In such problems, two cases are possible. Case 1 When battery is removed after charging If the battery is removed after charging, then the charge stored in the capacitor remains constant. q = constant First of all, find the change in capacitance and according to the formula find the change in other quantities. V

Example 2 An air capacitor is first charged through a battery. The charging battery is then removed and a dielectric slab of dielectric constant K = 4 is inserted between the plates. Simultaneously, the distance between the plates is reduced to half, then find change in C, E, V and U. Kε 0 A K ∝ d d d  ∴ Capacitance will become 8 times  K = 4, d′ =   2

Solution Change in capacitance,

C=

Change in electric field ∴ or or the electric field will become

E = E 0/K 1 E∝ K 1 times its initial value. 4

(if q = constant)

Chapter 25

Capacitors — 277

Change in potential difference, q C 1 V ∝ C V =

or Therefore, potential difference becomes

(if q = constant)

1 times of its initial value. 8

Alternate method V = Ed 1 Electric field has become times its initial value and d is reduced to half. Hence, V becomes 4 1 times. 8 Change in stored potential energy, 1 q2 2C 1 U ∝ C U =

or

( if q = constant )

Capacitance has become 8 times. Therefore, the stored potential energy U will become

1 times. 8

Case 2 When battery remains connected If the battery remains connected, the potential difference V becomes constant. So, in the above example, capacitance will become 8 times. The charge stored ( q = CV or q ∝ C ) will also increase to 8 times. The electric field V 1 1    or E ∝  becomes twice and the stored PE U = CV 2 or U ∝ C is 8 times. E =    d d 2

Type 3. To find self energy of a system of charges

Concept The self energy of a system of charges is Us =

q

∫0

Vdq

q2 in case of a capacitor or conductor. 2C A point charge does not have any self energy.

This comes out to be equal to

V

Example 3 Find the electric potential energy of a uniformly charged sphere. Solution Consider a uniformly charged sphere of radius R having a total charge q0. The volume charge density is ρ=

q0 3 q0 = 4 3 4 πR3 πR 3

278 — Electricity and Magnetism When the radius of the sphere is r, the charge contained in it is 4  q  q =  πr3  ρ =  03  r3 3  R  The potential at the surface is V =

q q0 = r2 4πε 0r 4πε 0R3

The charge needed to increase the radius from r to r + dr is dq = (4πr 2) dr ρ 3q = 30 ⋅ r 2 dr R ∴ The self energy of the sphere is R

U s = ∫ V dq 0

R   3q q0  =∫  r 2  30 ⋅ r 2 dr 0  4 πε R3   R  0

=

V

3q02 20 πε 0R

Ans.

Example 4 Find the electric potential energy of a uniformly charged, thin spherical shell. Solution Consider a uniformly charged thin spherical shell of radius R having a total charge q0. Suppose at some instant a charge q is placed on the shell. The potential at the surface is q V = 4πε 0R ∴ The self energy of the shell is q0

Us = ∫

0

=∫

0

=

q0

V dq  q    dq  4πε 0R

q02 8πε 0R

Ans.

Type 4. Based on flow of charge when position of a switch is changed

Concept From the flow of charge we mean that when a switch in a circuit is either closed or opened or it is shifted from one position to the other, then how much charge will flow through certain points of the circuit. Such problems can be solved by finding charges on different capacitors at initial and final positions and then by the difference we can find the charge flowing through a certain point. The following example will illustrate the theory.

Chapter 25 V

Capacitors — 279

Example 5 What charges will flow through A, B and C in the directions shown in the figure when switch S is closed? A 2 µF

30 V B S 60 V

3 µF C

Solution Let us draw two figures and find the charge on both the capacitors before closing the switch and after closing the switch. A + q 30 V

–

2 µF

+ q1

30 V

–

2 µF

B +

60 V

q

–

3 µF

+

60 V

q2

–

3 µF

C (a)

(b)

Refer Fig. (a), when switch is open capacitance is C eq =

Both capacitors are in series. Hence, their equivalent C1 C 2 (2) (3) 6 = = µF C1 + C 2 2 + 3 5

Therefore, charge on both capacitors will be same. Hence, using q = CV , we get  6 q = (30 + 60)   µC = 108 µC  5 Refer Fig. (b), when switch is closed Let q1 and q2 be the charges (in µC ) on two capacitors. Then, applying second law in upper and lower loops, we have q or 30 – 1 = 0 q1 = 60 µC 2 q or 60 – 2 = 0 q2 = 180 µC 3 Charges q1 and q2 can be calculated alternatively by seeing that upper plate of 2 µF capacitor is connected with positive terminal of 30 V battery. Therefore, they are at the same potential. Similarly, the lower plate of this capacitor is at the same potential as that of the negative terminal of 30 V battery. So, we can say that PD across 2 µF capacitor is also 30 V. ∴

q1 = (C ) (PD) = (2) (30) µC = 60 µC Similarly, PD across 3 µF capacitor is same as that between 60 V battery. Hence, q2 = (3) (60) µC = 180 µC

280 — Electricity and Magnetism Now, let qA charge flows from A in the direction shown. This charge goes to the upper plate of 2 µF capacitor. Initially, it had a charge + q and final charge on it is + q1. Hence, q1 = q + qA or qA = q1 – q = 60 – 108 Ans. = – 48 µC Similarly, charge qB goes to the upper plate of 3 µF capacitor and lower plate of 2 µF capacitor. Initially, both the plates had a charge + q – q or zero. And finally they have a charge (q2 – q1 ). Hence, (q2 – q1 ) = qB + 0 ∴ qB = q2 – q1 = 180 – 60 Ans. = 120 µC Charge qC goes to the lower plate of 3 µF capacitor. Initially, it had a charge – q and finally – q2. Hence, – q2 = (– q) + qC ∴ qC = q – q2 = 108 – 180 Ans. = – 72 µC So, the charges will flow as shown below –

48 µC

48 µC – 30 V 48 µC +

2 µF

120 µC +

72 µC + 72 µC –

3 µF

60 V

–

72 µC

Type 5. Based on heat generation or loss of energy during shifting of switch

Concept By heat generation (or loss of energy), we mean that when a switch is shifted from one position to the other, what amount of heat will be generated (or loss will be there) in the circuit. Such problems can be solved by simple energy conservation principle. For this, remember that when a charge + q flows from negative terminal to the positive terminal inside a battery of emf V is supplied an energy,

E = qV V

V

+ q

+ q

Energy supplied = qV

Energy consumed = qV

Chapter 25

Capacitors — 281

and if opposite is the case, i.e. charge + q flows in opposite direction, then it consumes energy by the same amount. Now, from energy conservation principle we can find the heat generated (or loss of energy) in the circuit in shifting the switch. Heat generated or loss of energy = energy supplied by the battery/batteries – energy consumed by the battery/batteries + ΣU i – ΣU f. Here,

ΣUi = energy stored in all the capacitors initially and ΣU f = energy stored in all the capacitors finally

V

Example 6 Find loss of energy in example 5. Solution In the above example, energy is supplied by 60 V battery and consumed by 30 V battery. Using E = qV , we have Energy supplied = (72 × 10–6 ) (60) = 4.32 × 10−3 J Energy consumed = (48 × 10–6 ) (30) = 1.44 × 10−3 J 1 6 ΣUi = × × 10–6 × (90)2 2 5 ΣU f

and

= 4.86 × 10−3 J 1 1 = × 2 × 10–6 × (30)2 + × 3 × 10–6 × (60)2 2 2 = 6.3 × 10−3 J

Loss of energy = (4.32 − 1.44 + 4.86 − 6.3) × 10−3 J

∴

= 1.44 × 10−3 J V

Ans.

Example 7 Prove that in charging a capacitor half of the energy supplied by the battery is stored in the capacitor and remaining half is lost during charging. Solution When switch S is closed, q = CV charge is stored in the capacitor. Charge transferred from the battery is also q. Hence, energy supplied by the battery = qV = (CV ) (V ) = CV 2. Half of its energy, i.e. 1 CV 2 is stored in the capacitor and the remaining 50 % or 1 CV 2 is lost. 2

2

C

+

C

S q

⇒ V V

+ q

–

282 — Electricity and Magnetism Type 6. Two or more than two capacitors are charged from different batteries and then connected in parallel

Concept In parallel, they come to a common potential given by V =

Total charge Total capacity

Moreover, total charge on them distributes in direct ratio of their capacity or we can also find the final charges on the capacitors by using the equation, q = CV V

Example 8 Three capacitors of capacities 1 µF , 2 µF and 3 µF are charged by 10 V , 20 V and 30 V respectively. Now, positive plates of first two capacitors are connected with the negative plate of third capacitor on one side and negative plates of first two capacitors are connected with positive plate of third capacitor on the other side. Find (a) common potential V (b) final charges on different capacitors Solution 10 µC + – S1 − 40 µC S2

q1 – +

1 µF, 10 V 40 µC + –

+ 40 µC

2 µF, 20 V

q2 – +

90 µC – +

q3 – +

3 µF, 30 V

V

Total charge on all three capacitors = 40 µC Total capacity = (1 + 2 + 3) µF = 6 µF (a) Common potential, V = = (b)

Total charge Total capacity 40 µC 20 = volt 6 µF 3

 20 20 q1 = C1V = (1 µF)   = µC 3 3  20 40 q2 = C 2V = (2 µF)   = µC 3 3  20 q3 = C3V = (3 µF)   = 20 µC 3

Chapter 25

Capacitors — 283

Type 7. To find final charges on different capacitors when position of switch is changed (opened, closed or shifted from one position to other position)

Concept (i) To find current in a C -R circuit at any time t, a capacitor may be assumed a battery of q emf E or V = . C (ii) Difference between a normal battery and a capacitor battery is, emf of a normal battery remains constant while emf of a capacitor battery keeps on changing with q. (iii) Before changing the position of switch, every loop of the circuit may not be balanced by Kirchhoff's second equation of potential. So, in a single loop problem rotate a charge q, either clockwise or anti-clockwise. From this charge q, some of the charges on capacitors may increase and others may decrease. To check this, always concentrate on positive plate of each capacitors. If positive charge comes towards this plate, then charge on this capacitor will increase. (iv) With these charges, apply Kirchhoff's loop equation and find the final charges on them. (v) In this redistribution of charges, there is some loss of energy as discussed in type 5. (vi) If there are only capacitors in the circuit, then redistribution of charges is immediate and if there are resistors in the circuit, then redistribution is exponential. (vii) If there are only capacitors, then loss is in the form of electromagnetic waves and if there are resistors in the circuit, then loss is in the form of heat. Further, this loss is proportional to R if resistors are in series (H = i 2 Rt or H ∝ R as i is same in series) and this loss is inversely proportional to R if resistors are in parallel V 2t 1 or H ∝ as V is same in parallel). (H = R R V

Example 9 In the circuit shown in figure, switch S is closed at time t = 0. Find

1F + – 40 V

10 V

2F – + 20 V

(a) Initial current at t = 0 and final current at t = ∞ in the S 1Ω loop. 4F 2Ω (b) Total charge q flown from the switch. (c) Final charges on capacitors in steady state at time 20V t = ∞. (d) Loss of energy during redistribution of charges. (e) Individual loss across 1 Ω and 2 Ω resistance. Solution (a) At t = 0 Three capacitors may be assumed like batteries of emf 40 V, 20 V and 0 V. ∴

Net emf Total resistance 40 + 20 − 10 − 20 = 1+2

i=

(anti-clockwise) = 10 A At t = ∞ When charge redistribution is complete and loop is balanced by Kirchhoff's second equation of potential, current in the loop becomes zero, as insulator is filled between the capacitors.

284 — Electricity and Magnetism (b) Redistribution current was anti-clockwise. So, we can assume that + q charge rotates anti-clockwise in the loop (between time t = 0 and t = ∞). After this rotation of charges, final charges on different capacitors are as shown below. a

(40– q) + – 1F + 4F –

q

(40 +q) – + b

10 V

2F

rq

1Ω i = 0 at t = ∞

2Ω c

d

20 V

Applying loop equation in loop abcda, (40 − q) (40 + q) q + 10 + − 20 + = 0 1 2 4 120 Solving this equation, we get q= C 7 (c) Final charges 120 160 q1F = 40 − q = 40 − = C 7 7 120 400 q 2F = 40 + q = 40 + = C 7 7 120 q 4F = q = C 7 −

Ans.

(d) Total loss of energy during redistribution 1 1 ΣU i = × (1) (40)2 + × (2) (20)2 = 1200 J 2 2 2 1 (160 / 7) 1 (400 / 7)2 1 (120 / 7)2 ΣU f = × + × + × 2 1 2 2 2 4 = 1114.3 J ∆U = ΣU f − ΣU i = − 85.7 J 20 V battery will supply energy but 10 V battery will consume energy. So, 120 120 Total energy supplied = 20 × − 10 × = 171.4 J 7 7 Total heat produced = Energy supplied − ∆U = 171.4 − (−85.7) = 257.1 J

Ans.

(e) Resistors are in series. Hence, H ∝ R or ∴

Note

H 1 R1 1 Ω 1 = = = H 2 R2 2 Ω 2

 1  H1 =   (257.1) = 85.7 J  1 + 2

Ans.

 2  H2 =   (257.1) = 171.4 J  1 + 2

Ans.

(i) For making the calculations simple, we have taken capacities in Farad, otherwise Farad is a large unit. (ii) For two loop problems, we will rotate two charges q1 and q2 .

Chapter 25

Capacitors — 285

Type 8. Shortcut method of finding time varying functions in a C-R circuit like q or i etc.

Concept (i) At time t = 0, when capacitor is uncharged it offers maximum current passing through it. So, it may be assumed like a conducting wire of zero resistance. With this concept, find initial values of q or i etc. (ii) At time t = ∞, when capacitor is fully charged it does not allow current through it, as insulator is filled between the plates. So, its resistance may be assumed as infinite. With this concept, find steady state values at time t = ∞ of q or i etc. (iii) Equivalent time constant To find the equivalent time constant of a circuit, the following steps are followed : (a) Short-circuit the battery. (b) Find net resistance across the capacitor (suppose it is Rnet ) (c) τC = ( Rnet )C (iv) In C - R circuit, increase or decrease is always exponential. So, first make exponential graph and then write exponential equation corresponding to this graph with the time constant obtained by the method discussed above. V

Example 10 Switch S is closed at time t = 0 in the circuit shown in figure. 3Ω 3Ω

S 15V

6Ω

2F

(a) Find the time varying quantities in the circuit. (b) Find their values at time t = 0. (c) Find their values at time t = ∞ (d) Find time constant of all time varying functions. (e) Make their exponential graphs and write their exponential equations. (f) Just write the equations to solve them to find different time varying functions. Solution (a) i3 3 Ω i1 i2 3Ω

15 V

6Ω

+ 2 F– q

There are four times variable functions i1 , i 2, i3 and q. (b) At t = 0, equivalent resistance of capacitor is zero. So, the simple circuit is as shown below 3Ω i1 15 V

i2 i3 3Ω

6Ω

286 — Electricity and Magnetism 3 ×6 =5Ω 3+6

Rnet = 3 +

∴

i2 6 2 = = i3 3 1

i1 =

15 =3 A 5

i2 =

2 (3 A ) = 2 A 2+1

⇒

 1  i3 =   (3 A ) = 1 A  2 + 1

and q = 0

(c) At t = ∞, equivalent resistance of capacitor is infinite. So, equivalent circuit is as shown below 15 5 i1 = i3 = = A 3Ω i3 3+6 3 i1

i2 = 0 V 2F = V 6 Ω

 5 = iR =   (6) = 10 volt  3

15 V

3Ω

6Ω

∴ q = CV = (2) (10) = 20 C (d) By short-circuiting the battery, the simplified circuit is as shown below 3Ω

3Ω

6Ω

a b

Net resistance across capacitor or ab is Rnet = 3 +

3 ×6 =5Ω 3+6

∴ τC = CRnet = (2) (5) = 10 s (e) Exponential graphs and their exponential equations are as under. i1(A) 3

i2(A)

i3(A)

q(C)

5/3 20

2

1

5/3

t(s)

t(s)

t(s)

−t

i1 =

t

5  5 5 4 − + 3 −  e τC = + e 10 3  3 3 3 −

i 2 = 2e

t τC

−

= 2e

t 10 t

t

− − 2 5  i3 = 1 +  − 1 (1 − e τC ) = 1 + (1 − e 10 ) 3  3 −

q = 20 (1 − e

t τC

−

) = 20 (1 − e

t 10 )

t(s)

Chapter 25

Capacitors — 287

(f) Unknowns are four: i1 , i 2, i3 and q. So, corresponding to the figure of part (a), four equations are i1 = i 2 + i3 dq i2 = dt Applying loop equation in left hand side loop, +15 − 3i1 − 3i 2 −

…(i) …(ii)

q =0 2

…(iii)

Applying loop equation in right hand side loop, q + + 3i 2 − 6i3 = 0 …(iv) 2 Solving these equations (with some integration), we can find same time functions as we have obtained in part (e).

Type 9. To find current and hence potential difference between two points in a wire having a capacitor

Concept If charge on capacitor is constant, then current through capacitor wire is zero. If charge is variable, then current is non-zero. Magnitude of this current is dq i =   dt  and direction of this current is towards the positive plate if charge is increasing and away from the positive plate if charge is decreasing. V

Example 11 In the circuit shown in figure, find V ab at 1 s. a

2F – +

4Ω

10 V

b

q =2t

Solution Charge on capacitor is increasing. So, there is a current in the circuit from right to left. This current is given by i=

dq =2 A dt

At 1 s, q = 2 C. So, at 1 s, circuit is as shown in figure. a

2F – + 2C

4Ω

2A

b

10 V

2 V a + + (2) (4) + 10 = Vb 2 ∴ or

V a − Vb V ab = − 19 volt

Ans.

288 — Electricity and Magnetism Type 10. To find capacitance of a capacitor filled with two or more than two dielectrics

Concept PQ and MN are two metallic plates. If we wish to find net capacitance between a and b, then

a S

P

V PS = V SQ = V1 ( say) V MT = VTN = V 2 ( say)

Q K2

K1

Hence, V PS − V MT = V SQ − VTN = V1 − V 2 Therefore, there are two capacitors, one on right hand side and other on left hand side which are in parallel.

K3 M

T

N b

∴ C = CRHS + CLHS For CRHS , we can use the formula, ε0 A C= t t d − t1 − t 2 + 1 + 2 K1 K 2 V

Example 12 What is capacitance of the capacitor shown in figure? A/2

2d

A/2 K2

d

K3

d

K1

Solution C = C LHS + C RHS =

K 1 ε 0 ( A /2) ε 0 ( A /2) + 2d (2d − d − d ) + (d /K 2) + (d /K 3 )

=

ε0 A  K1 K 2K 3  +   2d  2 K 2 + K3 

Type 11. To find charge on different capacitors in a C-R circuit

Concept In a C - R circuit, charge on different capacitors is normally asked either at t = 0, t = ∞ or t = t. If nothing is given in the question, then we have to find charges on capacitors at t = ∞ or steady state charges. In steady state, no current flows through a wire having capacitor. But, if there is any other closed circuit then current can flow through that circuit. So, first find this current and then steady state potential difference (say V 0 ) across two plates of capacitor. Now, q 0 = CV 0 = steady state charge

Chapter 25 V

Capacitors — 289

Example 13 Find potential difference across the capacitor (obviously in steady state) V

R

C

V

2R

2V

Solution In steady state condition, no current will flow through

R

VVVV

the capacitor C. Current in the outer circuit, 2V −V V i= = 2R + R 3R

V A

V A − V + V + iR = VB V V  VB − V A = iR =   R =  3 R 3

∴

V 2R

i

VVVV

i

Potential difference between A and B,

B i=0

2V

Note In this problem, charge stored in the capacitor can also be asked, which is equal to q = C

V with positive 3

charge on B side and negative on A side because VB > VA. V

Example 14 Find the charge stored in the capacitor. Insulator is filled between the plates of the capacitor. Therefore, a capacitor does not allow current flow through it after charging is over. Hence, in the circuit current will flow through 3 Ω and 5 Ω resistances and it will not flow through the capacitor. To find the charge stored in the capacitor, we need the PD across it. So, first we will find PD across the capacitor and then apply,

2 µF

HOW TO PROCEED

5Ω

3Ω

24 V

q = CV where, V = PD across the capacitor. Solution As we said earlier also, current will flow in loop ABCDA when charging is over. And this current is Net emf 24 i= = =3 A Total resistance 5 + 3 2 µF + – q 5Ω A

B i

D

3Ω

C 24 V

290 — Electricity and Magnetism Now, PD across the capacitor is equal to the PD across the 5 Ω resistance. Hence, V = V A – VB = iR = (3) (5) = 15 V ∴

q = CV = (2 × 15) µC = 30 µC

Ans.

Note VA – VB = 15 V , therefore VA > VB, i.e. the positive charge will be collected on the left plate of the capacitor and negative on the right plate.

Type 12. To find distribution of charges on different faces on parallel conducting plates

Concept

1

2

3

4

5

6

q Total 2 (ii) q 2 = − q3 and q 4 = − q5 (i) q1 = q 6 =

Note The above two results are proved in example 15.

(iii) Electric field between 2 and 3 is due to the charges q 2 and q3 . This electric field is given by q/A σ E= = [|q 2| = |q3| = q ] ε0 ε0 (iv) This electric field is uniform, so potential difference between any two points is given by V = Ed (v) If two plates are connected to each other, then distance between the plates is required, otherwise there is no requirement of that. V

Example 15 Three parallel metallic plates each of area A are kept as shown in figure and charges q1 , q2 and q3 are given to them. Find the resulting charge distribution on the six surfaces, neglecting edge effects as usual. q1

a

q2

b

P

c

q3

d

Q

e

f

R

Solution The plate separations do not affect the distribution of charge in this problem. In the figure, and

qb = q1 – qa, qf = q3 – qe.

qd = q2 – qc

Chapter 25

Capacitors — 291

Electric field at point P is zero because this point is lying inside a conductor. EP = 0 At P, charge qa will give an electric field towards right. All other charges qb , qc … , etc., will give the electric field towards left. So, 1 [qa – (q1 – qa ) – qc – (q2 – qc ) – qe – (q3 – qe )] = 0 2 Aε 0

qa

2qa – q1 – q2 – q3 = 0 q + q2 + q3 qa = 1 2

or or

qb qc

P

qd qe

Q

qf

R

ER = 0

Similarly the condition, will give the result,

q1 + q2 + q3 2 From here we may conclude that, half of the sum of all charges appears on each of the two outermost surfaces of the system of plates. Further we have a condition, qf =

EQ = 0 1 [qa + (q1 – qa ) + qc – (q2 – qc ) – qe – (q3 – qe )] = 0 2 Aε 0 q1 + 2qc – q2 – q3 = 0 q + q3 – q1 ∴ qc = 2 2 q – q2 – q3 qb = q1 – qa = 1 = – qc 2 Similarly, we can show that qd = – qe. From here we can find another important result that the pairs of opposite surfaces like b, c and d,e carry equal and opposite charges. or

V

Example 16 Three identical metallic plates are kept parallel to one another at a separation of a and b. The outer plates are connected by a thin conducting wire and a charge Q is placed on the central plate. Find final charges on all the six plate’s surfaces. 1

3

2

a

5

4

6

b

Solution Let the charge distribution in all the six faces be as shown in figure. While distributing the charge on different faces, we have used the fact that two opposite faces have equal and opposite charges on them.

292 — Electricity and Magnetism Net charge on plates A and C is zero. Hence, or

q2 – q1 + q3 + q1 – Q = 0 q2 + q3 = Q (Q –q 1)

–q 1 q 1

q2

…(i)

q3

(q1 – Q)

A

E1

B

E2 b

a

C

Further A and C are at same potentials. Hence, VB – VA = VB – VC E1a = E 2b q1 Q – q1 ⋅a = ⋅b Aε 0 Aε 0

or ∴ ∴

q1a = (Q – q1 ) b Qb q1 = a+b

∴

(A = Area of plates)

…(ii)

Electric field inside any conducting plate (say inside C) is zero. Therefore, q2 q1 q1 Q – q1 q1 – Q q3 – + + + – =0 2 Aε 0 2 Aε 0 2 Aε 0 2 Aε 0 2 Aε 0 2 Aε 0 ∴

q2 – q3 = 0 Qb Q Solving these three equations, we get q1 = , q2 = q3 = a+b 2

Hence, charge on different faces are as follows.

Table 25.3 Face

Charge

1

q2 =

2

– q1 = –

Q 2 Qb a+ b

Qb a+ b

3

q1 =

4

Q – q1 =

5

q1 − Q = –

6

q3 =

Qa a+ b Qa a+ b

Q 2

…(iii)

Chapter 25 V

Capacitors — 293

Example 17 Area of each plate is A. The conducting plates are connected to a battery of emf V volts. Find charges q1 to q 6 . 2d

d

1 A 2

3 B 4

5 C 6

V

Solution Net charge drawn from the battery is zero or qTotal = 0 qTotal =0 2 V AB = V with V A > VB  ε A |q2| = |q3| = CV =  0  V  d 

∴

q1 = q6 =

∴

q2 = +

ε 0 AV ε AV and q3 = − 0 d d

Similarly, VBC = V with VC > VB  ε A |q4| = |q5| = CV =  0  V  2d 

∴

q5 = +

with

ε 0 AV ε AV and q4 = − 0 2d 2d

Type 13. To find total electrostatic potential energy due to spherical charged shells

Concept (i) Capacity of a spherical capacitor is given by 4πε 0 C= 1 1 − a b

a

b

(ii) U =

1 q2 2 C

294 — Electricity and Magnetism V

Example 18 In the figure shown, q 2q 4q 12 34 5 6

R

3R 2R

(a) Find q1 to q6. (b) Total electrostatic potential energy. Solution (a) q1 = 0 q2 = 4q q3 = − q2 = − 4q q4 = 2q − q3 = 6q q5 = − q4 = − 6q q6 = q − q5 = 7q (b) U Total = U 1 + U 2 + U 3

U1 U2

U3

4q –4q 6q –6q 7q

Here, where,

where,

and where,

U1 =

1 (4q)2 2 C1

4πε 0 1 1 − R 2R 1 (6q)2 U2 = 2 C2 C1 =

4πε 0 1 1 − 2R 3R 1 (7q)2 U3 = 2 C3 C2 =

C3 =

4πε 0 = 4πε 0 ( 3R) 1 1 − 3R ∞

Miscellaneous Examples V

Example 19 In the circuit shown in figure switch S is closed at time t = 0. Find the current through different wires and charge stored on the capacitor at any time t. 6R S 3R

R V C

Solution Calculation of τC Equivalent resistance across capacitor after short-circuiting the battery is 6R R

Rnet = R +

3R

(6R) (3R) = 3R 6R + 3R

∴ τC = (C )Rnet = 3RC Calculation of steady state charge q0 At t = ∞, capacitor is fully charged and no current flows through it. 6R

R +

V

– i=

q0

V 9R

PD across capacitor = PD across 3R V  =   (3R)  9 R V 3 CV q0 = 3 =

∴

3R

296 — Electricity and Magnetism Now, let charge on the capacitor at any time t be q and current through it is i1. Then, q = q0 (1 – e– t/ τC ) dq q0 – t/ τC i1 = = e dt τC

and

…(i)

Applying Kirchhoff’s second law in loop ACDFA, we have 6R

A

i2

B

C

i

3R

R + V

q

– i1 F

D

E

– 6iR – 3i 2R + V = 0 V or 2i + i 2 = 3R Applying Kirchhoff’s junction law at B, we have

…(ii)

i = i1 + i 2 Solving Eqs. (i), (ii) and (iii), we have V 2 V 2q0 – t/ τC V q and i = i2 = – i1 = – e + 0 e– t/ τC 9R 3 9R 3τC 9R 3τC V

…(iii)

Example 20 In the circuit shown in figure, find the steady state charges on both the capacitors. 10 V 2 Ω A

H 3Ω

B 3 µF

G 6 µF

4Ω C

D

F

20 V 6 Ω

E

In steady state a capacitor offers an infinite resistance. Therefore, the two circuits ABGHA and CDEFC have no relation with each other. Hence, the battery of emf 10 V is not going to contribute any current in the lower circuit. Similarly, the battery of emf 20 V will not contribute to the current in the upper circuit. So, first we will calculate the current in the two circuits, then find the potential difference VBG and VCF and finally we can connect two batteries of emf VBG and VCF across the capacitors to find the charges stored in them.

HOW TO PROCEED

Solution Current in the upper circuit, i1 = ∴

10 =2 A 3+2

VBG = VB – VG = 3i1 = 3 × 2 = 6 V

Capacitors — 297

Chapter 25 Current in the lower circuit,

i2 =

20 =2 A 4+6

∴ VCF = VC – VF = 4i 2 = 4 × 2 = 8 V Charge on both the capacitors will be same. Let it be q. Applying Kirchhoff’s second law in loop BGFCB, 10 V

2Ω

A

6V

H i1

B

B

G

3 µF

6 µF

4Ω

C

+

⇒

q

–

3 µF

+

q

–

6 µF

F i2

C

D 20 V

8V

F

E

6Ω

–6–

q q +8– =0 6 × 10–6 3 × 10–6 (106 ) q =2 2

or

q = 4 × 10–6 C q = 4 µC

or or V

G

3Ω

Ans.

Example 21 An isolated parallel plate capacitor has circular plates of radius 4.0 cm. If the gap is filled with a partially conducting material of dielectric constant K and conductivity 5.0 × 10 –14 Ω –1 m –1 . When the capacitor is charged to a surface charge density of 15 µC / cm2 , the initial current between the plates is 1.0 µA? (a) Determine the value of dielectric constant K. (b) If the total joule heating produced is 7500 J, determine the separation of the capacitor plates. Solution (a) This is basically a problem of discharging of a capacitor from inside the capacitor. Charge at any time t is q = q0e– t/ τC Here, q0 = (area of plates) (surface charge density)  – dq q0 – t/ τC and discharging current, i= ⋅e = i 0e– t/ τC  =  dt  τC Here,

i0 =

Kε 0 A d Kε 0 CR = σ C=

∴

q0 q = 0 τC CR and R =

d σA

298 — Electricity and Magnetism i0 =

Therefore,

q0 σq0 = Kε 0 Kε 0 σ

⇒

K =

σq0 i 0ε 0

Substituting the values, we have (5.0 × 10–14 ) (π ) (4.0)2 (15 × 10–6 ) = 4.25 K = (1.0 × 10–6 ) (8.86 × 10–12)

Ans.

1 q02 1 q02 = 2 C 2 Kε 0 A d 2Kε 0 AU d= q02

U =

(b)

∴

=

2 × 4.25 × 8.86 × 10–12 × π × (4.0 × 10–2)2 × 7500 (15 × 10–6 × π × 4.0 × 4.0)2

= 5.0 × 10–3 m = 5.0 mm V

Ans.

Example 22 Three concentric conducting shells A, B and C of radii a, b and c are as shown in figure. A dielectric of dielectric constant K is filled between A and B. Find the capacitance between A and C.

C B A a b

When the dielectric is filled between A and B, the c electric field will change in this region. Therefore, the potential difference and hence the capacitance of the system will change. So, first find the electric field E(r ) in the region a ≤ r ≤ c. Then, find the PD (V ) between A and C and finally the capacitance of the system will be q C= V

HOW TO PROCEED

Here, q = charge on A E (r ) =

Solution

= Using, dV = –

q 4πε 0Kr 2 q 4πε 0r 2

a≤r≤b

for for

b≤r≤c

∫ E ⋅ dr

the PD between A and C is ∴

V = V A – VC = –

b

q

c

q

∫a 4πε 0Kr 2 ⋅ dr – ∫b 4πε 0r 2 dr

 1  1 1   1 1  q  (b – a ) (c – b)   K  a – b +  b – c  = 4πε  Kab + bc     0  q = [c (b – a ) + Ka (c – b)] 4πε 0Kabc =

q 4πε 0

∴ The desired capacitance is C=

q 4πε 0Kabc = V Ka (c – b) + c (b – a )

Ans.

Exercises LEVEL 1 Assertion and Reason Directions :

Choose the correct option.

(a) If both Assertion and Reason are true and the Reason is correct explanation of the Assertion. (b) If both Assertion and Reason are true but Reason is not the correct explanation of Assertion. (c) If Assertion is true, but the Reason is false. (d) If Assertion is false but the Reason is true. q . We can say that, if more charge q is given to a V conductor, its capacitance should increase. q Reason : Ratio will remain constant for a given conductor. V

1. Assertion : From the relation C =

2. Assertion : A parallel plate capacitor is first charged and then distance between the plates is increased. In this process, electric field between the plates remains the same, while potential difference gets decreased. q q Reason : E = and V = . Since, q remains same, E will remain same while V will A ε0 Ad ε 0 decrease.

3. Assertion : When an uncharged capacitor is charged by a battery, only 50% of the energy supplied by a battery is stored in the capacitor. Reason : Rest 50% is lost.

4. Assertion : Discharging graphs of two C-R circuits having the same value of C is shown in figure. From the graph we can say that τC1 > τC 2 . q

1 2 t

Reason :

R1 > R2.

5. Assertion : In series combination, charges on two capacitors are always equal. Reason : If charges are same, the total potential difference applied across two capacitors will be distributed in inverse ratio of capacities.

300 — Electricity and Magnetism 6. Assertion : Two capacitors are charged from the same battery and then connected as shown. A current will flow in anti-clockwise direction as soon as switch is closed. +

–

1 µF S +

– 2 µF

Reason :

In steady state charges on two capacitors are in the ratio 1 : 2..

7. Assertion : In the circuit shown in figure no charge will be stored in the capacitor.

R2 R1

Reason :

Current through R2 will be zero.

8. Assertion : In the circuit shown in figure, time constant of charging of capacitor is

CR . 2

R R

E C

Reason : battery.

In the absence of capacitor in the circuit, two resistors are in parallel with the

9. Assertion : Two capacitors are connected in series with a battery. Energy stored across them is in inverse ratio of their capacity. 1 Reason : U = qV or U ∝ qV . 2

10. Assertion : In the circuit shown in figure, when a dielectric slab is inserted in C2, the potential difference across C2 will decrease. C1

Reason :

C2

By inserting the slab a current will flow in the circuit in clockwise direction.

Chapter 25

Capacitors — 301

Objective Questions 1. The separation between the plates of a charged parallel- plate capacitor is increased. The force between the plates (b) decreases (d) first increases then decreases

(a) increases (c) remains same

2. If the plates of a capacitor are joined together by a conducting wire, then its capacitance (a) remains unchanged (c) becomes zero

(b) decreases (d) becomes infinite

3. Two metal spheres of radii a and b are connected by a thin wire. Their separation is very large compared to their dimensions. The capacitance of this system is (a) 4πε 0 (ab)

(b) 2πε 0 (a + b)  a 2 + b2 (d) 4πε 0    2 

(c) 4πε 0 (a + b)

4. n identical capacitors are connected in parallel to a potential difference V . These capacitors are then reconnected in series, their charges being left undisturbed. The potential difference obtained is (a) zero

(b) (n − 1) V

(d) n 2V

(c) nV

5. In the circuit shown in figure, the ratio of charge on 5 µF and 2 µF capacitor is 3 µF

2 µF

5 µF 6V

(a) 5 / 4

(b) 5 / 3

(c) 3 / 8

(d) None of these

6. In the circuit shown, a potential difference of 60 V is applied across AB. The potential difference between the points M and N is 2C M

A

60 V

C

C

N

B 2C

(a) 10 V

(b) 15 V

(c) 20 V

(d) 30 V

7. In Milikan’s oil drop experiment, an oil drop of radius r and charge q is held in equilibrium between the plates of a charged parallel-plate capacitor when the potential difference is V . To keep a drop of radius 2r and with a charge 2q in equilibrium between the plates the potential difference V required is (a) V

(b) 2 V

(c) 4 V

(d) 8 V

302 — Electricity and Magnetism σ

8. Two large parallel sheets charged uniformly with surface charge

−σ

density σ and − σ are located as shown in the figure. Which one of the following graphs shows the variation of electric field along a A line perpendicular to the sheets as one moves from A to B ?

B

E

(a)

(b) x E

(c)

(d)

x

9. When the switch is closed, the initial current through the 1 Ω resistor is 1Ω

6Ω

12 V

2 µF 3Ω

S

(a) 2 A

(b) 4 A

(c) 3 A

(d) 6 A

10. A capacitor of capacitance C carrying charge Q is connected to a source of emf E. Finally, the charge on capacitor would be (a) Q

(b) Q + CE

(c) CE

(d) None of these

11. In the circuit, the potential difference across the capacitor is 10 V. Each resistance is of 3 Ω. The cell is ideal. The emf of the cell is C = 3 µF

R R

R

R R

R

E

(a) 14 V

(b) 16 V

(c) 18 V

(d) 24 V

12. Four identical capacitors are connected in series with a 10 V

+ –

battery as shown in the figure. The point N is earthed. The potentials of points A and B are

10 V

(a) (b) (c) (d)

10 V, 0 V 7.5 V, – 2.5 V 5 V, – 5 V 7.5 V, 2.5 V

C A

C

C

N

C B

Chapter 25

Capacitors — 303

13. A capacitor of capacity 2 µF is charged to 100 V. What is the heat generated when this capacitor is connected in parallel to an another capacitor of same capacity? (a) 2.5 mJ (c) 10 mJ

(b) 5.0 mJ (d) 4 mJ

14. A charged capacitor is discharged through a resistance. The time constant of the circuit is η . Then, the value of time constant for the power dissipated through the resistance will be (a) η (c) η/2

(b) 2η (d) zero

15. A capacitor is charged by a cell of emf E and the charging battery is then removed. If an identical capacitor is now inserted in the circuit in parallel with the previous capacitor, the potential difference across the new capacitor is (a) 2E (c) E /2

(b) E (d) zero

16. The potential difference V A − V B between points A and B for the circuit segment shown in figure at the given instant is 3A

9 µC 6 V 2Ω – +

3Ω

– + 1 µF

A

(a) 12 V (c) 6 V

B

(b) – 12 V (d) – 6 V

17. For the circuit arrangement shown in figure, in the steady state condition charge on the capacitor is 2Ω

12 V 2 µF

4Ω

6Ω

(b) 14 µC (d) 18 µC

(a) 12 µC (c) 20 µC

18. In the circuit as shown in figure if all the symbols have their usual meanings, then identify the correct statement, C2 + – q2 V2

C1 + – q1 V1

C3 + q3

– V3

V

(a) q2 = q3 ; V 2 = V3 (c) q1 = q2 + q3 ; V = V1 + V 2 + V3

(b) q1 = q2 + q3 ; V 2 = V3 (d) q1 + q2 + q3 = 0 ; V 2 = V3 = V − V1

304 — Electricity and Magnetism –

19. An electron enters the region between the plates of a parallel-plate

capacitor at an angle θ to the plates. The plate width is l. The plate separation is d. The electron follows the path shown, just missing the d upper plate. Neglect gravity. Then,

(a) (b) (c) (d)

tan θ = 2d /l tan θ = 4d /l tan θ = 8d /l The data given is insufficient to find a relation between d , l and θ

θ l

+

20. An infinite sheet of charge has a surface charge density of 10−7 C/ m 2. The separation between two equipotential surfaces whose potentials differ by 5 V is (a) 0.64 cm (c) 0.32 cm

(b) 0.88 mm (d) 5 × 10−7 m

21. Find the equivalent capacitance across A and B for the arrangement shown in figure. All the capacitors are of capacitance C A

B

3C 14 3C (c) 16 (a)

(b)

C 8

(d) None of these

22. The equivalent capacitance between X and Y is 1 µF

1 µF

1 µF 2 µF

X

(a) 5 / 6 µF (c) 8 / 3 µF

Y

(b) 7 / 6 µF (d) 1 µF

23. In the arrangement shown in figure, dielectric constant K 1 = 2 and K 2 = 3. If the capacitance across P and Q are C1 and C2 respectively, then C1 / C2 will be (the gaps shown are negligible) P A/2

K1

K2

Q

(a) 1 : 1 (c) 9 : 5

P

A/2

A/2

A/2 K1

d/2

K2

d/2

Q

(b) 2 : 3 (d) 25 : 24

Chapter 25

Capacitors — 305

24. Six equal capacitors each of capacitance C are connected as shown in the figure. The equivalent capacitance between points A and B, is (a) (b) (c) (d)

1.5 C C 2C 0.5 C

C A C C

C C

C

B

25. Four ways of making a network of five capacitors of the same value are shown in four choices. Three out of four are identical. The one which is different is

(a)

A

A

B

B

(b)

A

B

A

(c)

(d)

B

26. The equivalent capacitance of the arrangement shown in figure, if A is the area of each plate, is

d

K2

d/2

K3

d/2

K1

(a) C =

ε0 A  K1 K 2 + K3  + d  2 K 2K 3 

(b) C =

ε0 A  K1 K 2K 3  + d  2 K 2 + K 3 

(c) C =

ε0 A  K 2K 3  K1 + 2d  K 2 + K 3 

(d) C =

ε0 A  K 2K 3  K1 + d  K 2 + K 3 

27. Find equivalent capacitance between points A and B. [Assume each conducting plate is having same dimensions and neglect the thickness of the plate,

ε0 A = 7 µF, where A is area of plates] d

d d d 2d B

(a) 7 µF (c) 12 µF

d

(b) 11 µF (d) 15 µF

A

306 — Electricity and Magnetism Subjective Questions Note You can take approximations in the answers.

1. Two metallic plates are kept parallel to one another and charges are given to them as shown in figure. Find the charge on all the four faces.

– 4 µC

10 µC

2. Charges 2 q and – 3 q are given to two identical metal plates of area of cross-section A. The distance between the plates is d. Find the capacitance and potential difference between the plates.

2q

–3q

3. Find the charge stored in all the capacitors.

1 µF

2 µF

10 V

4. Find the charge stored in the capacitor. 4 µF 6Ω

10 V

5. Find the charge stored in the capacitor. 2 µF

2Ω

4Ω 6Ω 30 V

3 µF

Capacitors — 307

Chapter 25

6. A 1 µF capacitor and a 2 µF capacitor are connected in series across a 1200 V supply line. (a) Find the charge on each capacitor and the voltage across them. (b) The charged capacitors are disconnected from the line and from each other and reconnected with terminals of like sign together. Find the final charge on each and the voltage across them.

7. A 100 µF capacitor is charged to 100 V. After the charging, battery is disconnected. The capacitor is then connected in parallel to another capacitor. The final voltage is 20 V. Calculate the capacity of second capacitor.

8. An uncharged capacitor C is connected to a battery through a resistance R. Show that by the time the capacitor gets fully charged, the energy dissipated in R is the same as the energy stored in C.

9. How many time constants will elapse before the current in a charging R-C circuit drops to half of its initial value?

10. A capacitor of capacitance C is given a charge q0. At time t = 0 it is connected to an uncharged capacitor of equal capacitance through a resistance R. Find the charge on the first capacitor and the second capacitor as a function of time t. Also plot the corresponding q-t graphs.

11. A capacitor of capacitance C is given a charge q0. At time t = 0, it is connected to a battery of emf E through a resistance R. Find the charge on the capacitor at time t.

12. Determine the current through the battery in the circuit shown in figure. E

S C1 R2

C2

R3

R1

(a) immediately after the switch S is closed (b) after a long time.

13. For the circuit shown in figure, find (a) (b) (c) (d)

the initial current through each resistor steady state current through each resistor final energy stored in the capacitor time constant of the circuit when switch is opened.

R1

E

R2 C

14. Find equivalent capacitance between points A and B, B A A

C

2C

2C

C

C

2C

2C

C C

C

A

C

C

C

4

B

C C B

(a)

(b)

(c)

308 — Electricity and Magnetism 15. A 4.00 µF capacitor and a 6.00 µF capacitor are connected in parallel across a 660 V supply line. (a) Find the charge on each capacitor and the voltage across each. (b) The charged capacitors are disconnected from the line and from each other, and then reconnected to each other with terminals of unlike sign together. Find the final charge on each and the voltage across each.

16. A 5.80 µF parallel-plate air capacitor has a plate separation of 5.00 mm and is charged to a potential difference of 400 V. Calculate the energy density in the region between the plates, in J/ m3 .

17. The dielectric to be used in a parallel-plate capacitor has a dielectric constant of 3.60 and a

dielectric strength of 1.60 × 107 V/ m. The capacitor is to have a capacitance of 1.25 × 10−9 F and must be able to withstand a maximum potential difference of 5500 V. What is the minimum area the plates of the capacitor may have?

18. Two condensers are in parallel and the energy of the combination is 0.1 J, when the difference of potential between terminals is 2 V. With the same two condensers in series, the energy is 1.6 × 10−2 J for the same difference of potential across the series combination. What are the capacities?

19. A circuit has section AB as shown in figure. The emf of the source equals E = 10 V, the capacitor capacitances are equal to C1 = 1.0 µF and C2 = 2.0 µF, and the potential difference V A − V B = 5.0 V. Find the voltage across each capacitor. B

A C1

E

C2

20. Several 10 pF capacitors are given, each capable of withstanding 100 V. How would you construct : (a) a unit possessing a capacitance of 2 pF and capable of withstanding 500 V? (b) a unit possessing a capacitance of 20 pF and capable of withstanding 300 V?

21. Two, capacitors A and B are connected in series across a 100 V supply and it is observed that the potential difference across them are 60 V and 40 V. A capacitor of 2 µF capacitance is now connected in parallel with A and the potential difference across B rises to 90 V. Determine the capacitance of A and B.

22. A 10.0 µF parallel-plate capacitor with circular plates is connected to a 12.0 V battery. (a) What is the charge on each plate? (b) How much charge would be on the plates if their separation were doubled while the capacitor remained connected to the battery? (c) How much charge would be on the plates if the capacitor were connected to the 12.0 V battery after the radius of each plate was doubled without changing their separation?

23. A 450 µF capacitor is charged to 295 V. Then, a wire is connected between the plates. How many joule of thermal energy are produced as the capacitor discharges if all of the energy that was stored goes into heating the wire?

24. The plates of a parallel-plate capacitor in vacuum are 5.00 mm apart and 2.00 m 2 in area. A potential difference of 10,000 V is applied across the capacitor. Compute (a) the capacitance (b) the charge on each plate, and (c) the magnitude of the electric field in the space between them.

Chapter 25

Capacitors — 309

25. Three capacitors having capacitances of 8.4 µF , 8.2 µF and 4.2 µF are connected in series across a 36 V potential difference. (a) What is the charge on 4.2 µF capacitor? (b) What is the total energy stored in all three capacitors? (c) The capacitors are disconnected from the potential difference without allowing them to discharge. They are then reconnected in parallel with each other, with the positively charged plates connected together. What is the voltage across each capacitor in the parallel combination? (d) What is the total energy now stored in the capacitors?

26. Find the charges on 6 µF and 4 µF capacitors. 3 µF

5V

2 µF

6 µF

5V

4 µF

10 V

27. In figure, C1 = C5 = 8.4 µF and C2 = C3 = C4 = 4.2 µF. The applied potential is Vab = 220 V. C3

C1 a C2 b C5

C4

(a) What is the equivalent capacitance of the network between points a and b? (b) Calculate the charge on each capacitor and the potential difference across each capacitor.

28. Two condensers A and B each having slabs of dielectric constant K = 2 are connected in series. When they are connected across 230 V supply, potential difference across A is 130 V and that across B is 100 V. If the dielectric in the condenser of smaller capacitance is replaced by one for which K = 5, what will be the values of potential difference across them?

29. A capacitor of capacitance C1 = 1.0 µF charged upto a voltage V = 110 V is connected in parallel to the terminals of a circuit consisting of two uncharged capacitors connected in series and possessing the capacitance C2 = 2.0 µF and C3 = 3.0 µF. What charge will flow through the connecting wires?

30. In figure, the battery has a potential difference of 20 V. Find

C2 = 2 µF

2 µF 20 V

4 µF C3 = 4 µF

+ – 3 µF

C 1 = 3 µF

310 — Electricity and Magnetism (a) the equivalent capacitance of all the capacitors across the battery and (b) the charge stored on that equivalent capacitance. Find the charge on (c) capacitor 1, (d) capacitor 2, and (e) capacitor 3.

31. In figure, battery B supplies 12 V. Find the charge on each capacitor C1

C3

+

S2

C2 –

C4 S1

B

(a) first when only switch S1 is closed and (b) later when S 2 is also closed. (Take C1 = 1.0 µF, C 2 = 2.0 µF, C3 = 3.0 µF and C 4 = 4.0 µF )

32. When switch S is thrown to the left in figure, the plates of capacitor 1 acquire a potential difference V 0. Capacitors 2 and 3 are initially uncharged. The switch is now thrown to the right. What are the final charges q1 , q2 and q3 on the capacitors? C2

+

S V0

–

C3

C1

33. A parallel-plate capacitor has plates of area A and separation d and is charged to a potential difference V. The charging battery is then disconnected, and the plates are pulled apart until their separation is 2d. Derive expression in terms of A, d and V for (a) the new potential difference (b) the initial and final stored energies, Ui and U f and (c) the work required to increase the separation of plates from d to 2d.

34. In the circuit shown in figure E1 = 2E2 = 20 V , R1 = R2 = 10 k Ω and C = 1 µF. Find the current through R1 , R2 and C when

R1 S A

E1

B R2 E2

(a) S has been kept connected to A for a long time. (b) The switch is suddenly shifted to B.

C

Chapter 25

Capacitors — 311

35. (a) What is the steady state potential of point a with respect to point b in figure when switch S is open? V = 18.0 V 6.00 Ω

6.00 µF

a

b S

3.00 µF

3.00 Ω

(b) Which point, a or b, is at the higher potential? (c) What is the final potential of point b with respect to ground when switch S is closed? (d) How much does the charge on each capacitor change when S is closed?

36. (a) What is the potential of point a with respect to point b in figure, when switch S is open? V = 18.0 V 6.00 Ω

6.00 µF

a

b

S

3.00 µF

3.00 Ω

(b) Which point, a or b, is at the higher potential? (c) What is the final potential of point b with respect to ground when switch S is closed? (d) How much charge flows through switch S when it is closed?

37. In the circuit shown in figure, the battery is an ideal one with emf V. The capacitor is initially uncharged. The switch S is closed at time t = 0.

R

A

R

V

R

C

B

(a) Find the charge Q on the capacitor at time t. (b) Find the current in AB at time t. What is its limiting value as t → ∞?

LEVEL 2 Single Correct Option 1. Two

very large thin conducting plates having same Y cross-sectional area are placed as shown in figure. They are carrying charges Q and 3 Q, respectively. The variation of electric field as a function at x (for x = 0 to x = 3d) will be best represented by E

Q

E

(a)

3Q

(d, 0) (2d, 0)

(3d, 0)

X

(b) 2d

d

3d

d

X

2d

3d

X

2d

3d

X

E

E

(c)

(d) 3d

X

2d d

d

2. The electric field on two sides of a thin sheet of charge is shown in the figure. The charge density on the sheet is E1 = 8 V/m

(a) 2 ε 0 (c) 10 ε 0

+ + + + + + + +

E2 = 12 V/m

(b) 4 ε 0 (d) zero

3. In the circuit shown in figure, the capacitors are initially

10 V, 2 Ω

uncharged. The current through resistor PQ just after closing the switch is (a) (b) (c) (d)

2 A from P to Q 2 A from Q to P 6 A from P to Q zero

2 µF P 6Ω

4 µF

5Ω Q

6Ω

Chapter 25

Capacitors — 313

4. A graph between current and time during charging of a capacitor by a

Log l

battery in series with a resistor is shown. The graphs are drawn for two circuits. R1, R2 , C1 , C2 and V1 , V 2 are the values of resistance, capacitance and EMF of the cell in the two circuits. If only two parameters (out of resistance, capacitance, EMF) are different in the two circuits. What may be the correct option(s)? (a) V1 = V 2, R1 > R2, C1 > C 2

(b) V1 > V 2, R1 > R2, C1 = C 2

(c) V1 < V 2, R1 < R2, C1 = C 2

(d) V1 < V 2, R1 = R2, C1 < C 2

2 1 t

5. A capacitor of capacitance C is charged by a battery of emf E and internal resistance r. A resistance 2r is also connected in series with the capacitor. The amount of heat liberated inside the battery by the time capacitor is 50% charged is 3 2 E C 8 E 2C (c) 12 (a)

E 2C 6 E 2C (d) 24

(b)

3V

6. For the circuit shown in the figure, determine the

3V

charge on capacitor in steady state. (a) (b) (c) (d)

4 µC 6 µC 1 µC Zero

10 V 1Ω

1Ω

5Ω

2Ω 2Ω

1 µF

6V

1Ω

7. For the circuit shown in the figure, find the charge stored on capacitor in steady state. RC E R + R0 RC (b) (E − E 0 ) R0

(a)

(c) zero RC (d) (E − E 0 ) R + R0

E

R

E

C

E0

Rn

8. Two similar parallel-plate capacitors each of capacity C0 are connected in series. The combination is connected with a voltage source of V 0. Now, separation between the plates of one capacitor is increased by a distance d and the separation between the plates of another capacitor is decreased by the distance d/ 2. The distance between the plates of each capacitor was d before the change in separation. Then, select the correct choice. (a) (b) (c) (d)

The new capacity of the system will increase The new capacity of the system will decrease The new capacity of the system will remain same data insufficient

9. The switch shown in the figure is closed at t = 0. The charge on the

C

capacitor as a function of time is given by (a) (b) (c) (d)

CV (1 − e− t/RC ) 3 CV (1 − e−t/RC ) CV (1 − e−3 t/RC ) CV (1 − e− t/3RC )

k R

R

R

V

314 — Electricity and Magnetism 10. A 2 µF capacitor C1 is charged to a voltage 100 V and a 4 µF capacitor C2 is charged to a voltage 50 V. The capacitors are then connected in parallel. What is the loss of energy due to parallel connection? (b) 0.17 J (d) 1.7 × 10−3 J

(a) 1.7 J (c) 1.7 × 10−2 J

11. The figure shows a graph of the current in a discharging circuit of a capacitor through a resistor of resistance 10 Ω.

i(A) 10

(a) The initial potential difference across the capacitor is 100 V 1 (b) The capacitance of the capacitor is F 10 ln 2 500 (c) The total heat produced in the circuit will be J ln 2

2.5 2

t(s)

(d) All of the above

12. Four capacitors are connected in series with a battery of emf 10 V as shown in the figure. The point P is earthed. The potential of point A is equal in magnitude to potential of point B but opposite in sign if 10 V C1

C2

C3

P

C4

A

(a) C1 + C 2 + C3 = C 4 (c)

C1C 2C3 = C4 C12 + C 22 + C32

B

(b)

1 1 1 1 + + = C1 C 2 C3 C 4

(d) It is never possible

13. A capacitor of capacity C is charged to a potential difference V and another capacitor of capacity 2C is charged to a potential difference 4 V . The charging batteries are disconnected and the two capacitors are connected with reverse polarity (i.e. positive plate of first capacitor is connected to negative plate of second capacitor). The heat produced during the redistribution of charge between the capacitors will be (a)

125 CV 2 3

(c) 2 CV 2

50 CV 2 3 25 CV 2 (d) 3

(b)

14. A capacitor of capacitance 2 µF is charged to a potential difference of 5 V.

2 µF

Now, the charging battery is disconnected and the capacitor is connected in parallel to a resistor of 5 Ω and another unknown resistor of resistance R as shown in figure. If the total heat produced in 5 Ω resistance is 10 µJ , then the unknown resistance R is equal to

5Ω

(a) 10 Ω (b) 15 Ω 10 (c) Ω 3 (d) 7.5 Ω

R

Chapter 25

Capacitors — 315

15. In the circuit shown in figure switch S is thrown to position 1 at t = 0. When the current in the resistor is 1 A, it is shifted to position 2. The total heat generated in the circuit after shifting to position 2 is 10 V 1

5Ω

S 2 µF

2

5V

(a) zero

(b) 625 µJ

(c) 100 µJ

(d) None of these

16. The flow of charge through switch S if it is closed is 2 µF + – q

(a) zero

q

q

+ – 6 µF

+ – 3 µF

S

(c) 2q/3

(b) q/4

(d) q/3

17. Consider the arrangement of three plates X , Y and Z each of the area A and separation d. The energy stored when the plates are fully charged is X d

Y

V

d

Z

(b) ε 0 AV 2 /d (d) 3ε 0 AV 2 /d

(a) ε 0 AV 2 /2d (c) 2ε 0 AV 2 /d

18. Consider a capacitor – charging circuit. Let Q1 be the charge given to the capacitor in time interval of 20 ms and Q2 be the charge given in the next time interval of 20 ms. Let 10 µCcharge be deposited in a time interval t1 and the next 10 µCcharge is deposited in the next time interval t2. Then,

(a) Q1 > Q2, t1 > t2 (c) Q1 < Q2, t1 > t2

(b) Q1 > Q2, t1 < t2 (d) Q1 < Q2, t1 < t2

19. The current in 1 Ω resistance and charge stored in the capacitor are 1Ω

5V

2Ω

2V

4Ω

2 µF

6V

3Ω

(a) 4 A , 6 µC

(b) 7 A, 12 µC

(c) 4 A , 12 µC

(d) 7 A, 6 µC

316 — Electricity and Magnetism 20. A capacitor C is connected to two equal resistances as shown in the

S

R

figure. Consider the following statements. (i) At the time of charging of capacitor time constant of the circuit is R E 2 CR (ii)At the time of discharging of the capacitor the time constant of the circuit is CR (iii)At the time of discharging of the capacitor the time constant of the circuit is 2 CR (iv) At the time of charging of capacitor the time constant of the circuit is CR (a) Statements (i) and (ii) only are correct (b) Statements (ii) and (iii) only are correct (c) Statements (iii) and (iv) only are correct (d) Statements (i) and (iii) only are correct

C

21. Two capacitors C1 = 1 µF and C2 = 3 µF each are charged to a potential difference of 100 V but with opposite polarity as shown in the figure. When the switch S is closed, the new potential difference between the points a and b is a

S

+ –

– C1

C2

+

b

(a) 200 V

(b) 100 V

(c) 50 V

(d) 25 V

22. Four capacitors are connected as shown in figure to a 30 V battery. The potential difference between points a and b is 1.0 µF

2.5 µF +

1.5 µF

a

b

0.5 µF –

30 V

(b) 9 V (d) 13 V

(a) 5 V (c) 10 V

23. Three uncharged capacitors of capacitance C1 , C2 and C3 are connected to one another as shown in figure. The potential at O will be +6 V C1 = 60 µF C2 = 20 µF +2 V

(a) 3 V

(b)

49 V 11

O

C3 = 30 µF +3 V

(c) 4 V

(d)

3 V 11

Chapter 25

Capacitors — 317

24. In the circuit shown in figure, the potential difference between the points A and B in the steady state is 3 µF

1 µF

B

3 µF

2 µF 1 µF

20 Ω

10 Ω

10 V A

(a) zero

(b) 6 V

(c) 4 V

(d)

10 V 3

25. Two cells, two resistors and two capacitors are connected as shown in figure. The charge on 2 µF capacitor is 3 µF

5Ω

18 V 1Ω

15 V 2Ω

4Ω

2 µF

(a) 30 µC

(b) 20 µC

(c) 25 µC

(d) 48 µC

26. In the circuit shown in figure, the capacitor is charged with a cell of 5 V. If the switch is closed at t = 0, then at t = 12 s, charge on the capacitor is 2 µF +

3 MΩ

–

S

(a) (0.37) 10 µC

(b) (0.37)2 10 µC

(c) (0.63) 10 µC

(d) (0.63)2 10 µC

27. The potential difference between points a and b of circuits shown in the figure is E1 a

C1

C2

b E2

 E + E 2 (a)  1  C2  C1 + C 2 

 E − E 2 (b)  1  C2  C1 + C 2 

 E + E 2 (c)  1  C1  C1 + C 2 

 E − E 2 (d)  1  C1  C1 + C 2 

318 — Electricity and Magnetism 28. A capacitor C1 is charged to a potential V and connected to another capacitor in series with a resistor R as shown. It is observed that heat H 1 is dissipated across resistance R, till the circuit reaches steady state. Same process is repeated using resistance of 2R. If H 2 is heat dissipated in this case, then C1 R C2

H2 =1 H1 H2 1 (c) = H1 4

H2 =4 H1 H2 (d) =2 H1

(b)

(a)

29. In the circuit diagram, the current through the battery immediately after the switch S is closed is E

S C1 C2

R1 R2 R3

(a) zero

(b)

E R1

(c)

E R1 + R2

(d)

E R2R3 R1 + R2 + R3

30. In the circuit shown, switch S is closed at t = 0. Let i1 and i2 be the current at any finite time t, then the ratio i1 / i2

i1

3C

2R

C

R

i2

V

(a) is constant (c) decreases with time

S

(b) increases with time (d) first increases and then decreases

31. A leaky parallel capacitor is filled completely with a material having dielectric constant K = 5 and electrical conductivity σ = 7.4 × 10−12 Ω −1 m −1. Charge on the plate at instant t = 0 is q = 8.885 µC. Then, time constant of leaky capacitor is (a) 3 s (c) 5 s

(b) 4 s (d) 6 s

Chapter 25

Capacitors — 319

32. A charged capacitor is allowed to discharge through a resistor by closing the key at the instant

t = 0. At the instant t = (ln 4) µs, the reading of the ammeter falls half the initial value. The resistance of the ammeter is equal to C = 0.5 F +

–

K A

2Ω

(a) 0.5 Ω (c) 2 Ω

(b) 1 Ω (d) 4 Ω

33. Five identical capacitor plates are arranged such that they make four capacitors each of 2 µF. The plates are connected to a source of emf 10 V. The charge on plate C is 10 V

A B C D E

(a) + 20 µC (c) + 60 µC

(b) + 40 µC (d) + 80 µC

34. A capacitor of capacitance C is charged to a potential difference V from a cell and then disconnected from it. A charge + Q is now given to its positive plate. The potential difference across the capacitor is now Q C Q (d) V − , if Q < CV C

(b) V +

(a) V (c) V +

Q 2C

More than One Correct Options 1. X and Y are large, parallel conducting plates close to each other. Each face has an area A. X is given a charge Q. Y is without any charge. Points A, B and C are as shown in the figure. X

A

Y

B

C

Q 2 ε0 A Q (b) The field at B is ε0 A

(a) The field at B is

(c) The fields at A , B and C are of the same magnitude (d) The fields at A and C are of the same magnitude, but in opposite directions

320 — Electricity and Magnetism 2. In the circuit shown in the figure, switch S is closed at time t = 0. Select the correct statements. 2R

C

R

2C

S

E

(a) (b) (c) (d)

Rate of increase of charge is same in both the capacitors Ratio of charge stored in capacitors C and 2C at any time t would be 1 : 2 Time constants of both the capacitors are equal Steady state charges on capacitors C and 2C are in the ratio of 1 : 2

3. An electrical circuit is shown in the given figure. The resistance of

each voltmeter is infinite and each ammeter is 100 Ω. The charge on the capacitor of 100 µF in steady state is 4 mC. Choose correct statement(s) regarding the given circuit.

(a) (b) (c) (d)

Reading of voltmeter V 2 is 16 V Reading of ammeter A1 is zero and A2 is 1/25 A Reading of voltmeter V1 is 40 V Emf of the ideal cell is 66 V

V2 200 Ω V1

900 Ω

C

100 Ω

A2 A1

4. In the circuit shown, A and B are equal resistances. When S is closed, the capacitor C charges from the cell of emf ε and reaches a steady state. C

B S −

+

A

ε

(a) During charging, more heat is produced in A than in B (b) In steady state, heat is produced at the same rate in A and B 1 (c) In the steady state, energy stored in C is Cε 2 4 1 2 (d) In the steady state energy stored in C is Cε 8

5. A parallel-plate capacitor is charged from a cell and then isolated from it. The separation between the plates is now increased (a) (b) (c) (d)

The force of attraction between the plates will decrease The field in the region between the plates will not change The energy stored in the capacitor will increase The potential difference between the plates will decrease

Chapter 25

Capacitors — 321

6. In the circuit shown, each capacitor has a capacitance C. The emf of the

S

C

cell is E. If the switch S is closed, then (a) positive charge will flow out of the positive terminal of the cell (b) positive charge will enter the positive terminal of the cell 1 (c) the amount of the charge flowing through the cell will be CE 3  4 (d) the amount of charge flowing through the cell is   CE  3

C C –

+

E

7. Two capacitors of 2 µF and 3 µF are charged to 150 V and 120 V,

1.5 µF

A

respectively. The plates of capacitor are connected as shown in the figure. An uncharged capacitor of capacity 1.5 µF falls to the free end of the wire. Then, (a) (b) (c) (d)

+

charge on 1.5 µF capacitor is 180 µC charge on 2 µF capacitor is 120 µC positive charge flows through A from right to left positive charge flows through A from left to right

–

+

2 µF

–

3 µF

8. A parallel plate capacitor is charged and then the battery is disconnected. When the plates of the capacitor are brought closer, then (a) (b) (c) (d)

energy stored in the capacitor decreases the potential difference between the plates decreases the capacitance increases the electric field between the plates decreases

9. A capacitor of 2 F (practically not possible to have a capacity of 2 F) is

–

charged by a battery of 6 V. The battery is removed and circuit is made as shown. Switch is closed at time t = 0. Choose the correct options. (a) At time t = 0 current in the circuit is 2 A (b) At time t = ( 6 ln 2) second, potential difference across capacitor is 3 V (c) At time t = ( 6 ln 2) second, potential difference across 1 Ω resistance is 1V (d) At time t = ( 6 ln 2) second, potential difference across 2 Ω resistance is 2 V.

10. Given that potential difference across 1 µF capacitor is 10 V. Then, 6 µF

1 µF

4 µF 3 µF

E

(a) (b) (c) (d)

potential difference across 4 µF capacitor is 40 V potential difference across 4 µF capacitor is 2.5 V potential difference across 3 µF capacitor is 5 V value of E is 50 V

6V

+

2F

1Ω

S 2Ω

322 — Electricity and Magnetism Comprehension Based Questions Passage I (Q. No. 1 and 2) The capacitor C1 in the figure shown initially carries a charge q0. When the switches S1 and S 2 are closed, capacitor C1 is connected in series to a resistor R and a second capacitor C2, which is initially uncharged. S1

C1

+ q – 0

R

C2 S2

1. The charge flown through wires as a function of time t is (a) q0e− t/RC +

C q0 C2

(b)

C − t/CR e C1 C1C 2 where, C = C1 + C 2

q0C × [1 − e− t/RC ] C1

(d) q0e− t/RC

(c) q0

2. The total heat dissipated in the circuit during the discharging process of C1 is q02 ×C 2 C12 q2C (c) 0 22 2 C1

q02 2C q02 (d) 2 C1C 2 (b)

(a)

Passage II (Q. No. 3 and 4) Figure shows a parallel plate capacitor with plate area A and plate separation d. A potential difference is being applied between the plates. The battery is then disconnected and a dielectric slab of dielectric constant K is placed in between the plates of the capacitor as shown. + + + + + + + + + + t d − − − − − − − − − −

Now, answer the following questions based on above information.

3. The electric field in the gaps between the plates and the dielectric slab will be (a)

ε 0 AV d

(b)

V d

(c)

KV d

(d)

V d−t

(c)

V d

(d)

KV t

4. The electric field in the dielectric slab is (a)

V Kd

(b)

KV d

Chapter 25

Capacitors — 323

Match the Columns 1. In the figure shown, C1 = 4 µF (without dielectric) and

Column I (a) (b) (c) (d)

C2

C1

C2 = 4 µF (with a dielectric slab of dielectric constant K = 2) . Now, the same slab after removing from C2 is filled in C1. Then, match the following two columns. Column II

Charge on C 2 Energy stored in C 2 Potential difference across C 2 Electric field between the plates of C 2

(p) (q) (r) (s)

will increase will decrease will remain same data insufficient

2. In the circuit shown in figure, match the following two columns for Column I (a) (b) (c) (d)

From the battery From 2 µF capacitor From 3 µF From 4 µF capacitor

4 µF

S

the flow of charge when switch is closed. Column II (p) (q) (r) (s)

3 µF

40 µC 100 µC 60 µC None of these

2 µF

30 V

3. Three identical capacitors are connected in three different configurations as shown in Column II. Points a and b are connected with a battery. Match the two columns. Column I

Column II

(a) Maximum charge on C1 (p)

a

C1

C2

C3

b

C1

(b) Minimum charge on C 2

(q)

C2

a

b

C3

C2

(c) Maximum potential difference across C1

(r)

a

C1

b

C3 C1

(d) Minimum potential difference across C1

(s)

a

C2

b

C3

324 — Electricity and Magnetism 4. A capacitor C is charged by a battery of V volts. Then, it is connected to an uncharged capacitor of capacity 2C as shown in figure. Now, match the following two columns. C +

–

S 2C

Column I

Column II

1 (a) After closing the switch energy stored (p) CV 2 9 in C. 1 (b) After closing the switch energy stored (q) CV 2 6 in 2C. 1 (c) After closing the switch loss of energy (r) CV 2 18 during redistribution of charge. (s)

None of these

5. Two identical sized capacitors C1 and C2 are connected with a battery as shown in figure. Capacitor plates are square plates. A dielectric slab of dielectric constant K = 2, is filled in half the region of the two capacitors as shown :

C1

C2

C → capacity, q → charge stored, U → energy stored. Match the following two columns. Column I

Column II

(a) C1 / C 2

(p) 9 / 4

(b) q1 / q2

(q) 4 / 9

(c) U 1 /U 2

(r) 4 / 3 (s) None of these

6. Four large parallel identical conducting plates are arranged as

4Q

2Q

Q

7Q

shown. Column I

Column II

(a) Surfaces having charges of same magnitude and sign

(p) 1 and 8

(b) Surfaces charges

(q) 3 and 5

having

positive

(1) (2) (3) (4) (5) (6) (7) (8)

d

(c) Uncharged surfaces

(r) 2 and 3

(d) Charged surfaces

(s) 6 and 7

d

d

Chapter 25

Capacitors — 325

Subjective Questions 1. Five identical conducting plates, 1, 2, 3,4 and 5 are fixed parallel plates equidistant from each other (see figure). A conductor connects plates 2 and 5 while another conductor joins 1 and 3. The junction of 1 and 3 and the plate 4 are connected to a source of constant emf V 0. Find 5 4

–

3 2 +

1

(a) the effective capacity of the system between the terminals of source. (b) the charges on the plates 3 and 5. Given, d = distance between any two successive plates and A = area of either face of each plate.

2. A 8 µF capacitor C1 is charged to V 0 = 120 V. The charging battery is then removed and the capacitor is connected in parallel to an uncharged + 4µF capacitor C2. S

C2

C1

(a) what is the potential difference V across the combination? (b) what is the stored energy before and after the switch S is closed?

3. Condensers with capacities C , 2 C , 3 C and 4C are charged to the voltage, V, 2 V, 3 V and 4 V correspondingly. The circuit is closed. Find the voltage on all condensers in the equilibrium. CV –

+

– +

4C 4V

2C 2V

+ – +

–

3C 3V

4. In the circuit shown, a time varying voltage V = 2000t volt is applied where t is in second. At time t = 5 ms, determine the current through the resistor R = 4 Ω and through the capacitor C = 300 µF. + V

R –

C

326 — Electricity and Magnetism 5. A capacitor of capacitance 5 µF is connected to a source of constant emf of 200 V. Then, the switch was shifted to contact 2 from contact 1. Find the amount of heat generated in the 400 Ω resistance. 5 µF

400 Ω 2 S 500 Ω

1

200 V

6. Analyze the given circuit in the steady state condition. Charge on the capacitor is q0 = 16 µC. 2Ω

+

1Ω

–

A

3Ω

B

C = 4 µF

3Ω C

4Ω

4Ω –

+

E

(a) Find the current in each branch (b) Find the emf of the battery. (c) If now the battery is removed and the points A and C are shorted. Find the time during which charge on the capacitor becomes 8 µC.

7. Find the potential difference between points M and N of the system shown in figure, if the emf is equal to E = 110 V and the capacitance ratio C1

C1 is 2. C2 C2

M

C2

N

E

E

8. In the given circuit diagram, find the charges which flow through directions 1 and 2 when switch S is closed. S

C1

C2

E 1

2

E

Capacitors — 327

Chapter 25

2 µF

9. Two capacitors A and B with capacities 3 µF and 2 µF are charged to a

potential difference of 100 V and 180 V, respectively. The plates of the capacitors are connected as shown in figure with one wire of each C + capacitor free. The upper plate of A is positive and that of B is negative. An 3 µF uncharged 2 µF capacitor C with lead wires falls on the free ends to A complete the circuit. Calculate

–

2 µF

B

(i) the final charge on the three capacitors, (ii) the amount of electrostatic energy stored in the system before and after completion of the circuit.

10. The capacitor C1 in the figure initially carries a charge q0. When the switch S1 and S 2 are closed, capacitor C1 is connected to a resistor R and a second capacitor C2, which initially does not carry any charge. (a) Find the charges deposited on the capacitors in steady state and the current through R as a function of time. (b) What is heat lost in the resistor after a long time of closing the switch?

S1 +

C1

R

–

C2 S2

11. A leaky parallel plate capacitor is filled completely with a material having dielectric constant K = 5 and electrical conductivity σ = 7.4 × 10–12 Ω –1m –1. If the charge on the capacitor at the instant t = 0 is q0 = 8.55 µC, then calculate the leakage current at the instant t = 12 s.

12. A parallel plate vacuum capacitor with plate area A and separation x has charges +Q and −Q on its plates. The capacitor is disconnected from the source of charge, so the charge on each plate remains fixed. (a) What is the total energy stored in the capacitor? (b) The plates are pulled apart an additional distance dx. What is the change in the stored energy? (c) If F is the force with which the plates attract each other, then the change in the stored energy must equal the work dW = Fdx done in pulling the plates apart. Find an expression for F. (d) Explain why F is not equal to QE , where E is the electric field between the plates.

13. A spherical capacitor has the inner sphere of radius 2 cm and the outer one of 4 cm. If the inner sphere is earthed and the outer one is charged with a charge of 2 µC and isolated. Calculate

(a) the potential to which the outer sphere is raised. (b) the charge retained on the outer surface of the outer sphere.

14. Calculate the charge on each capacitor and the potential difference across it in the circuits shown in figure for the cases : A 6 µF

3 µF S

100 Ω 100 Ω 90 V

6 µF

2 µF

1 µF 100 Ω

20 Ω

S

10 Ω

20 Ω 100 V

(a)

(i) switch S is closed and (ii) switch S is open. (iii) In figure (b), what is the potential of point A when S is open?

(b)

328 — Electricity and Magnetism 15. In the shown network, find the charges on capacitors of capacitances 5 µF and 3 µF, in steady state. 5 µF

1Ω

2Ω

3Ω 3 µF

10 V 4Ω

C = 10 µF, R1 = 4 MΩ, R2 = 6 MΩ , R3 = 3 MΩ. With C completely uncharged, switch S is suddenly closed (at t = 0).

16. In the circuit shown, E = 18 kV,

R1 I1

S R2

R3

I2

C I3

E

(a) Determine the current through each resistor for t = 0 and t = ∞. (b) What are the values of V 2 (potential difference across R2) at t = 0 and t = ∞ ? (c) Plot a graph of the potential difference V 2 versus t and determine the instantaneous value of V 2.

17. The charge on the capacitor is initially zero. Find the charge on the capacitor as a function of time t. All resistors are of equal value R. C

R2 E

R3

R1

18. The capacitors are initially uncharged. In a certain time the capacitor of capacitance 2 µF gets a charge of 20 µC. In that time interval find the heat produced by each resistor individually. 2Ω

20 V

3Ω 6Ω 1 µF

2 µF

Capacitors — 329

Chapter 25

19. A capacitor of capacitance C has potential difference E / 2 and another capacitor of capacitance 2C is uncharged. They are joined to form a closed circuit as shown in the figure. R + E/2 + –

C

–

E 2C

C

2C

(a) Find the current in the circuit at t = 0. (b) Find the charge on C as a function of time.

20. The capacitor shown in figure has been charged to a potential difference of V volt, so that it carries a charge CV with both the switches S1 and S 2 remaining open. Switch S1 is closed at t = 0. At t = R1C switch S1 is opened and S 2 is closed. Find the charge on the capacitor at t = 2R1C + R2C. +

C

– R1

S1 R2 E S2

21. The switch S is closed at t = 0. The capacitor C is uncharged but C0 has a charge Q0 = 2 µC at t = 0. If R = 100 Ω ,C = 2 µF,C0 = 2 µF, E = 4 V. Calculate i( t ) in the circuit. C0

R S

+

E

– C

22. A time varying voltage is applied to the clamps A and B such that voltage across the capacitor plates is as shown in the figure. Plot the time dependence of voltage across the terminals of the resistance E and D. VAB A

E C

B

R D t

23. In the above problem if given graph is between VAB and time. Then, plot graph between VED and time.

330 — Electricity and Magnetism 24. Initially, the switch is in position 1 for a long time. At t = 0, the switch is moved from 1 to 2. Obtain expressions for VC and V R for t > 0.

1

S

2

5 kΩ

100 V 1 µF

50 V

25. For the arrangement shown in the figure, the switch is closed at t = 0. C2 is initially uncharged while C1 has a charge of 2 µC.

30 Ω

C2 2 µF

60 Ω 30 Ω 1 µF

C1

S

9V

(a) Find the current coming out of the battery just after the switch is closed. (b) Find the charge on the capacitors in the steady state condition.

26. In the given circuit, the switch is closed in the position 1 at t = 0 and then moved to 2 after 250 µs. Derive an expression for current as a function of time for t > 0. Also plot the variation of current with time. 1 500 Ω

2

20 V

40 V

0.5 µF

27. A charged capacitor C1 is discharged through a resistance R by putting switch S in position 1 of circuit shown in figure. When discharge current reduces to I 0 , the switch is suddenly shifted to position 2. Calculate the amount of heat liberated in resistance R starting from this instant. 1

2 +

C1 R

–

C2

Answers Introductory Exercise 25.1 1. [M–1 L–2 T 4A2 ]

3. (a) −10 V (b) −10 µC , −20 µC (c) 3.0 × 10 −4 J

2. False

Introductory Exercise 25.2 1. ±182 µC

2. (a) 604 V (b) 90.8 cm2

(c) 16.3 µC/m2

3. (a) 1.28

(b) 6.2 × 10 –7 C/m2

Introductory Exercise 25.3 1. q3µF = 30 µC, q 4µF = 20 µC, q 2µF = 10 µC

2. q 4µF = 120 µC, q 9µF = 90 µC, q3µF = 30 µC

Exercises LEVEL 1 Assertion and Reason 1. (d)

2. (a)

3. (a,b)

4. (a,b)

5. (d)

6. (d)

7. (b)

8. (d)

9. (b)

10. (b)

Objective Questions 1. (c)

2. (d)

3. (c)

4. (c)

5. (d)

6. (d)

7. (c)

8. (b)

9. (b)

10. (c)

11. (a)

12. (b)

13. (b)

14. (c)

15. (c)

16. (a)

17. (d)

18. (b)

19. (b)

20. (b)

21. (a)

22. (c)

23. (d)

24. (c)

25. (d)

26. (b)

27. (b)

Subjective Questions ε0 A 5dq , d 2Aε0

1. Starting from the left face the charges are, 3 µC, 7 µC, –7 µC, 3 µC

2.

3. 10 µC, 20 µC, 30 µC

5. 24 µC

4. 40 µC

1600 3200 1600 6. (a) 800 µC, 800 V, 800 µC, 400 V (b) µC, µC, V 7. 400 µF 3 3 3 q q q 9. 0.69 10. q1 = 0 + 0 e –2t /RC , q2 = 0 (1 – e –2t /RC ) 2 2 2 q1 q2 q0 2

q0

q0 2 t

t

11. CE (1 – e – t /CR ) + q0 e – t /CR 12. (a) E /R1 (b) E /(R1 + R3 ) 13. (a) i1 = E /R1, i2 = E /R 2

(b) i1 = E /R1, i2 = 0 (c)

5C 4C (b) (c) 2C 3 3 –3 15. (a) 4.0 µF : 2.64 × 10 C, 660 V, 6.0 µF : 3.96 × 10 –3 C, 660 V

14. (a)

(b) 4.0 µF : 5.28 × 10 –4C, 132 V, 6.0 µF : 7.92 × 10 –4C, 132 V

1 CE 2 2

(d) C (R1 + R 2 )

332 — Electricity and Magnetism 16. 2.83 × 10 –2 J/m3

17.

0.0135 m2

18.

19.

40 mF, 10 mF

20. (a) Five capacitors in series (b) Six rows of three capacitors in each row. 21. 0.16 µF, 0.24 µF 22. (a) 120 µC (b) 60 µC (c) 480 µC 24. (a) 3.54 × 10

–9

10 V, 5 V

23. 19.6 J

F (b) ±35.4 µC (c) 2.0 × 10 N/C 6

25. (a) 76 µC (b) 1.4 mJ (c) 11 V (d) 1.2 mJ

26. 10 µC,

40 µC 3

27. (a) 2.5 µF (b) Q1 = 5.5 × 10 –4 C, V1 = 66 V, Q2 = 3.7 × 10 –4C, V2 = 88 V, Q3 = Q4 = 1.8 × 10 –4 C,V3 = V4 = 44 V, Q5 = Q1,V5 = V1 28. 78.68 V, 151.32 V

29. 60 µC

30. (a) 3 µF (b) 60 µC (c) 30 µC (d) 20 µC (e) 20 µC

31. (a) q1 = q3 = 9 µC, q2 = q4 = 16 µC (b) q1 = 8.64 µC, q2 = 17.28 µC, q3 = 10.08 µC, q4 = 13.44 µC 32. q2 = q3 =

1+

CV CV 1 0 1 0 , q1 = CV 1 0 − C1 C C1 C 1+ + 1 + 1 C2 C3 C2 C3

33. (a) 2 V (b) Ui =

1  ε0A  2  ε0A  V 2   V , Uf =    d  2 d 

(c) W =

1  ε0A  2  V 2 d 

35. (a) 18 V (b) a is at higher potential (c) 6 V

34. (a) 1 mA, 1 mA, 0 (b) 2 mA, 1 mA, 3 mA

(d) –36 µC on both the capacitors 36. (a) –6.0 V (b) b (c) 6.0 V (d) –54.0 µC 37. (a)

2 CV V V −αt V Here α = − (1 − e – αt ) (b) e , 2 2R 6R 2R 3RC

LEVEL 2 Single Correct Option 1.(c) 11.(d) 21.(c) 31.(d)

2.(b) 12.(b) 22.(d) 32.(c)

3.(d) 13.(d) 23.(b) 33.(b)

4.(c) 14.(c) 24.(d) 34.(c)

5.(d) 15.(c) 25.(a)

6.(d) 16.(a) 26.(b)

7.(d) 17.(b) 27.(c)

8.(b) 18.(b) 28.(a)

9.(c) 19.(b) 29.(b)

10.(d) 20.(c) 30.(b)

More than One Correct Options 1. (a,c,d) 2. (b,c,d) 3. (b,c)

4. (a,b,d) 5. (b,c)

6. (a,d)

Comprehension Based Questions 1.(b)

2.(a)

3.(b)

4.(a)

Match the Columns 1. (a) → q

(b) → p

(c) → p

(d) → p

2. (a) → s

(b) → p

(c) → r

(d) → s

3. (a) → q

(b) → p,r

(c) → q

(d) → p,s

4. (a) → r

(b) → p

(c) → s

5. (a) → s

(b) → s

(c) → s

6. (a) → p

(b) → p,q

(c) → s

(d) → p,q,r

7. (a,b,d)

8. (a,b,c) 9. (a,b,c,d)

10. (b)

Capacitors — 333

Chapter 25 Subjective Questions 5  ε0A  4  ε0 AV0  2  ε0 AV0    (b) q3 =   , q5 =   3 d  3 d  3 d  19 2 7 14 3. − V, − V, V, V 5 5 5 5 1. (a)

6. (a) 3 A, 2.67A (b) 24 V (c) 11.1 µs

2. (a) 80 V

(b) 57.6 mJ, 38.4 mJ

4. 2.5 A, 0.6 A 7. VN − VM =

5. 44.4 mJ

110 volt 3

8. q1 = EC 2 , q2 =

−EC1C 2 (C1 + C 2 )

9. (i) 90 µC, 210 µC,150 µC (ii) (a) 47.4 mJ (b) 18 mJ  C1  10. (a) q1 =   q0  C1 + C 2  11. 0.193 µA

12. (a)

and

 C2  q2 =   q0 ,  C1 + C 2 

i=

q0 − t /RC e RC1

 Q2  Q 2x Q2 (b)   ⋅ dx (c) 2ε0 A 2ε0 A  2ε0 A 

14.

C1C 2 q02C 2 , here C = C1 + C 2 2C1(C1 + C 2 )

(b) ∆H =

13. (a) 2.25 × 10 5 V (b) +1 µC

Fig. (a)

(i)

PD (volts) charge (µC)

(ii)

(iii)

Fig. (b)

6 µF

3µF

1µF

6 µF

2µF

30

30

0

10

30

180

90

0

60

60

PD (volts)

0

90

100

25

75

charge (µC)

0

270

100

150

150

VA = 75 volt

15. 15 µC, 15 µC 16. (a) At t = 0,

i1 = 3 mA, i2 = 1 mA, i3 = 2 mA At t = ∞,

V2(kV) 10.8

i1 = i2 = 1.8 mA, i3 = 0 (b) At t = 0, V2 = 6 kV At t = ∞, V2 = 10.8 kV (c) V2 = (10.8 – 4.8 e – t /54 ) kV CE 17. q = 2 19. (a)

E 2R

2t   −  1 − e CR     

(b)

6

18. H 2 = 0.075 mJ, H3 = 0.05 mJ, H 6 = 0.025 mJ

CE [ 5 – 2e –3t /2RC ] 6

VC 1 20. EC  1 –  + 2  e e

VR

4

21. (0.03 e–10 t ) A

VED

22.

23.

O

t

t

7 11 A or A (b) Q1 = 9 µC , Q2 = 0 50 50 = – (0.11 e –4000t ) A for t ≥ 250 µs

24. VC = 50 (3 e –200t – 1) , VR = 150 e –200t 26. i = (0.04e –4000t ) A

for t ≤ 250 µs,

For i - t graph, see the hints. 27.

O

(I0R )2C1C 2 2(C1 + C 2 )

25. (a)

t

Magnetics Chapter Contents 26.1

Introduction

26.2

Magnetic force on a moving charge(Fm)

26.3

Path of a charged particle in Uniform magnetic field

26.4

Magnetic force on a current carrying conductor

26.5

Magnetic dipole

26.6

Magnetic dipole in uniform magnetic field

26.7

Biot savart law

26.8

Applications of Biot savart law

26.9

Ampere's circuital law

26.10

Force between parallel current carrying wires

26.11

Magnetic poles and Bar magnets

26.12

Earth's magnetism

26.13

Vibration magnetometer

26.14

Magnetic induction and Magnetic materials

26.15

Some important terms used in magnetism

26.16

Properties of magnetic materials

26.17

Explanation of paramagnetism, Diamagnetism and Ferromagnetism

26.18

Moving coil galvanometer

336 — Electricity and Magnetism

26.1 Introduction The fascinating attractive properties of magnets have been known since ancient times. The word magnet comes from ancient Greek place name Magnesia (the modern town Manisa in Western Turkey), where the natural magnets called lodestones were found. The fundamental nature of magnetism is the interaction of moving electric charges. Unlike electric forces which act on electric charges whether they are moving or not, magnetic forces act only on moving charges and current carrying wires. We will describe magnetic forces using the concept of a field. A magnetic field is established by a permanent magnet, by an electric current or by other moving charges. This magnetic field, in turn, exerts forces on other moving charges and current carrying conductors. In this chapter, first we study the magnetic forces and torques exerted on moving charges and currents by magnetic fields, then we will see how to calculate the magnetic fields produced by currents and moving charges.

26.2 Magnetic Force on a Moving Charge ( Fm ) An unknown electric field can be determined by magnitude and direction of the force on a test charge q 0 at rest. To explore an unknown magnetic field (denoted by B), we must measure the magnitude and direction of the force on a moving test charge. The magnetic force ( Fm ) on a charge q moving with velocity v in a magnetic field B is given, both in magnitude and direction, by Fm = q ( v × B)

…(i)

Following points are worthnoting regarding the above expression. (i) The magnitude of Fm is Fm = Bqv sin θ where, θ is the angle between v and B. (ii) Fm is zero when, (b) q = 0, i.e. particle is neutral. (a) B = 0, i.e. no magnetic field is present. (c) v = 0, i.e. charged particle is at rest or (d) θ = 0° or 180°, i.e. v ↑↑ B or v ↑↓ B (iii) Fm is maximum at θ = 90° and this maximum value is Bqv. (iv) The units of B must be the same as the units of F qv. Therefore, the SI unit of B is equivalent to N-s . This unit is called the tesla (abbreviated as T), in honour of Nikola Tesla, the prominent C-m Serbian-American scientist and inventor. Thus, 1 tesla = 1T =

1 N-s 1 N = C-m A-m

The CGS unit of B, the gauss (1G = 10 –4 T ) is also in common use. (v) In equation number (i) q is to be substituted with sign. If q is positive, magnetic force is along v × B and if q is negative, magnetic force is in a direction opposite to v × B.

Chapter 26

Magnetics — 337

(vi) Direction of Fm From the property of cross product we can infer that Fm is perpendicular to both v and B or it is perpendicular to the plane formed by v and B. The exact direction of Fm can be given by any of the following methods: (a) Direction of Fm = (sign of q) (direction of v × B) or, as we stated earlier also, Fm ↑↑ v × B if q is positive and Fm ↑↓ v × B if q is negative. (b) Fleming's left hand rule According to this rule, the forefinger, the central finger and the thumb of the left hand are stretched in such a way that they are mutually perpendicular to each other. If the central finger shows the direction of velocity of positive charge ( v +q ) and forefinger shows the direction of magnetic field ( B), then the thumb will give the direction of magnetic force ( Fm ). If instead of positive charge we have the negative charge, then Fm is in opposite direction. Thumb Fm

B Forefinger v+q

Central finger

Fig. 26.1

(c) Right hand rule Wrap the fingers of your right hand around the line perpendicular to the plane of v and B as shown in figure, so that they curl around with the sense of rotation from v to B through the smaller angle between them. Your thumb then points in the direction of the force Fm on a positive charge. (Alternatively, the direction of the force Fm on a positive charge is the direction in which a right hand thread screw would advance if turned the same way). Fm = v × B

Fm v B

Right hand rule

B

v+q

Fig. 26.2

(vii) Fm ⊥ v or Fm ⊥

ds . Therefore, Fm ⊥ d s or the work done by the magnetic force in a static magnetic dt

field is zero. WFm = 0 So, from work energy theorem KE and hence the speed of the charged particle remains constant in magnetic field. The magnetic force can change the direction only. It cannot increase or decrease the speed or kinetic energy of the particle.

338 — Electricity and Magnetism Note By convention the direction of magnetic field B perpendicular to the paper going inwards is shown by and the direction perpendicular to the paper coming out is shown by

.

Fig. 26.3 V

Example 26.1 A charged particle is projected in a magnetic field B = ( 3$i + 4$j) × 10 –2 T The acceleration of the particle is found to be a = ( x$i + 2$j) m /s2 Find the value of x. As we have read, Fm ⊥ B i.e. the acceleration a ⊥ B or a ⋅ B = 0 $ $ or ( xi + 2j ) ⋅ ( 3$i + 4$j ) × 10–2 = 0 Solution

V

or

( 3x + 8) × 10–2 = 0

∴

x=–

8 m/s 2 3

Ans.

Example 26.2 When a proton has a velocity v = (2$i + 3$j) × 10 6 m/s, it experiences a force F = – (1.28 × 10 –13 k$ ) N . When its velocity is along the z-axis, it experiences a force along the x-axis. What is the magnetic field? HOW TO PROCEED In the second part of the question, it is given that magnetic force is along x-axis when velocity is along z-axis. Hence, magnetic field should be along negative y-direction. As in case of positive charge (here proton) Fm ↑↑ v × B So, let

B = – B 0 $j

where, B 0 = positive constant. Now, applying Fm = q ( v × B) we can find value of B 0 from the first part of the question. Solution Substituting proper values in, Fm = q ( v × B ) We have, – (1.28 × 10–13 k$ ) = (1.6 × 10–19 ) [( 2i$ + 3$j ) × (– B 0 $j )] × 106 ∴ or Therefore, the magnetic field is

1.28 = 1.6 × 2 × B 0 1.28 B0 = = 0.4 3.2 B = (– 0.4 $j ) T

Ans.

Chapter 26 V

Magnetics — 339

Example 26.3 A magnetic field of (4.0 × 10 –3 k$ ) T exerts a force ( 4.0 $i + 3.0 $j) × 10 –10 N on a particle having a charge 10 –9 C and moving in the x-y plane. Find the velocity of the particle. Solution

Given,

B = ( 4 × 10–3 k$ ) T, q = 10–9 C

and magnetic force Fm = ( 4.0 $i + 3.0 $j ) × 10–10 N Let velocity of the particle in x-y plane be v = v x i$ + v y $j Then, from the relation Fm = q ( v × B ) We have ( 4.0 i$ + 3.0 $j ) × 10–10 = 10–9 [( v x i$ + v y $j ) × ( 4 × 10–3 k$ )]

= ( 4v y × 10 –12 $i – 4v x × 10 –12 $j) Comparing the coefficients of $i and $j , we have 4 × 10–10 = 4v y × 10–12 ∴

v y = 102 m/s = 100 m/s

and ∴

3.0 × 10–10 = – 4v x × 10–12 v x = – 75 m/s v = (– 75 i$ + 100 $j ) m/s

∴

INTRODUCTORY EXERCISE

Ans.

26.1

1. Write the dimensions of E/B. Here, E is the electric field and B the magnetic field. 2. In the relation F = q ( v × B), which pairs are always perpendicular to each other. 3. If a beam of electrons travels in a straight line in a certain region. Can we say there is no magnetic field?

4. A charge q = – 4 µC has an instantaneous velocity v = (2 $i – 3$j + k$ ) × 106 m /s in a uniform magnetic field B = (2 $i + 5$j – 3 k$ ) × 10–2 T. What is the force on the charge?

5. A particle initially moving towards south in a vertically downward magnetic field is deflected toward the east. What is the sign of the charge on the particle?

6. An electron experiences a magnetic force of magnitude 4.60 × 10−15 N, when moving at an angle of 60° with respect to a magnetic field of magnitude 3.50 × 10−3 T. Find the speed of the electron.

7. He2+ ion travels at right angles to a magnetic field of 0.80 T with a velocity of 105 m /s. Find the magnitude of the magnetic force on the ion.

340 — Electricity and Magnetism

26.3 Path of a Charged Particle in Uniform Magnetic Field The path of a charged particle in uniform magnetic field depends on the angle θ (the angle between v and B ). Depending on the different values of θ, the following three cases are possible. When θ is 0° or 180° As we have seen in Art. 26.2, Fm = 0, when θ is either 0° or180°. Hence, path of the charged particle is a straight line (undeviated) when it enters parallel or antiparallel to magnetic field. Case 1

B

B or

q + –

v

v

+ – q

Fm = 0

Fig. 26.4

Case 2

When θ = 90°

When θ = 90°, the magnetic force is Fm = Bqv sin 90° = Bqv. This magnetic force is perpendicular to the velocity at every instant. Hence, path is a circle. The necessary centripetal force is provided by the magnetic force. Hence, if r be the radius of the circle, then mv 2 = Bqv r mv r= Bq

or

This expression of r can be written in the following different ways r=

2qVm p mv 2Km = = = Bq Bq Bq Bq

Here, p = momentum of particle p2 or p = 2Km 2m We also know that if the charged particle is accelerated by a potential difference of V volts, it acquires a KE given by K = qV Further, time period of the circular path will be K =KE of particle =

or

T=

2π r = v

T=

2πm Bq

 mv  2π    Bq  v

=

2π m Bq

Chapter 26 or The angular speed (ω ) of the particle is ∴

Magnetics — 341

2π Bq = T m Bq ω= m

ω=

1 T Bq f = 2πm f =

Frequency of rotation is or

The following points are worthnoting regarding a circular path: (i) The plane of the circle is perpendicular to magnetic field. If the magnetic field is along z-direction, the circular path is in x-y plane. The speed of the particle does not change in magnetic field. Hence, if v 0 be the speed of the particle, then velocity of particle at any instant of time will be v = v $i + v $j x

y

v x2 + v y2 = v 02

where,

×

×

×

×

×

×

× ×

×

× ×

×

×

×

+ q, m

v2

×

×

×

v1

×

×

× ×

×

× ×

(ii) T , f and ω are independent of v while the radius is directly proportional to v.

+ q, m

Fig. 26.5

Hence, if two charged particles of equal mass and charge enter in a magnetic field B with different speeds v1 and v 2 ( > v1 ) at right angles, then T1 = T2 r2 > r1

but as shown in figure.

q is known as specific charge. It is sometimes denoted by α. So, in terms of α, the m above formulae can be written as v 2π Bα , T= , f = and ω = Bα r= Bα Bα 2π

Note Charge per unit mass

Case 3

When θ is other than 0° , 180° or 90°

In this case velocity can be resolved into two components, one along B and another perpendicular to B. Let the two components be v| | and v ⊥ .

342 — Electricity and Magnetism Then, v| | = v cos θ v ⊥ = v sin θ

and

B v sin θ

v

q, m +

θ

v cos θ

Fig. 26.6

The component perpendicular to field ( v ⊥ ) gives a circular path and the component parallel to field ( v| | ) gives a straight line path. The resultant path is a helix as shown in figure. The radius of this helical path is r=

Fig. 26.7

mv ⊥ mv sin θ = Bq Bq

Time period and frequency do not depend on velocity and so they are given by T=

2πm Bq

and

f =

Bq 2πm

There is one more term associated with a helical path, that is pitch (p) of the helical path. Pitch is defined as the distance travelled along magnetic field in one complete cycle. i.e. p = v| | T 2π m or p = ( v cos θ ) Bq ∴

p= V

2πmv cos θ Bq

Example 26.4 Two particles A and B of masses m A and mB respectively and having the same charge are moving in a plane. A uniform magnetic field exists perpendicular to this plane. The speeds of the particles are v A and vB respectively and the trajectories are as shown in the figure. Then, (JEE 2001) A B

Fig. 26.8

(a) m A v A < m B v B (c) m A < m B and v A < v B

(b) m A v A > m B v B (d) m A = m B and v A = v B

Chapter 26 Solution

Radius of the circle =

Magnetics — 343

mv Bq

or radius ∝ mv if B and q are same. ( Radius ) A > ( Radius ) B ∴

m A v A > mB v B

∴ Correct option is (b). V

Example 26.5 A proton, a deuteron and an α-particle having the same kinetic energy are moving in circular trajectories in a constant magnetic field. If rp , rd and rα denote respectively the radii of the trajectories of these particles, then

(a) rα = r p < rd (c) rα = rd > r p Solution Radius of the circular path is given by r=

(b) rα > rd > r p (d) r p = rd = rα

(JEE 1997)

2Km Bq

mv = Bq

Here, K is the kinetic energy to the particle. Therefore, r ∝

m if K and B are same. q

∴

rp : rd : rα =

1 2 4 : : = 1: 2 : 1 1 1 2

rα = rp < rd

Hence, ∴ Correct option is (a). V

Example 26.6 Two particles X and Y having equal charges, after being accelerated through the same potential difference, enter a region of uniform magnetic field and describe circular paths of radii R1 and R2 , respectively. The ratio of the mass of X to that of Y is (JEE 1988) (a) ( R1/R2 )1/ 2 (c) ( R1/R2 ) 2 Solution

R=

(b) R2/R1 (d) R1/R2 2qVm Bq

or or

or ∴ Correct option is (c).

R∝ m R1 = R2

mX mY

m X  R1  =  mY  R2 

2

344 — Electricity and Magnetism INTRODUCTORY EXERCISE

26.2

1. A neutron, a proton, an electron and an α - particle enter a region of constant magnetic field with equal velocities. The magnetic field is along the inward normal to the plane of the paper. The tracks of the particles are labeled in figure. The electron follows track…… and the α - particle follows track…… (JEE 1984) C B

A

D

Fig. 26.9

2. An electron and a proton are moving with the same kinetic energy along the same direction. When they pass through a uniform magnetic field perpendicular to the direction of their motion, they describe circular path of the same radius. Is this statement true or false? (JEE 1985)

3. A charged particle enters a region of uniform magnetic field at an angle of 85° to the magnetic line of force. The path of the particle is a circle. Is this statement true or false?

(JEE 1983)

4. Can a charged particle be accelerated by a magnetic field. Can its speed be increased? 5. An electron beam projected along positive x-axis deflects along the positive y-axis. If this deflection is caused by a magnetic field, what is the direction of the field?

6. An electron and a proton are projected with same velocity perpendicular to a magnetic field. (a) Which particle will describe the smaller circle? (b) Which particle will have greater frequency?

7. An electron is accelerated through a PD of 100 V and then enters a region where it is moving perpendicular to a magnetic field B = 0.2 T. Find the radius of the circular path. Repeat this problem for a proton.

26.4 Magnetic Force on a Current Carrying Conductor A charged particle in motion experiences a magnetic force in a magnetic field. Similarly, a current carrying wire also experiences a force when placed in a magnetic field. This follows from the fact that the current is a collection of many charged particles in motion. Hence, the resultant force exerted by the field on the wire is the vector sum of the individual forces exerted on all the charged particles making up the current. The force exerted on the particles is transmitted to the wire when the particles collide with the atoms making up the wire.

×

×

×

× Fm ×

×

×

×B A

vd ×

×

×

i ×

× l

Fig. 26.10

×

×

×

Chapter 26

Magnetics — 345

Suppose a conducting wire carrying a current i is placed in a magnetic field B. The length of the wire is l and area of cross-section is A. The free electrons drift with a speed v d opposite to the direction of current. The magnetic force exerted on the electron is d Fm = – e ( v d × B) If n be the number of free electrons per unit volume of the wire, then total number of electrons in volume Al of the wire are, nAl. Therefore, total force on the wire is Fm = – e ( nAl) ( v d × B) If we denote the length l along the direction of the current by l, then the above equation becomes Fm = i ( l × B)

…(i)

where, neAv d = i The following points are worthnoting regarding the above expression : (i) Magnitude of Fm is, Fm = ilB sin θ, here θ is the angle between l and B. Fm is zero for θ = 0° or 180° and maximum for θ = 90°. (ii) Here, l is a vector that points in the direction of the current i and has a magnitude equal to the length. (iii) The above expression applies only to a straight segment of wire in a uniform magnetic field. (iv) For the magnetic force on an arbitrarily shaped wire segment, let us consider the magnetic force exerted on a small segment of vector length dl. D dl

C

B

B

i

i

A (b)

(a)

Fig. 26.11

…(ii) d Fm = i ( d l × B) To calculate the total force Fm acting on the wire shown in figure, we integrate Eq. (ii) over the length of the wire. Fm = i∫ ( d l × B) D

…(iii)

A

Now, let us consider two special cases involving Eq. (iii). In both cases, the magnetic field is taken to be constant in magnitude and direction. Case 1 A curved wire ACD as shown in Fig. (a) carries a current i and is located in a uniform magnetic field B. Because the field is uniform, we can take B outside the integral in Eq. (iii) and we obtain, D …(iv) Fm = i  ∫ d l  × B A D

But, the quantity ∫ d l represents the vector sum of all length elements from A to D. From the A

polygon law of vector addition, the sum equals the vector l directed from A to D. Thus,

346 — Electricity and Magnetism Fm = i ( l × B) or we can write F ACD = F AD = i ( AD × B) in uniform field. An arbitrarily shaped closed loop carrying a current i is placed in a uniform magnetic field as shown in Fig. (b). We can again express the force acting on the loop in the form of Eq. (iv), but this time we must take the vector sum of the length elements dl over the entire loop,

Case 2

Fm = i ( ∫ d l ) × B Because the set of length elements forms a closed polygon, the vector sum must be zero. ∴ Fm = 0 Thus, the net magnetic force acting on any closed current carrying loop in a uniform magnetic field is zero. F (v) The direction of Fm can be given by Fleming's left hand rule as Thumb m discussed in Art. 26.2. According to this rule, the forefinger, the central finger and the thumb of the left hand are stretched in such B a way that they are mutually perpendicular to each other. If the Forefinger central finger shows the direction of current (or l) and forefinger shows the direction of magnetic field ( B), then the thumb will i or l Central finger give the direction of magnetic force ( Fm ). Fig. 26.12

V

Example 26.7 A horizontal rod 0.2 m long is mounted on a balance and carries a current. At the location of the rod a uniform horizontal magnetic field has magnitude 0.067 T and direction perpendicular to the rod. The magnetic force on the rod is measured by the balance and is found to be 0.13 N. What is the current? Solution F = ilB sin 90° F 0.13 ∴ i= = lB 0.2 × 0.067 = 9.7 A

V

Example 26.8 A square of side 2.0 m is placed in a uniform magnetic field B = 2.0 T in a direction perpendicular to the plane of the square inwards. Equal current i = 3.0 A is flowing in the directions shown in figure. Find the magnitude of magnetic force on the loop. Solution

∴ or

Force on wire ACD = Force on AD = Force on AED Net force on the loop = 3 ( FAD ) Fnet = 3 ( i ) ( AD ) ( B ) = ( 3) ( 3.0) ( 2 2 ) ( 2.0) N = 36 2 N

Direction of this force is towards EC.

Ans. ×

×

×

C

B × D

×

×

×

×

×

×

×

×

×

A

× × Fig. 26.13

E

×

Chapter 26 V

Magnetics — 347

Example 26.9 In the figure shown a semicircular wire loop is placed in a uniform magnetic field B = 1.0 T . The plane of the loop is perpendicular to the magnetic field. Current i = 2 A flows in the loop in the directions shown. Find the magnitude of the magnetic force in both the cases (a) and (b). The radius of the loop is 1.0 m × B ×

× i = 2 A×

×

×

×

×

×

×

×

×

×

×

1m ×

×

×

×

1m ×

×

×

×

×

×

×

×

×

×

(a)

i = 2A

(b)

Fig. 26.14

Refer figure (a) It forms a closed loop and the current completes the loop. Therefore, net force on the loop in uniform field should be zero. Ans.

×

×

×

×B

×

×

×

×

×

×

Solution

Refer figure (b) In this case although it forms a closed loop, but current does not complete the loop. Hence, net force is not zero. ∴ ∴

×

C

A ×

FACD = FAD Floop = FACD + FAD = 2FAD | Floop | = 2 | FAD | = 2ilB sin θ

×

×

×

Fig. 26.15

[l = 2r = 2.0 m]

= ( 2) ( 2) ( 2) (1) sin 90° = 8 N

INTRODUCTORY EXERCISE

× D

Ans.

26.3

1. A wire of length l carries a current i along the x-axis. A magnetic field B = B0 ( $j + k$ ) exists in the space. Find the magnitude of the magnetic force acting on the wire.

2. In the above problem will the answer change if magnetic field becomes B = B0 ( $i + $j + k$ ). 3. A wire along the x-axis carries a current of 3.50 A in the negative direction. Calculate the force (expressed in terms of unit vectors) on a 1.00 cm section of the wire exerted by these magnetic fields (a) B = – (0.65 T )$j (b) B = + (0.56 T ) k$ (c) B = – (0.31 T )$i (d) B = + (0.33 T ) $i − (0.28 T ) k$ (e) B = + (0.74 T ) $j − (0.36 T ) k$

4. Find net force on the equilateral loop of side 4 m carrying a current of 2 A kept in a uniform magnetic field of 2 T as shown in figure. ×

×

×

×

×

×

×

×

×

×

× Fig. 26.16

×

348 — Electricity and Magnetism

26.5 Magnetic Dipole Every current carrying loop is a magnetic dipole. It has two poles: south ( S ) and north ( N ). This is similar to a bar magnet. Magnetic field lines emanate from the north pole and after forming a closed path terminate on south pole. Each magnetic dipole has some magnetic moment ( M). The magnitude of M is | M | = NiA N = number of turns in the loop i = current in the loop and A = area of cross-section of the loop. For the direction of M any one of the following methods can be used: (i) As in case of an electric dipole, the dipole moment p has a direction from negative charge to positive charge. In the similar manner, direction of M is from south to north pole. The south and north poles can be identified by the sense of current. The side from where the current seems to be clockwise becomes south pole and the opposite side from where it seems anti-clockwise becomes north pole.

Here,

i

i a

b

M × M

Fig. 26.17

M

a

R

i (a)

a (b)

(c)

Fig. 26.18

Now, let us find the direction and magnitude of M in the three loops shown in Fig. 26.18. Refer figure (a) In this case, current appears to be clockwise from outside the paper, so this side becomes the south pole. From the back of the paper it seems anti-clockwise. Hence, this side becomes the north pole. As the magnetic moment is from south to north pole. It is directed perpendicular to paper inwards. Further, | M | = NiA = πR 2 i Refer figure (b) Here, opposite is the case. South pole is into the paper and north pole is outside the paper. Therefore, magnetic moment is perpendicular to paper in outward direction. The magnitude of M is | M | = a 2i Refer figure (c) In this case, south pole is on the right side of the loop and north pole on the left side. Hence, M is directed from right to left. The magnitude of magnetic moment is | M | = abi (ii) Vector M is along the normal to the plane of the loop. The orientation (up or down along the normal) is given by the right hand rule. Wrap your fingers of the right hand around the perimeter

Chapter 26

Magnetics — 349

of the loop in the direction of current as shown in figure. Then, extend your thumb so that it is perpendicular to the plane of the loop. The thumb points in the direction of M. M

i

Fig. 26.19

.

Extra Points to Remember ˜

In addition to the method discussed above for finding M here are two more A methods for calculating M. Method 1 This method is useful for calculating M for a rectangular or square loop. The magnetic moment (M) of the rectangular loop shown in figure is D M = i (AB × BC ) = i (BC × CD ) = i (CD × DA ) = i (DA × AB)

B i

C

Fig 26.20

Here, the cross product of any two consecutive sides (taken in order) gives the area as well as the correct direction of M also.

Note

If coordinates of vertices are known. Then, vector of any side can be written in terms of coordinates, e.g. AB = ( xB – xA ) $i + ( yB – yA ) $j + ( zB – zA ) k$

Method 2 Sometimes, a current carrying loop does not lie in a single plane. But by assuming two equal and opposite currents in one branch (which obviously makes no change in the given circuit) two (or more) closed loops are completed in different planes. Now, the net magnetic moment of the given loop is the vector sum of individual loops. z B

B

C

i A

F

A

D y

H

E

x

C

i

F

G

D H

E

(a)

G (b)

Fig. 26.21

For example, in Fig. (a), six sides of a cube of side l carry a current i in the directions shown. By assuming two equal and opposite currents in wire AD, two loops in two different planes (xy and yz) are completed. M = – il 2 k$ ABCDA

M ADGFA = – il 2 $i ∴

M net = – il 2 ($i + k$ )

350 — Electricity and Magnetism V

Example 26.10 A square loop OABCO of side l carries a current i. It is placed as shown in figure. Find the magnetic moment of the loop. z A B i y

O 60° C x

Fig. 26.22

Solution

As discussed above, magnetic moment of the loop can be written as M = i ( BC × CO )

Here, BC = – l k$ ,

V

l 3l $ CO = – l cos 60° $i – l sin 60° $j = – $i – j 2 2

∴

  l 3l $  M = i (– l k$ ) ×  – $i – j  2   2 

or

M=

il 2 $ (j – 2

3 $i )

Ans.

Example 26.11 Find the magnitude of magnetic moment of the current carrying loop ABCDEFA. Each side of the loop is 10 cm long and current in the loop is i = 2.0 A. C

i D

B A

E F Fig. 26.23

By assuming two equal and opposite currents in BE, two current carrying loops ( ABEFA and BCDEB) are formed. Their magnetic moments are equal in magnitude but perpendicular to each other. Hence, Solution

Mnet = M + M 2

where, ∴

2

C

D

B

E

= 2M

M = iA = ( 2.0) ( 0.1) ( 0.1) = 0.02 A-m 2

A

F Fig. 26.24

M net = ( 2 ) ( 0.02) A-m 2 = 0.028 A-m 2

Ans.

Chapter 26

Magnetics — 351

26.6 Magnetic Dipole in Uniform Magnetic Field Let us consider a rectangular ( a × b) current carrying loop OACDO placed in xy - plane. A uniform magnetic field B = B x $i + B y $j + B z k$ z

D

O

y

a i b

A

C

x

Fig. 26.25

exists in space. We are interested in finding the net force and torque in the loop. Force : Net force on the loop is F = FOA + F AC + FCD + FDO = i [ (OA × B) + ( AC × B) + (CD × B) + ( DO × B)] = i [(OA + AC + CD + DO) × B] = null vector or | F | = 0, as OA + AC + CD + DO forms a null vector. Torque : Using F = i ( l × B), we have FOA = i (OA × B) = i [( a$i ) × ( B x $i + B y $j + B z k$ )] = ia [ B y k$ – B z $j] F = i ( AC × B) = i [(b$j) × ( B $i + B $j + B k$ )] = ib [– B k$ + B $i ] AC

x

y

x

z

z

FCD = i (CD × B) = i [(– a$i ) × ( B x $i + B y $j + B z k$ )] = ia [– B y k$ + B z $j] FDO = i ( DO × B) = i [(– b$j) × ( B x $i + B y $j + B z k$ )] = ib [ B x k$ – B z $i ] All these forces are acting at the centre of the wires. For example, FOA will act at the centre of OA. When the forces are in equilibrium, net torque about any point remains the same. Let us find the torque about O. H O

D

E

G

A

C F Fig. 26.26

E , F , G and H are the mid-points of OA, AC , CD and DO, respectively.

352 — Electricity and Magnetism Using

τ = r × F, we have τ O = (OE × FOA ) + (OF × F AC ) + (OG × FCD ) + (OH × FDO )  a     b  =   $i  × {ia (B y k$ – B z $j)} +   a$i + $j × {ib (– B x k$ + B z $i )}     2  2      b   a   +   $i + b$j × {ia (– B y k$ + B z $j)} +   $j × ib ( B x k$ – B z $i )      2 2     = iab B x $j – iabB y $i

This can also be written as τ O = ( iabk$ ) × ( B x $i + B y $j + B z k$ ) iabk$ = magnetic moment of the dipole M

Here,

B x $i + B y $j + B z k$ = B

and ∴

τ =M×B

Note that although this formula has been derived for a rectangular loop, it comes out to be true for any shape of loop. The following points are worthnoting regarding the torque acting on the loop in uniform magnetic field. (i) Magnitude of τ is MB sin θ or NiAB sin θ. Here, θ is the angle between M and B. Torque is zero when θ = 0° or 180° and it is maximum at θ = 90°. (ii) If the loop is free to rotate in a magnetic field, the axis of rotation becomes an axis parallel to τ passing through the centre of mass of the loop. The above equation for the torque is very similar to that of an electric dipole in an electric field. The similarity between electric and magnetic dipoles extends even further as illustrated in the table below. Table 26.1 S.No. Field of similarity

Electric dipole

Magnetic dipole

1.

Magnitude

|p| = q (2d )

| M | = NiA

2.

Direction

from –q to +q

from S to N

3.

Net force in uniform field

zero

zero

4.

Torque

τ =p×E

τ =M×B

5.

Potential energy

U = – p⋅E

U = – M ⋅B

6.

Work done in rotating the dipole

Wθ1 – θ 2 = pE (cos θ1 – cos θ2 )

Wθ1 – θ 2 = MB (cos θ1 – cos θ2 )

7

Field along axis

E=

8.

Field perpendicular to axis

E=–

Note In last two points r >> size of loop.

1 2p ⋅ 4 πε0 r 3 1 p ⋅ 4 πε0 r 3

B=

µ 0 2M ⋅ 4π r3

B=–

µ0 M ⋅ 4π r3

Chapter 26

Magnetics — 353

Note that the expressions for the magnetic dipole can be obtained from the expressions for the electric 1 dipole by replacing pby Mand ε 0 by . Here, µ 0 is called the permeability of free space. It is related µ0 with ε 0 and speed of light c as 1 c= ε 0µ 0 and it has the value, µ 0 = 4π × 10 –7 T-m /A Dimensions of

ε 0µ 0

are that of speed or [LT –1 ].  1   ε 0µ 0

Hence, V

1

 –1  = [ LT ] 

Example 26.12 A circular loop of radius R = 20 cm is placed in a uniform magnetic field B = 2 T in xy-plane as shown in figure. The loop carries a current i = 1.0 A in the direction shown in figure. Find the magnitude of torque acting on the loop. y B i 45°

x

Fig. 26.27

Solution

Magnitude of torque is given by | τ | = MB sin θ

Here,

M = NiA = (1) (1.0) ( π ) ( 0.2)

2

= ( 0.04 π ) A-m 2 and ∴

B = 2T θ = angle between M and B = 90° | τ | = ( 0.04 π ) ( 2) sin 90° = 0.25 N-m

Note M is along negative z-direction (perpendicular to paper inwards) while B is in xy-plane. So, the angle between M and B is 90° not 45°. If the direction of torque is also desired, then we can write B = 2 cos 45 ° $i + 2 sin 45 ° $j = 2 ( $i + $j) T

Ans.

354 — Electricity and Magnetism M = – (0.04 π ) k$ A- m2 τ = M × B = (0.04 2π ) (– $j + $i ) τ = 0.18 ( $i – $j)

and ∴ or

INTRODUCTORY EXERCISE

Ans.

26.4

1. A charge q is uniformly distributed on a non-conducting disc of radius R. It is rotated with an angular speed ω about an axis passing through the centre of mass of the disc and perpendicular to its plane. Find the magnetic moment of the disc.  q  [Hint : For any charge distribution : Magnetic moment =   (angular momentum)]  2m 

2. A circular loop of wire having a radius of 8.0 cm carries a current of 0.20 A. A vector of unit length and parallel to the dipole moment M of the loop is given by 0.60 $i − 0.80 $j . If the loop is located in uniform magnetic field given by B = (0.25 T ) $i + (0.30 T ) k$ find, (a) the torque on the loop and (b) the magnetic potential energy of the loop.

3. A length L of wire carries a current i. Show that if the wire is formed into a circular coil, then the maximum torque in a given magnetic field is developed when the coil has one turn only, and that maximum torque has the magnitude τ = L2iB / 4π .

4. A coil with magnetic moment 1.45 A -m 2 is oriented initially with its magnetic moment antiparallel to a uniform 0.835 T magnetic field. What is the change in potential energy of the coil when it is rotated 180° so that its magnetic moment is parallel to the field?

26.7 Biot Savart Law In the preceding articles, we discussed the magnetic force exerted on a charged particle and current carrying conductor in a magnetic field. To P complete the description of the magnetic interaction, this and the next article r θ deals with the origin of the magnetic field. As in electrostatics, there are two r dl methods of calculating the electric field at some point. One is Coulomb's law which gives the electric field due to a point charge and the another is Gauss's i law which is useful in calculating the electric field of a highly symmetric configuration of charge. Similarly, in magnetics, there are basically two methods of calculating magnetic field at some point. One is Biot Savart law Fig. 26.28 which gives the magnetic field due to an infinitesimally small current carrying wire at some point and the another is Ampere's law, which is useful in calculating the magnetic field of a highly symmetric configuration carrying a steady current. We begin by showing how to use the law of Biot and Savart to calculate the magnetic field produced at some point in space by a small current element. Using this formalism and the principle of superposition, we then calculate the total magnetic field due to various current distributions. From their experimental results, Biot and Savart arrived at a mathematical expression that gives the magnetic field at some point in space in terms of the current that produces the field. That expression is based on the following experimental observations for the magnetic field dB at a point P associated with a length element dl of a wire carrying a steady current i.

Chapter 26

Magnetics — 355

(i) The vector dB is perpendicular to both dl (which points in the direction of the current) and the unit vector r$ directed from dl to P. (ii) The magnitude of dB is inversely proportional to r 2 , where r is the distance from dl to P. (iii) The magnitude of dB is proportional to the current and to the magnitude dl of the length element dl. (iv) The magnitude of dB is proportional to sin θ where θ is the angle between dl and r$ . These observations are summarized in mathematical formula known today as Biot Savart law µ i ( d l × r$ ) …(i) dB = 0 4π r2 Here,

µ0 T-m = 10 –7 4π A

It is important to note that dB in Eq. (i) is the field created by the current in only a small length element dl of the conductor. To find the total magnetic field B created at some point by a current of finite size, we must sum up contributions from all current elements that make up the current. That is, we must evaluate B by integrating Eq. (i). B=

µ 0 i d l × r$ 4π ∫ r 2

where, the integral is taken over the entire current distribution. This expression must be handled with special care because the integrand is a cross product and therefore, a vector quantity. The following points are worthnoting regarding the Biot Savart law. (i) Magnitude of dB is given by µ idl sin θ | dB| = 0 4π r 2 | dB | is zero at θ = 0° or 180° and maximum at θ = 90°. (ii) For the direction of dB either of the following methods can be employed.

dl

×

×

×

×

×

×

×

×

×

×

dB = 0

Fig. 26.29

(a) d B ↑↑ d l × r$ . So, dB is along dl × r. (b) If dl is in the plane of paper. dB = 0 at all points lying on the straight line passing through dl. The magnetic field to the right of this line is in ⊗ direction and to the left of this line is in u direction.

356 — Electricity and Magnetism

26.8 Applications of Biot Savart Law A

Let us now consider few applications of Biot Savart law.

Magnetic Field Surrounding a Thin, Straight Conductor

dy

According to Biot Savart law, µ id l × r$ …(i) B= 0 ∫ 4π r2 As here every element of the wire contributes to B in the same direction (which is here ⊗). Eq. (i) for this case becomes, µ idl sin θ µ 0 i dy sin θ B= 0∫ = 4π 4π ∫ r 2 r2 y = d tan φ or

θ

r α

y

φ

d β

P

i B

Fig. 26.30

dy = ( d sec 2φ ) dφ

r = d sec φ and θ = 90° – φ

{

}

2 µ 0 i φ = α ( d sec φ ) dφ sin (90° − φ ) B= 4π ∫φ = –β ( d sec φ ) 2

or

B=

µ 0i α cos φ ⋅ dφ or 4πd ∫–β

B=

µ0 i (sin α + sin β) 4π d

Note down the following points regarding the above equation. (i) For an infinitely long straight wire, α = β = 90° ∴

sin α + sin β = 2 or

B=

µ0 i 2π d

(ii) The direction of magnetic field at a point P due to a long i straight wire can be found by the right hand thumb rule. B If we stretch the thumb of the right hand along the B current and curl our fingers to pass through P, the i direction of the fingers at P gives the direction of magnetic field there. Fig. 26.31 1 (iii) B ∝ , i.e. B-d graph for an infinitely long straight wire is a rectangular hyperbola as shown in d the figure. B

d

Fig. 26.32

Chapter 26

Magnetics — 357

Magnetic Field on the Axis of a Circular Coil Suppose a current carrying circular loop has a radius R. Current in the loop is i. We want to find the magnetic field at a point P on the axis of the loop a distance z from the centre. We can take the loop in xy-plane with its centre at origin and point P on the z-axis. P (0, 0, z)

P z

y

y θ x

O

Q (R cos θ, R sin θ, 0)

O

i

x

i

Fig. 26.33

Fig. 26.34

Let us take a small current element at angle θ as shown. y Rd θ

θ θ

x

Q

θ

O

Fig. 26.35

P ≡ (0, 0, z ) Q ≡ ( R cos θ, R sin θ, 0) d l = – ( Rdθ ) sin θ$i + ( Rdθ ) cos θ$j r$ = unit vector along QP (– R cos θ$i – R sin θ$j + zk$ ) = r Here,

r = distance QP = R 2 + z 2

Now, magnetic field at point P, due to current element d l at Q is µ i d B = 0 2 ( d l × r$ ) 4π r = or

µ0 i [(– R sin θ dθ$i + R cos θ dθ$j) × (– R cos θ$i – R sin θ$j + zk$ )] 4π r 3

µ0 i [( zR cos θ dθ )$i + ( zR sin θ dθ ) $j + ( R 2 dθ ) k$ ] 4π r 3 = dB $i + dB $j + dB k$

dB =

x

Here,

dB x =

y

z

µ0 i ( zR cos θdθ ) 4π r 3

358 — Electricity and Magnetism µ0 4π µ dB z = 0 4π

dB y = and

i

( zR sin θdθ )

r3 i

( R 2 dθ )

3

r Integrating these differentials from θ = 0° to θ = 2π for the complete loop, we get µ ziR 2π B x = 0 3 ∫ cos θ dθ = 0 4π r 0

and

By =

µ 0 ziR 4π r 3

∫0

Bz =

µ 0 iR 2 4π r 3

∫0

Substituting r = ( R 2 + z 2 )1/ 2 , we get

sin θ dθ = 0

2π

BP = B z = B=

For N number of loops,

2π

dθ =

µ 0 iR 2 2 r3

µ 0 iR 2 2 ( R 2 + z 2 ) 3/ 2

µ 0 NiR 2 2 ( R 2 + z 2 ) 3/ 2

Note down the following points regarding a circular current carrying loop. (i) At the centre of the loop, z = 0 B (centre) =

and

µ 0 Ni 2R

(ii) For z >> R , z 2 + R 2 ≈ z 2 ∴

B=

µ 0 NiR 2 2z 3

 µ  (2 NiπR 2 )  µ 0   2M  = 0 =   3   4π   4π   z  z3

Here, M = magnetic moment of the loop = NiA = NiπR 2 . µ 0 2M . 4π r 3 (iii) Direction of magnetic field on the axis of a circular loop can be obtained using the right hand thumb rule. If the fingers are curled along the current, the stretched thumb will point towards the magnetic field. This result was expected as the magnetic field on the axis of a dipole is

B i

i B

Fig. 26.36

Chapter 26

Magnetics — 359

(iv) The magnetic field at a point not on the axis is mathematically difficult to calculate. We have shown qualitatively in figure the magnetic field lines due to a circular current.

Fig. 26.37

(v) Magnetic field is maximum at the centre and decreases as we move away from the centre (on the axis of the loop). The B - z graph is somewhat like shown in figure. B µ0Ni 2R

–z

z

O

Fig. 26.38 i

(vi) Magnetic field due to an arc of a circle at the centre is  θ µ i µ  i B =  0 = 0  θ  2π  2R 4π  R  or

θ

µ   i  B = 0  θ  4π   R 

R Inwards

O

Here, θ is to be substituted in radians.

Fig. 26.39

Field Along the Axis of a Solenoid The name solenoid was first given by Ampere to a wire wound in a closely spaced spiral over a hollow cylindrical non-conducting core. If n is the number of turns per unit length, each carries a current i uniformly wound round a cylinder of radius R, the number of turns in length dx are ndx. Thus, the magnetic field at the axial point O due to this element dx is dx θ2

dθ

R θ1

O x

L

Fig. 26.40

θ

360 — Electricity and Magnetism dB =

µ 0 ( indx ) R 2 2 ( R 2 + x 2 ) 3/ 2

…(i)

Its direction is along the axis of the solenoid. From the geometry, we know dx Number of turns = ndx r R θ O dB

x

Fig. 26.41

x = R cot θ dx = – R cosec 2 θ ⋅ dθ Substituting these values in Eq. (i), we get 1 dB = – µ 0 ni sin θ ⋅ dθ 2 Total field B due to the entire solenoid is θ2 1 B = µ 0 ni∫ (– sin θ ) dθ θ1 2 µ ni ∴ B = 0 (cos θ 2 – cos θ 1 ) 2 If the solenoid is very long ( L >> R ) and the point O is chosen at the middle, i.e. if θ 1 = 180° and θ 2 = 0°, then we get B (centre ) = µ 0 ni

[For L >> R ]

1 B (end ) = µ 0 ni 2

[For L >> R ]

At the end of the solenoid, θ 2 = 0°, θ 1 = 90° and we get

Fig. 26.42

Thus, the field at the end of a solenoid is just one half at the centre. The field lines are as shown in Fig. 26.42.

Chapter 26 V

Magnetics — 361

Example 26.13 In a high tension wire electric current runs from east to west. Find the direction of magnetic field at points above and below the wire. B

i

(a) N

E

W

S (b)

Fig. 26.43

When the current flows from east to west, magnetic field lines are circular round it as shown in figure (a). And so, the magnetic field above the wire is towards north and below the wire towards south. Solution

V

Example 26.14 A current path shaped as shown in figure produces a magnetic field at P, the centre of the arc. If the arc subtends an angle of 30° and the radius of the arc is 0.6 m, what are the magnitude and direction of the field produced at P if the current is 3.0 A. A C

i

30° P D

E

Fig. 26.44

The magnetic field at P due to the straight segments AC and DE is zero. CD is arc of circle.  θ   µ i ∴ B=  0   2π   2R  Solution

or

µ   i  B =  0  θ  4π   R 

∴

 3.0 B = (10–7 )    0.6

( N = 1)

 π    6

= 2.62 × 10–7 T

Ans.

362 — Electricity and Magnetism V

Example 26.15 Figure shows a current loop having two circular arcs joined by two radial lines. Find the magnetic field B at the centre O. i

C

D A

B

θ

b a O

Fig. 26.45

2

Solution Magnetic field at point O, due to wires CB and AD will be zero. Magnetic field due to wire BA will be  θ   µ i B1 =    0   2π   2a 

Direction of field B1 is coming out of the plane of the figure. Similarly, field at O due to arc DC will be  θ   µ i B2 =    0   2π   2b  Direction of field B 2 is going into the plane of the figure. The resultant field at O is B = B1 – B 2 =

µ 0 i θ (b – a ) 4πab

Ans.

Coming out of the plane. V

Example 26.16 The magnetic field B due to a current carrying circular loop of radius 12 cm at its centre is 0.5 × 10 –4 T . Find the magnetic field due to this loop at a point on the axis at a distance of 5.0 cm from the centre. Solution

Magnetic field at the centre of a circular loop is µ i B1 = 0 2R µ 0 iR 2

and that at an axial point,

B2 =

Thus,

B2 R3 = 2 B1 ( R + x 2 ) 3/ 2

or

  R3 B 2 = B1  2 2 3/ 2   (R + x ) 

Substituting the values, we have

  (12) 3 B 2 = ( 0.5 × 10–4 )  3/ 2   (144 + 25) 

2 ( R 2 + x 2 ) 3/ 2

= 3.9 × 10–5 T

Ans.

Chapter 26

INTRODUCTORY EXERCISE

Magnetics — 363

26.5

1. (a) A conductor in the shape of a square of edge length l = 0.4 m carries a current i = 10.0 A. Calculate the magnitude and direction of magnetic field i at the centre of the square. (b) If this conductor is formed into a single circular turn and carries the same current, what is the value of the magnetic field at the centre. l

Fig. 26.46

2. Determine the magnetic field at point P located a distance x from the corner of an infinitely long wire bent at right angle as shown in figure. The wire carries a steady currenti . x

i

P

i

Fig. 26.47

3. A conductor consists of a circular loop of radius R = 10 cm and two

i = 7.0 A

straight, long sections as shown in figure. The wire lies in the plane of the paper and carries a current of i = 7.00 A. Determine the magnitude and direction of the magnetic field at the centre of the loop. Fig. 26.48

4. The segment of wire shown in figure carries a current of i = 5.0 A, where the radius of the circular arc is R = 3.0 cm. Determine the magnitude and direction of the magnetic field at the origin. (Fig. 26.49) i

R O

Fig. 26.49

5. Consider the current carrying loop shown in figure formed of radial lines and segments of circles whose centres are at point P. Find the magnitude and direction of B at point P. (Fig. 26.50)

b 60° P a Fig. 26.50

364 — Electricity and Magnetism

26.9 Ampere’s Circuital Law The electrical force on a charge is related to the electric field (caused by other charges) by the equation, Fe = qE Just like the gravitational force, the static electrical force is a conservative force. This means that the work done by the static electric force around any closed path is zero. q ∫ E⋅ dl = 0 J

∫ E⋅ dl = 0 V

Hence, we have

In other words, the integral of the static (time independent) electric field around a closed path is zero. What about the integral of the magnetic field around a closed path? That is, we want to determine the value of ∫ B⋅ dl Here, we have to be careful. The quantity B ⋅ d l does not represent some physical quantity, and certainly not work. Although the static magnetic force does no work on a moving charge, we cannot conclude that the path integral of the magnetic field around a closed path is zero. We are just curious about what this analogous line integral amounts to. The line integral ∫ B ⋅ d l of the resultant magnetic field along a closed, plane curve is equal to µ 0 times the total current crossing the area bounded by the closed curve provided the electric field inside the loop remains constant. Thus,

∫ B ⋅ d l = µ 0 (i net )

…(i)

This is known as Ampere's circuital law. Eq. (i) in simplified form can be written as Bl = µ 0 ( i net )

…(ii)

But this equation can be used only under the following conditions. (i) At every point of the closed path B || d l. (ii) Magnetic field has the same magnitude B at all places on the closed path. If this is not the case, then Eq. (i) is written as B1 dl1 cos θ 1 + B 2 dl2 cos θ 2 +… = µ 0 ( i net ) Here, θ 1 is the angle between B1 and dl1 , θ 2 the angle between B 2 and dl 2 and so on. Besides the Biot Savart law, Ampere’s law gives another method to calculate the magnetic field due to a given current distribution. Ampere’s law may be derived from the Biot Savart law and Bio Savart law may be derived from the Ampere’s law. However, Ampere's law is more useful under certain symmetrical conditions. To illustrate the theory now let us take few applications of Ampere ’s circuital law.

Magnetic Field Created by a Long Current Carrying Wire A long straight wire of radius R carries a steady current i that is uniformly distributed through the cross-section of the wire.

Chapter 26 For finding the behaviour of magnetic field due to this wire, let us divide the whole region into two parts. One is r ≥ R and the another is r < R . Here, r is the distance from the centre of the wire. For r ≥ R : Let us choose for our path of integration circle 1. From symmetry B must be constant in magnitude and parallel to dl at every point on this circle. Because the total current passing through the plane of the circle is i. Ampere’s law gives

∫ B ⋅ d l = µ 0 i net

∴

2 R r r

dl

[simplified form]

B (2πr ) = µ 0 i µ i B= 0 2π r

or

i 1

Fig. 26.51

Bl = µ 0 i

or

Magnetics — 365

[for r ≥ R ] …(iii)

For r < R : Here, the current i ′passing through the plane of circle 2 is less than the total current i. Because the current is uniform over the cross-section of the wire, the fraction of the current enclosed by circle 2 must equal the ratio of the area πr 2 enclosed by circle 2 to the cross-sectional area πR 2 of the wire. i ′ πr 2 = i πR 2

 r2  i′=  2  i R 

⇒

Then, the following procedure same as for circle 1, we apply Ampere’s law to circle 2.

∫ B ⋅ d l = µ 0 i net Bl = µ 0 i ′

[simplified form]

∴

r  B (2πr ) = µ 0  2  i R 

∴

 µ i  B = 0 2 r  2π R 

2

[For r < R ] …(iv)

This result is similar in the form to the expression for the electric field inside a uniformly charged sphere. The magnitude of the magnetic field versus r for this configuration is plotted in figure. Note that inside the wire B → 0 as r → 0. Note also that Eqs. (iii) and (iv) give the same value of the magnetic field at r = R , demonstrating that the magnetic field is continuous at the surface of the wire.

B

r

B

O

B

R Fig. 26.52

1 r

r

366 — Electricity and Magnetism Magnetic Field of a Solenoid A solenoid is a long wire wound in the form of a helix. With this configuration, a reasonably uniform magnetic field can be produced in the space surrounded by the turns of wire, which we shall call the interior of the solenoid, when the solenoid carries a current. When the turns are closely spaced, each can be approximated as a circular loop, and the net magnetic field is the vector sum of the fields resulting from all the turns (as done in Art. 26.9). If the turns are closely spaced and the solenoid is of infinite length, the magnetic field lines are as shown in Fig. 26.53. S

N

Fig. 26.53

One end of the solenoid behaves like the north pole ( ) and the opposite end behaves like the south pole ( ). As the length of the solenoid increases, the interior field becomes more uniform and the exterior field becomes weaker. An ideal solenoid is approached when the turns are closely spaced and the length is much greater than the radius of the turns. In this case, the external field is zero, and the interior field is uniform over a great volume. We can use Ampere’s law to obtain an expression for the interior magnetic field in an ideal solenoid. Fig. 26.54 shows a longitudinal cross-section of part of such a solenoid carrying a current i. Because the solenoid is ideal, B in the interior space is uniform and parallel to the axis, and B in the exterior space is zero. ×

w

× ×

2

×

1

×

3 l

× × ×

4

× × ×

Fig. 26.54

Consider the rectangular path of length l and width w as shown in figure. We can apply Ampere’s law to this path by evaluating the line integral B ⋅ d l over each side of the rectangle.

Chapter 26

∫ (B ⋅ d l ) side 3 = 0

B =0

as

∫ (B ⋅ d l ) side 2 and 4 = 0 ∫ (B ⋅ d l ) side1 = Bl

Magnetics — 367

as

B ⊥ d l or B = 0 along these paths

as

B is uniform and parallel to dl

The integral over the closed rectangular path is therefore,

∫ B ⋅ d l = Bl The right side of Ampere’s law involves the total current passing through the area bounded by the path of integration. In this case, i net = (number of turns inside the area) (current through each turn) (n = number of turns per unit length) = ( nl) ( i)

∫ B ⋅ d l = µ 0 i net

Using Ampere’s law,

Bl = (µ 0 ) ( nli) or

or

B = µ 0 ni

…(v)

This result is same as obtained in Art. 26.9. Eq. (v) is valid only for points near the centre (that is far from the ends) of a very long solenoid. The field near each end is half the value given by Eq. (v). V

Example 26.17 A closed curve encircles several conductors. The line integral −7 ∫ B ⋅ d l around this curve is 3.83 × 10 T - m. (a) What is the net current in the conductors? (b) If you were to integrate around the curve in the opposite directions, what would be the value of the line integral? Solution

(a)

∫ B ⋅ d l = µ 0 i net

∴

i net =

∫ B ⋅ d l = 3.83 × 10−7 4π × 10−7

µ0

= 0.3A

(b) In opposite direction, line integral will be negative. V

Example 26.18 An infinitely long hollow conducting cylinder with inner radius R/2 and outer radius R carries a uniform current density along its length. The magnitude of the magnetic field,|B| as a function of the radial distance r from the axis is best represented by (JEE 2012) (a)

|B|

| B|

(b) R/2

(c)

R

r

|B|

R/2

(d) R/2

R

r

R

r

|B|

R/2

R

r

368 — Electricity and Magnetism Solution ×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×R/2×

×

×

×

×

×

×

× × R×

×

× ×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

Fig. 26.55

r = distance of a point from centre For r ≤ R/2 Using Ampere’s circuital law,

∫ B ⋅ d l = µ 0 i net or

Bl = µ 0 ( I in )

or

B ( 2πr ) = µ 0 ( I in ) µ I B = 0 in 2π r

or

I in = 0

Since, ∴ For

…(i)

B=0 2   R  I in = πr 2 − π    σ  2   

R ≤ r≤ R 2

Here, σ = current per unit area Substituting in Eq. (i), we have  2 R2  πr − π σ 4  µ0  B= 2π r = At At

µ 0σ  2 R 2  r −  2r  4 

R ,B=0 2 3µ 0σR r = R, B = 8

r=

For r ≥ R

I in = I Total = I (say)

Therefore, substituting in Eq. (i), we have B= ∴ The correct graph is (d).

µ0 I . 2π r

or

B∝

1 r

Chapter 26 V

Magnetics — 369

Example 26.19 A device called a toroid (figure) is often used to create an almost uniform magnetic field in some enclosed area. The device consists of a conducting wire wrapped around a ring (a torus) made of a non-conducting material. For a toroid having N closely spaced turns of wire, calculate the magnetic field in the region occupied by the torus, a distance r from the centre. Solution

To calculate this field, we must evaluate

∫ B⋅ dl

over the circle of radius r. By

symmetry we see that the magnitude of the field is constant on this circle and tangent to it.

∫ B ⋅ d l = Bl = B ( 2πr )

So,

B

r

i i

Fig. 26.56

Furthermore, the circular closed path surrounds N loops of wire, each of which carries a current i. Therefore, right side of Eq. (i) is µ 0 Ni in this case. ∴

∫ B ⋅ d l = µ 0 i net

or

B ( 2πr ) = µ 0 Ni µ Ni B= 0 2πr

or

1 This result shows that B ∝ and hence is non-uniform in the region occupied by torus. However, r if r is very large compared with the cross-sectional radius of the torus, then the field is approximately uniform inside the torus. In that case, × ×

×

×

×

×

× ×

×

Fig. 26.57

N = n = number of turns per unit length of torus 2πr ∴ B = µ 0 ni

370 — Electricity and Magnetism For an ideal toroid, in which turns are closely spaced, the external magnetic field is zero. This is because the net current passing through any circular path lying outside the toroid is zero. Therefore, from Ampere’s law we find that B = 0, in the regions exterior to the torus.

INTRODUCTORY EXERCISE

26.6

1. Figure given in the question is a cross-sectional view of a coaxial cable. The centre conductor is surrounded by a rubber layer, which is surrounded by an outer conductor, which is surrounded by another rubber layer. The current in the inner conductor is 1.0 A out of the page, and the current in the outer conductor is 3.0 A into the page. Determine the magnitude and direction of the magnetic field at points a and b .

×

3A

×

× 1A

a ×

×

b

× ×

×

1mm1mm1mm

Fig. 26.58

2. Figure shows, in cross-section, several conductors that carry currents through the plane of the figure. The currents have the magnitudes I1 = 4.0 A , I 2 = 6.0 A, and I 3 = 2.0 A, in the directions shown. Four paths labelled a to d, are shown. What is the line integral ∫ B ⋅ d l for each path? Each integral involves going around the path in the counter-clockwise direction.

a

I1

I3 b I2 c

d

Fig. 26.59

3. A current I flows along the length of an infinitely long, straight, thin-walled pipe. Then, (JEE 1993) (a) (b) (c) (d)

the magnetic field at all points inside the pipe is the same, but not zero the magnetic field at any point inside the pipe is zero the magnetic field is zero only on the axis of the pipe the magnetic field is different at different points inside the pipe

Chapter 26

Magnetics — 371

26.10 Force Between Parallel Current Carrying Wires Consider two long wires 1 and 2 kept parallel to each other at a distance r and carrying currents i1 and i2 respectively in the same direction. 1

2

i1

i2 F

×

dl

r Fig. 26.60

Magnetic field on wire 2 due to current in wire 1 is, B =

µ 0 i1 ⋅ 2π r

[in ⊗ direction]

Magnetic force on a small element dl of wire 2 due to this magnetic field is d F = i2 ( d l × B) Magnitude of this force is

dF = i2 [( dl) ( B ) sin 90° ] µ i  µ i i = i2 ( dl)  0 1  = 0 ⋅ 1 2 ⋅ dl  2π r  2π r

Direction of this force is along dl × B or towards the wire 1. The force per unit length of wire 2 due to wire 1 is dF µ 0 i1 i2 = dl 2π r The same force acts on wire 1 due to wire 2. The wires attract each other if currents in the wires are flowing in the same direction and they repel each other if the currents are in opposite directions. V

Example 26.20 Two long parallel wires are separated by a distance of 2.50 cm. The force per unit length that each wire exerts on the other is 4.00 × 10 −5 N / m , and the wires repel each other. The current in one wire is 0.600 A. (a) What is the current in the second wire? (b) Are the two currents in the same direction or in opposite directions? F  µ 0  i1 i 2 Solution (a) =  l  2π  r ∴

4 × 10−5 =

( 2 × 10−7 ) (0.6) i 2 2.5 × 10−2

∴ i 2 = 8.33 A (b) Wires repel each other if currents are in opposite directions.

372 — Electricity and Magnetism V

Example 26.21 Consider three long straight parallel wires as shown in figure. Find the force experienced by a 25 cm length of wire C. D

C

G

3 cm

30 A

5 cm

10 A

20 A

Fig. 26.61

Solution

Repulsion by wire D ,

[towards right]

µ i i l F1 = 0 1 2 2π r =

( 2 × 10−7 ) ( 30 × 10) 3 × 10−2

(0.25)

= 5 × 10−4 N Repulsion by wire G,

[towards left] F2 =

( 2 × 10

−7

) ( 20 × 10)

5 × 10−2

(0.25)

= 2 × 10−4 N ∴

F net = F1 − F2 = 3 × 10−4 N

[towards right]

26.11 Magnetic Poles and Bar Magnets In electricity, the isolated charge q is the simplest structure that can exist. If two such charges of opposite sign are placed near each other, they form an electric dipole characterized by an electric dipole moment p. In magnetism isolated magnetic ‘poles’ which would correspond to isolated electric charges do not exist. The simplest magnetic structure is the magnetic dipole, characterized by a magnetic dipole moment M. A current loop, a bar magnet and a solenoid of finite length are examples of magnetic dipoles. When a magnetic dipole is placed in an external magnetic field B, a magnetic torque τ acts on it, which is given by τ =M×B Alternatively, we can measure B due to the dipole at a point along its axis a (large) distance r from its centre by the expression, µ 2M B= 0⋅ 3 4π r

Chapter 26

Magnetics — 373

A bar magnet might be viewed as two poles (North and South) separated by some distance. However, all attempts to isolate these poles fail. If a magnet is broken, the fragments prove to be dipoles and not isolated poles. If we break up a magnet into the electrons and nuclei that make up its atoms, it will be found that even these elementary particles are magnetic dipoles. N

N Fig. 26.62

S

S

N

S

N

S

If a bar magnet is broken, each fragment becomes a small dipole.

Each current carrying loop is just like a magnetic dipole, whose magnetic dipole moment is given by M = niA i S

N

Fig. 26.63

Here, n is the number of turns in the loop, i is the current and A represents the area vector of the current loop. The behaviour of a current loop can be described by the following hypothetical model: (i) There are two magnetic charges; positive magnetic charge and negative magnetic charge. We call the positive magnetic charge a north pole and the negative magnetic charge as the south pole. Every pole has a pole strength m. The unit of pole strength is A-m. (ii) A magnetic charge placed in a magnetic field experiences a force, F = mB The force on positive magnetic charge is along the field and a force on a negative magnetic charge is opposite to the field. (iii) A magnetic dipole is formed when a negative magnetic charge −m and a positive magnetic charge +m are placed at a small separation d. The magnetic dipole moment is M = md The direction of M is from −m to +m.

Geometrical Length and Magnetic Length In case of a bar magnet, the poles appear at points which are slightly inside the two ends. The distance between the locations of the assumed poles is called the magnetic length of the magnet. The distance between the ends is called the geometrical length. The magnetic length of a bar magnet is written as 2l. If m be the pole strength and 2l the magnetic length of a bar magnet, then its magnetic moment is M = 2ml

Geometrical length S

N

Magnetic length Fig. 26.64

374 — Electricity and Magnetism Extra Points to Remember Current carrying loop, solenoid etc. are just like magnetic dipoles, whose dipole moment M is equal to NiA. Direction of M is from south pole (S) to north pole (N ).

˜

The behaviour of a magnetic dipole (may be a bar magnet also) is similar to the behaviour of an electric dipole. The only difference is that the electric dipole moment p is replaced by magnetic dipole moment M and the µ 1 constant is replaced by 0 . 4 πε0 4π

˜

Table given below makes a comparison between an electric dipole and a magnetic dipole.

˜

Table 26.2 S.No.

Physical quality to be compared

Electric dipole

Magnetic dipole

p = q (2 l )

M = m (2 l )

1.

Dipole moment

2.

Direction of dipole moment

3.

Net force in uniform field

4.

Net torque in uniform field

5.

Field at far away point on the axis

1 2p ⋅ 4 πε0 r 3

6.

Field at far away point on perpendicular bisector

1 p ⋅ 4 πε0 r 3

7.

Potential energy

U θ = − p ⋅ E = − pE cos θ

8.

Work done in rotating the dipole

From negative charge to the positive charge

From south to north pole

0

0

τ =p×E

τ =M×B

(along p)

(opposite to p)

Wθ1 − θ 2 = pE (cos θ1 − cos θ2 )

µ 0 2M ⋅ (along M) 4π r3 µ0 M ⋅ 4π r3

(opposite to M )

U θ = − M ⋅ B = − MB cos θ Wθ1 − θ 2 = MB (cos θ1 − cos θ2 )

Note In the above table, θ is the angle between field ( E or B) and dipole moment ( p or M ). V

Example 26.22 Calculate the magnetic induction (or magnetic field) at a point 1 Å away from a proton, measured along its axis of spin. The magnetic moment of the proton is 1.4 × 10 −26 A- m2 . On the axis of a magnetic dipole, magnetic induction is given by µ 2M B= 0 ⋅ 3 4π r Substituting the values, we get (10−7 ) (2) (1.4 × 10−26 ) B= (10−10 ) 3 Solution

= 2.8 × 10−3 T = 2.8 mT

Ans.

Chapter 26 V

Magnetics — 375

Example 26.23 A bar magnet of magnetic moment 2.0 A-m2 is free to rotate about a vertical axis through its centre. The magnet is released from rest from the east-west position. Find the kinetic energy of the magnet as it takes the north-south position. The horizontal component of the earth’s magnetic field is B = 25 µT . Earth’s magnetic field is from south to north. Solution

Gain in kinetic energy = loss in potential energy

Thus, KE = U i − U f U = − MB cos θ

As,

 π KE = − MB cos   − ( − MB cos 0° )  2

∴

= MB Substituting the values, we have KE = ( 2. 0) ( 25 × 10−6 ) J = 50 µJ

Ans.

26.12 Earth’s Magnetism Our earth behaves as it has a powerful magnet within it. The value of magnetic field on the surface of earth is a few tenths of a gauss (1 G = 10 −4 T ). The earth’s south magnetic pole is located near the north geographic pole and the earth’s north magnetic pole is located near the south geographic pole. In fact, the configuration of the earth’s magnetic field is very much like the one that would be achieved by burying a gigantic bar magnet deep in the interior of the earth. Axis of rotation of the earth Geographical North pole Magnetic South pole S

11.5°

Magnetic equator

North

N (b)

Magnetic North pole

Geographical South pole (a) Fig. 26.65

The axis of earth’s magnet makes an angle of 11.5° with the earth’s rotational axis.

376 — Electricity and Magnetism Theories Regarding the Origin of Earth’s Magnetism First Theory : Gilbert for the first time in 1600, gave the idea that there is a powerful magnet within the earth at its centre. Later on this theory was denied because the temperature in the interior of the earth is so high that it is impossible to retain its magnetism. Second Theory : The second theory was put forward by Grover in 1849. He put the view that the earth magnetism is due to electric currents flowing near the outer surface of the earth. Hot air, rising from the region near equator, goes towards north and south hemispheres and become electrified. These currents magnetise the ferromagnetic material near the outer surface of the earth. Third Theory : There are many conducting materials including iron and nickel in the molten state within the central core of the earth. Conventional currents are generated in this semifluid core due to earth’s rotation about its axis. Due to these currents, magnetism is generated within the earth. Till date not a single theory can explain all events regarding earth’s magnetism.

Elements of Earth’s Magnetism There are three elements of earth’s magnetism. (i) Angle of Declination ( α ) At any point (say P) on earth’s surface the longitude determines the north-south direction. The vertical plane in the direction of longitude or the vertical plane passing through the line joining the geographical north and south poles is called the ‘geographical meridian’. At point P, there also exists the magnetic field B. A vertical plane in the direction of B is called ‘magnetic meridian’. At any place the acute angle between the magnetic meridian and the geographical meridian is called ‘angle of declination ‘α’. (ii) Angle of Dip ( θ ) ‘The angle of dip (θ ) at a place is the angle between the direction of earth’s magnetic field and the horizontal at that place.’ Angle of dip at some place can be measured from a magnetic needle free to rotate in a vertical plane about a horizontal axis passing through centre of gravity of the needle. At earth’s magnetic poles the magnetic field of earth is vertical, i.e. angle of dip is 90°, the freely suspended magnetic needle is vertical there. At magnetic equator field is horizontal, or angle of dip is 0°. The needle is horizontal. In northern hemisphere, the north pole of the magnetic needle inclines downwards, whereas in the southern hemisphere the south pole of the needle inclines downwards. (iii) Horizontal Component of Earth’s Magnetic Field Let B e be the net magnetic field at some point. H and V be the horizontal and vertical components of B e . Let θ is the angle of dip at the same place, then we can see that Geographical North

L P

Magnetic North

α H

Geographical meridian Magnetic meridian

O θ S

N M BE Q Fig. 26.66

V R

Chapter 26 H = B e cos θ and V = B e sin θ Squaring and adding Eqs. (i) and (ii), we get

Magnetics — 377 …(i) …(ii)

Be = H 2 + V 2 Further, dividing Eq. (ii) by Eq. (i), we get V  θ = tan −1   H By knowing H and θ at some place we can find B e and V at that place.

Neutral Points When a magnet is placed at some point on earth’s surface, there are points where horizontal component of earth’s magnetic field is just equal and opposite to the field due to the magnet. Such points are called neutral points. If a magnetic compass is placed at a neutral point, no force acts on it and it may set in any direction. Suppose a small bar magnet is placed such that north pole of the magnet is towards the magnetic south pole of the earth then neutral points are obtained both sides on the axis of the magnet. If distance of each neutral point from the middle point of a magnet be r, and the magnitude of the magnetic moment of the magnet be M, then µ 0 2M ⋅ =H 4π r 3 When north pole of bar magnet is towards the magnetic north pole of the earth, the neutral points are obtained on perpendicular bisectors of the magnet. Let r be the distance of neutral points from centre, then µ0 M ⋅ =H 4π r 3 V

Example 26.24 In the magnetic meridian of a certain place, the horizontal component of earth’s magnetic field is 0.26 G and the dip angle is 60°. Find (a) vertical component of earth’s magnetic field. (b) the net magnetic field at this place. Solution

Given,

(a) tan θ =

V H

∴

H = 0.26 G and

θ = 60°

V = H tan θ = ( 0.26) tan 60° = 0.45 G

Ans.

(b) H = B e cos θ ∴

Be =

H 0.26 = cos θ cos 60°

= 0.52 G

Ans.

378 — Electricity and Magnetism V

Example 26.25 A magnetic needle suspended in a vertical plane at 30° from the magnetic meridian makes an angle of 45° with the horizontal. Find the true angle of dip. Solution

In a vertical plane at 30° from the magnetic meridian, the horizontal component is H ′ = H cos 30° H Magnetic meridian

30° 0° os 3 Hc

O

V

Fig. 26.67

While vertical component is still V. Therefore, apparent dip will be given by V V tan θ′ = = H ′ H cos 30° But, ∴ ∴

V = tan θ H tan θ tan θ′ = cos 30°

(where, θ = true angle of dip)

θ = tan −1 [tan θ′ cos 30° ] = tan −1 [(tan 45° ) (cos 30° )] ≈ 41°

Ans.

26.13 Vibration Magnetometer Vibration magnetometer is an instrument which is used for the following two purposes: (i) To find magnetic moment of a bar magnet. (ii) To compare magnetic fields of two magnets. The construction of a vibration magnetometer is as shown in figure.The magnet shown in figure is free to rotate in a horizontal plane. The magnet stays parallel to the horizontal component of earth’s magnetic field. If the magnet is now displaced through an angle θ, a restoring torque of magnitude MH sin θ acts on it and the magnet starts oscillating. From the theory of simple harmonic motion, we can find the time period of oscillations of the magnet. Restoring torque in displaced position is τ = − MH sin θ

Torsion head Screw Glass tube

S1

S2

S

N Plane mirror

Magnetic meridian Fig. 26.68

…(i)

Chapter 26

Magnetics — 379

Here, M = Magnetic moment of the magnet and H = Horizontal component of earth’s magnetic field. Negative sign shows the restoring nature of torque. Now since, τ = Iα and sin θ ≈ θ for small angular displacement. Thus, Eq. (i) can be written as Iα = − MHθ Since, α is proportional to −θ. Therefore, motion is simple harmonic in nature, time period of which will be given by θ T = 2π   = 2π α ∴

T = 2π

I MH

I MH …(ii)

In the expression of T, I is the moment of inertia of the magnet about its axis of vibration. (i) Measurement of Magnetic Moment : By finding time period T of vibrations of the given magnet, we can calculate magnetic moment M by the relation, 4π 2 I M= 2 T H (ii) Comparison of Two Magnetic Fields : Suppose we wish to compare the magnetic fields B1 and B 2 at some point P due to two magnets. For this, vibration magnetometer is so placed that the centre of its magnet lies on P. Now, one of the given magnets is placed at some known distance from P in the magnetic meridian, such that point P lies on its axial line and its north pole points north. In this position, the field B1 at P produced by the magnet will be in the direction of H. Hence, the magnet suspended in the magnetometer will vibrate in the resultant magnetic field ( H + B1 ). Its period of vibration is noted, say it is T1 , then I T1 = 2π M ( H + B1 ) Now, the first magnet is replaced by the second magnet and the second magnet is placed in the same position and again the time period is noted. If the field produced at P due to this magnet be B 2 and the new time period be T2 , then I T2 = 2π M (H + B 2 ) Finally, the time period of the magnetometer under the influence of the earth’s magnetic field alone is determined. Let it be T, then I T = 2π MH Solving above three equations for T , T1 and T2 , we can show that B1 (T 2 − T12 ) T22 = B 2 (T 2 − T22 ) T12

380 — Electricity and Magnetism V

Example 26.26 A short bar magnet is placed with its north pole pointing north. The neutral point is 10 cm away from the centre of the magnet. If H = 0.4 G, calculate the magnetic moment of the magnet. When north pole of the magnet points towards magnetic north, null point is obtained on perpendicular bisector of the magnet. Simultaneously, magnetic field due to the bar magnet should be equal to the horizontal component of earth’s magnetic field H. Solution

H=

Thus,

µ0 M ⋅ 4π r 3

or

M=

Hr 3 (µ 0 /4π )

Substituting the values, we have M= V

( 0.4 × 10−4 ) (10 × 10−2 ) 3 10

−7

= 0.4 A-m 2

Ans.

Example 26.27 A magnetic needle performs 20 oscillations per minute in a horizontal plane. If the angle of dip be 30°, then how many oscillations per minute will this needle perform in vertical north-south plane and in vertical east-west plane? Solution

In horizontal plane, the magnetic needle oscillates in horizontal component H.

∴

T = 2π

I MH

In the vertical north-south plane (magnetic meridian), the needle oscillates in the total earth’s magnetic field B e , and in vertical east-west plane (plane perpendicular to the magnetic meridian) it oscillates only in earth’s vertical component V. If its time period be T1 and T2 , then I MB e

and T2 = 2π

2

=

H Be

or

n 22 2

=

V H

T1 = 2π

I MV

From above equations, we can find T12 T Similarly, Further, and ∴

n12 n

2

=

Be H

n Be 2 = sec θ = sec 30° = H 3 V 1 = tan θ = tan 30° = H 3 B  2   n12 = ( n ) 2  e  = ( 20) 2   H  3

or

n1 = 21.5 oscillations/min

and

 1 V  n 22 = ( n ) 2   = ( 20) 2    H  3

∴

n 2 = 15.2 oscillations/min

Ans.

Ans.

Chapter 26

Magnetics — 381

26.14 Magnetic Induction and Magnetic Materials We know that the electric lines of force change when a dielectric is placed between the parallel plates of a capacitor. Experiments show that magnetic lines also get modified due to the presence of certain materials in the magnetic field. Few substances such as O 2 , air, platinum, aluminium etc., show a very small increase in the magnetic flux passing through them, when placed in a magnetic field. Such substances are called paramagnetic substances. Few other substances such as H 2 , H 2O, Cu, Zn, Sb etc. show a very small decrease in flux and are said to be diamagnetic. There are other substances like Fe, Co etc. through which the flux increases to a larger value and are known as ferromagnetic substances.

Magnetisation of Matter A material body is consisting of large number of atoms and thus large number of electrons. Each electron produces orbital and spin magnetic moments and can be assumed as magnetic dipoles. In the absence of any external magnetic field, the dipoles of individual atoms are randomly oriented and the magnetic moments thus, cancel. When we apply an external magnetic field to a substance, two processes may occur. (i) All atoms which have non-zero magnetic moment are aligned along the magnetic field. (ii) If the atom has a zero magnetic moment, the applied magnetic field distorts the electron orbit and thus, induces magnetic moment in opposite directions. In diatomic substances, the individual atoms do not have a magnetic moment by its own. When an external field is applied, the second process occurs. The induced magnetic moment is thus set up in the direction opposite to B. In this case, the magnetic flux density in the interior of the body will be less than that of the external field B. In paramagnetic substances, the constituent atoms have intrinsic magnetic moments. When an external magnetic field is applied, both of the above processes occur and the resultant magnetic moment is always in the direction of magnetic field B as the first effect predominates over the second.

26.15 Some Important Terms Used in Magnetism Magnetic Induction ( B) When a piece of any substance is placed in an external magnetic field, the substance becomes magnetised. If an iron bar is placed in a uniform S N S N magnetic field, the magnetised bar produces its own magnetic field in the same direction as those of the original field inside the bar, but in opposite direction outside the bar. This results in a concentration of the lines of force within the bar. (a) (b) The magnetic flux density within the bar is Fig. 26.69 increased whereas it becomes weak at certain places outside the bar. “The number of magnetic lines of induction inside a magnetic substance crossing unit area normal to their direction is called the magnitude of magnetic induction, or magnetic flux density inside the

382 — Electricity and Magnetism substance. It is denoted by B. The SI unit of B is tesla (T) or weber/metre 2 (Wb/m 2 ). The CGS unit is gauss (G). 1 Wb/ m 2 = 1T = 10 4 G

Intensity of Magnetisation ( I ) “Intensity of magnetisation ( I ) is defined as the magnetic moment per unit volume of the magnetised substance.” This basically represents the extent to which the substance is magnetised. Thus, M I= V The SI unit of I is ampere/metre (A/m).

Magnetic Intensity or Magnetic Field Strength (H ) When a substance is placed in an external magnetic field, the actual magnetic field inside the substance is the sum of the external field and the field due to its magnetisation. The capability of the magnetising field to magnetise the substance is expressed by means of a vector H, called the ‘magnetic intensity’ of the field. It is defined through the vector relation, B H= −I µ0 The SI unit of H is same as that of I, i.e. ampere/metre (A/m). The CGS unit is oersted.

Magnetic Permeability (µ) “It is defined as the ratio of the magnetic induction B inside the magnetised substance to the magnetic intensity H of the magnetising field, i.e. B µ= H It is basically a measure of conduction of magnetic lines of force through it. The SI unit of magnetic permeability is weber/ampere-metre (Wb/A-m).

Relative Magnetic Permeability (µ r ) It is the ratio of the magnetic permeability µ of the substance to the permeability of free space. µ Thus, µr = µ0 µ r is a pure ratio, hence, dimensionless. For vacuum its value is 1. µ r can also be defined as the ratio of the magnetic field B in the substance when placed in magnetic field B 0 . Thus, B µr = B0 For paramagnetic substance, µ r >1, For diamagnetic substance, µ r >1.

Chapter 26

Magnetics — 383

Magnetic Susceptibility ( χ m ) We know that both diamagnetic and paramagnetic substances develop a magnetic moment depending on the applied field. Magnetic susceptibility is a measure of how easily a substance is magnetised in a magnetising field. For paramagnetic and diamagnetic substances, I, H and χ m are related by the equation, I = χ mH I or χm = H Thus, the magnetic susceptibility χ m may be defined as the ratio of the intensity of magnetisation to the magnetic intensity of the magnetising field. Since, I and H have the same units, χ m is unitless. It is a pure number. By doing simple calculation, we can prove that µ r and χ m are related by µ r =1 + χ m For paramagnetic substances χ m is slightly positive. For diamagnetic substances, it is slightly negative and for ferromagnetic substances, χ m is positive and very large.

26.16 Properties of Magnetic Materials As discussed earlier, all substances (whether solid, liquid or gaseous) may be classified into three categories in terms of their magnetic properties. (i) paramagnetic, (ii) diamagnetic and (iii) ferromagnetic.

Paramagnetic Substances Examples of such substances are platinum, aluminium, chromium, manganese, CuSO 4 solution etc. They have the following properties: (i) The substances when placed in a magnetic field, acquire a feeble magnetisation in the same sense as the applied field. Thus, the magnetic inductance inside the substance is slightly greater than outside to it. (ii) In a uniform magnetic field, these substances rotate until their longest axes are parallel to the field. (iii) These substances are attracted towards regions of stronger magnetic field when placed in a non-uniform magnetic field.

N

S

Fig. 26.70

Figure shows a strong electromagnet in which one of the pole pieces is sharply pointed while the other is flat. Magnetic field is much stronger near the pointed pole than near the flat pole. If a small piece of paramagnetic material is suspended in this region, a force can be observed in the direction of arrow.

384 — Electricity and Magnetism (iv) If a paramagnetic liquid is filled in a narrow U-tube and one limb is placed in between the pole pieces of an electromagnet such that the level of the liquid is in line with the field, then the liquid will rise in the limb as the field is switched on.

Fig. 26.71

(v) For paramagnetic substances, the relative permeability µ r is slightly greater than one. (vi) At a given temperature the magnetic susceptibility χ m does not change with the magnetising field. However, it varies inversely as the absolute temperature. As temperature increases, χ m decreases. At some higher temperature, χ m becomes negative and the substance becomes diamagnetic.

Diamagnetic Substances Examples of such substances are bismuth, antimony, gold, quartz, water, alcohol etc. They have the following properties: (i) These substances when placed in a magnetic field, acquire feeble magnetisation in a direction opposite to that of the applied field. Thus, the lines of induction inside the N S substance is smaller than that outside to it. (ii) In a uniform field, these substances rotate until their longest axes are normal to the field. Fig. 26.72 (iii) In a non-uniform field, these substances move from stronger to weaker parts of the field. (iv) If a diamagnetic liquid is filled in a narrow U-tube and one limb is placed in between the pole of an electromagnet, the level of liquid depresses when the field is switched on. (v) The relative permeability µ r is slightly less than 1. (vi) The susceptibility χ m of such substances is always negative. It is constant and does not vary with field or the temperature.

Ferromagnetic Substances Examples of such substances are iron, nickel, steel, cobalt and their alloys. These substances resemble to a higher degree with paramagnetic substances as regard their behaviour. They have the following additional properties: (i) These substances are strongly magnetised by even a weak magnetic field. (ii) The relative permeability is very large and is of the order of hundreds and thousands. (iii) The susceptibility is positive and very large.

Chapter 26

Magnetics — 385

(iv) Susceptibility remains constant for very small values of H, increases for larger values of H and then decreases for very large values of H. (v) Susceptibility decreases steadily with the rise of temperature. Above a certain temperature known as Curie Temperature, the ferromagnetic substances become paramagnetic. It is 1000°C for iron, 770°C for steel, 360°C for nickel and 1150°C for cobalt.

26.17 Explanation of Paramagnetism, Diamagnetism

and Ferromagnetism There are three properties of atoms that give rise to magnetic dipole moment. 1. The electrons moving around the nucleus in the orbits act as small current loops and contribute

magnetic moments. 2. The spinning electron has an intrinsic magnetic dipole moment. 3. The nucleus contribute to magnetic moment due to the motion of charge within the nucleus. The magnitude of nuclear moments is about 10 − 3 times that of electronic moments or the spin magnetic moments, as the later two are of the same order. Still most of the magnetic moment of an atom is produced by electron spin, the net contribution of the orbital revolution is very small. This is because most of the electrons pair off in such a way that they produce equal and opposite orbital magnetic moment and they cancel out. Although, the electrons also try to pair up with their opposite spins but in case of spin motion of an electron it is not always possible to form equal and opposite pairs. Tiny bar magnets

In the absence of external magnetic field Fig. 26.73

Paramagnetism The property of paramagnetism is found in those substances whose atoms or molecules have an excess of electrons spinning in the same direction. Hence, atoms of paramagnetic substances have a permanent magnetic moment and behave like tiny bar magnets. In the absence of external magnetic field, the atomic magnets are randomly oriented and net magnetic Fig. 26.74 moment is thus, zero. When paramagnetic substance is placed in an external magnetic field, then each atomic magnet experiences a torque which tends to turn the magnet in the direction of the field. The atomic magnets are thus, aligned in the direction of the field. Thus, the whole substance is magnetised in the direction of the external magnetic field. As the temperature of substance is increased, the thermal agitation disturbs the magnetic alignment of the atoms. Thus, we can say that paramagnetism is temperature dependent.

386 — Electricity and Magnetism Curie’s law According to Curie’s law, magnetic susceptibility of a paramagnetic substance is inversely proportional to absolute temperature T. 1 χm ∝ T The exact law is beyond the scope of our course.

Diamagnetism The property of diamagnetism is generally found in those substances whose atoms (or molecules) have even number of electrons which form pairs. “The net magnetic moment of an atom of a diamagnetic substance is thus zero.” When a diamagnetic substance is placed in an external magnetic field, the spin motion of electrons is so modified that the electrons which produce the magnetic moments in the direction of external field slow down while the electrons which produce magnetic moments in opposite direction get accelerated. Thus, a net magnetic moment is induced in the opposite directions of applied magnetic field. Hence, the substance is magnetised opposite to the external field. Note That diamagnetism is temperature independent.

Ferromagnetism Iron like elements and their alloys are known as ferromagnetic substances. The susceptibility of these substances is in several thousands. Like paramagnetic substances, atoms of ferromagnetic substances have a permanent magnetic moment and behave like tiny magnets. But in ferromagnetic substances the atoms form innumerable small effective regions called ‘domains’.

Unmagnetised Fig. 26.75 −6

The size of the domain vary from about 10 cm 3 to 10 −2 cm 3 . Each domain has 1017 to 10 21 atoms whose magnetic moments are aligned in the same direction. In an unmagnetised ferromagnetic specimen, the domains are oriented randomly, so that their resultant magnetic moment is zero. External field

(a)

(b) Fig. 26.76

Chapter 26

Magnetics — 387

When the specimen is placed in a magnetic field, the resultant magnetisation may increase in two different ways. (a) The domains which are oriented favourably with respect to the field increase in size. Whereas those oriented opposite to the external field are reduced. (b) The domains rotate towards the field direction. Note That if the external field is weak, specimen gets magnetised by the first method and if the field is strong they get magnetised by the second method.

Hysteresis : Retentivity and Coercivity The distinguishing characteristics of a ferromagnetic material is not that it can be strongly magnetised but that the intensity of magnetisation I is not directly proportional to the magnetising field H. If a gradually increasing magnetic field H is applied to an unmagnetised piece of iron, its magnetisation increases non-linearly until it reaches a maximum. I A

B Retentivity C O

F

H

E D Coercivity Fig. 26.77

If I is plotted against H, a curve like OA is obtained. This curve is known as magnetisation curve. At this stage all the dipoles are aligned and I has reached to a maximum or saturated value. If the magnetic field H is now decreased, the I does not return along magnetisation curve but follows path AB. At H = 0, I does not come to its zero value but its value is still near the saturated value. The value of I at this point (i.e. OB) is known as remanence, remanent magnetisation or retentivity. The value of I at this point is known as residual induction. On applying a reverse field the value of I finally becomes zero. The abscissa OC represents the reversed magnetic field needed to demagnetise the specimen. This is known as coercivity of the material. If the reverse field is further increased, a reverse magnetisation is set up which quickly reaches the saturation value. This is shown as CD. If H is now taken back from its negative saturation value to its original positive saturation value, a similar curve DEFA will be traced. The whole graph ABCDEFA thus, forms a closed loop, usually known as hysteresis loop. The whole process described above and the property of the iron characterized by it are called hysteresis. The energy lost per unit volume of a substance in a complete cycle is equal to the area. Thus, we can conclude the following three points from the above discussion: (i) The retentivity of a substance is a measure of the magnetisation remaining in the substance when the magnetising field is removed. (ii) The coercivity of a substance is a measure of the reverse magnetising field required to destroy the residual magnetism of the substance.

388 — Electricity and Magnetism (iii) The energy loss per unit volume of a substance in a complete cycle of magnetisation is equal to the area of the hysteresis loop.

Demagnetisation I It is clear from the hysteresis loop that the intensity of magnetisation I does not reduce to zero on removing the magnetising field H. Further, I is zero when the magnetising field H is equal to the coercive field. At these points the magnetic induction is not zero, and the specimen is not demagnetised. To demagnetise a substance, it is subjected to several cycles is magnetisation, each time with decreasing magnetising field and finally the field is reduced to zero. In this way, the size of the hysteresis curve goes on decreasing and the area finally reduces to zero. Demagnetisation is obtained by placing the specimen in an alternating Fig. 26.78 field of continuously diminishing amplitude. It is also obtained by heating. Ferromagnetic materials become practically non-magnetic at sufficiently high temperatures.

H

Magnetic Properties of Soft Iron and Steel A comparison of the magnetic properties of ferromagnetic substances can be made by the comparison of the shapes and sizes of their hysteresis loops. Following three conclusions can be drawn from their hysteresis loops: (i) Retentivity of soft iron is more than the retentivity of steel. (ii) Coercivity of soft iron is less than the coercivity of steel. (iii) Area of hysteresis loop (i.e. hysteresis loss) in soft iron is smaller than that in steel.

Choice of Magnetic Materials The choice of a magnetic material for different uses is decided from the hysteresis curve of a specimen of the material. (i) Permanent Magnets The materials for a permanent magnet should have (a) high retentivity (so that the magnet is strong) and (b) high coercivity (so that the magnetising is not wiped out by stray magnetic fields). As the material in this case is never put to cyclic changes of magnetisation, hence, hysteresis is immaterial. From the point of view of these facts steel is more suitable for the construction of permanent magnets than soft iron. Modern permanent magnets are made of ‘cobalt-steel’, alloys ‘ticonal’. (ii) Electromagnets The materials for the construction of electromagnets should have (a) high initial permeability (b) low hysteresis loss From the view point of these facts, soft iron is an ideal material for this purpose. (iii) Transformer Cores and Telephone Diaphragms As the magnetic material used in these cases is subjected to cyclic changes. Thus, the essential requirements for the selection of the material are (a) high initial permeability (b) low hysteresis loss to prevent the breakdown

Chapter 26

Magnetics — 389

Electromagnet As we know that a current carrying solenoid behaves like a bar magnet. If we place a soft iron rod in the solenoid, the magnetism of the solenoid increases hundreds of times and the solenoid is called an ‘electromagnet’. It is a temporary magnet.

N

S N

S

(a)

(b) Fig. 26.79

An electromagnet is made by winding closely a number of turns of insulated copper wire over a soft iron straight rod or a horse shoe rod. On passing current through this solenoid, a magnetic field is produced in the space within the solenoid.

Applications of Electromagnets (i) Electromagnets are used in electric bell, transformer, telephone diaphragms etc. (ii) In medical field, they are used in extracting bullets from the human body. (iii) Large electromagnets are used in cranes for lifting and transferring big machines and parts.

26.18 Moving Coil Galvanometer The moving coil galvanometer is a device used to measure an electric current.

Principle Action of a moving coil galvanometer is based upon the principle that when a current carrying coil is placed in a magnetic field, it experiences a torque whose magnitude depends on the magnitude of current. T M W North

Core South

N

F S

× F

Coil

S

Soft iron core Fig. 26.80

390 — Electricity and Magnetism Construction The main parts of a moving coil galvanometer are shown in figure. The galvanometer consists of a coil, with many turns free to rotate about a fixed vertical axis in a uniform radial magnetic field. There is a cylindrical soft iron core, which not only makes the field radial but also increases the strength of magnetic field.

Theory The current to be measured is passed through the galvanometer. As the coil is in the magnetic field (of constant magnitude) it experiences a torque given by τ = MB sin θ = ( NiA ) B sin θ

...(i)

As shown in the figure, the pole pieces are made cylindrical, the magnetic field always remains parallel to the plane of the coil. Or angle between B and M always remains 90 o . Therefore, Eq. (i) can be written as

B

M

(as sin θ = sin 90 o = 1)

τ = NiAB Here, N = total number of turns of the coil

Fig. 26.81

i = current passing through the coil A = area of cross-section of the coil and B = magnitude of radial magnetic field.

This torque rotates the coil. The spring S shown in figure provides a counter torque kφ that balances the above torque NiAB. In equilibrium, ...(ii) kφ = NiAB Here, k is the torsional constant of the spring. With rotation of coil a small light mirror M (attached with phosphor bronze wire W ) also rotates and equilibrium deflection φ can be measured by a lamp and scale arrangement. The above Eq. (ii) can be written as  k  i= φ  NAB 

...(iii)

Hence, the current i is proportional to the deflection φ .

Galvanometer Constant The constant

k in Eq. (iii) is called galvanometer constant. NAB

Hence, Galvanometer constant =

k NAB

...(iv)

This constant may be found by passing a known current through the coil. Measuring the deflection φ and putting these values in Eq. (iii), we can find galvanometer constant.

Chapter 26

Magnetics — 391

Sensitivity of Galvanometer Deflection per unit current (φ i) is called sensitivity of galvanometer. From Eq. (iii), we can see that φ i = NAB k . Hence, φ NAB Sensitivity = = ...(v) i k The sensitivity of a galvanometer can be increased by (i) increasing the number of turns in the coil N or (ii) increasing the magnitude of magnetic field. V

Example 26.28 A rectangular coil of area 5.0 × 10 −4 m2 and 60 turns is pivoted about one of its vertical sides. The coil is in a radial horizontal magnetic field of 9 × 10 −3 T. What is the torsional constant of the spring connected to the coil if a current of 0 .20 mA produces an angular deflection of 18 o ? Solution

From the equation,  k  i= φ  NAB 

We find that torsional constant of the spring is given by NABi k= φ Substituting the values in SI units, we have k=

( 60)( 5.0 × 10−4 )( 9 × 10−3 )( 0.2 × 10−3 ) 18

= 3 × 10−9 N-m/degree

INTRODUCTORY EXERCISE

Ans.

26.7

1. A coil of a moving coil galvanometer twists through 90o when a current of one microampere is passed through it. If the area of the coil is 10−4 m 2 and it has 100 turns, calculate the magnetic field of the magnet of the galvanometer. Given, k = 10−8 N -m /degree. 2. A galvanometer coil 5 cm × 2 cm with 200 turns is suspended vertically in a field of 5 × 10−2 T. The suspension fibre needs a torque of 0.125 × 10−7 N -m to twist it through one radian. Calculate the strength of the current required to be maintained in the coil if we require a deflection of 6°.

392 — Electricity and Magnetism

Final Touch Points 1. Sometimes, a non-conducting charged body is rotated with some angular speed. In this case, the ratio of magnetic moment and angular momentum is constant which is equal to q /2m, where q is the charge and m the mass of the body. ω + + +

R + + +

+ + +

+

+

+ + +

+ + + + + +

+

+

e.g. In case of a ring of mass m, radius R and charge q distributed on its circumference. Angular momentum, L = Iω = (mR 2 ) (ω )

…(i)

2

Magnetic moment, M = iA = (qf ) ( πR ) ω Here, f = frequency = 2π ωR 2 ω M = (q )   ( πR 2 ) = q  2π  2

∴

…(ii)

From Eqs. (i) and (ii), we get M q = L 2m Although this expression is derived for simple case of a ring, it holds good for other bodies also. For example, for a disc or a sphere.

2. Determination of e/m of an Electron (Thomson Method) JJ Thomson in 1897, devised an experiment for the determination of e/m (specific charge) of the electron by using electric and magnetic fields in mutually perpendicular directions. –

+

S

+ A

E

G

C –

O A1 A2

Q

O F

P R

O′

O′ D

The discharge is maintained by the application of high PD between the cathode C and anode A of a discharge tube containing air at a very low pressure (~ 10–2 mm of Hg). The electrons so produced are allowed to pass through slits A1 and A2 also kept at the potential of A. The beam then passes along the axis of the tube and produces a spot of light at O on the fluorescent screen S. The electric field E is applied between two horizontal plates P and Q. The magnetic field B is applied in the direction perpendicular to the paper plane by passing the current through coils, in the region within the dotted

Chapter 26

Magnetics — 393

circle. It is clear from Fleming’s left hand rule, that F due to E is in upward direction, while due to B in downward direction. Hence, fields E and B can be adjusted so that the electrons suffer no deflection and strike at point O on the screen. In this case, or

eE = evB or v = E /B

…(i)

Now, electric field is switched off, the electrons thus, describe the circular arc and fall at O′ on the screen. In this case, the force F = Bev bends the electron beam in a circular arc, such that it is balanced by the centripetal force mv 2 /R . ∴

Bev = mv 2 /R

or

v = BRe /m

…(ii)

e /m = E /RB 2

…(iii)

Combining Eqs. (i) and (ii), we get As E and B are known. To find R, consider arc EF of the circular path in the magnetic field region. From the geometry, we get OO ′ /GO = EF /R or

…(iv)

R = EF × GO /OO ′

Practically, EF is replaced by the width of the magnetic flux region and G is taken at the middle of the region. Thomson’s value for e/m was 1.7 × 1011 C/kg, which is in excellent agreement with the modern value of 175890 . × 1011 C/kg.

3. Cyclotron In 1932, Lawrence developed a machine named cyclotron, for the acceleration of charged particles, such as protons or deuterons. These particles (ions) are caused to move in circular orbits by magnetic field and are accelerated by the electric field. In its simplest form, it consists of two flat semicircular metal boxes, called dees because of their shape. These hollow chambers have their diametric edges parallel and slightly separated from each other. An alternating potential (with frequency of the order of megacycles per second) is applied between the dees. The dees are placed between the poles of a strong electromagnet which provides a magnetic field perpendicular to the plane of the dees. D2

D1

Magnet S D2

D1 S F Magnet

Suppose that at any particular instant the alternating potential is in the direction which makes D1 positive and D2 negative. A positive ion of mass m, charge q starting from the source S (of positive ion) will be attracted by the dee D2. Let its velocity while entering in dee D2 is v. Due to magnetic field B, it will move in a circular path of radius r inside the dee D2, where

or

r =

mv Bq

v =

Bqr m

394 — Electricity and Magnetism In the interior of the dee, the speed of the ion remains constant. After it has traversed half a cycle, the ion comes to the edge of D2. If in the meantime, the potential difference between D1 and D2 has changed direction so that D2 is now positive and D1 negative, the positive ion will receive an additional acceleration, while going across the gap between the dees and speed of ion will increase. Then, it travels in a circular path of larger radius inside D1 under the influence of magnetic field (because r ∝ v ). After traversing a half cycle in D1, it will reach the edge of D1 and receive an additional acceleration between the gaps because in the meantime the direction of potential difference between the dees has changed. The ion will continue travelling in a semicircle of increasing radii, the direction of potential difference changes every time the ion goes from D1 to D2 and from D2 to D1. The time taken by the charged particle to traverse the semicircular path in the dee is given by πm …(i) t = πr /v = Bq This relation indicates that time t is independent of the velocity of the particle and of the radius. For any given value of m /q , it is determined by the magnetic field intensity. By adjusting the magnetic field intensity the time can be made the same as that required to change the potentials. On the other hand, the oscillator frequency (of alternating potential) can also be adjusted to the nature of a given ion and to the strength of the magnetic field. The frequency of the oscillations required to keep the ion in phase is given by the relation 1 1 Bq …(ii) f = = = T 2 t 2πm If the oscillation frequency is adjusted to keep the charged ion always in phase, each time the ion crosses the gap it receives an additional energy and at the same time it describes a flat spiral of increasing radius. Eventually, the ion reaches the periphery of the dee, where it can be brought out of the chamber by means of a deflecting plate charged to a high negative potential. This attractive force draws the ion out of its spiral path and thus can be used easily. If R is the radius of the dee, kinetic energy of the ion emerging from the cyclotron is thus given by 1 1 K = mv 2 = m (BqR /m )2 2 2 K = B 2R 2q 2 / 2m

…(iii)

This relation indicates that the maximum energy attained by the ion is limited by the radius R, magnetic field B or the frequency of the alternating potential f. It is independent of the alternating voltage. It can be explained by the fact that when the voltage is low, the ion makes a large number of turns before reaching the periphery, but when the voltage is high the number of turns is small. The total energy remains same in both the cases provided B and R are unchanged.

Note The cyclotron is used to bombard nuclei with energetic particles and study the resulting nuclear reactions. It is also used in hospitals to produce radioactive substances which can be used in diagnosis and treatment. Cyclotron is suitable only for accelerating heavy particles like proton, deuteron, α-particle etc. Electrons cannot be accelerated by the cyclotron because the mass of the electron is small and a small increase in energy of the electron makes the electrons move with a very high speed. The uncharged particles (e.g., neutrons) cannot be accelerated by cyclotron.

Solved Examples TYPED PROBLEMS Type 1. Based on deviation of charged particle in uniform magnetic field when θ = 90° or path is uniform circular

Concept Suppose a charged particle ( q , m) enters a uniform magnetic field B at right angles with speed v as shown in figure. The magnetic field extends upto a length x. The path of the particle is a circle of radius r, where mv r= Bq ×θ

r

×

×

x

q, m

×

×

×

×

×

×

+ ×

v

×

×

×

×

×

v θ

B

x

The speed of the particle in magnetic field does not change. But, it gets deviated in the magnetic field. The deviation θ can be found in two ways (i) After time t, deviation will be  Bq  θ = ωt =   t  m (ii) In terms of the length of the magnetic field (i.e. when the particle leaves the magnetic field) the deviation will be  x θ = sin –1   r But, since, sin θ | > 1, this relation can be used only when x < r. For x ≥ r, the deviation will be 180° as shown in figure. V

Bq   as ω = m  ×

×

×

×

×

×

×

×

×

v ×

×

×

v

r

x>r

Example 1 The region between x = 0 and x = L is filled with uniform steady magnetic field −B 0 k$ . A particle of mass m, positive charge q and velocity v0 $i travels along x-axis and enters the region of the magnetic field. [JEE 1999]

396 — Electricity and Magnetism Neglect the gravity throughout the question. (a) Find the value of L if the particle emerges from the region of magnetic field with its final velocity at an angle 30° to its initial velocity. (b) Find the final velocity of the particle and the time spent by it in the magnetic field, if the magnetic field now extends upto 2.1 L. Solution (a) θ = 30° L R mv0 R= B0q

sin θ = Here, ∴

y C

or

1 B0qL = 2 mv0

∴

L=

θ

v0 L

x

mv0 2B0q

(b) In part (a)

y

L sin 30° = R

1 L or = 2 R

L = R /2

or Now, when L ′ = 2.1 L

2.1 R ⇒ 2

or

L′ > R

O

x L'

Therefore, deviation of the particle is θ = 180° as shown in figure. ∴ v = − v i$ f

0

tAB = T /2 =

and

πm B0q

Type 2. To find coordinates and velocity of particle at any time t in circular path V

θ

L

L mv0 B0q

sin 30° =

v0 R

Example 2 A particle of specific charge α enters a uniform magnetic field $ with velocity v = v $i from the origin. Find the time dependence of B = – B0 k 0 velocity and position of the particle. y C

v0 θ

r x

Fm O

v0

θ P y x

OC = CP = radius of circle

B = –B0k

Chapter 26

Magnetics — 397

In such type of problems first of all see the angle between v and B. Because only this angle decides the path of the particle. Here, the angle is 90°. Therefore, the path is a circle. If it is a circle, see the plane of the circle (perpendicular to the magnetic field). Here, the plane is xy. Then, see the sense of the rotation. Here, it will be anti-clockwise as shown in figure, because at origin the magnetic force is along positive y-direction (which can be seen from Fleming's left hand rule). Find the deviation and radius of the particle. v θ = ωt = B0αt and r = 0 B0α

HOW TO PROCEED

Now, according to the figure, find v( t ) and r ( t ) . Solution Velocity of the particle at any time t is v (t ) = vx $i + vy $j = v0 cos θ $i + v0 sin θ $j v (t ) = v cos (B αt ) $i + v sin (B αt ) $j

or

0

0

0

0

Ans.

Position of particle at time t is r (t ) = x$i + y$j = r sin θ i$ + (r – r cos θ )$j Substituting the values of r and θ, we have v r (t ) = 0 [sin (B0αt )i$ + { 1 – cos (B0αt )}$j ] B0α

Ans.

Type 3. To find coordinates and velocity of particle at any time t in helical path V

Example 3 A particle of specific charge α is projected from origin with velocity $ in a uniform magnetic field B = – B k $ v = v0 $i – v0 k 0 . Find time dependence of velocity and position of the particle. Here, the angle between v and B is B0 v 0 v⋅B  1 θ = cos –1 = cos –1 = cos –1   |v|| B |  2 2v 0 ⋅ B0

HOW TO PROCEED

or

θ = 45°

Hence, the path is a helix. The axis of the helix is along z-axis (parallel to B ) and plane of the circle of helix is xy (perpendicular to B ). So, in xy-plane, the velocity components and x and y-coordinates are same as that of the above problem. The only change is along z-axis. Velocity component in this direction will remain unchanged while the z-coordinate of particle at time t would be v z t. Solution Velocity of particle at time t is $ v (t ) = vx $i + vy $j + vz k = v0 cos (B0αt )$i + v0 sin (B0αt )$j – v0k$ vx and vy can be found in the similar manner as done in Example 2. The position of the particle at time t would be r (t ) = x$i + y$j + zk$ Here,

z = vz t = – v0t

Ans.

398 — Electricity and Magnetism and x and y are same as in Example 2. v Hence, r (t ) = 0 [sin (B0αt )$i + {1 – cos (B0αt )}$j] – v0t k$ B0α

Ans.

Type 4. To find the time spent in magnetic field, deviation etc. if a charged particle enters from the e outside in uniform magnetic field (which extends upto large distance from point of entering) V

Example 4 A charged particle ( q, m) enters a uniform magnetic field B at angle α as shown in figure with speed v0 . Find β

x

x

x

x

x

x

x

x

x

x

x

x

x×

Px

x

x

x

x

x

x

x

x

x

x

C

A

v0

α

B

(q, m)

(a) the angle β at which it leaves the magnetic field. (b) time spent by the particle in magnetic field and (c) the distance AC. Solution (a) Here, velocity of the particle is in the plane of paper while the v0 magnetic field is perpendicular to the paper inwards,. i.e. angle between v and mv0 B is 90°. So, the path is a circle. The radius of the circle is r = Bq O is the centre of the circle. In ∆AOC, ∠OCD = ∠OAD 90° – β = 90° – α

∴

β =α

(b) ∠COD = ∠DOA = α

(as ∠OCD = ∠OAD = 90° – α)

∴

∠AOC = 2 α

or

length APC = r (2 α ) =

∴

tAPC

2mv0 .α Bq APC 2mα = = v0 Bq

C

r O

or

β 90° α α

P D

r 90°

Ans. v0

A

α

Ans.

Alternate method T α tAPC =   (2α ) =   ⋅ T  2π   π  α   2πm 2 αm =   =  π   Bq  Bq

Ans.

(c) Distance, AC = 2 ( AD ) = 2 (r sin α ) =

2mv0 sin α Bq

Ans.

Chapter 26 V

Magnetics — 399

Example 5 A particle of mass m = 1.6 × 10 −27 kg and charge q = 1.6 × 10 −19 C enters a region of uniform magnetic field of strength 1 T along the direction shown in figure. The speed of the particle is 10 7 m/s. (JEE 1984) θ F E 45°

x x x x x x x x x x x

x x x x x x x x x x x

x x x x x x x x x x x

x x x x x x x x x x x

x x x x x x x x x x x

(a) The magnetic field is directed along the inward normal to the plane of the paper. The particle leaves the region of the field at the point F. Find the distance EF and the angle θ. (b) If the direction of the field is along the outward normal to the plane of the paper, find the time spent by the particle in the region of the magnetic field after entering it at E. Solution Inside a magnetic field, speed of charged particle does not change. Further, velocity is perpendicular to magnetic field in both the cases hence path of the particle in the magnetic field will be circular. Centre of circle can be obtained by drawing perpendiculars to velocity (or tangent to the circular path) at E and F. Radius and angular speed of circular path would be mv Bq and ω = r= Bq m v

×

×

×

×

F×

×

×

×

×

×

×

×

×

v

×

×

×

×

×

×

θ

r C

45° 45°

G

45° E 90°

C

r 45°

E

45°

×

×

(i)

×

F

×

(ii)

(a) Refer figure (i) Since, ∴ or Further, ∴

∠CFG = 90° − θ and ∠CEG = 90° − 45° = 45° CF = CE ∠CFG = ∠CEG 90° − θ = 45° or θ = 45° FG = GE = r cos 45° 2mv cos 45° EF = 2FG = 2r cos 45° = Bq  1 2 (1.6 × 10−27 ) (107 )    2 = = 0.14 m (1) (1.6 × 10−19 )

400 — Electricity and Magnetism Note That in this case particle completes 1 / 4th of circle in the magnetic field because the angle rotated is 90°. (b) Refer figure (ii) In this case, particle will complete

3 th of circle in the magnetic field. 4

Hence, the time spent in the magnetic field : 3 t = (time period of circular motion) 4 3  2πm 3πm =  = 4  Bq  2Bq =

(3π ) (1.6 × 10−27 ) (2) (1) (1.6 × 10−19 )

= 4.712 × 10−8 s

Ans.

Note From the above examples, we can see that particle never completes circular path if it enters from outside in uniform magnetic field at right angles (as in Examples 1, 4 and 5). Circle is completed if magnetic field extends all around (Example-2). Following figures explain these points more clearly. In all figures, particle is positively charged. × v (a)

v

×

× v ×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

× v ×

× v ×

×

×

×

×

× v ×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

× ×

v

× ×

C

×

v

× ×

×

v

×

×

(b) C

(c)

× C

v

×

In figure (a) Centre of circular path is lying on the boundary line of magnetic field. Deviation of the T particle is 180° and time spent in magnetic field t = . 2 In figure (b) Centre of circular path lies outside the magnetic field. Deviation of the particle is less than T 180° and time spent in magnetic field t < . 2 In figure (c) Centre of circular path lies inside the magnetic field. Deviation of the particle is more than T 180° and time spent in magnetic field t > . 2

Type 5. Based on the concept of helical path

Concept Following points are worthnoting in case of a helical path. (i) The plane of the circle of the helix is perpendicular to the magnetic field. (ii) The axis of the helix is parallel to magnetic field.

Chapter 26

Magnetics — 401

y (iii) The particle while moving in helical path in magnetic field touches the line passing through the starting point B parallel to the magnetic field after every pitch. v For example, a charged particle is projected from origin in a magnetic field (along x-direction) at angle θ from the θ x-axis as shown. As the velocity vector v makes an angle θ O with B,its path is a helix. The plane of the circle of the helix is yz (perpendicular to magnetic field) and axis of the helix is parallel to x-axis. The particle while moving in helical path touches the x-axis after every pitch, i.e. it will touch the x-axis at a distance

x = np V

x

where, n = 0, 1, 2 …

Example 6 An electron gun G emits electrons of energy 2 keV travelling in the positive x-direction. The electrons are required to hit the spot S where GS = 0.1 m, and the line GS makes an angle of 60° with the x-axis as shown in figure. A uniform magnetic field B parallel to GS exists in the region outside the electron gun. S

B 60°

v

G

X

Find the minimum value of B needed to make the electrons hit S.

(JEE 1993)

Solution Kinetic energy of electron,

∴

S

1 K = mv2 = 2 keV 2 2K Speed of electron, v = m 2 × 2 × 1.6 × 10−16 m/s v= 9.1 × 10−31

B 60° G

= 2.65 × 107 m/s Since, the velocity (v) of the electron makes an angle of θ = 60° with the magnetic field B, the path will be a helix. So, the particle will hit S if GS = np Here, n = 1, 2, 3 .............. 2πm p = pitch of helix = v cos θ qB But for B to be minimum, n = 1 Hence,

2πm v cos θ qB 2πmv cos θ B = Bmin = q(GS )

GS = p =

v

402 — Electricity and Magnetism Substituting the values, we have Bmin =

 1 (2π )(9.1 × 10−31 )(2.65 × 107 )    2 (1.6 × 10−19 ) (0.1)

Bmin = 4.73 × 10−3 T

or

Ans.

Type 6. Based on calculation of magnetic field due to current carrying wires V

Example 7 A wire shaped to a regular hexagon of side 2 cm carries a current of 2 A. Find the magnetic field at the centre of the hexagon. Solution Q θ = 30° BC = tan θ OC 1 1 = tan 30° = r 3

∴ ∴

(BC = 1 cm ) O

r = 3 cm

θ θ

Net magnetic field at O is 6 times the magnetic field due to one side. µ i ∴ B = 6  0 (sin θ + sin θ )  4π r =

r

i

A

6 (10–7 ) (2)  1 1  +  3 × 10–2  2 2

= 6.9 × 10–5 T V

C

B

Ans.

Example 8 Find the magnetic field B at the point P in figure. a

a P

2a

I a

Solution Magnetic field at P due to SM and OQ is zero. Due to QR and RS are equal and outwards. Due to MN and NO are equal and inwards. O P

Q

N

M

I R

S

Chapter 26

Magnetics — 403

Due to QR and RS, µ I (sin 0° + sin 45° ) B1 = 2  0  4π 2a =

µ 0I 4 2 πa

[outwards]

Due to MN and NO, µ I (sin 0° + sin 45° ) B2 = 2  0  4π a =

Bnet = B2 − B1 =

∴ V

µ 0I 2 2 πa

[inwards] µ 0I 4 2 πa

[inwards]

Example 9 A long insulated copper wire is closely wound as a spiral of N turns. The spiral has inner radius a and outer radius b. The spiral lies in the xy-plane and a steady current I flows through the wire. The z-component of the magnetic field at the centre of the spiral is (JEE 2011) y

a

I

x b

(a) (c)

µ 0 NI  b ln   2 (b − a)  a µ 0 NI 2b

 b ln    a

(b)

 b + a µ 0 NI ln   2( b − a )  b − a 

(d)

 b + a µ 0 NI ln   2b  b − a

Solution (a) If we take a small strip of dr at distance r from centre, then number of turns in this strip would be  N  dN =   dr  b − a Magnetic field due to this element at the centre of the coil will be µ (dN )I µ NI dr dB = 0 = 0 2r (b − a ) 2 r ∴ ∴ Correct answer is (a).

B=∫

r =b r=a

dB =

µ 0NI  b ln    a 2(b − a )

404 — Electricity and Magnetism V

Example 10 An infinitely long conductor PQR is bent to form a right angle as shown in figure. A current I flows through PQR. The magnetic field due to this current at the point M is H 1 . Now, another infinitely long straight conductor QS is connected at Q, so that current is I/2 in QR as well as in QS, the current in PQ remaining unchanged. The magnetic field at M is now H 2 . The ratio H 1 / H 2 is given by (JEE 2000) M

−∞

90° P I

∞ S

90°

Q R −∞

(a) 1/2 (b) 1 (c) 2/3 (d) 2 Solution H 1 = Magnetic field at M due to PQ + Magnetic field at M due to QR But magnetic field at M due to QR = 0 ∴ Magnetic field at M due to PQ (or due to current I in PQ) = H1 Now, H 2 = Magnetic field at M due to PQ (current I) + magnetic field at M due to QS (current I/2) + magnetic field at M due to QR H 3 = H1 + 1 + 0 = H1 2 2 H1 2 = H2 3

Note Magnetic field at any point lying on the current carrying straight conductor is zero.

Type 7. Based on the magnetic force on current carrying wire V

Example 11 A long horizontal wire AB, which is free to move in a vertical plane and carries a steady current of 20 A, is in equilibrium at a height of 0.01 m over another parallel long wire CD which is fixed in a horizontal plane and carries a steady current of 30 A, as shown in figure. Show that when AB is slightly depressed, it executes simple harmonic motion. Find the period of oscillations. (JEE 1994) A

B

C

D

Solution Let m be the mass per unit length of wire AB. At a height x above the wire CD, magnetic force per unit length on wire AB will be given by µ ii Fm = 0 1 2 2π x

(upwards)

…(i)

Weight per unit length of wire AB is Fg = mg

(downwards)

Magnetics — 405

Chapter 26 Here, m = mass per unit length of wire AB At x = d, wire is in equilibrium, i.e. µ0 2π µ0 2π

or or

Fm = Fg i1i 2 = mg d i1i 2 mg = d d2

…(ii)

Fm B i1 = 20 A

A Fg C

x = d = 0.01 m D

When AB is depressed, x decreases therefore, Fm will increase, while Fg remains the same. Change in magnetic force will become the net restoring force, Let AB is displaced by dx downwards. Differentiating Eq. (i) w.r.t. x, we get µ ii …(iii) dFm = − 0 1 22 . dx 2π x i.e. restoring force, F = dFm ∝ − dx Hence, the motion of wire is simple harmonic. From Eqs. (ii) and (iii), we can write  mg  dFm = −   ⋅ dx  d 

[Q x = d ]

 g ∴ Acceleration of wire, a = –   . dx  d Hence, period of oscillation

V

T = 2π

| displacement | dx = 2π | acceleration | a

or

T = 2π

d 0.01 = 2π g 9.8

or

T = 0.2 s

Example 12 A straight segment OC (of length L) of a circuit carrying a current I is placed along the x-axis . Two infinitely long straight wires A and B, each extending from z = − ∞ to + ∞, are fixed at y = − a and y = + a respectively, as shown in the figure. If the wires A and B each carry a current I into the plane of the paper, obtain the expression for the force acting on the segment OC. What will be the force on OC if the current in the wire B is reversed?

Ans. y B O

I

C

x

A z

(JEE 1992)

406 — Electricity and Magnetism Solution (a) Let us assume a segment of wire OC at a point P, a distance x from the centre of length dx as shown in figure. y B θ dx O

I

x

P θ

BB

A

Net

BA

Magnetic field at P due to current in wires A and B will be in the directions perpendicular to AP and BP respectively as shown. µ I |B|= 0 2π AP Therefore, net magnetic force at P will be along negative y-axis as shown below Bnet = 2|B|cos θ µ  I  x  = 2  0    2π  AP  AP   µ  Ix Bnet =  0   π  ( AP )2 µ Ix Bnet = 0 . 2 π (a + x2) y B BB I

90°

O

Net

x

90° BA A

Therefore, force on this element will be µ Ix  dx dF = I  0 2 2 π a +x 

[in negative z-direction]

∴ Total force on the wire will be µ 0I 2 L xdx x=0 π ∫0 x2 + a 2 2 2  µ I L + a 2 = 0 ln   2π  a2 

F =∫

Hence,

F=−

x =L

dF =

[in negative z-axis]

 L2 + a 2 $ µ 0I 2 ln  k 2π  a2 

(b) When direction of current in B is reversed net magnetic field is along the current. Hence, force is zero.

V

Magnetics — 407 A

6 1.

Find (JEE 1997) (a) the magnitude and direction of the current in B. B (b) the magnitude of the magnetic field of induction at the point S. (c) the force per unit length on the wire B. Solution (a) Direction of current at B should be perpendicular to paper

S

1.2 10 m 11

A × B2

iB 10 = i A 32 iB =

or (b) Since, AS + BS = AB 2

2

m

P

outwards. Let current in this wire be iB . Then, µ0 iA µ iB = 0 10 2π (10 / 11) 2π  2 +   11 or

m

Example 13 Two long straight parallel wires are 2 m apart, perpendicular to the plane of the paper. The wire A carries a current of 9.6 A, directed into the plane of the paper. The wire B carries a current such that the magnetic field of induction at the point P, at a distance of 10 / 11 m from the wire B, is zero.

2m

Chapter 26

90°

10 10 × iA = × 9.6 = 3 A 32 32

S

B1

2

B

∴

∠ ASB = 90°

At S : B1 = Magnetic field due to i A =

µ 0 i A (2 × 10−7 ) (9.6) = 2π 1.6 1.6

= 12 × 10−7 T B2 = Magnetic field due to iB µ i = 0 B 2π 1.2 =

(2 × 10−7 ) (3) 1.2

= 5 × 10−7 T Since, B1 and B2 are mutually perpendicular. Net magnetic field at S would be B = B12 + B22 = (12 × 10−7 )2 + (5 × 10−7 )2 = 13 × 10−7 T (c) Force per unit length on wire B : F µ 0 i AiB = l 2π r =

(2 × 10−7 ) (9.6 × 3) 2

= 2.88 × 10−6 N/m

[Q r = AB = 2 m]

408 — Electricity and Magnetism V

Example 14 A current of 10 A flows around a closed path in a circuit which is in the horizontal plane as shown in the figure. The circuit consists of eight alternating arcs of radii r1 =0.08 m and r2 =0.12 m. Each subtends the same angle at the centre. (JEE 2001) D r2

C A

r1 i

(a) Find the magnetic field produced by this circuit at the centre. (b) An infinitely long straight wire carrying a current of 10 A is passing through the centre of the above circuit vertically with the direction of the current being into the plane of the circuit. What is the force acting on the wire at the centre due to the current in the circuit? What is the force acting on the arc AC and the straight segment CD due to the current at the centre? Solution (a) Given, i = 10 A, r1 = 0.08 m and r2 = 0.12 m. Straight portions, i.e. CD etc, will produce zero magnetic field at the centre. Rest eight arcs will produce the magnetic field at the centre in the same direction, i.e. perpendicular to the paper outwards or vertically upwards and its magnitude is B = Binner arcs + Bouter arcs =

1 µ 0 i  1 µ 0 i    +   2 2 r1  2 2 r2 

r + r  µ  =  0  (πi )  1 2  4π   r1 r2  Substituting the values, we have B=

(10−7 )(3.14)(10)(0.08 + 0.12) (0.08 × 0.12)

B = 6.54 × 10−5 T (vertically upward or outward normal to the paper) (b) Force on AC Force on circular portions of the circuit, i.e. AC etc, due to the wire at the centre will be zero because magnetic field due to the central wire at these arcs will be tangential (θ = 180°). Force on CD Current in central wire is also i = 10 A. Magnetic field at distance x due to central wire µ i B= 0 . 2π x ∴ Magnetic force on element dx due to this magnetic field i µ  µ  dx dF = (i )  0 .  ⋅ dx =  0  i 2  2π x  2π  x

[F = ilB sin 90° ]

Chapter 26

Magnetics — 409

Therefore, net force on CD is F =∫

x = r2 x = r1

dF =

µ 0i 2 2π

−7

0.12 dx

∫ 0. 08

x

=

µ 0 2  3 i ln    2 2π

Substituting the values, F = (2 × 10 ) (10) ln (1 . 5) 2

or F = 8.1 × 10−6 N (inwards) Force on wire at the centre Net magnetic field at the centre due to the circuit is in vertical direction and current in the wire in centre is also in vertical direction. Therefore, net force on the wire at the centre will be zero. ( θ = 180°). Hence, (i) Force acting on the wire at the centre is zero. (ii) Force on arc AC = 0. (iii) Force on segment CD is 8.1 × 10−6 N (inwards).

Type 8. Based on the magnetic force on a charged particle in electric and (or) magnetic field V

Example 15 Consider the motion of a positive point charge in a region where there are simultaneous uniform electric and magnetic fields E = E 0 $j and B = B 0 $j. At time t = 0, this charge has velocity v in the xy-plane making an angle θ with the (JEE 2012) x-axis. Which of the following option(s) is(are) correct for time t > 0? (a) If θ = 0°, the charge moves in a circular path in the xz-plane. (b) If θ = 0°, the charge undergoes helical motion with constant pitch along the y-axis (c) If θ = 10°, the charge undergoes helical motion with its pitch increasing with time along the y-axis. (d) If θ = 90°, the charge undergoes linear but accelerated motion along the y-axis. Solution Magnetic field will rotate the particle in a circular path (in xz-plane or perpendicular to B). Electric field will exert a constant force on the particle in positive y-direction. Therefore, resultant path is neither purely circular nor helical or the options (a) and (b) both are wrong. (c) v⊥ and B will rotate the particle in a circular path in xz- plane (or perpendicular to B ). Further, v|| and E will move the particle (with increasing speed) along positive y-axis (or along the axis of above circular path). Therefore, the resultant path is helical with increasing pitch along the y-axis (or along B and E ). Therefore, option (c) is correct. (d) B

y

E

v

x

Magnetic force is zero, as θ between B and v is zero. But, electric force will act in y-direction. Therefore, motion is 1-D and uniformly accelerated (towards positive y-direction). Therefore, option (d) is also correct.

410 — Electricity and Magnetism V

Example 16 A particle of charge +q and mass m moving under the influence of a uniform electric field E $i $ follows a trajectory from and uniform magnetic field B k P to Q as shown in figure. The velocities at P and Q are v $i and −2 $j . Which of the following statement(s) is/are correct ? (JEE 1991) (a) E =

y P

E

v

B a 2a

Q 2v

x

3  mv 2    4  qa 

(b) Rate of work done by the electric field at P is

3  mv3    4 a 

(c) Rate of work done by the electric field at P is zero (d) Rate of work done by both the fields at Q is zero Solution Magnetic force does not do work. From work-energy theorem : WFe = ∆KE 1 (qE ) (2a ) = m [4v2 − v2] 2 3  mv2 E=   4  qa 

or or ∴ Option (a) is correct.

At P, rate of work done by electric field = Fe ⋅ v = (qE ) (v) cos 0°  3 mv2 3  mv3  =q v=   4 a   4 qa  Therefore, option (b) is also correct. Rate of work done at Q : of electric field = Fe ⋅ v = (qE ) (2v) cos 90° = 0 and of magnetic field is always zero. Therefore, option (d) is also correct. Note that F = qE$i e

V

Example 17 A proton moving with a constant velocity passes through a region of space without any change in its velocity. If E and B represent the electric and magnetic fields, respectively. Then, this region of space may have (JEE 1985) (a) E = 0, B = 0 (c) E ≠ 0, B = 0

(b) E = 0, B ≠ 0 (d) E ≠ 0, B ≠ 0

Solution If both E and B are zero, then Fe and Fm both are zero. Hence, velocity may remain constant. Therefore, option (a) is correct. If E = 0, B ≠ 0 but velocity is parallel or antiparallel to magnetic field, then also Fe and Fm both are zero. Hence, option (b) is also correct. If E ≠ 0, B ≠ 0 but Fe + Fm = 0, then again velocity may remain constant or option (d) is also correct.

Chapter 26

Example 18 A wire loop carrying a current I is placed in the xy-plane as shown in figure. (JEE 1991) y M

v +Q

120°

V

Magnetics — 411

I

x

O

P

a N

(a) If a particle with charge +Q and mass m is placed at the centre P and given a velocity v along NP (see figure), find its instantaneous acceleration. (b) If an external uniform magnetic induction field B = B i$ is applied, find the force and the torque acting on the loop due to this field. Solution (a) Magnetic field at P due to arc of circle, y

M

a

x

r 60° 60° P

I

+Q y

v

a 60° x

N

Subtending an angle of 120° at centre would be 1 1 µ I B1 = (field due to circle) =  0  3 3  2a  µ I = 0 6a 0.16 µ 0I = a 0.16 µ 0I $ or B1 = k a Magnetic field due to straight wire NM at P , µ I B2 = 0 (sin 60° + sin 60° ) 4π r Here, ∴ or

r = a cos 60° µ I B2 = 0 (2 sin 60° ) 4π a cos 60° B2 =

µ0 I tan 60° 2π a

[outwards] [ outwards]

412 — Electricity and Magnetism 0.27 µ 0I a 0.27 µ 0I $ or B2 = − k a 0.11 µ 0 I $ ∴ Bnet = B1 + B2 = − k a Now, velocity of particle can be written as v = v cos 60° i$ + v sin 60° $j =

=

(inwards)

v$ 3v $ i+ j 2 2

Magnetic force, Fm = Q (v × B) =

0.11 µ 0IQv $ 0.11 3 µ 0IQv $ j− i 2a 2a

∴ Instantaneous acceleration,

Fm 0.11 µ 0IQv $ = ( j − 3 i$ ) m 2am (b) In uniform magnetic field, force on a current loop is zero. Further, magnetic dipole moment of the loop will be $ M = (IA ) k a=

Here, A is the area of the loop. 1 1 (πa 2) − [2 × a sin 60° ] [a cos 60° ] 3 2 2 2 πa a = − sin 120° 3 2

A=

= 0.61 a 2 $ M = (0.61 Ia 2) k B = B $i

∴ Given,

τ = M × B = (0.61 Ia 2B) $j

∴ V

Example 19 Two long parallel wires carrying currents 2.5 A and I (ampere) in the same direction (directed into the plane of the paper) are held at P and Q respectively such that they are perpendicular to the plane of paper. The points P and Q are located at a distance of 5 m and 2 m respectively from a collinear point R (see figure). (JEE 1990) Q

P 2.5 A

R X

IA 2m 5m

(a) An electron moving with a velocity of 4 × 105 m/s along the positive x-direction experiences a force of magnitude 3.2 × 10−20 N at the point R. Find the value of I. (b) Find all the positions at which a third long parallel wire carrying a current of magnitude 2.5 A may be placed, so that the magnetic induction at R is zero.

Chapter 26

Magnetics — 413

Solution (a) Magnetic field at R due to both the wires P and Q will be downwards as shown in figure. ×

×

P

Q

R

v

BP BQ

Therefore, net field at R will be sum of these two. B = BP + BQ µ I µ IQ µ 0  2.5 I  = 0 P + 0 = +   2π 5 2π 2 2π  5 2 µ0 = (I + 1) = 10−7 (I + 1) 4π Net force on the electron will be M

R

N ×

1m

1m

Fm = Bqv sin 90° or or

(3.2 × 10

−20

) = (10−7 ) (I + 1) (1.6 × 10−19 ) (4 × 105 )

I + 1 =5

∴ I =4A (b) Net field at R due to wires P and Q is B = 10−7 (I + 1) T = 5 × 10−7 T Magnetic field due to third wire carrying a current of 2.5 A should be 5 × 10−7 T in upward direction, so that net field at R becomes zero. Let distance of this wire from R be r. Then, µ 0 2.5 = 5 × 10−7 2π r or

(2 × 10−7 ) (2.5) = 5 × 10−7 m r

or r =1m So, the third wire can be put at M or N as shown in figure. If it is placed at M, then current in it should be outwards and if placed at N, then current be inwards.

Type 9. Path of charged particle in both electric and magnetic fields

Concept Here, normally two cases are popular. In the first case, E ↑↑ B and particle velocity is perpendicular to both of these fields. In the second case, E ⊥ B and the particle is released from rest. Let us now consider both the cases separately.

414 — Electricity and Magnetism V

Example 20 fields.

When E ↑↑ B and particle velocity is perpendicular to both of these

Solution Consider a particle of charge q and mass m released from the origin with velocity

v = v0i$ into a region of uniform electric and magnetic fields parallel to y-axis, i.e. E = E 0$j and B = B0$j. The electric field accelerates the particle in y-direction, i.e. y-component of velocity goes on increasing with acceleration, Fy Fe qE 0 …(i) ay = = = m m m The magnetic field rotates the particle in a circle in xz-plane (perpendicular to magnetic field). The resultant path of the particle is a helix with increasing pitch. The axis of the plane is parallel to y-axis. Velocity of the particle at time t would be $ v (t ) = v $i + v $j + v k x

Here, and

y

z

qE 0 vy = a y t = t m vx2 + vz2 = constant = v02 Bq θ = ωt = t m  Bqt  vx = v0 cos θ = v0 cos    m 

and

 Bqt  vz = v0 sin θ = v0 sin    m   Bqt  $  qE 0  $  Bqt  $ v (t ) = v0 cos  t j + v0 sin   i+  k  m   m   m 

∴

Similarly, position vector of particle at time t can be given by $ r (t ) = x$i + y$j + z k Here,

y=

z v0 θ

1 1  qE 0  2 ayt2 =   t 2 2 m  Fm

Note

t=0

z v0

x

 mv    Bqt   z = r (1 – cos θ ) =  0  1 – cos    m    Bq    mv   Bqt  $ 1  qE 0  2 $  mv0  r (t ) =  0  sin   i+   t j+    m   Bq   Bq  2 m 

∴

θ x

 mv   Bqt  x = r sin θ =  0  sin    m   Bq  and

r

  Bqt   $ 1 – cos  m   k   

(i) While moving in helical path the particle touches the y-axis after every T or after, Here,

t = nT , 2πm T= Bq

where n = 0 , 1 , 2…

(ii) At t = 0, velocity is along positive x-axis and magnetic field is along y-axis. Therefore, magnetic force is along positive z-axis and the particle rotates in xz-plane as shown in figure.

Chapter 26 V

Magnetics — 415

Example 21 When E ⊥ B and the particle is released at rest from origin. Solution Consider a particle of charge q and mass m emitted at origin with zero initial velocity into a region of uniform electric and magnetic fields. The field E is acting along x-axis and field B along y-axis, i.e. E = E i$ 0

and B = B0$j Electric field will provide the particle an acceleration (and therefore a velocity component) in x-direction and the magnetic field will rotate the particle in xz-plane (perpendicular to B). Hence, at any instant of time its velocity (and hence, position) will have only x and z components. Let at time t its velocity be $ v = v i$ + v k x

z

Net force on it at this instant is F = Fe + Fm = qE + q (v × B) $ ) × (B $j)] = q [E 0i$ + (vx i$ + vz k 0 $ $ F = q (E – v B ) i + qv B k

or

z

0

x

0

0

F $ ∴ a= = a x i$ + a z k m q where, ax = (E 0 – vz B0 ) m q and az = vx B0 m Differentiating Eq. (i) w.r.t. time, we have

…(i) …(ii)

d 2vx qB0  dvz  =–   m  dt  dt 2 dvz qB0 = az = vx dt m

But,

d 2vx =– dt 2

∴

2

 qB0    v  m  x

…(iii)

 d 2y  Comparing this equation with the differential equation of SHM  2 = – ω 2y , we get  dt  ω=

qB0 m

and the general solution of Eq. (iii) is vx = A sin (ωt + φ ) At time t = 0,

vx = 0,

hence, φ = 0 dvx = Aω cos ωt dt

Again, From Eq. (i), a x =

qE 0 m

at t = 0, as vz = 0 at t = 0 Aω =

∴ Substituting ω =

…(iv)

qB0 , we get m

qE 0 m A=

or E0 B0

A=

qE 0 ωm

(as φ = 0)

416 — Electricity and Magnetism Therefore, Eq. (iv) becomes vx =

E0 sin ωt B0

ω=

where,

qB0 m

Now substituting value of vx in Eq. (ii), we get dvz qE 0 = sin ωt dt m vz qE 0 t ∴ ∫0 dvz = m ∫0 sin ωt dt qE 0 or vz = (1 – cos ωt ) ωm qB0 Substituting ω = , we get m E vz = 0 (1 – cos ωt ) B0 On integrating equations for vx and vz and knowing that at t = 0, x = 0 and z = 0, we get

and

x=

E0 (1 – cos ωt ) B0ω

z=

E0 (ωt – sin ωt ) B0ω

These equations are the equations for a cycloid which is defined as the path generated by the point on the circumference of a wheel rolling on a ground. x

2E0 B0ω z

In the present case, the radius of the rolling wheel is along x-direction is

E0 , the maximum displacement B0ω

2E 0 . The x-displacement becomes zero at t = 0, 2π /ω, 4π /ω, etc. B0ω

Note Path of a charged particle in uniform electric and magnetic field will remain unchanged if Fnet = 0 or

Fe + Fm = 0

or

qE + q ( v × B) = 0

or

E = − (v × E ) = (E × v )

Miscellaneous Examples V

Example 22 A cyclotron’s oscillator frequency is 10 MHz. What should be the operating magnetic field for accelerating protons? If the radius of its dees is 60 cm, what is the kinetic energy (in MeV) of the proton beam produced by the accelerator? ( e = 1.60 × 10 −19 C , m p = 1.67 × 10 −27 kg, 1 MeV = 1.6 × 10 −13 J ) Solution Magnetic field Cyclotron’s oscillator frequency should be same as the proton’s revolution frequency (in circular path) Bq ∴ f = 2πm 2πmf or B= q Substituting the values in SI units, we have B=

(2)(22 7)(1.67 × 10−27 )(10 × 106 ) 1.6 × 10−19

= 0.67 T

Ans.

Kinetic energy Let final velocity of proton just after leaving the cyclotron is v. Then, radius of dee should be equal to mv BqR or v = R= Bq m ∴

Kinetic energy of proton, 2

K =

2 2 2 1 1  BqR B q R mv2 = m   =  m  2 2 2m

Substituting the values in SI units, we have K =

(0.67)2(1.6 × 10−19 )2(0.60)2 2 × 1.67 × 10−27

= 1.2 × 10−12 J =

1.2 × 10−12 MeV (1.6 × 10−19 ) (106 )

= 7 . 5 MeV V

Ans.

Example 23 A charged particle carrying charge q = 1 µC moves in uniform magnetic field with velocity v1 = 10 6 m/s at angle 45° with x-axis in the xy-plane and experiences a force F1 = 5 2 mN along the negative z-axis. When the same particle moves with velocity v2 = 10 6 m/ s along the z-axis, it experiences a force F2 in y-direction. Find (a) the magnitude and direction of the magnetic field (b) the magnitude of the force F2.

418 — Electricity and Magnetism Solution F2 is in y-direction when velocity is along z-axis. Therefore, magnetic field should be along x-axis. So let, B = B0i$ v1 =

(a) Given,

$ F1 = – 5 2 × 10−3 k

and From the equation, We have

∴

106 $ 106 $ i+ j 2 2

F = q (v × B) 6 6 $ = (10–6 )  10 $i + 10 $j × (B $i ) (– 5 2 × 10–3 ) k  0  2   2  B0 $ =– k 2 B0 = 5 2 × 10–3 2 B0 = 10–2 T

or Therefore, the magnetic field is

B = (10–2 $i ) T (b)

F2 = B0qv2 sin 90° As the angle between B and v in this case is 90°. ∴

V

Ans.

F2 = (10–2) (10–6 ) (106 ) = 10–2 N

Ans.

Example 24 A wire PQ of mass 10 g is at rest on two parallel metal rails. The separation between the rails is 4.9 cm. A magnetic field of 0.80 T is applied perpendicular to the plane of the rails, directed downwards. The resistance of the circuit is slowly decreased. When the resistance decreases to below 20 Ω, the wire PQ begins to slide on the rails. Calculate the coefficient of friction between the wire and the rails. P x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

6V

4.9 cm

x Q

Solution Wire PQ begins to slide when magnetic force is just equal to the force of friction, i.e. Here, ∴

µ mg = ilB sin θ E 6 i= = = 0.3 A R 20 il B (0.3) (4.9 × 10–2) (0.8) µ= = mg (10 × 10–3 ) (9.8) = 0.12

( θ = 90° )

Ans.

Chapter 26

Magnetics — 419

Example 25 What is the value of B that can be set up at the equator to permit a proton of speed 107 m/ s to circulate around the earth? [R = 6.4 × 10 6 m, m p = 1.67 × 10 –27 kg] . Solution From the relation mv , Bq mv B= qr r=

We have Substituting the values, we have B=

(1.67 × 10–27 ) (107 ) (1.6 × 10–19 ) (6.4 × 106 )

= 1.6 × 10–8 T V

Ans.

Example 26 Deuteron in a cyclotron describes a circle of radius 32.0 cm. Just before emerging from the D’s. The frequency of the applied alternating voltage is 10 MHz. Find (a) the magnetic flux density (i.e. the magnetic field). (b) the energy and speed of the deuteron upon emergence. Solution (a) Frequency of the applied emf = Cyclotron frequency or

f =

Bq 2π m

∴

B=

2πmf q

=

(2) (3.14) (2 × 1.67 × 10–27 ) (10 × 106 ) 1.6 × 10–19

= 1.30 T (b) The speed of deuteron on the emergence from the cyclotron, 2 πR v= = 2πfR T

∴

= (2) (3.14) (10 × 106 ) (32 × 10–2) = 2.01 × 107 m/s 1 Energy of deuteron = mv2 2 1 = × (2 × 1.67 × 10–27 ) (2.01 × 107 )2 J 2 = 4.22 MeV

Note V

Ans.

Ans.

1 MeV = 1.6 × 10 −13 J

Example 27 In the Bohr model of the hydrogen atom, the electron circulates around the nucleus in a path of radius 5 × 10 –11 m at a frequency of 6.8 × 1015 Hz. (a) What value of magnetic field is set up at the centre of the orbit? (b) What is the equivalent magnetic dipole moment?

420 — Electricity and Magnetism Solution (a) An electron moving around the nucleus is equivalent to a current, i = qf Magnetic field at the centre, B=

µ 0i µ 0qf = 2R 2R

Substituting the values, we have (4π × 10–7 ) (1.6 × 10–19 ) (6.8 × 1015 ) B= 2 × 5.1 × 10–11 Ans. = 13.4 T (b) The current carrying circular loop is equivalent to a magnetic dipole with magnetic dipole moment, M = NiA = (Nqf πR2) Substituting the values, we have M = (1) (1.6 × 10–19 ) (6.8 × 1015 ) (3.14) (5.1 × 10–11 )2 = 8.9 × 10–24 A -m2 V

Ans.

Example 28 A flat dielectric disc of radius R carries an excess charge on its surface. The surface charge density is σ. The disc rotates about an axis perpendicular to its plane passing through the centre with angular velocity ω. Find the torque on the disc if it is placed in a uniform magnetic field B directed perpendicular to the rotation axis. Solution Consider an annular ring of radius r and of thickness dr on this disc. Charge within this ring, ω B dr

r

dq = (σ ) (2πrdr ) As ring rotates with angular velocity ω, the equivalent current is i = (dq) (frequency) ω = (σ ) (2πrdr )   or  2π 

i = σωrdr

Magnetic moment of this annular ring, M = iA = (σωrdr ) (πr 2)

(along the axis of rotation)

Torque on this ring, dτ = MB sin 90° = (σωπr3 B) dr ∴ Total torque on the disc is

R

R

0

0

τ = ∫ dτ = (σωπB)∫ r3 dr =

σωπBR4 4

Ans.

Chapter 26 V

Magnetics — 421

Example 29 Three infinitely long thin wires, each carrying current i in the same direction, are in the xy-plane of a gravity free space. The central wire is along the y-axis while the other two are along x = ± d . (i) Find the locus of the points for which the magnetic field B is zero. (ii) If the central wire is displaced along the z-direction by a small amount and released, show that it will execute simple harmonic motion. If the linear mass density of the wires is λ, find the frequency of oscillation. Solution (i) Magnetic field will be zero on the y-axis, i.e. x = 0 = z. I

II

y

i

i

x = –d

O

III

IV

i x x = +d

Magnetic field cannot be zero in region I and region IV because in region I magnetic field will be along positive z-direction due to all the three wires, while in region IV magnetic field will be along negative z-axis due to all the three wires. It can be zero only in region II and III. Let magnetic field is zero on line z = 0 and x = x (shown as dotted). The magnetic field on this line due to wires 1 and 2 will be along negative z-axis and due to wire 3 along positive z-axis. Thus, B=0

i

i

y

B=0

i x

O

O x

d –x

d+x 1

or

2

3

x= d 3 z=0

B=0

x=– d 3 z=0

B1 + B2 = B3 µ0 i µ i µ i + 0 = 0 2π (d + x) 2πx 2π (d – x)

or This equation gives Hence, there will be two lines and where, magnetic field is zero.

1 1 1 + = d+x x d–x d 3 d x= 3 d x=– 3

x=±

(z = 0) Ans.

422 — Electricity and Magnetism (ii) In this part, we change our coordinate axes system, just for better understanding. z

1 x

x y-axis

2

3 x

x

x = –d

x=0

x

x=d

There are three wires 1, 2 and 3 as shown in figure. If we displace the wire 2 towards the z-axis, then force of attraction per unit length between wires (1 and 2) and (2 and 3) will be given by 2 x F θ θ

F

r

r z

1 x

3 x d

F =

d

µ0 i2 2π r

The components of F along x-axis will be cancelled out. Net resultant force will be towards negative z-axis (or mean position) and will be given by µ i 2  z Fnet = 2F cos θ = 2 0  2π r  r Fnet =

µ0 i2 .z 2 π (z + d 2)

(r 2 = z 2 + d 2)

If z R, measured from the axis.

LEVEL 2 Single Correct Option 1. A uniform current carrying ring of mass m and radius R is connected by a massless string as shown. A uniform magnetic field B0 exists in the region to keep the ring in horizontal position, then the current in the ring is

B0

(a)

mg πRB0

(b)

mg RB0

(c)

mg 3πRB0

(d)

mg πR2B0

2. A wire of mass 100 g is carrying a current of 2 A towards increasing x in the form of $ tesla. The y = x 2( −2m ≤ x ≤ + 2m ). This wire is placed in a magnetic field B = − 0.02 k 2 acceleration of the wire (in m/ s ) is (a) − 1.6 $j

(b) − 3.2 $j

(c) 1.6 $j

(d) zero

3. A conductor of length l is placed perpendicular to a horizontal uniform magnetic field B. Suddenly, a certain amount of charge is passed through it, when it is found to jump to a height h. The amount of charge that passes through the conductor is m gh Bl m 2 gh (c) Bl (a)

(b)

m gh 2Bl

(d) None of these

440 — Electricity and Magnetism 4. A solid conducting sphere of radius R and total charge q rotates about its diametric axis with constant angular speed ω. The magnetic moment of the sphere is

(a)

1 qR2ω 3

(b)

2 qR2ω 3

(c)

1 qR2ω 5

(d)

2 qR2ω 5

5. A charged particle moving along positive x-direction with a velocity v enters a region where $ , from x = 0 to x = d. The particle gets deflected at an there is a uniform magnetic field B = − B k angle θ from its initial path. The specific charge of the particle is v tan θ Bd v sin θ (d) Bd

Bd v cos θ B sin θ (c) vd

(b)

(a)

6. A current carrying rod AB is placed perpendicular to an infinitely long current carrying wire as shown in figure. The point at which the conductor should be hinged so that it will not rotate ( AC = CB) C

A l

l

(a) A (c) C

B l

(b) somewhere between B and C (d) somewhere between A and C

7. The segment AB of wire carrying current I1 is placed perpendicular to a long straight wire carrying current I 2 as shown in figure. The magnitude of force experienced by the straight wire AB is

µ 0I1I 2 ln 3 2π 2 µ 0I1I 2 (c) 2π

µ 0I1I 2 ln 2 2π µ II (d) 0 1 2 2π

(b)

(a)

I1 A 2a

I2 B a

8. A straight long conductor carries current along the positive x-axis. Identify the correct statement related to the four points A ( a , a , 0), B ( a , 0, a ), C ( a , − a , 0) and D ( a , 0, − a ).

(a) (b) (c) (d)

The magnitude of magnetic field at all points is same Fields at A and B are mutually perpendicular Fields at A and C are antiparallel All of the above

9. The figure shows two coaxial circular loops 1 and 2, which forms same solid angle θ at point O. If B1 and B2 are the magnetic fields produced at the point O due to loop 1 and 2 respectively, then 2 1 O

θ

I

x

l 2x

(a)

B1 =1 B2

(b)

B1 =2 B2

(c)

B1 =8 B2

(d)

B1 =4 B2

Chapter 26

Magnetics — 441

10. In the figure shown, a charge q moving with a velocity v along the x-axis enter into a region of uniform magnetic field. The minimum value of v so that the charge q is able to enter the region x> b y

q

x

v

a b

qBb m qB(b − a ) (c) m

qBa m qB(b + a ) (d) 2m (b)

(a)

11. An insulating rod of length l carries a charge q uniformly distributed on it. The rod is pivoted at one of its ends and is rotated at a frequency f about a fixed perpendicular axis. The magnetic moment of the rod is πqfl2 12 πqfl2 (c) 6

πqfl2 2 πqfl2 (d) 3

(a)

(b)

12. A wire carrying a current of 3 A is bent in the form of a parabola y 2 = 4 − x as shown in figure, $ tesla. The where x and y are in metre. The wire is placed in a uniform magnetic field B = 5 k force acting on the wire is y (m)

x (m)

O

(a) 60 $i N (c) 30 $i N

(b) − 60$i N (d) −30 $i N

13. An equilateral triangle frame PQR of mass M and side a is kept under

Q

P

the influence of magnetic force due to inward perpendicular magnetic field B and gravitational field as shown in the figure. The magnitude and direction of current in the frame so that the frame remains at √3 a rest, is 4 2Mg ; anti-clockwise aB Mg ; anti-clockwise (c) I = aB (a) I =

2Mg ; clockwise aB Mg (d) I = ; clockwise aB (b) I =

g R

442 — Electricity and Magnetism 14. A tightly wound long solenoid has n turns per unit length, radius r and carries a current i. A particle having charge q and mass m is projected from a point on the axis in the direction perpendicular to the axis. The maximum speed for which particle does not strike the solenoid will be

µ 0qrni 2m 2 µ 0qrni (c) 3m

(b)

(a)

µ 0qrni m

(d) None of these

15. If the acceleration and velocity of a charged particle moving in a constant magnetic region is $ v = b i$ + b k. $ [a , a , b and b are constants], then choose the wrong given by a = a1i$ + a2k, 1 2 1 2 1 2 statement.

(a) (b) (c) (d)

Magnetic field may be along y-axis a1b1 + a 2b2 = 0 Magnetic field is along x-axis Kinetic energy of particle is always constant

16. A simple pendulum with a charged bob is oscillating as shown in the figure. Time period of

oscillation is T and angular amplitude is θ. If a uniform magnetic field perpendicular to the plane of oscillation is switched on, then B θ θ

(a) T will decrease but θ will remain constant (c) Both T and θ will remain the same

(b) T will remain constant but θ will decrease (d) Both T and θ will decrease

$ Two loops each of side a is placed in this 17. Magnetic field in a region is given by B = B0 x k. magnetic region in the xy-plane with one of its sides on x-axis. If F1 is the force on loop 1 and F2 be the force on loop 2, then y 1

2

I

I x

(a) F1 = F2 = 0 (c) F2 > F1

(b) F1 > F2 (d) F1 = F2 ≠ 0

18. Consider a coaxial cable which consists of an inner wire of radius a surrounded by an outer shell of inner and outer radii b and c, respectively. The inner wire carries a current I and outer shell carries an equal and opposite current. The magnetic field at a distance x from the axis where b < x < c is (a)

µ 0I (c2 − b2) 2πx(c2 − a 2)

(b)

µ 0I (c2 − x2) 2πx(c2 − a 2)

(c)

µ 0I (c2 − x2) 2πx(c2 − b2)

(d) zero

Chapter 26

Magnetics — 443

19. A particle of mass 1 × 10−26 kg and charge +1.6 × 10−19 C travelling with a velocity of 1.28 × 106 m/ s along positive direction of x-axis enters a region in which a uniform electric field E and a uniform magnetic field B are present such that Ez = − 102.4 kV/ m and By = 8 × 10−2 Wb/ m 2 The particle enters this region at origin at time t = 0. Then, (a) (b) (c) (d)

net force acts on the particle along the +ve z-direction net force acts on the particle along –ve z-direction net force acting on particle is zero net force acts in xz-plane

20. A wire lying along y-axis from y = 0 to y = 1 m carries a current of 2 mA in the negative y-direction. The wire lies in a non-uniform magnetic field $ The magnetic force on the entire wire is + ( 0.4 Tm ) yj.

(a) −3 × 10−4 $j N (c) −3 × 10−4 k$ N

given

by B = ( 0.3 T/ m ) y$i

(b) 6 × 10−3 k$ N (d) 3 × 10−4 k$ N

21. A particle having a charge of 20 µC and mass 20 µg moves along a circle of

v

radius 5 cm under the action of a magnetic field B = 0.1 tesla. When the particle is at P, uniform transverse electric field is switched on and it is found that the particle continues along the tangent with a uniform velocity. Find the electric field

(a) 2 V/m (c) 5 V/m

P

(b) 0.5 V/m (d) 1.5 V/m

22. Two circular coils A and B of radius

5

cm and 5 cm carry currents 5 A and 5 2 A, 2 respectively. The plane of B is perpendicular to plane of A and their centres coincide. Magnetic field at the centre is

(a) 0

(b) 4π 2 × 10−5 T

(c) 4π × 10−5 T

(d) 2π 2 × 10−5 T

23. A charged particle with specific charge s moves undeflected through a region of space containing mutually perpendicular and uniform electric and magnetic fields E and B. When the electric field is switched off, the particle will move in a circular path of radius (a)

E Bs

(b)

Es B

(c)

Es B2

(d)

E B2s

24. Two long parallel conductors are carrying currents in the same direction as shown in the figure. The upper conductor ( A) carrying a current of 100 A is held firmly in position. The lower conductor ( B) carries a current of 50 A and free to move up and down. The linear mass density of the lower conductor is 0.01 kg/m. 100 A A

B 50 A

(a) (b) (c) (d)

Conductor B will be in equilibrium if the distance between the conductors is 0.1 m Equilibrium of conductor B is unstable Both (a) and (b) are wrong Both (a) and (b) are correct

444 — Electricity and Magnetism 25. Equal currents are flowing in three infinitely long wires along positive x , y and z-directions. The magnetic field at a point ( 0, 0, − a ) would be (i = current in each wire)

µ 0i 2 πa µ i (c) 0 2 πa

(a)

µ 0i $ $ (i − j) 2 πa µ i (d) 0 (− $i − $j) 2 πa

($j − i$ )

(b)

($i + $j)

26. In the figure, the force on the wire ABC in the given uniform magnetic field will be ( B = 2 tesla ) C

4m I = 2A

A

B

3m

(a) 4(3 + 2π ) N (c) 30 N

(b) 20 N (d) 40 N

27. A uniformly charged ring of radius R is rotated about its axis with constant linear speed v of each of its particles. The ratio of electric field to magnetic field at a point P on the axis of the ring distant x = R from centre of ring is (c is speed of light) v

P

O x=R

c2 v v (c) c (a)

v2 c c (d) v (b)

More than One Correct Options 1. Two circular coils of radii 5 cm and 10 cm carry currents of 2 A. The coils have 50 and 100 turns respectively and are placed in such a way that their planes as well as their centres coincide. Magnitude of magnetic field at the common centre of coils is (a) (b) (c) (d)

8π × 10−4 T if currents in the coils are in same sense 4π × 10−4 T if currents in the coils are in opposite sense zero if currents in the coils are in opposite sense 8π × 10−4 T if currents in the coils are in opposite sense

2. A charged particle enters into a gravity free space occupied by an electric field E and magnetic field B and it comes out without any change in velocity. Then, the possible cases may be (a) E = 0 and B ≠ 0 (c) E ≠ 0 and B ≠ 0

(b) E ≠ 0 and B = 0 (d) E = 0, B = 0

Chapter 26

Magnetics — 445

3. A charged particle of unit mass and unit charge moves with velocity v = ( 8i$ + 6j$ ) m/s in a magnetic field of B = 2k$ T. Choose the correct alternative (s).

(a) (b) (c) (d)

The path of the particle may be x2 + y2 − 4x − 21 = 0 The path of the particle may be x2 + y2 = 25 The path of the particle may be y2 + z 2 = 25 The time period of the particle will be 3.14 s

4. When a current carrying coil is placed in a uniform magnetic field with its magnetic moment anti-parallel to the field, then (a) (b) (c) (d)

torque on it is maximum torque on it is zero potential energy is maximum dipole is in unstable equilibrium

5. If a long cylindrical conductor carries a steady current parallel to its length, then (a) (b) (c) (d)

the electric field along the axis is zero the magnetic field along the axis is zero the magnetic field outside the conductor is zero the electric field outside the conductor is zero

6. An infinitely long straight wire is carrying a current I1. Adjacent to it there is another equilateral triangular wire having current I 2. Choose the wrong options. b I2

I1 a

(a) Net force on loop is leftwards (c) Net force on loop is upwards

c

(b) Net force on loop is rightwards (d) Net force on loop is downwards

7. A charged particle is moving along positive y-axis in uniform electric and magnetic fields $ E = E0k B = B $i

and 0 Here, E0 and B0 are positive constants. Choose the correct options. (a) (b) (c) (d)

particle may be deflected towards positive z-axis particle may be deflected towards negative z-axis particle may pass undeflected kinetic energy of particle may remain constant

8. A charged particle revolves in circular path in uniform magnetic field after accelerating by a potential difference of V volts. Choose the correct options if V is doubled. (a) (b) (c) (d)

kinetic energy of particle will become two times radius in circular path will become two times radius in circular path will become 2 times angular velocity will remain unchanged

446 — Electricity and Magnetism 9. abcd is a square. There is a current I in wire efg as shown.

e

b

Choose the correct options. (a) (b) (c) (d)

c I

Net magnetic field at a is inwards Net magnetic field at b is zero Net magnetic field at c is outwards Net magnetic field at d is inwards

g f d

a

10. There are two wires ab and cd in a vertical plane as shown in figure. Direction of current in wire ab is rightwards. Choose the correct options. c

a

d

i1

b

(a) If wire ab is fixed, then wire cd can be kept in equilibrium by the current in cd in leftward direction (b) Equilibrium of wire cd will be stable equilibrium (c) If wire cd is fixed, then wire ab can be kept in equilibrium by flowing current in cd in rightward direction (d) Equilibrium of wire ab will be stable equilibrium

Match the Columns 1. An electron is moving towards positive x-direction. Match the following two columns for deflection of electron just after the fields are switched on. (E0 and B0 are positive constants) Column I

Column II

(a) If magnetic field B = B0$j is switched on (p) Negative y-axis (b) If magnetic field B = B0k$ is switched on (q) Positive y-axis (c) If magnetic field B = B0$i and electric field E = E 0$j is switched on

(r) Negative z-axis

(d) If electric field E = E 0k$ is switched on

(s) Positive z-axis

2. Four charged particles, ( −q , m ), ( −3q , 4m ), ( +q , m ) and ( + 2q , m ) enter in uniform magnetic field (in inward direction) with same kinetic energy as shown in figure. Inside the magnetic field their paths are shown. Match the following two columns. Column I

Column II y

x

(a) Particle (−q, m)

(p) w

(b) Particle (−3q, 4m)

(q) x

(c) Particle (+ q, m)

(r) y

(d) Particle (+ 2q, m)

(s) z

w

z

Chapter 26

Magnetics — 447

3. In Column I, a current carrying loop and a uniform magnetic field are shown. Match this with Column II. Column I

Column II

B

(p) Force = 0

(a)

(b)

(q) Maximum torque

B

(c)

(r) Minimum potential energy B B

(d)

(s) Positive potential energy

y

4. Equal currents are flowing in two infinitely long wires lying along x and y-axes in the directions shown in figure. Match the following two columns. Column I

Column II

(a) Magnetic field at (a , a )

O

x

(p) along positive y-axis

(b) Magnetic field at (− a , − a ) (q) along positive z-axis (c) Magnetic field at (a , − a )

(r) along negative z-axis

(d) Magnetic field at (−a, a)

(s) zero

5. Equal currents are flowing in four infinitely long wires. Distance between two wires is same and directions of currents are shown in figure. Match the following two columns. 1

2

3

Column I (a) (b) (c) (d)

Force on wire-1 Force on wire-2 Force on wire-3 Force on wire-4

4

Column II (p) (q) (r) (s)

inwards leftwards rightwards zero

448 — Electricity and Magnetism 6. A square loop of uniform conducting wire is as shown in figure. A current I (in ampere) enters the loop from one end and exits the loop from opposite end as shown in figure. y I/2 I/2

I/2 x I/2

I

The length of one side of square loop is l metre. The wire has uniform cross-section area and uniform linear mass density. Column I (a) B = B0$i in tesla (b) B = B0$j in tesla (c) B = B0 ($i + $j) in tesla (d) B = B0k$ in tesla

Column II (p) magnitude of net force on loop is 2B0Il newton (q) magnitude of net force on loop is zero (r) magnitude of force on loop is 2 B0Il (s) magnitude of net force on loop is B0Il newton

Subjective Questions 1. An equilateral triangular frame with side a carrying a current I is placed at a distance a from an infinitely long straight wire carrying a current I as shown in the figure. One side of the frame is parallel to the wire. The whole system lies in the xy-plane. Find the magnetic force F acting on the frame. y x 2

I

I

I

1

3 I

a

2. Find an expression for the magnetic dipole moment and magnetic field induction at the centre of a Bohr’s hypothetical hydrogen atom in the n th orbit of the electron in terms of universal constants.

3. A square loop of side 6 cm carries a current of 30 A. Calculate the magnitude of magnetic field B at a point P lying on the axis of the loop and a distance 7 cm from centre of the loop.

4. A positively charged particle of charge 1 C and mass 40 g, is revolving along a circle of radius 40 cm with velocity 5 m/ s in a uniform magnetic field with centre at origin O in xy-plane. At t = 0, the particle was at ( 0, 0.4 m, 0 ) and velocity was directed along positive x-direction. Another particle having charge 1 C and mass 10 g moving uniformly parallel to z-direction 40 with velocity m/ s collides with revolving particle at t = 0 and gets stuck with it. Neglecting π gravitational force and colombians force, calculate x , y and z-coordinates of the combined π particle at t = sec. 40

Magnetics — 449

Chapter 26

5. A proton beam passes without deviation through a region of space where there are uniform kV and m B = 50 mT. Then, the beam strikes a grounded target. Find the force imparted by the beam on the target if the beam current is equal to I = 0.80 mA.

transverse mutually perpendicular electric and magnetic fields with E = 120

y

6. A positively charged particle having charge q is accelerated by a potential difference V. This particle moving along the x-axis enters a region where an electric field E exists. The direction of the electric field is along positive y-axis. The electric field exists in the region bounded by the lines x = 0 and x = a. Beyond the line x = a (i.e. in the region x ≥ a) there exists a magnetic field of strength B, directed along the positive y-axis. Find

E

B

q m

O

x=a

x

(a) at which point does the particle meet the line x = a (b) the pitch of the helix formed after the particle enters the region x ≥ a. Mass of the particle is m.

7. A charged particle having charge 10−6 C and mass of 10−10 kg is fired from the middle of the plate making an angle 30° with plane of the plate. Length of the plate is 0.17 m and it is separated by 0.1 m. Electric field E = 10−3 N/C is present between the plates. Just outside the plates magnetic field is present. Find the velocity of projection of charged particle and magnitude of the magnetic field perpendicular to the plane of the figure, if it has to graze the plate at C and A parallel to the surface of the plate. (Neglect gravity)

C 30°

E A

8. A uniform constant magnetic field B is directed at an angle of 45° to the x-axis in xy-plane. PQRS is a rigid square wire frame carrying a steady current I 0, with its centre at the origin O. At time t = 0, the frame is at rest in the position shown in the figure with its sides parallel to x and y-axis. Each side of the frame has mass M and length L. Y I0

S

R O

P

X

Q

(a) What is the magnitude of torque τ acting on the frame due to the magnetic field? (b) Find the angle by which the frame rotates under the action of this torque in a short interval of time ∆t, and the axis about which the rotation occurs (∆t is so short that any variation in the torque during this interval may be neglected). Given : The moment of inertia of the frame 4 about an axis through its centre perpendicular to its plane is ML2. 3

450 — Electricity and Magnetism 9. A ring of radius R having uniformly distributed charge Q is mounted on a rod suspended by two identical strings. The tension in strings in equilibrium is T0. Now, a vertical magnetic field is switched on and ring is rotated at constant angular velocity ω. Find the maximum value of ω 3T0 . with which the ring can be rotated if the strings can withstand a maximum tension of 2 D

R ω

B

10. Figure shows a cross-section of a long ribbon of width ω that is carrying a uniformly distributed total current i into the page. Calculate the magnitude and direction of the magnetic field B at a point P in the plane of the ribbon at a distance d from its edge. P

x x x x x x x x x x x ω

d

11. A particle of mass m having a charge q enters into a circular region of radius R with velocity v directed towards the centre. The strength of magnetic field is B. Find the deviation in the path of the particle. x x x x x x x x x R x

x

x x

x

x x x x x x x x x x x x x x x x x x x

v

12. A thin, uniform rod with negligible mass and length 0.2 m is attached to the floor by a

frictionless hinge at point P. A horizontal spring with force constant k = 4.80 N/ m connects the other end of the rod with a vertical wall. The rod is in a uniform magnetic field B = 0.340 T directed into the plane of the figure. There is current I = 6.50 A in the rod, in the direction shown. x

k x

x

x I

x

x

x P

x

x

x B x x

x

53° x x

x

Chapter 26

Magnetics — 451

(a) Calculate the torque due to the magnetic force on the rod, for an axis at P. Is it correct to take the total magnetic force to act at the centre of gravity of the rod when calculating the torque? (b) When the rod is in equilibrium and makes an angle of 53.0° with the floor, is the spring stretched or compressed? (c) How much energy is stored in the spring when the rod is in equilibrium?

13. A rectangular loop PQRS made from a uniform wire has length a, width b and mass m. It is free to rotate about the arm PQ, which remains hinged along a horizontal line taken as the y-axis (see figure). Take the vertically upward direction as the z-axis. A uniform magnetic field B = ( 3 $i + 4 k$ )B0 exists in the region. The loop is held in the xy-plane and a current I is passed through it. The loop is now released and is found to stay in the horizontal position in equilibrium. z

P

Q a

x

S

R b

(a) What is the direction of the current I in PQ? (b) Find the magnetic force on the arm RS. (c) Find the expression for I in terms of B0 , a, b and m.

y

Answers Introductory Exercise 26.1 $ $)N 4. (– 0.16 $i – 0.32 j – 0.64 k

1. [LT –1]

2. ( F , v ), ( F , B)

3. No

5. Positive

6. 9.47 × 10 6 m /s

7. 2.56 × 10 –14 N

Introductory Exercise 26.2 1. D , B

2. False

3. False

6. (a) electron (b) electron

4. Yes, No

5. Along positive z-direction

7. 0.0167 cm, 0.7 cm

Introductory Exercise 26.3 1.

2. No

2 B 0 il

$ (b) (0.02 N) $j 3. (a) (0.023 N) k

$ (c) zero (d) (– 0.0098 N) j

$ $ (e) (– 0.013 N) j + (– 0.026 N) k

4. 32 N upwards

Introductory Exercise 26.4 1.

qR 2ω

$ $ ) × 10 −4 N- m (b) U = − (6.0 × 10 −4 ) J 2. (a) τ = (–9.6 $i − 7.2 j + 8.0 k

4 4. –2.42 J

Introductory Exercise 26.5 µ 0i 4 πx µ i 5. 0 12

1. (a) 28.3 µT into the page (b) 24.7 µT into the page 3. 58.0 µT into the page

2.

4. 26.2 µT into the page

into the page  1 – 1  out of the page    a b

Introductory Exercise 26.6 1. 200 µT toward the top of the page, 133 µT toward the bottom of the page 2. (a) zero (b) −5.0 × 10 –6 T - m (c) 2.5 × 10 –6 T - m (d) 5.0 × 10 –6 T - m 3. (b) the magnetic field at any point inside the pipe is zero

Introductory Exercise 26.7 1. 90 T

2. 1.3 × 10 −7 A

Exercises LEVEL 1 Assertion and Reason 1. (c) 11. (d)

2. (c)

3. (b)

4. (d)

5. (c)

6. (b)

7. (a)

8. (c)

9. (a)

10. (d)

Objective Questions 1. (a) 11. (c)

2. (c) 12. (d)

3. (c) 13. (c)

4. (c) 14. (a)

5. (c) 15. (c)

6. (b) 16. (a)

7. (d) 17. (d)

8. (c) 18. (c)

9. (d) 19. (c)

10. (a) 20. (c)

Magnetics — 453

Chapter 26 21. (c) 31. (a)

22. (b) 32. (a)

23. (d) 33. (b)

24. (c) 34. (d)

25. (a) 35. (a)

26. (b)

27. (c)

28. (a)

29. (b)

30. (a)

Subjective Questions $ $ (b) − (6.24 × 10 – 14 N) k 1. (a) (6.24 × 10 –14 N) k 3. (a) B x = (–0.175)T, B z = (−0.256)T $ (b) + qvB $j 4. (a) −qvB k 6. (a) 8.35 × 10 5 m /s

2. B x = – 2.0 T

(b) Yes, B y

(c) zero (d)

−qvB $ j 2

(c) zero, 90°

 qvB  $ $ (e)  −  ( j + k)  2 

(b) 2.62 × 10 –8 s (c) 7.26 kV

7. (a) –q

8. (a) 1.6 × 10 –4 T into the page (b) 1.11 × 10 –7 s $ $ , Here ω = Be 9. v = vx $i + vy cos ωt j − vy sin ωt k m 2qE 0 z 2 πmE tan θ 11. vy = 0, v = 12. m qB 2 14. (a) 817.5 V 16.

πR 3 2µ 0

(b) 112.8 m /s 2

πm Bq

$ 10. E = (−0.1 V /m) k 13. 0.47 A from left to right

18. (a) 1.5 × 10 –16 s (b) 1.1 mA (c) 9.3 × 10 –24 A - m2 22. 2 × 10 −6 T

21. 2.0 µT

20. zero

(b)

$ , F = (0.04$i + 0.04k $)N 15. Fab = 0,Fbc = (– 0.04 N) $i, Fcd = (– 0.04 N) k da

17. (a) 76.7° (b) 76.7°

$ 19. M = 2Il 2 j

5. 8.38 × 10 –4 T

I  23. y =  1  x  I2 

24. 2 rad

25. (a) Between the wires, 30.0 cm from wire carrying 75.0 A (b) 20.0 cm from wire carrying 25.0 A and 60.0 cm from wire carrying 75.0 A πD  28. (a) 2.77 A (b) 0.0184 m 26. 69 27. I1 =   I , towards right  R  2 29. 2.0 A toward bottom of page π π µ 0 in2 sin   tan    n  n 31. (a) 2 π 2r 32. (a)

µ 0I πr

−µ 0 qv0 I $ k (b) F1 = F2 = 2B 0 IR$i , F = 4B 0 IR$i 4R

µ 0i 2r

 2r 2 − a 2  µ 0I  2   4r − a2  to the left (b) πr  

 2r 2 + a 2   2   4r + a2  towards the top of the page  

µ 0 I1I2L , zero 2 πa  2 π m v0 sin θ  38.  , 0 , 0 Bq  

33. BP = 0, BQ = µ 0 λ 37 v0 ,

(b)

30. (a)

34.

mv0 qE 0

36. (a) π

(b) π

(c)

π 6

39. (0.04 $j – 0.07 k$ ) A - m2

40. 9.98 N-m, clockwise as seen looking down from above. 41. 20.0 µT toward the bottom of the square

42. (a)

µ 0 b r12 3

(b)

µ 0b R 3 3r2

LEVEL 2 Objective Questions 1.(a)

2.(c)

3.(c)

4.(c)

5.(d)

6.(d)

7.(b)

8.(d)

9.(b)

10.(c)

11.(d)

12.(a)

13.(b)

14.(a)

15.(c)

16.(c)

17.(d)

18.(c)

19.(c)

20.(d)

21.(b)

22.(c)

23.(d)

24.(d)

25.(a)

26.(b)

27.(a)

454 — Electricity and Magnetism More than One Correct Options 1.(a,c)

2.(a,c,d) 3.(a,b,d) 4.(b,c,d) 5.(b,d)

6.(a,b,c,d) 7.(a,b,c,d) 8.(a,c,d) 9.(a,c,d)

Match the Columns 1. (a) → r

(b) → q

(c) → p

(d) → r

2. (a) → r

(b) → s

(c) → q

(d) → p

3. (a) → p,s

(b) → p,q

(c) → p,r

(d) → p,s

4. (a) → q

(b) → r

(c) → s

(d) → s

5. (a) → q

(b) → r

(c) → q

(d) → r

6. (a) → s

(b) → s

(c) → q

(d) → p

Subjective Questions 1. F =

 2 + 3 $ µ 0I 2  1 1   ( i) ln   − 2   π 2 3   

3. 2.7 × 10 –4 T 5. 2 × 10

−5

N

7. 2.0 m/s, 3.46 mT 9. ωmax =

DT 0 BQR 2

2. M =

µ πm2e7 neh ,B = 0 3 5 5 4π m 8 ε0h n

4. (0.2 m, 0.2 m, 0.2 m) 6. (a) y =

Ea2 4V

(b) p =

πEa B

2m qV

3 I0B (∆t )2 4 M  d + ω µ i 10. B = 0 ln   (upwards) 2π ω  d  8. (a) I0L2B

BqR  11. 2 tan−1    mv  12. (a) 0.0442 N-m, clockwise, yes (b) stretched (c) 7.8 × 10 –3 J mg $ $ 13. (a) P to Q (b) IbB 0 (3 k − 4 i ) (c) 6bB 0

(b)

10.(a,b,c)

Electromagnetic Induction Chapter Contents 27.1

Introduction

27.2

Magnetic field lines and magnetic flux

27.3

Faraday's law

27.4

Lenz's law

27.5

Motional electromotive force

27.6

Self inductance and inductors

27.7

Mutual inductance

27.8

Growth and decay of current in an L-R circuit

27.9

Oscillations in L-C circuit

27.10 Induced electric field

456 — Electricity and Magnetism

27.1 Introduction Almost every modern device has electric circuits at its heart. We learned in the chapter of current electricity that an electromagnetic force (emf) is required for a current to flow in a circuit. But for most of the electric devices used in industry the source of emf is not a battery but an electrical generating station. In these stations other forms of energy are converted into electric energy. For example, in a hydroelectric plant gravitational potential energy is converted into electric energy. Similarly, in a nuclear plant nuclear energy is converted into electric energy. But how this conversion is done? Or what is the physics behind this? The branch of physics, known as electromagnetic induction gives the answer to all these queries. If the magnetic flux ( φB ) through a circuit changes, an emf and a current are induced in the circuit. Electromagnetic induction was discovered in 1830. The central principle of electromagnetic induction is Faraday’s law. This law relates induced emf to change in magnetic flux in any loop, including a closed circuit. We will also discuss Lenz’s law, which helps us to predict the directions of induced emf and current.

27.2 Magnetic Field Lines and Magnetic Flux Let us first discuss the concept of magnetic field lines and magnetic flux. We can represent any magnetic field by magnetic field lines. Unlike the electric lines of force, it is wrong to call them magnetic lines of force, because they do not point in the direction of the force on a charge. The force on a moving charged particle is always perpendicular to the magnetic field (or magnetic field lines) at the particle’s position. The idea of magnetic field lines is same as for the electric field lines as discussed in the chapter of electrostatics. The magnetic field at any point is tangential to the field line at that point. Where the field lines are close, the magnitude of field is large, where the field lines are far apart, the field magnitude is small. Also, because the direction of Bat each point is unique, field lines never intersect. Unlike the electric field lines, magnetic lines form a closed loop.

Magnetic Flux The flux associated with a magnetic field is defined in a similar manner to that used to define electric flux. Consider an element of area dS on an arbitrary shaped surface as shown in figure. If the magnetic field at this element is B, the magnetic flux through the element is dS B θ

Fig. 27.1

dφB = B ⋅ dS = BdS cos θ Here, dS is a vector that is perpendicular to the surface and has a magnitude equal to the area dS and θ is the angle between B and dS at that element. In general, dφB varies from element to element. The total magnetic flux through the surface is the sum of the contributions from the individual area elements. ∴

φB = ∫ BdS cos θ = ∫ B ⋅ dS

Chapter 27

Electromagnetic Induction — 457

Note down the following points regarding the magnetic flux : (i) Magnetic flux is a scalar quantity (dot product of two vector quantities is a scalar quantity) (ii) The SI unit of magnetic flux is tesla-metre 2 (1 T-m 2 ). This unit is called weber (1Wb). 1 Wb = 1 T-m 2 = 1 N-m / A Thus, unit of magnetic field is also weber/m 2 (1Wb/m 2 ). 1 T = 1 Wb/ m 2

or

(iii) In the special case in which B is uniform over a plane surface with total area S, φB = BS cos θ B

B θ

S

φB = BS cosθ

S φB = BS

Fig. 27.2

If B is perpendicular to the surface, then cos θ =1 and φB = BS

Gauss’s Law for Magnetism In Gauss’s law, the total electric flux through a closed surface is proportional to the total electric charge enclosed by the surface. For example, if a closed surface encloses an electric dipole, the total electric flux is zero because the total charge is zero. By analogy, if there were such as thing as a single magnetic charge (magnetic monopole), the total magnetic flux through a closed surface would be proportional to the total magnetic charge enclosed. But as no magnetic monopole has ever been observed, we conclude that the total magnetic flux through a closed surface is zero.

∫ B ⋅ dS = 0 Unlike electric field lines that begin and end on electric charges, magnetic field lines never have end points. Such a point would otherwise indicate the existence of a monopole. For a closed surface, the vector area element dS always points out of the surface. However, for an open surface we choose one of the possible sides of the surface to be the positive and use that choice consistently.

27.3 Faraday’s Law This law states that, “the induced emf in a closed loop equals the negative of the time rate of change of magnetic flux through the loop.” dφ e=– B dt

458 — Electricity and Magnetism If a circuit is a coil consisting of N loops all of the same area and if φB is the flux through one loop, an emf is induced in every loop, thus the total induced emf in the coil is given by the expression, dφ e=– N B dt The negative sign in the above equations is of important physical significance, which we will discuss in article. 27.4. Note down the following points regarding the Faraday’s law: (i) As we have seen, induced emf is produced only when there is a change in magnetic flux passing through a loop. The flux passing through the loop is given by φ = BS cos θ This, flux can be changed in several ways: (a) The magnitude of B can change with time. In the problems if magnetic field is given a function of time, it implies that the magnetic field is changing. Thus, B = B (t ) (b) The current producing the magnetic field can change with time. For this, the current can be given as a function of time. Hence, i = i( t ) (c) The area of the loop inside the magnetic field can change with time.This can be done by pulling a loop inside (or outside) a magnetic field.

Fig. 27.3

(d) The angle θ between B and the normal to the loop (or s) can change with time. ω

Fig. 27.4

This can be done by rotating a loop in a magnetic field. (e) Any combination of the above can occur. (ii) When the magnetic flux passing through a loop is changed, an induced emf and hence, an induced current is produced in the circuit. If R is the resistance of the circuit, then induced current is given by e 1  – dφB  i= =   R R  dt 

Chapter 27

Electromagnetic Induction — 459

Current starts flowing in the circuit, means flow of charge takes place. Charge flown in the circuit in time dt will be given by 1 dq = idt = (– dφB ) R Thus, for a time interval ∆t we can write the average values as, e=–

∆φB , ∆t

i=

1  –∆φB  1   and ∆q = (– ∆φB ) R  ∆t  R

From these equations, we can see that e and i are inversely proportional to ∆t while ∆q is independent of ∆t. It depends on the magnitude of change in flux, not the time taken in it. This can be explained by the following example. V

Example 27.1 A square loop ACDE of area 20 cm2 and resistance 5 Ω is rotated in a magnetic field B = 2T through 180°, (a) in 0.01 s and (b) in 0.02 s Find the magnitudes of average values of e, i and ∆q in both the cases.

B A

C

E

D

Let us take the area vector S perpendicular to plane of loop inwards. So initially, S ↑↑ B and when it is rotated by 180°, S ↑↓ B. Hence, initial flux passing through the loop,

Solution

Fig. 27.5

φi = BS cos 0° = ( 2) ( 20 × 10–4 ) (1) = 4.0 × 10–3 Wb Flux passing through the loop when it is rotated by 180°, φ f = BS cos 180° = ( 2) ( 20 × 10–4 ) (–1) = – 4.0 × 10–3 Wb Therefore, change in flux, ∆φB = φ f – φi = – 8.0 × 10–3 Wb (a) Given, ∆t = 0.01s, ∴

and

R = 5Ω ∆φB 8.0 × 10–3 = = 0.8 V ∆t 0.01 | e | 0.8 i= = = 0.16 A R 5 ∆q = i∆t = 0.16 × 0.01 = 1.6 × 10–3 C |e | = –

(b) ∆t = 0.02 s ∴

and

∆φB 8.0 × 10–3 = = 0.4 V ∆t 0.02 | e | 0.4 i= = = 0.08 A R 5 ∆q = i∆t = ( 0.08) ( 0.02) = 1.6 × 10–3 C

|e | = –

Note Time interval ∆t in part (b) is two times the time interval in part (a), so e and i are half while ∆q is same.

460 — Electricity and Magnetism V

Example 27.2 A coil consists of 200 turns of wire having a total resistance of 2.0 Ω. Each turn is a square of side 18 cm, and a uniform magnetic field directed perpendicular to the plane of the coil is turned on. If the field changes linearly from 0 to 0.5 T in 0.80 s, what is the magnitude of induced emf and current in the coil while the field is changing? Solution From the Faraday’s law, N∆φ ∆B Induced emf, | e | = = ( NS ) ∆t ∆t =

( 200) (18 × 10−2 ) 2 ( 0.5 − 0) 0.8

= 4.05 V Induced current, i = V

Ans.

| e | 4.05 = ≈ 2.0 A R 2

Ans.

Example 27.3 The magnetic flux passing through a metal ring varies with time t as : φB = 3 ( at3 − bt2 ) T- m2 with a = 2.00 s −3 and b = 6.00 s −2 . The resistance of the ring is 3.0 Ω. Determine the maximum current induced in the ring during the interval from t = 0 to t = 2.0 s. Solution

Given,

φB = 3 ( at 3 − bt 2 )

∴

|e | =

∴ Induced current,

i=

dφB = 9at 2 – 6bt dt

| e | 9at 2 − 6bt = = 3at 2 − 2bt R 3

For current to be maximum, di =0 dt ∴

6 at − 2b = 0 b t= 3a

or i.e. at t =

b , current is maximum. This maximum current is 3a 2

 b  b i max = 3a   – 2b    3a   3a  =

b 2 2b 2 b 2 − = 3a 3a 3a

Substituting the given values of a and b, we have i max =

( 6) 2 = 6.0 A 3( 2)

Ans.

Chapter 27

Electromagnetic Induction — 461

27.4 Lenz’s Law The negative sign in Faraday's equations of electromagnetic induction describes the direction in which the induced emf drives current around a circuit. However, that direction is most easily determined with the help of Lenz’s law. This law states that: “The direction of any magnetic induction effect is such as to oppose the cause of the effect.” For different types of problems, Lenz’s law has been further subdivided into following concepts. 1. Attraction and repulsion concept If magnetic flux is changed by bringing a magnet and a loop (or solenoid etc.) closer to each other then direction of induced current is so produced, that the magnetic field produced by it always repels the two. Similarly, if they are moved away from each other then they are attracted towards each other. Following two examples will illustrate this. V

Example 27.4 A bar magnet is freely falling along the axis of a circular loop as shown in figure. State whether its acceleration a is equal to, greater than or less than the acceleration due to gravity g. S N

a

Fig. 27.6

Solution a < g. Because according to Lenz’s law, whatever may be the direction of induced current, it will oppose the cause. S N

(a)

(b)

Fig. 27.7

Here, the cause is, the free fall of magnet and so the induced current will oppose it and the acceleration of magnet will be less than the acceleration due to gravity g. This can also be explained in a different manner. When the magnet falls downwards with its north pole downwards. The magnetic field lines passing through the coil in the downward direction increase. Since, the induced current opposes this, the upper side of the coil will become north pole, so that field lines of coil’s magnetic field are upwards. Now, like poles repel each other. Hence, a < g. V

Example 27.5 A bar magnet is brought near a solenoid as shown in figure. Will the solenoid attract or repel the magnet?

S

N

Fig. 27.8

462 — Electricity and Magnetism When the magnet is brought near the solenoid, according to Lenz’s law, both repel each other. On the other hand, if the magnet is moved away from the solenoid, it attracts the magnet. When the magnet is brought near the solenoid, the nearer side becomes the same pole and when it is moved away it becomes the opposite pole as shown in figure. Solution

S

N

N

S

S

N

S

N

Fig. 27.9

It can also be explained by increasing or decreasing field lines as discussed in example 27.4. 2. Cross or dot magnetic field increasing or decreasing concept

If cross magnetic field passing through a loop increases then induced current will produce dot magnetic field. Similarly, if dot magnetic field passing through a loop decreases then dot magnetic field is produced by the induced current. Let us take some examples in support of it. V

Example 27.6 A circular loop is placed in magnetic field B = 2 t. Find the direction of induced current produced in the loop. x

x

x

x

x

x

x

x Fig. 27.10

x

Solution B = 2 t , means ⊗ magnetic field (or we can also say cross magnetic flux) passing through the loop is increasing. So, induced current will produce dot magnetic field. To produce magnetic field, induced current from our side should be anti-clockwise. V

Example 27.7 A rectangular loop is placed to the left of large current carrying straight wire as shown in figure. Current varies with time as I = 2 t. Find direction of induced current I in in the square loop.

I = 2t

Fig. 27.11

magnetic field passing through the loop. Current I is Current I will produce increasing, so dot magnetic field will also increase. Therefore, induced current should produce cross magnetic field. For producing cross magnetic field in the loop, induced current from our side should be clockwise.

Solution

3. Situations where flux passing through the loop is always zero or change in flux is zero.

Chapter 27 V

Electromagnetic Induction — 463

Example 27.8 A current carrying straight wire passes inside a triangular loop as shown in Fig. 27.12. The current in the wire is perpendicular to paper inwards. Find the direction of the induced current in the loop if current in the wire is increased.

i

Fig. 27.12

Magnetic field lines round the current carrying wire are as shown in Fig.27.13. Since, the lines are tangential to the loop (θ = 90° ) the flux passing through the loop is always zero, whether the current is increased or decreased. Hence, change in flux is also zero. Therefore, induced current in the loop will be zero. Solution

Fig. 27.13 V

.

Example 27.9 A rectangular loop is placed adjacent to a current carrying straight wire as shown in figure. If the loop is rotated about an axis passing through one of its sides, find the direction of induced current in the loop. ω

i

Fig. 27.14

Magnetic field lines around the straight wire are circular. So, same magnetic lines will pass through loop under all conditions. Solution

⇒ ⇒

∆φ = 0 emf = 0 i=0

4. Attraction or repulsion between two loops facing each other if current in one loop is

changed

464 — Electricity and Magnetism V

Example 27.10 Two loops are facing each other as shown in Fig. 27.15. State whether the loops will attract each other or repel each other if current I1 is increased. Solution If current I 1 is increased then induced current in loop-2 (say I 2 ) will be in opposite direction. Now, two wires having currents in opposite directions repel each other. So, the loops will repel each other.

INTRODUCTORY EXERCISE

I1 1

2

Fig. 27.15

27.1

1. Figure shows a conducting loop placed near a long straight wire carrying a current i as shown. If the current increases continuously, find the direction of the induced current in the loop.

i

Fig. 27.16

2. A metallic loop is placed in a non-uniform steady magnetic field. Will an emf be induced in the loop?

3. Write the dimensions of

dφ B . dt

4. A triangular loop is placed in a dot

magnetic field as shown in figure. Find the direction of induced current in the loop if magnetic field is increasing.

Fig. 27.17

5. Two circular loops lie side by side in the same plane. One is connected to a source that supplies an increasing current, the other is a simple closed ring. Is the induced current in the ring is in the same direction as that in the loop connected to the source or opposite? What if the current in the first loop is decreasing?

6. A wire in the form of a circular loop of radius 10 cm lies in a plane normal to a magnetic field of 100 T. If this wire is pulled to take a square shape in the same plane in 0.1 s, find the average induced emf in the loop.

7. A closed coil consists of 500 turns has area 4 cm 2 and a resistance of 50 Ω. The coil is kept with its plane perpendicular to a uniform magnetic field of 0.2 Wb/m 2. Calculate the amount of charge flowing through the coil if it is rotated through 180°. 8. The magnetic field in a certain region is given by B = (4.0 $i − 1.8 k$ ) × 10–3 T. How much flux passes through a 5.0 cm 2 area loop in this region if the loop lies flat on the xy -plane?

Chapter 27

Electromagnetic Induction — 465

27.5 Motional Electromotive Force Till now, we have considered the cases in which an emf is induced in a stationary circuit placed in a magnetic field, when the field changes with time. In this section, we describe what is called motional emf, which is the emf induced in a conductor moving through a constant magnetic field. The straight conductor of length l shown in figure is moving through a uniform magnetic field directed into the page. For simplicity we assume that the conductor is moving in a direction perpendicular to the field with constant velocity under the influence of some external agent. The electrons in the conductor experience a force Fm = – e ( v × B) ++ ++ l

Fe

v

–

Fm –– ––

Fig. 27.18

Under the influence of this force, the electrons move to the lower end of the conductor and accumulate there, leaving a net positive charge at the upper end. As a result of this charge separation, an electric field is produced inside the conductor. The charges accumulate at both ends untill the downward magnetic force evB is balanced by the upward electric force eE. At this point, electrons stop moving. The condition for equilibrium requires that, eE = evB

or

E = vB

The electric field produced in the conductor (once the electrons stop moving and E is constant) is related to the potential difference across the ends of the conductor according to the relationship, ∆V = El = Blv ∆V = Blv where the upper end is at a higher electric potential than the lower end. Thus, “a potential difference is maintained between the ends of a straight conductor as long as the conductor continues to move through the uniform magnetic field.” x x Now, suppose the moving rod slides along a stationary U-shaped x a conductor forming a complete circuit. We call this a motional electromagnetic force denoted by e, we can write e = Bvl

x

x

x

b x

v

x

If R is the resistance of the circuit, then current in the circuit is e Bvl i= = R R

Fig. 27.19

Va > Vb . Therefore, direction of current in the loop is anti-clockwise as shown in figure.

x

466 — Electricity and Magnetism Extra Points to Remember ˜

Induced current (Upper side of palm) The direction of motional emf or current can be given by right hand rule. Stretch your right hand. Stretched fingers The stretched fingers point in the direction of magnetic field. (B ) Thumb is along the velocity of conductor. The upper side of the palm is at higher potential and lower side on lower Velocity of conductor (Thumb) potential. If the circuit is closed, the induced current within the Fig. 27.20 conductor is along perpendicular to palm upwards.

˜

i

a

e = Bv l l

⇒

v

R

R r

b

Fig. 27.21

In the Fig. 27.20, we can replace the moving rod ab by a battery of emf Bvl with the positive terminal at a and the negative terminal at b. The resistance r of the rod ab may be treated as the internal resistance of the battery. Hence, the current in the circuit is e Bvl or i = i = R+ r R+ r ˜

Induction and energy transfers In the Fig. 27.21, if you move the conductor ab with a constant velocity v, the current in the circuit is Bvl i = (r = 0) R a F, v l

R

Fm

b

Fig. 27.22 2 2

B l v acts on the conductor in opposite direction of velocity. So, to move the R conductor with a constant velocity v an equal and opposite force F has to be applied in the conductor. Thus,

A magnetic force Fm = ilB =

B2 l 2 v R The rate at which work is done by the applied force is F = Fm =

Papplied = Fv =

B2 l 2 v 2 R

Electromagnetic Induction — 467

Chapter 27 and the rate at which energy is dissipated in the circuit is 2

B2 l 2 v 2 Bvl  Pdissipated = i 2 R =   R=  R  R This is just equal to the rate at which work is done by the applied force. ˜

Motional emf is not a different kind of induced emf, it is exactly the induced emf described by Faraday’s d φB law, in the case in which there is a conductor moving in a magnetic field. Equation, e = – is best dt applied to problems in which there is a changing flux through a closed loop while e = Bvl is applied to problems in which a conductor moves through a magnetic field. Note that, if a conductor is moving in a magnetic field but circuit is not closed, then only PD will be asked between two points of the conductor. If the circuit is closed, then current will be asked in the circuit. Now, let us see how these two are similar.

d

a

l

i=0

d

a

v

v c

d

a

c

b

v

b

c

x (a)

b x (c)

(b)

Fig. 27.23

A loop abcd enters a uniform magnetic field B at constant speed v.

Refer figure (a)

Using Faraday’s equation, |e| = –

dφB dx d (BS ) d (Blx) = = = Bl = Blv dt dt dt dt

For the direction of current, we can use Lenz’s law. As the loop enters the field, ⊗ magnetic field passing through the loop increases, hence, induced current should produce magnetic field or current in the loop is anti-clockwise. From the theory of motional emf, e = Bvl and using right hand rule also, current in the d φB circuit is anti-clockwise. Thus, we see that e = – and e = Bvl give the same result. In the similar dt manner, we can show that current in the loop in figure (b) is zero and in figure (c) it is clockwise. ˜

We can generalize the concept of motional emf for a conductor with any shape moving in any magnetic field uniform or not. For an element dl of conductor the contribution de to the emf is the magnitude dl multiplied by the component of v × B parallel to dl, that is de = (v × B) ⋅ d l For any two points a and b the motional emf in the direction from b to a is, e=

a

∫b(v × B) ⋅ d l

b c a

v

b c

⇒

θ a

Fig. 27.24

v

v⊥ = v cos θ

468 — Electricity and Magnetism In general, we can say that motional emf in wire acb in a uniform magnetic field is the motional emf in an imaginary straight wire ab. Thus, e acb = e ab = (length of ab ) (v ⊥ ) (B) Here, v ⊥ is the component of velocity perpendicular to both B and ab. From right hand rule we can see that b is at higher potential and a at lower potential. Hence, ˜

Vba = Vb – Va = (ab ) (v cos θ) (B)

Motional emf induced in a rotating bar : A conducting rod of length l rotates with a constant angular speed ω about a pivot at one end. A uniform magnetic field B is directed perpendicular to the plane of rotation as shown in figure. Consider a segment of rod of lengthdr at a distance r from O. This segment has a velocity, v = rω B dr

l

P v

ω

O

Fig. 27.25

The induced emf in this segment is

de = Bvdr = B (rω) dr

Summing the emfs induced across all segments, which are in series, gives the total emf across the rod. l

∴

e = ∫ de =

∴

e=

0

l

∫0 Brωdr =

B ωl 2 2

B ωl 2 2

From right hand rule we can see that P is at higher potential than O. Thus, VP – VO = ˜

˜

B ωl 2 2

Note that in the problems of electromagnetic induction whenever you see a conductor moving in a magnetic field use the motional approach. It is easier than the other approach. But, if the conductor (or d φB loop) is stationary, you have no choice. Use e = – . dt Now onwards, the following integrations will be used very frequently. x

t

a (1 – e – bct ) b x t dx ∫ x0 a – bx = ∫0 c dt x=

then, and if then,

dx

∫0 a – bx = ∫0 c dt

If

x=

Here a, b and c are positive constants.

a a –  – x0  e – bct  b b

Electromagnetic Induction — 469

Chapter 27

B ωl 2 and 2 then it can be solved with the help of Kirchhoff’s laws. The following example will illustrate this concept.

Note In an electrical circuit, a moving or rotating wire may be assumed as a battery of emf Bvl or

V

Example 27.11 Two parallel rails with negligible resistance are 10.0 cm apart. They are connected by a 5.0 Ω resistor. The circuit also contains two metal rods having resistances of 10.0 Ω and 15.0 Ω along the rails. The rods are pulled away from the resistor at constant speeds 4.00 m/ s and 2.00 m/ s respectively. A uniform magnetic field of magnitude 0.01T is applied perpendicular to the, plane of the rails. Determine the current in the 5.0 Ω resistor. a

c

e

5.0 Ω

4.0 m/s

b 10.0 Ω

2.0 m/s

f 15.0 Ω

d

Fig. 27.26

Here, two conductors are moving in a uniform magnetic field. So, we will use the motional approach. The rod ab will act as a source of emf, e1 = Bvl = ( 0.01)( 4.0)( 0.1) = 4 × 10 –3 V HOW TO PROCEED

and internal resistance r1 = 10.0 Ω Similarly, rod ef will also act as a source of emf, e2 = ( 0.01)( 2.0)( 0.1) = 2.0 × 10 –3 V and internal resistance r2 = 15.0 Ω. From right hand rule we can see that, V b > V a and V e > V f Now, either by applying Kirchhoff’s laws or applying principle of superposition (discussed in the chapter of current electricity) we can find current through 5.0 Ω resistor. We will here use the superposition principle. You solve it by using Kirchhoff’s laws. Solution In the figures R = 5.0 Ω, r1 = 10 Ω, r2 = 15 Ω, e1 = 4 × 10–3 V and e 2 = 2 × 10–3 V i ′2

i e2

e2

r1

r1

⇒

R r2

r1

R

R

+ i2

i1

e1

i ′1

r2 e1

(a)

(b)

Fig. 27.27

i′

(c)

r2

470 — Electricity and Magnetism Refer figure (b) ∴

Net resistance of the circuit = r2 + Current, i =

Rr1 10 × 5 55 = 15 + = Ω R + r1 10 + 5 3

e2 2 × 10–3 6 = = × 10–3 A Net resistance 55/ 3 55

∴ Current through R,  r   10  i1 =  1  i =    10 + 5  R + r1 

6 –3   × 10  A   55

4 4 × 10–3 A = mA 55 55 Rr2 Refer figure (c) Net resistance of the circuit = r1 + R + r2 =

= 10 + ∴

Current, i ′ = =

∴ Current through R,

5 × 15 55 = Ω 5 + 15 4

e1 Net resistance 4 × 10–3 16 = × 10–3 A 55/ 4 55

 r   15  i ′1 =  2  i ′ =    15 + 5  R + r2  =

 16 –3   × 10 A  55 12 mA 55

From superposition principle net current through 5.0 Ω resistor is 8 i ′1 – i1 = mA from d to c 55 V

Ans.

Example 27.12 Figure shows the top view of a rod that can slide without friction. The resistor is 6.0 Ω and a 2.5 T magnetic field is directed perpendicularly downward into the paper. Let l = 1.20 m. B

l

R

F

Fig. 27.28

(a) Calculate the force F required to move the rod to the right at a constant speed of 2.0 m/s. (b) At what rate is energy delivered to the resistor? (c) Show that this rate is equal to the rate of work done by the applied force.

Chapter 27

Electromagnetic Induction — 471

The motional emf in the rod, e = Bvl or e = (2.5) (2.0) (1.2) V = 6.0 V e 6.0 The current in the circuit, i= = = 1.0 A R 6.0 (a) The magnitude of force F required will be equal to the magnetic force acting on the rod, which opposes the motion. Solution

∴

F = Fm = ilB or

F = (1.0) (1.2) (2.5) N = 3 N

Ans.

(b) Rate by which energy is delivered to the resistor is P1 = i 2 R = (1) 2 (6.0) = 6 W

Ans.

(c) The rate by which work is done by the applied force is P2 = F ⋅ v = ( 3) (2.0) = 6 W P1 = P2

INTRODUCTORY EXERCISE

Hence proved.

27.2

1. A horizontal wire 0.8 m long is falling at a speed of 5 m / s perpendicular to a uniform magnetic field of 1.1 T, which is directed from east to west. Calculate the magnitude of the induced emf. Is the north or south end of the wire positive?

2. As shown in figure, a metal rod completes the circuit. The circuit area is perpendicular to a magnetic field with B = 0.15 T. If the resistance of the total circuit is 3 Ω, how large a force is needed to move the rod as indicated with a constant speed of 2 m / s? x x

x x x B = 0.15 T (into page) x x x

R

50 cm x x

x

x

x

x

x

x x Fig. 27.29

x

x

x

x

v = 2 m/s

x

x

x

x

x

3. A rod of length 3l is rotated with an angular velocity ω as shown in figure. The uniform magnetic field B is into the paper. Find (a) VA − VC (b) VA − VD

A l

2l C ω Fig. 27.30

D ⊗B

4. As the bar shown in figure moves in a direction perpendicular to the field, is an external force required to keep it moving with constant speed. x

x

x

x

x

v

x Fig. 27.31

x

x

472 — Electricity and Magnetism

27.6 Self-inductance and Inductors Consider a single isolated circuit. When a current is present in the circuit, it sets up a magnetic field that causes a magnetic flux through the same circuit. This flux changes as the current in the circuit is changed. According to Faraday’s law any change in flux in a circuit produces an induced emf in it. Such an emf is called a self-induced emf. The name is so called because the source of this induced emf is the change of current in the same circuit. According to Lenz’s law the self-induced emf always opposes the change in the current that caused the emf and so tends to make it more difficult for variations in current to occur. We will here like to define a term self-inductance L of a circuit which is of great importance in our proceeding discussions. It can be defined in the following two ways : First Definition Suppose a circuit includes a coil with N turns of wire. It carries a current i. The total flux ( NφB ) linked with the coil is directly proportional to the current ( i) in the coil, i.e. NφB ∝ i When the proportionality sign is removed a constant L comes in picture, which depends on the dimensions and number of turns in the coil. This constant is called self-inductance. Thus, NφB or L= NφB = Li i From here we can define self-inductance ( L) of any circuit as, the total flux per unit current. The SI unit of self-inductance is henry (1H). Second Definition If a current i is passed in a circuit and it is changed with a rate di / dt, the induced emf e produced in the circuit is directly proportional to the rate of change of current. Thus, di e∝ dt When the proportionality constant is removed, the same constant L again comes here. di Hence, e=– L dt The minus sign here is a reflection of Lenz’s law. It says that the self-induced emf in a circuit opposes any change in the current in that circuit. From the above equation, L=

–e di / dt

This equation states that, the self-inductance of a circuit is the magnitude of self induced emf per unit rate of change of current. A circuit or part of a circuit, that is designed to have a particular inductance is called an inductor. The usual symbol for an inductor is Fig. 27.32

Thus, an inductor is a circuit element which opposes the change in current through it. It may be a circular coil, solenoid etc.

Chapter 27

Electromagnetic Induction — 473

Significance of Self-inductance and Inductor Like capacitors and resistors, inductors are among the circuit elements of modern electronics. Their purpose is to oppose any variations in the current through the circuit. In a DC circuit, an inductor helps to maintain a steady state current despite fluctuations in the applied emf. In an AC circuit, an inductor tends to suppress variations of the current that are more rapid than desired. An inductor plays a dormant role in a circuit so far as current is constant. It becomes active when current changes in the circuit. Every inductor has some self-inductance which depends on the size, shape and the number of turns etc. For N turns close together, it is always proportional to N 2 . It also depends on the magnetic properties of the material enclosed by the circuit. When the current passing through it is changed, an emf of magnitude L di / dt is induced across it. Later in this article, we will explore the method of finding the self-inductance of an inductor.

Potential Difference Across an Inductor We can find the polarities of self-induced emf across an inductor from Lenz’s law. The induced emf is produced whenever there is a change in the current in the inductor. This emf always acts to oppose this change. Figure shows three cases. Assume that the inductor has negligible resistance, so the PD, Vab = Va – Vb between the inductor terminals a and b is equal in magnitude to the self-induced emf. Refer figure (a) The current is constant, and there is no self-induced emf. Hence, Vab = 0 di Refer figure (b) The current is increasing, so is positive. The induced emf e dt must oppose the increasing current, so it must be in the sense from b to a, a becomes the higher potential terminal and Vab is positive. The direction of the emf is analogous to a battery with a as its positive terminal. di Refer figure (c) The current is decreasing and is negative. The self-induced dt emf eopposes this decrease and Vab is negative. This is analogous to a battery with b as its positive terminal. In each case, we can write the PD, Vab as di Vab = – e = L dt i a

i (constant) a

b di = 0 dt e=0 Vab = 0 (a) i (increasing)

a +

b – e di > 0 dt Vab > 0 (b) i (decreasing)

a –

b + e di < 0 dt Vab < 0 (c)

Fig. 27.33

i b

a

R Vab = iR

b L di Vab = L dt (b)

(a)

Fig. 27.34

The circuit’s behaviour of an inductor is quite different from that of a resistor. While a resistor opposes the current i, an inductor opposes the change ( di / dt ) in the current.

474 — Electricity and Magnetism Kirchhoff’s potential law with an inductor In Kirchhoff’s potential law when we go through an inductor in the same direction as the assumed current, we encounter a voltage drop equal to L di / dt, where di / dt is to be substituted with sign. R

L

H

L

L

di Drop = L dt

Drop = iR

i

H

E Fig. 27.35

For example in the loop shown in figure, Kirchhoff’s second law gives the equation. di E – iR – L = 0 dt V

Example 27.13 The inductor shown in the figure has inductance 0.54 H and carries a current in the direction shown that is decreasing at a uniform rate di = – 0.03 A/s. dt i b

a L

Fig. 27.36

(a) Find the self-induced emf. (b) Which end of the inductor a or b is at a higher potential? Solution (a) Self-induced emf, di e = – L = (– 0.54 ) (– 0.03) V dt = 1.62 × 10–2 V di (b) Vba = L = – 1.62 × 10–2 V dt

Ans.

Since, Vba ( = Vb – Va ) is negative. It implies thatVa > Vb or a is at higher potential. V

Ans.

Example 27.14 In the circuit diagram shown in figure, R = 10 Ω, L = 5H , E = 20 V , i = 2 A. This current is decreasing at a rate of –1.0 A/s. Find V ab at this instant. a

R

i

L

E

Fig. 27.37

Solution

PD across inductor, VL = L

di = ( 5) (– 1.0) = – 5 V dt

b

Chapter 27 Now,

Electromagnetic Induction — 475

Va – iR – VL – E = Vb

∴

Vab = Va – Vb = E + iR + VL = 20 + ( 2) (10) – 5 = 35 V

Ans.

di = 5 V in such a dt manner that this emf supports the decreasing current, or it sends the current in the circuit in the same direction as the existing current. So, positive terminal of this source is towards b. Thus, the given circuit can be drawn as shown below,

Note As the current is decreasing, the inductor can be replaced by a source of emf e = L ⋅

e= L a

di dt = 5 V

R

E = 20 V b

i

Fig. 27.38

Now, we can find Vab .

Method of Finding Self-inductance of a Circuit We use the equation, L = NφB / i to calculate the inductance of given circuit. A good approach for calculating the self-inductance of a circuit consists of the following steps: (a) Assume that there is a current i flowing through the circuit (we can call the circuit as inductor). (b) Determine the magnetic field B produced by the current. (c) Obtain the magnetic flux φB . (d) With the flux known, the self-inductance can be found from L = NφB / i. To demonstrate this procedure, we now calculate the self-inductance of two inductors.

Inductance of a Solenoid Let us find the inductance of a uniformly wound solenoid having N turns and length l. Assume that l is much longer than the radius of the windings and that the core of the solenoid is air. We can assume that the interior magnetic field due to a current i is uniform and given by equation, N B = µ 0 ni = µ 0   i  l  where, n =

N is the number of turns per unit length. l

The magnetic flux through each turn is,

φB = BS = µ 0

NS i l

Here, S is the cross-sectional area of the solenoid. Now, L= ∴

NφB N  µ 0 NSi  µ 0 N 2 S =  = i i  l  l L=

µ 0 N 2S l

476 — Electricity and Magnetism This result shows that L depends on dimensions ( S , l) and is proportional to the square of the number of turns. L∝N2 Because N = nl, we can also express the result in the form, ( nl) 2 L =µ0 S = µ 0 n 2 Sl = µ 0 n 2V or l Here, V = Sl is the volume of the solenoid.

L = µ 0 n 2V

Inductance of a Rectangular Toroid A toroid with a rectangular cross-section is shown in figure. The inner and outer radii of the toroid are R1 and R 2 and h is the height of the toroid. Applying Ampere’s law for a toroid, we can show that magnetic field inside a rectangular toroid is given by R2 R1 r dr

dS

h

Fig. 27.39

µ 0 Ni 2π r where, r is the distance from the central axis of the toroid. Because the magnitude of magnetic field changes within the toroid, we must calculate the flux by integrating over the toroid’s cross-section. Using the infinitesimal cross-sectional area element dS = hdr shown in the figure, we obtain R2  µ Ni  φB = ∫ B dS = ∫  0  ( hdr ) R1  2 π r  B=

=

µ 0 Nhi  R 2  ln   2π  R1 

Now,

L=

NφB µ 0 N 2 h  R 2  = ln   i 2π  R1 

or

L=

µ 0 N 2h  R2  ln   2π  R1 

As expected, the self-inductance is a constant determined by only the physical properties of the toroid.

Chapter 27

Electromagnetic Induction — 477

Energy Stored in an Inductor The energy of a capacitor is stored in the electric field between its plates. Similarly, an inductor has the capability of storing energy in its magnetic field. i (increasing)

e=L e=L

di dt

di dt

Fig. 27.40

An increasing current in an inductor causes an emf between its terminals. The work done per unit time is power. dW di P= = – ei = – Li dt dt dW dU From dW = – dU or P = =– dt dt dU di We have, or dU = Li di = Li dt dt The total energy U supplied while the current increases from zero to a final value i is i 1 U = L∫ idi = Li 2 0 2 1 U = Li 2 ∴ 2 This is the expression for the energy stored in the magnetic field of an inductor when a current i flows through it. The source of this energy is the external source of emf that supplies the current. Note

(i) After the current has reached its final steady state value i, di / dt = 0 and no more energy is taken by the inductor. (ii) When the current decreases from i to zero, the inductor acts as a source that supplies a total amount of 1 energy Li 2 to the external circuit. If we interrupt the circuit suddenly by opening a switch, the current 2 decreases very rapidly, the induced emf is very large and the energy may be dissipated as a spark across the switch. (iii) If we compare the behaviour of a resistor and an inductor towards the current flow we can observe that energy flows into a resistor whenever a current passes through it. Whether the current is steady (constant) or varying this energy is dissipated in the form of heat. By contrast energy flows into an ideal, zero resistance inductor only when the current in the inductor increases. This energy is not dissipated, it is stored in the inductor and released when the current decreases. (iv) As we said earlier also, the energy in an inductor is actually stored in the magnetic field within the coil. We can develop relations of magnetic energy density u (energy stored per unit volume) analogous to those we obtained in electrostatics. We will concentrate on one simple case of an ideal long cylindrical solenoid. For a long solenoid its magnetic field can be assumed completely of within the solenoid. The energy U stored in the solenoid when a current i is 1 1 U = Li 2 = ( µ 0 n 2V ) i 2 as L = µ 0 n 2V 2 2

478 — Electricity and Magnetism U V U 1 1 (µ 0 ni ) 2 1 B2 u = = µ 0 n 2i 2 = = V 2 2 µ0 2 µ0

The energy per unit volume is u = ∴ as

B = µ 0 ni

Thus,

u=

1 B2 2 µ0

1 ε 0E 2 used in electrostatics. Although, we have derived it for one 2 special situation, it turns out to be correct for any magnetic field configuration.

This expression is similar to u =

V

Example 27.15 (a) Calculate the inductance of an air core solenoid containing 300 turns if the length of the solenoid is 25.0 cm and its cross-sectional area is 4.00 cm2 . (b) Calculate the self-induced emf in the solenoid if the current through it is decreasing at the rate of 50.0 A/s. Solution

(a) The inductance of a solenoid is given by L=

µ 0N 2S l

Substituting the values, we have L=

( 4π × 10–7 ) ( 300) 2 ( 4.00 × 10–4 ) ( 25.0 × 10–2 )

= 1.81 × 10–4 H (b) e = – L

H Ans.

di dt di = – 50.0 A/s dt

Here, ∴

e = – (1.81 × 10–4 ) (–50.0) = 9.05 × 10–3 V e = 9.05 mV

or V

Ans.

Example 27.16 What inductance would be needed to store 1.0 kWh of energy in a coil carrying a 200 A current. (1 kWh = 3.6 × 10 6 J ) Solution

We have, i = 200 A

and

U = 1 kWh = 3.6 × 106 J

∴

L= =

U = 1 Li 2      2

2U i

2

2 ( 3.6 × 106 ) ( 200) 2

= 180 H

Ans.

Chapter 27 V

Electromagnetic Induction — 479

Example 27.17 (a) What is the magnetic flux through one turn of a solenoid of self-inductance 8.0 × 10 −5 H when a current of 3.0 A flows through it? Assume that the solenoid has 1000 turns and is wound from wire of diameter 1.0 mm. (b) What is the cross-sectional area of the solenoid? Solution

Given,

L = 8.0 × 10−5 H, i = 3.0 A and N = 1000 turns

Nφ i The flux linked with one turn,

(a) From the relation, L =

φ=

Li ( 8.0 × 10−5 ) ( 3.0) = N 1000

= 2.4 × 10−7 Wb (b) This φ = BS = (µ 0 ni ) ( S ) l x x x x x x x x x x x x x x x x x

d

Fig. 27.41

Here, n = number of turns per unit length N N 1 = = = l Nd d µ 0 iS ∴ φ= d S =

or

φd ( 2.4 × 10–7 ) (1.0 × 10−3 ) = µ 0i ( 4π × 10−7 ) ( 3.0)

= 6.37 × 10−5 m 2 V

Ans.

Example 27.18 A 10 H inductor carries a current of 20 A. How much ice at 0°C could be melted by the energy stored in the magnetic field of the inductor? Latent heat of ice is 22.6 × 103 J / kg . Solution

Energy stored is

1 2 Li . 2

This energy is completely used in melting the ice. 1 2 Hence, Li = mL f 2 Here,

L f = latent heat of fusion

480 — Electricity and Magnetism m=

Hence, mass of ice melted,

Li 2 2L f

Substituting the values, we have m=

(10) ( 20) 2 2 ( 2.26 × 103 )

= 0.88 kg

Ans.

27.3

INTRODUCTORY EXERCISE

1. The current through an inductor of 1H is given by i = 3 t sin t . Find the voltage across the inductor.

2. In the figure shown i = 10e −4t A. Find VL and Vab. R = 4Ω

L = 2H

a

i

b

Fig. 27.42

3. The current (in Ampere) in an inductor is given by I = 5 + 16 t , where t is in seconds. The self-induced emf in it is 10 mV. Find (a) the self-inductance, and (b) the energy stored in the inductor and the power supplied to it at t = 1s 4. (a) Calculate the self-inductance of a solenoid that is tightly wound with wire of diameter 0.10 cm, has a cross-sectional area 0.90 cm 2 and is 40 cm long. (b) If the current through the solenoid decreases uniformly from 10 A to 0 A in 0.10 s, what is the emf induced between the ends of the solenoid?

27.7 Mutual Inductance We have already discussed in Chapter 26, the magnetic interaction between two wires carrying steady currents. The current in one wire causes a magnetic field, which exerts a force on the current in the second wire. An additional interaction arises between two circuits when there is a changing current in one of the circuits. 1

i1

2

i1

Fig. 27.43

Chapter 27

Electromagnetic Induction — 481

Consider two neighbouring coils of wire as shown in Fig. 27.42. A current flowing in coil 1 produces magnetic field and hence, a magnetic flux through coil 2. If the current in coil 1 changes, the flux through coil 2 changes as well. According to Faraday’s law this induces an emf in coil 2. In this way, a change in the current in one circuit can induce a current in a second circuit. This phenomenon is known as mutual induction. Like the self-inductance ( L), two circuits have mutual inductance ( M ). It also have two definitions as under: First Definition Suppose the circuit 1 has a current i1 flowing in it. Then, total flux N 2φB2 linked with circuit 2 is proportional to the current in 1. Thus, N 2 φB2 ∝ i1 or N 2φB2 = Mi1 Here, the proportionality constant M is known as the mutual inductance M of the two circuits. Thus,

M=

N 2φB2 i1

From this expression, M can be defined as the total flux N 2φB 2 linked with circuit 2 per unit current in circuit 1. Second Definition If we change the current in circuit 1 at a rate di1 / dt, an induced emf e2 is developed in circuit 1, which is proportional to the rate di1 / dt. Thus, e2 ∝ di1 / dt or e2 = – Mdi1 / dt Here, the proportionality constant is again M. Minus sign indicates that e2 is in such a direction that it opposes any change in the current in circuit 1. From the above equation, M=

– e2 di1 / dt

This equation states that, the mutual inductance of two circuits is the magnitude of induced emf e 2 per unit rate of change of current di 1 / dt . Note down the following points regarding the mutual inductance: 1. The SI unit of mutual inductance is henry (1H). 2. M depends upon closeness of the two circuits, their orientations and sizes and the number of turns etc. 3. Reciprocity theorem : M 21 = M 12 = M and

and

e2 = – M ( di1 /dt ) e1 = – M ( di2 / dt ) N 2φB2 M 12 = i1 M 21 =

N 1φB1 i2

482 — Electricity and Magnetism 4. A good approach for calculating the mutual inductance of two circuits consists of the following

steps: (a) Assume any one of the circuits as primary (first) and the other as secondary (second). (b) Pass a current i1 through the primary circuit. (c) Determine the magnetic field B produced by the current i1 . (d) Obtain the magnetic flux φB2 . (e) With this flux, the mutual inductance can be found from, N 2 φB2 M= i1

Mutual Inductance of a Solenoid Surrounded by a Coil Figure shows a coil of N 2 turns and radius R 2 surrounding a long solenoid of length l1 , radius R1 and number of turns N 1 . l1 R2 R1

Fig. 27.44

To calculate M between them, let us assume a current i1 in solenoid. There is no magnetic field outside the solenoid and the field inside has magnitude, N  B = µ 0  1  i1  l1  and is directed parallel to the solenoid’s axis. The magnetic flux φB2 through the surrounding coil is, therefore, µ N i φB2 = B ( πR12 ) = 0 1 1 πR12 l1 Now, ∴

M=

N 2φB2 i1

µ N N πR 2  N  µ N i  =  2   0 1 1  πR12 = 0 1 2 1 l1  i1   l1 

M=

µ 0 N 1 N 2 πR12 l1

Note that M is independent of the radius R 2 of the surrounding coil. This is because solenoid’s magnetic field is confined to its interior. In principle, we can also calculate M by finding the magnetic flux through the solenoid produced by the current in the surrounding coil. This approach is much more difficult, because φB1 is so complicated. However, since M 12 = M 21 , we do know the result of this calculation.

Electromagnetic Induction — 483

Chapter 27

Combination of Inductances In series If several inductances are in series so that there are no interactions through mutual inductance. a i

L1

c

L2

d

L3

⇒

b

L

a

b

i

(a)

(b)

Fig. 27.45

Refer figure (a) di dt di Vc – Vd = L2 dt di Vd – Vb = L3 dt Va – Vc = L1

and Adding all these equations, we have

Va – Vb = ( L1 + L2 + L3 )

di dt

…(i)

Refer figure (b) Va – Vb = L

di dt

…(ii)

Here, L = equivalent inductance. From Eqs. (i) and (ii), we have L = L1 + L2 + L3 In parallel Refer figure (a) i

i1

L1

i2

L2

i3

L3

a

b

⇒

i

L

a

(a)

b

(b)

Fig. 27.46

or or

i = i1 + i2 + i3 di di1 di2 di3 = + + dt dt dt dt di Va – Vb Va – Vb Va – Vb = + + dt L1 L2 L3

…(i)

484 — Electricity and Magnetism Refer figure (b) di Va – Vb = dt L 1 1 1 1 = + + L L1 L2 L3

From Eqs. (i) and (ii), V

…(ii)

Example 27.19 A straight solenoid has 50 turns per cm in primary and total 200 turns in the secondary. The area of cross-section of the solenoids is 4 cm2 . Calculate the mutual inductance. Primary is tightly kept inside the secondary. Solution The magnetic field at any point inside the straight solenoid of primary with n1 turns per unit length carrying a current i1 is given by the relation, B = µ 0 n1 i1 The magnetic flux through the secondary of N 2 turns each of area S is given as N 2 φ2 = N 2 ( BS ) = µ 0 n1 N 2 i1 S N φ ∴ M = 2 2 = µ 0 n1 N 2 S i1 Substituting the values, we get  50  M = ( 4π × 10–7 )  –2  ( 200) ( 4 × 10–4 )  10  = 5.0 × 10–4 H

V

Ans.

Example 27.20 Two solenoids A and B spaced close to each other and sharing the same cylindrical axis have 400 and 700 turns, respectively. A current of 3.50 A in coil A produced an average flux of 300 µT- m2 through each turn of A and a flux of 900 . µT- m2 through each turn of B. (a) Calculate the mutual inductance of the two solenoids. (b) What is the self-inductance of A? (c) What emf is induced in B when the current in A increases at the rate of 0.5 A/s? N φ Solution (a) M= B B iA =

( 700) ( 90 × 10–6 ) 3.5

= 1.8 × 10–2 H (b) L A =

Ans.

N A φA iA =

( 400) ( 300 × 10–6 ) 3.5

= 3.43 × 10–2 H

Ans.

Chapter 27

Electromagnetic Induction — 485

 di  eB = M  A   dt 

(c)

= (1.8 × 10–2 ) ( 0.5) = 9.0 × 10–3 V

INTRODUCTORY EXERCISE

Ans.

27.4

1. Calculate the mutual inductance between two coils when a current of 4 A changes to 12 A in 0.5 s in primary and induces an emf of 50 mV in the secondary. Also, calculate the induced emf in the secondary if current in the primary changes from 3 A to 9 A is 0.02 s.

2. A coil has 600 turns which produces 5 × 10−3 Wb / turn of flux when 3 A current flows in the wire. This produced 6 × 10−3 Wb/turn in 1000 turns secondary coil. When the switch is opened, the current drops to zero in 0.2 s in primary. Find (a) mutual inductance, (b) the induced emf in the secondary, (c) the self-inductance of the primary coil.

3. Two coils have mutual inductance M = 3.25 × 10– 4 H. The current i1 in the first coil increases at a uniform rate of 830 A /s. (a) What is the magnitude of the induced emf in the second coil? Is it constant? (b) Suppose that the current described is in the second coil rather than the first. What is the induced emf in the first coil?

27.8 Growth and Decay of Current in an L-R Circuit Growth of Current Let us consider a circuit consisting of a battery of emf E, a coil of self-inductance L and a resistor R. The resistor R may be a separate circuit element, or it may be the resistance of the inductor windings. By closing switch S 1 , we connect R and L in series with constant emf E. Let i be the current at some time t after switch S 1 is closed and di / dt be its rate of increase at that time. Applying Kirchhoff’s loop rule starting at the negative terminal and proceeding counterclockwise around the loop di E – Vab – Vbc = 0 or E – iR – L = 0 dt ∴

i

∫0

di = E – iR

t

∫0

dt L

Rt

or

– E i = (1 – e L ) R

E S1

S2 i a

R

b

L

c

Fig. 27.47

By letting E / R = i0 and L / R = τ L , the above expression reduces to i = i0 (1 – e – t / τ L ) Here, i0 = E / R is the current at t = ∞. It is also called the steady state current or the maximum current in the circuit.

486 — Electricity and Magnetism L is called time constant of the L-R circuit. At a time R equal to one time constant the current has risen to (1 – 1/ e) or about 63% of its final value i0 .

And τ L =

The i-t graph is as shown in figure.

i i0 =

E/R

0.63 i0

Note that the final current i0 does not depend on the inductance L, it is the same as it would be if the resistance R alone were connected to the source with emf E.

τL

t

Fig. 27.48

Let us have an insight into the behaviour of an L - R circuit from energy considerations. The instantaneous rate at which the source delivers energy to the circuit ( P = Ei) is equal to the instantaneous rate at which energy is dissipated in the resistor ( = i 2 R ) plus the rate at which energy is di  d 1 2 di  stored in the inductor  = iVbc = Li  or  Li  = Li ⋅     dt dt 2 dt Ei = i 2 R + Li

Thus,

di dt

Decay of Current Now suppose switch S 1 in the circuit shown in figure has been closed for a long time and that the current has reached its steady state value i0 . Resetting our stopwatch to redefine the initial time we close switch S 2 at time t = 0 and at the same time we should open the switch S 1 to by pass the battery. The current through L and R does not instantaneously go to zero but decays exponentially. To apply Kirchhoff’s loop rule and to find current in the circuit at time t, let us draw the circuit once more.

i a

R

i b

L

c

Fig. 27.49

Applying loop rule we have, (Va – Vb ) + (Vb – Vc ) = 0  di  iR + L   = 0  dt 

or Note Don’t bother about the sign of

di . dt

∴ ∴ ∴

(as Va = Vc )

i

∫i

0

di R = – dt i L di R t = – ∫ dt i L 0 i = i0 e – t / τ L

L , is the time for current to decrease to 1/ e or about 37% of its original value. The i-t graph R is as shown in Fig. 27.49.`

where, τ L =

Chapter 27

Electromagnetic Induction — 487

The energy that is needed to maintain the current during this decay is provided by energy stored in the magnetic field. Thus, the rate at which energy is dissipated in the resistor = rate at which the stored energy decreases in magnetic field of inductor dU d 1 2  di  or =– i2R = –  Li  = Li  –     dt  dt dt 2  – di  i 2 R = Li    dt 

or i i0

0.37 i0 tL

t

Fig. 27.50 V

Example 27.21 A coil of resistance 20 Ω and inductance 0.5 H is switched to DC 200 V supply. Calculate the rate of increase of current (a) at the instant of closing the switch and (b) after one time constant. (c) Find the steady state current in the circuit. Solution (a) This is the case of growth of current in an L-R circuit. Hence, current at time t is given by i = i 0 (1 – e – t / τ L ) Rate of increase of current, di i 0 – t / τ L = e dt τ L E/ R E di i 0 = = = dt τ L L/ R L

At t = 0, Substituting the value, we have

di 200 = = 400 A/s dt 0.5

Ans.

(b) At t = τ L , di = ( 400) e –1 = ( 0.37) ( 400) dt = 148 A/s

Ans.

(c) The steady state current in the circuit, i0 =

E 200 = = 10 A R 20

Ans.

488 — Electricity and Magnetism V

Example 27.22 A 5 H inductor is placed in series with a 10 Ω resistor. An emf of 5 V being suddenly applied to the combination. Using these values prove the principle of conservation of energy, for time equal to the time constant. Solution At any instant t, current in L-R circuit is given as i = i 0 (1 – e – t / τ L ) E L and τ L = R R After one time constant ( t = τ L ), current in the circuit is E 1 5  1 i = 1 –  = 1 –  = 0.316 A R  e 10  e i0 =

Here,

The rate at which the energy is delivered by the battery is P1 = Ei = ( 5) ( 0.316) = 1.58 W At this time rate by which energy is dissipated in the resistor is

…(i)

P2 = i 2 R = ( 0.316) 2 (10) = 0.998 W

…(ii)

The rate at which energy is stored in the inductor is d 1 di P3 =  Li 2  = Li     dt  dt  2 di i 0 –1 E = e = dt τ L eL

Here,

(after one time constant)

Substituting the values, we get E Ei P3 = ( L ) ( i )   =  eL e =

5 × 0.316 = 0.582 W 2.718

…(iii)

From Eqs. (i), (ii) and (iii), we have P1 = P2 + P3 It is the same as required by the principle of conservation of energy.

INTRODUCTORY EXERCISE 1. Show that

27.5

L has units of time. R

2. A coil of inductance 2 H and resistance 10 Ω are in a series circuit with an open key and a cell of constant 100 V with negligible resistance. At time t = 0, the key is closed. Find (a) the time constant of the circuit. (b) the maximum steady current in the circuit. (c) the current in the circuit at t = 1s.

3. In the simple L-R circuit, can the emf induced across the inductor ever be greater than the emf of the battery used to produce the current?

Chapter 27

Electromagnetic Induction — 489

27.9 Oscillations in L-C Circuit If a charged capacitor C is short-circuited through an inductor L, the charge and current in the circuit start oscillating simple harmonically. If the resistance of the circuit is zero, no energy is dissipated as heat. We also assume an idealized situation in which energy is not radiated away from the circuit. With these idealizations-zero resistance and no radiation, the oscillations in the circuit persist indefinitely and the energy is transferred from the capacitor’s electric field to the inductor’s magnetic field and back. The total energy associated with the circuit is constant. This is analogous to the transfer of energy in an oscillating mechanical system from potential energy to kinetic energy and back, with constant total energy. Later, we will see that this analogy goes much further. Let us now derive an equation for the oscillations in an L-C circuit. i t=0 C

t=t

b

+ q – 0

+ q –

L

L

a S

d S

(a)

(b)

Fig. 27.51

Refer figure (a)

c

]

A capacitor is charged to a PD, V0 = q 0C

Here, q 0 is the maximum charge on the capacitor. At time t = 0, it is connected to an inductor through a switch S. At time t = 0, switch S is closed. Refer figure (b) When the switch is closed, the capacitor starts discharging. Let at time t charge on the capacitor is q ( < q 0 ) and since, it is further decreasing there is a current i in the circuit in the direction shown in figure. Later we will see that, as the charge is oscillating there may be a situation when q will be increasing, but in that case direction of current is also reversed and the equation remains unchanged. The potential difference across capacitor = potential difference across inductor, or Vb – Va = Vc – Vd q  di  …(i) ∴ =L   dt  C  – dq  i=   dt 

Now, as the charge is decreasing, or

d 2q di =– 2 dt dt

Substituting in Eq. (i), we get

 d 2q  q =–L 2  C  dt 

or

d 2q dt

2

 1  =– q  LC 

…(ii)

490 — Electricity and Magnetism  d 2x  This is the standard equation of simple harmonic motion  2 = – ω 2 x  .  dt  1 Here, ω= LC

…(iii)

The general solution of Eq. (ii), is q = q 0 cos (ωt ± φ ) For example in our case φ = 0 as q = q 0 at t = 0. q = q 0 cos ωt

Hence,

…(iv)

Thus, we can say that charge in the circuit oscillates simple harmonically with angular frequency given by Eq. (iii). Thus, 1 ω 1 1 and T = = 2π LC ω= , f = = 2π 2π LC f LC The oscillations of the L-C circuit are an electromagnetic analog to the mechanical oscillations of a block-spring system. Table below shows a comparison of oscillations of a mass-spring system and an L-C circuit. Table 27.1 S.No.

Mass spring system

Inductor-capacitor circuit

1.

Displacement ( x)

Charge (q)

2.

Velocity (v )

Current (i)

3.

Acceleration (a)

di Rate of change of current    dt 

4.

d2x dt

2

= – ω2 x, where ω =

k m

d 2q dt

2

1 LC

= – ω2q , where ω =

5.

x = A sin (ωt ± φ) or x = A cos (ωt ± φ)

q = q 0 sin (ωt ± φ) or q = q 0 cos (ωt ± φ)

6.

dx = ω A 2 – x2 dt dv a= = – ω2 x dt 1 Kinetic energy = mv 2 2 1 Potential energy = kx2 2

i =

10.

1 1 1 1 2 mv 2 + kx2 = constant = kA 2 = mvmax 2 2 2 2

1 q 02 1 2 1 2 1q2 = constant = Li + = Limax 2 2 C 2 C 2

11.

|vmax| = Aω

imax = q 0ω

12

|amax| = ω A

7. 8. 9.

v=

2

dq = ω q 02 – q 2 dt

Rate of change of current = Magnetic energy =

1 Li 2

Potential energy =

1q2 2 C

 di  = ω2q 0    dt  max

13.

1 k

C

14.

m

L

di = – ω2q dt

2

Chapter 27

Electromagnetic Induction — 491

A graphical description of the energy transfer between the inductor and the capacitor in an L-C circuit is shown in the figure. The right side of the figure shows the analogous energy transfer in the oscillating block-spring system. 0 t=0

0 max

T=

T 4

0

0 t=

T 2 0 i = imax

t=

3 T 4

0

0

Fig. 27.52

di all oscillate simple harmonically with same angular frequency ω. But the dt di π di phase difference between q and i or between i and is , while that between i and is π. Their dt 2 dt 2 amplitudes are q0 , q0ω and ω q0 respectively. So, now suppose q = q0 cos ωt , then dq i= = – q0ω sin ωt and dt di = – q0ω 2 cos ωt dt

Note In L-C oscillations, q, i and

492 — Electricity and Magnetism Similarly, potential energy across capacitor (UC ) and across inductor (UL) also oscillate with double the frequency 2ω but not simple harmonically. The different graphs are as shown in Fig. 27.52. q

UC

q0

q 2max 2C t

t

UL

i i0

0

Li 2max 2 t

t T 2

T

3T 2

0

T 4

2T

T 2

3T 4

T

Fig. 27.53 V

Example 27.23 A capacitor of capacitance 25 µF is charged to 300 V . It is then connected across a 10 mH inductor. The resistance in the circuit is negligible. (a) Find the frequency of oscillation of the circuit. (b) Find the potential difference across capacitor and magnitude of circuit current 1.2 ms after the inductor and capacitor are connected. (c) Find the magnetic energy and electric energy at t = 0 and t = 1.2 ms. Solution (a) The frequency of oscillation of the circuit is 1 f= 2π LC Substituting the given values, we have f=

1 2π (10 × 10 ) ( 25 × 10–6 ) –3

(b) Charge across the capacitor at time t will be q = q 0 cos ωt and Now, charge in the capacitor after t = 1.2 × 10

Ans.

i = – q 0ω sin ωt

Here, q 0 = CV0 = ( 25 × 10 ) ( 300) = 7.5 × 10 –6

= 318.3 Hz

–3

C

–3

s is

q = ( 7.5 × 10–3 ) cos ( 2π × 318.3) (1.2 × 10–3 ) C = – 5.53 × 10–3 C ∴ PD across capacitor, V =

| q| 5.53 × 10–3 = = 221.2 volt C 25 × 10–6

Ans.

The magnitude of current in the circuit at t = 1.2 × 10–3 s is | i | = q 0ω sin ωt = ( 7.5 × 10–3 ) ( 2π ) ( 318.3) sin ( 2π × 318.3) (1.2 × 10–3 ) A = 10.13 A

Ans.

Chapter 27 (c) At t = 0

Electromagnetic Induction — 493

Current in the circuit is zero.

Hence, U L = 0 Charge in the capacitor is maximum. Hence,

UC =

1 q 02 2 C

or

UC =

1 ( 7.5 × 10–3 ) 2 × 2 ( 25 × 10–6 )

= 1.125 J

Ans.

∴ Total energy, E = U L + UC = 1.125 J At t = 1.2 ms UL = =

1 2 Li 2 1 (10 × 10–3 ) (10.13) 2 2

= 0.513 J ∴

UC = E – U L = 1.125 – 0.513 = 0.612 J

Ans.

Otherwise UC can be calculated as UC = =

1 q2 2 C 1 ( 5.53 × 10–3 ) 2 × 2 ( 25 × 10–6 )

= 0.612 J

INTRODUCTORY EXERCISE

27.6

1. Show that LC has units of time. 2. While comparing the L-C oscillations with the oscillations of spring-block system, with whom the magnetic energy can be compared and why?

3. In an L-C circuit, L = 0.75 H and C = 18 µF, (a) At the instant when the current in the inductor is changing at a rate of 3.40 A/s, what is the charge on the capacitor? (b) When the charge on the capacitor is 4.2 × 10–4 C, what is the induced emf in the inductor?

4. An L-C circuit consists of a 20.0 mH inductor and a 0.5 µF capacitor. If the maximum instantaneous current is 0.1 A, what is the greatest potential difference across the capacitor?

494 — Electricity and Magnetism

27.10 Induced Electric Field When a conductor moves in a magnetic field, we can understand the induced emf on the basis of magnetic forces on charges in the conductor as described in Art. 27.5. But an induced emf also occurs when there is a changing flux through a stationary conductor. What is it that pushes the charges around the circuit in this type of situation? x

x

x

x

x

x

x

x

x

x x Fig. 27.54

x

As an example, let’s consider the situation shown in figure. A conducting circular loop is placed in a magnetic field which is directed perpendicular to the paper inwards. When the magnetic field changes with time (suppose it increases with time) the magnetic flux φB also changes and according dφ to Faraday’s law the induced emf e = – B is produced in the loop. If the total resistance of the loop dt is R, the induced current in the loop is given by e i= R But what force makes the charges move around the loop? It can’t be the magnetic force, because the charges are not moving in the magnetic field. x

x E x x x

x

x

x x

x x

x

x

x E x x

x xr

x

x x

x

x x

x

x E x B in

x

x

x xE x

x

Fig. 27.55

Actually, there is an induced electric field in the conductor caused by the changing magnetic flux. This electric field has the following important properties: 1. It is non-conservative in nature. The line integral of E around a closed path is not zero. This line

integral is given by

∫ E⋅ dl = –

dφB dt

Note that this equation is valid only if the path around which we integrate is stationary.

…(i)

Chapter 27

Electromagnetic Induction — 495

2. Because of symmetry, the electric field E has the same magnitude at every point on the circle and

is tangent to it at each point. The directions of Eat several points on the loop are shown in figure. 3. Being a non-conservative field, the concept of potential has no meaning for such a field. 4. This field is different from the electrostatic field produced by stationary charges (which is

conservative in nature). 5. The relation F = qE is still valid for this field. 6. This field can vary with time. So, a changing magnetic field acts as a source of electric field of a sort that we cannot produce with any static charge distribution. This may seen strange but its the way nature behaves. 1. For symmetrical situations (as shown in figure) Eq. (i), in simplified form can be written as

Note

El =

dφB dB =S dt dt

Here, l is the length of closed loop in which electric field is to be calculated and S is the area in which magnetic field is changing. 2. Direction of electric field is the same as the direction of induced current. V

Example 27.24 The magnetic field at all points within the cylindrical region whose cross-section is indicated in the accompanying figure start increasing at a constant rate α T / s . Find the magnitude of electric field as a function of r, the distance from the geometric centre of the region.

R

Fig. 27.56

Solution

For r ≤ R

Using El = S

dB dt

or

E ( 2πr ) = ( πr 2 ) α

r R

Fig. 27.57

∴

E=

rα 2

496 — Electricity and Magnetism ∴ E ∝ r, i.e. E - r graph is a straight line passing through origin. Rα At r = R , E = 2 For r ≥ R E

r

R

Fig. 27.58

Using El = S

dB , dt

∴

E ( 2πr ) = ( πR 2 ) (α ) E=

∴

αR 2 2r

1 E ∝ , i.e. E-r graph is a rectangular hyperbola. r The E-r graph is as shown in figure.

∴

E

Rα 2

E ∝r

E ∝1 r r

R Fig. 27.59

The direction of electric field is shown in above figure. V

Example 27.25 A long thin solenoid has 900 turns/metre and radius 2.50 cm. The current in the solenoid is increasing at a uniform rate of 60 A / s. What is the magnitude of the induced electric field at a point? (a) 0.5 cm from the axis of the solenoid. (b) 1.0 cm from the axis of the solenoid. Solution Q B = µ 0 ni dB di ∴ = µ 0n dt dt = ( 4π × 10− 7 ) ( 900) ( 60) = 0.068 T/s

Chapter 27

Electromagnetic Induction — 497

Using the result of electric field derived in above problem (as both points lie inside the solenoid). r dB E =   2  dt   0.5 × 10−2  −4 (a) E =   ( 0.068) = 1.7 × 10 V/ m 2   (b) E =

(1.0 × 10−2 ) ( 0.068) = 3.4 × 10− 4 V/ m 2

INTRODUCTORY EXERCISE

27.7

1. A long solenoid of cross-sectional area 5.0 cm 2 is wound with 25 turns of wire per centimetre. It is placed in the middle of a closely wrapped coil of 10 turns and radius 25 cm as shown. 25cm

Fig. 27.60

(a) What is the emf induced in the coil when the current through the solenoid is decreasing at a rate –0.20 A/s? (b) What is the electric field induced in the coil?

2. For the situation described in figure, the magnetic field changes with time according to B = (2.00 t 3 – 4.00 t 2 + 0.8) T and r2 = 2R = 5.0 cm

× × × ×

× × × × × ×

× × × × × ×

××× r1 P1 ×××× × R× ×× × ×××

r2

P2

Bin

Fig. 27.61

(a) Calculate the force on an electron located at P2 at t = 2.00 s (b) What are the magnitude and direction of the electric field at P1 when t = 3.00 s and r1 = 0.02 m. Hint : For the direction, see whether the field is increasing or decreasing at given times.

498 — Electricity and Magnetism

Final Touch Points 1. Eddy currents When a changing magnetic flux is applied to a piece of conducting material, circulating currents called eddy currents are induced in the material. These eddy currents often have large magnitudes and heat up the conductor. When a metal plate is allowed to swing through a strong magnetic field, then in entering or leaving the field the eddy currents are set up in the plate which opposes the motion as shown in figure. The kinetic energy dissipates in the form of heat. The slowing down of the plate is called the electromagnetic damping.

v

v

F F

The electromagnetic damping is used to damp the oscillations of a galvanometer coil or chemical balance and in braking electric trains. Otherwise, the eddy currents are often undesirable. To reduce the eddy currents some slots are cut into moving metallic parts of machinery. These slots intercept the conducting paths and decreases the magnitudes of the induced currents.

2. Back EMF of Motors An electric motor converts electrical energy into mechanical energy and is based on the fact that a current carrying coil in a uniform magnetic field experiences a torque. As the coil rotates in the magnetic field, the flux linked with the rotating coil will change and hence, an emf called back emf is produced in the coil. When the motor is first turned on, the coil is at rest and so there is no back emf. The ‘start up’ current can be quite large. To reduce ‘start up’ current a resistance called ‘starter’ is put in series with the motor for a short period when the motor is started. As the rotation rate increases the back emf increases and hence, the current reduces.

3. Electric Generator or Dynamo A dynamo converts mechanical energy (rotational kinetic energy) into electrical energy. It consists of a coil rotating in a magnetic field. Due to rotation of the coil magnetic flux linked with it changes, so an emf is induced in the coil.

ω

Suppose at time t = 0, plane of coil is perpendicular to the magnetic field.

Chapter 27

Electromagnetic Induction — 499

The flux linked with it at any time t will be given by φ = NBA cos ωt dφ e=– = NBAω sin ωt dt

∴

(N = number of turns in the coil)

e = e 0 sin ωt

or

e 0 = NBA ω

where,

4. Transformer It is a device which is either used to increase or decrease the voltage in AC circuits through mutual induction. A transformer consists of two coils wound on the same core.

Load

Laminated sheets

Input

Output

Iron core

The coil connected to input is called primary while the other connected to output is called secondary coil. An alternating current passing through the primary creates a continuously changing flux through the core. This changing flux induces an alternating emf in the secondary. As magnetic lines of force are closed curves, the flux per turn of primary must be equal to flux per turn of the secondary. Therefore, φP φ = S NP NS or

1 d φP 1 d φS ⋅ = ⋅ NP dt NS dt

∴

eS NS = eP NP

d φ  as e ∝   dt 

In an ideal transformer, there is no loss of power. Hence, ∴

ei = constant eS NS iP = = eP NP iS

Regarding a transformer, the following are few important points. In step-up transformer, NS > NP . It increases voltage and reduces current In step-down transformer, NP > NS . It increases current and reduces voltage It works only on AC A transformer cannot increase (or decrease) voltage and current simultaneously. As, ei = constant (v) Some power is always lost due to eddy currents, hysteresis, etc.

(i) (ii) (iii) (iv)

Solved Examples TYPED PROBLEMS Type 1. Based on Faraday’s and Lenz’s law

Concept Problems of induced emf or induced current can be solved by the following two methods. Method 1 Magnitudes are given by N dφB   and |i| = |e| |e| =  dt R   Direction is given by Lenz’s law. Method 2

Magnitudes are given by |e| = |Bvl| or |e| =

Bωl 2 2

and |i | =

|e| R

Direction is given by right hand rule. Note In the first method, we have to first find the magnetic flux passing through the loop and then differentiate it with respect to time. Second method is simple but it can be applied if and only if some conductor is either in translational or rotational motion. V

Example 1 Current in a long current carrying wire is I = 2t A conducting loop is placed to the right of this wire. Find (a) magnetic flux φB passing through the loop. (b) induced emf|e| produced in the loop. (c) if total resistance of the loop is R, then find induced current I in in the loop.

I = 2t

c a

b

Solution Here, no conductor is in motion. So, we can apply only method-1. Further, magnetic field of straight wire is non-uniform. Therefore, magnetic flux can be obtained by integration. x x x dx x x x

I

a

b

c

Chapter 27

Electromagnetic Induction — 501

(a) At a distance x from the straight wire, magnetic field is µ I B= 0 2π x Let us take a small strip of width dx. ∴ Area of this strip is

[ in ⊗ direction ]

dS = c (dx) Now, dS can also be assumed inwards. Or, angle between B and dS may be assumed to be 0°. Therefore, small magnetic flux passing through the loop is dφB = BdS cos 0° µ I = 0 cdx 2π x Total magnetic flux is φB = ∫ =∫ =

x=a+b x=a

dφB

a + b µ a

Ic dx  0   2π  x

µ 0Ic  a + b ln    a  2π

Substituting the values of I, we get φB = (b)

µ 0ct  a + b ln    a  π

dφ d µ 0ct |e| =  B  = ln  dt  dt  π µ c a + = 0 ln   a π

 a + b     a   b  

Ans.

Ans.

(c) Induced current, I in =

|e| µ 0c  a + b = ln    a  R πR

Note The main current I ( = 2 t ) is increasing with time. Hence, ⊗ magnetic field passing through the loop will also increase. So, induced current Iin will produce V

magnetic field. Or, induced current is anti-clockwise.

Example 2 A constant current I flows through a long straight wire as shown in figure. A square loop starts moving towards right with a constant speed v.

I

v

a

x

a

(a) Find induced emf produced in the loop as a function of x. (b) If total resistance of the loop is R, then find induced current in the loop.

502 — Electricity and Magnetism Note In this problem, loop is in motion therefore both methods can be applied. Solution Method 1 (a) Using the result of magnetic flux obtained in Example-1, we have µ Ic  a + b φB = 0 ln    a  2π Here, a = x, b = c = a Substituting the values, we get

|e| =

Now, Putting

φB =

µ 0Ia  x + a ln    x  2π

=

µ 0Ia a  ln 1 +   2π x

dφB µ Ia  x   a  dx = 0     dt 2π  x + a   x2 dt

dx = v, we have dt µ 0Ia 2 v 2πx (x + a )

Ans.

|e| µ 0Ia 2v = R 2πRx (x + a )

Ans.

|e| = (b) Induced current, I in =

Note Near the wire (towards right) value of ⊗ magnetic field is high. So, the loop is moving from higher magnetic

field to lower magnetic field. or, ⊗ magnetic field passing through the loop is decreasing. Hence, induced current will produce ⊗ magnetic field or it should be clockwise.. Method 2

v e2

e1

I x

x+a

µ0 I va 2π x µ I e2 = B2vl = 0 va 2π x + a e1 = B1vl =

e1 > e2 ∴

enet = e1 − e2 = =

µ 0Iva  1 1   −  2π  x x + a 

µ 0Iva 2 2πx (x + a )

This is the same result as was obtained in Method 1.

Chapter 27 V

Electromagnetic Induction — 503

Example 3 A conducting circular ring is rotated with angular velocity ω about point A as shown in figure. Radius of ring is a. Find

(a)

⊗B B

(a) potential difference between points A and C (b) potential difference between the points A and D. Solution Here, the loop is rotating. So, we can applying e =

C

Bωl2 2

D

A

ω

C ⊗ B l

A

ω

Using right hand rule, we can see that VC > V A ∴

VC − V A =

Bωl2 Bω (2a )2 = = 2Bωa 2 2 2

Ans.

(b) ⊗ B D l A

ω

Using right hand rule, we can see that VD > V A ∴

VD − V A =

Bωl2 Bω ( 2 a )2 = = Bωa 2 2 2

Ans.

Type 2. Based on potential difference across an inductor V

Example 4 Two different coils have self-inductances L1 = 8 mH and L2 = 2 mH. The current in one coil is increased at a constant rate. The current in the second coil is also increased at the same constant rate. At a certain instant of time, the power given to the two coils is the same. At that time, the current, the induced voltage and the energy stored in the first coil are i1 , V1 and W1 respectively. Corresponding values for the second coil at the same instant are i2 , V2 and W2 respectively. Then, (JEE 1994) i1 1 = i2 4 W 1 (c) 1 = W2 4

(a)

i1 =4 i2 V (d) 1 = 4 V2 (b)

504 — Electricity and Magnetism Solution Potential difference across an inductor : di   V ∝ L, if rate of change of current is constant V = − L   dt  V2 L2 2 1 = = = ∴ V1 L1 8 4 V1 =4 V2

or Power given to the two coils is same, i.e.

V1i1 = V 2i 2 i1 V 2 1 = = i 2 V1 4

or Energy stored, W =

1 2 Li 2

∴

W 2  L 2 =  W1  L1 

or

W1 1 = W2 4

2

 i 2  1 2   =   (4)  4 i  1

∴ The correct options are (a), (c) and (d). V

Example 5 In the figure shown, i1 = 10 e –2 t A, i2 = 4 A and VC = 3e –2 t V . Determine (JEE 1992) b + V – C i 2 R2 = 3 Ω

C = 2F a

R1 = 2 Ω i 1

c

O i L + L = 4H

VL –

d

(a) iL and V L

(b) V ac, V ab and V cd .

Solution (a) Charge stored in the capacitor at time t, q = CVC = (2) (3e

–2t

ic +

)

q –

= 6e C dq ic = = – 12e–2t A dt –2t

∴

(Direction of current is from b to O) Applying junction rule at O, iL = i1 + i 2 + i c = 10e–2t + 4 – 12e–2t = (4 – 2e–2t ) A = [2 + 2 (1 – e–2t )] A

Ans.

Chapter 27

Electromagnetic Induction — 505

iL versus time graph is as shown in figure. iL(A) 4 2 t

iL increases from 2 A to 4 A exponentially. di d VL = VOd = L L = (4) (4 – 2e–2t ) dt dt = 16e–2t V VL versus time graph is as shown in figure.

Ans.

VL(V) 16

t

VL decreases exponentially from 16 V to 0. (b) Vac = Va – Vc Va – i1R1 + i 2R2 = Vc ∴ Va – Vc = Vac = i1R1 – i 2R2 Substituting the values, we have Vac = (10e–2t ) (2) – (4) (3) Vac = (20e–2t – 12) V

Vac(V)

8

At t = 0, Vac = 8 V and at t = ∞, V ac = – 12V Therefore, Vac decreases exponentially from 8 V to –12 V.

t

Vab = Va – Vb

Va – i1R1 + VC = Vb ∴ V a – Vb = V ab = i1R1 – VC Substituting the values, we have or Thus, V ab

Vab = (10e–2t ) (2) – 3e–2t Vab = 17e–2t V decreases exponentially from 17 V to 0. Vab(V) 17

t

–12

Ans.

506 — Electricity and Magnetism Vab versus t graph is shown in figure. Vcd = Vc – Vd Vc – i 2R2 – VL = Vd ∴ V c – V d = V cd = i 2R2 + VL Substituting the values, we have V cd = (4) (3) + 16e–2t V cd = (12 + 16e–2t ) V At t = 0, V cd = 28 V and at t = ∞, V cd = 12 V i.e. V cd decreases exponentially from 28 V to 12 V. V cd versus t graph is shown in figure. or

Vcd (V )

Ans.

28

12 t

Type 3. Based on L - R circuit

Concept At time t = 0, when there is zero current in the circuit, an inductor offers infinite resistance and at t = ∞, when steady state is reached an ideal inductor (of zero resistance) offers zero resistance. R1

i1

i i2

S L

R2

E

Thus, in the circuit shown, if switch S is closed at time t = 0, then i2 = 0 E at t = 0 and i = i1 = R1 + R2 as initially the inductor offers infinite resistance and at t = ∞, E i1 = 0, while i = i 2 = R1 as in steady state the inductor offers zero resistance. V

Example 6 For the circuit shown in figure, E = 50 V , R1 = 10 Ω, R2 = 20 Ω, R3 = 30 Ω and L = 2.0 mH . Find the current through R1 and R2 . (a) (b) (c) (d)

Immediately after switch S is closed. A long time after S is closed. Immediately after S is reopened. A long time after S is reopened.

R3

R1 S E

R2

L

Chapter 27

Electromagnetic Induction — 507

Solution (a) Resistance offered by inductor immediately after switch is closed will be infinite. Therefore, current through R3 will be zero and current through R1 = current through R2 = =

E R1 + R2 50 5 = A 10 + 20 3

Ans.

(b) After long time of closing the switch, resistance offered by inductor will be zero. In that case R2 and R3 are in parallel, and the resultant of these two is then in series with R1. Hence, R2R3 Rnet = R1 + R2 + R3 = 10 +

(20) (30) = 22 Ω 20 + 30

Current through the battery (or through R1) E 50 = = A Rnet 22

Ans.

This current will distribute in R2 and R3 in inverse ratio of resistance. Hence,  50  R3  Current through R2 =      22  R2 + R3   50  30  15 =   A  =  22  30 + 20 11

Ans.

(c) Immediately after switch is reopened, the current through R1 will become zero. But current through R2 will be equal to the steady state current through R3 , which is equal to,  50 15 Ans. –  A = 0.91 A   22 11 (d) A long after S is reopened, current through all resistors will be zero. V

Example 7 An inductor of inductance L = 400 mH and resistors of resistances R1 = 2 Ω and R2 = 2 Ω are connected to a battery of emf E =12 V as shown in the figure. The internal resistance of the battery is negligible. The switch S is closed at time t = 0. E

L R1

S

R2

What is the potential drop across L as a function of time? After the steady state is reached, the switch is opened. What is the direction and the magnitude of current through R1 as a function of time? (JEE 2001) Solution (a) Given, R1 = R2 = 2 Ω, E = 12 V and L = 400 mH = 0.4 H. Two parts of the circuit are in parallel with the applied battery.

508 — Electricity and Magnetism So, the given circuit can be broken as :

L

E R1 S

E

R1

E

L

+

S

VVVV

VVVV

R2

S

(a)

R2

(b)

Now refer Fig. (b) This is a simple L-R circuit, whose time constant 0.4 τL = L /R2 = = 0.2 s 2 and steady state current E 12 i0 = = =6A R2 2 Therefore, if switch S is closed at time t = 0, then current in the circuit at any time t will be given by i (t ) = i 0 (1 − e− t/ τL ) i (t ) = 6(1 − e−t/ 0. 2) = 6(1 − e−5 t ) = i Therefore, potential drop across L at any time t is di V =  L  = L (30e−5 t ) = (0.4)(30) e−5 t or V = 12 e−5 t volt  dt  (b) The steady state current in L or R2 is

(say)

i0 = 6 A Now, as soon as the switch is opened, current in R1 is reduced to zero immediately. But in L and R2 it decreases exponentially. The situation is as follows : 6A

i

6A

t=0 S is open (d)

VVVV

VVVV

i=0

i0 R1

VVVV

R2

Steady state condition (c)

L

VVVV

R1

VVVV

E

VVVV

L

t=t (e)

R2

Chapter 27

Electromagnetic Induction — 509

Refer figure (e) Time constant of this circuit would be τL′ =

L 0.4 = = 0.1 s R1 + R2 (2 + 2)

∴ Current through R1 at any time t is i = i 0e− t/ τL′ = 6 e − t/ 0.1 or i = 6 e−10t A Direction of current in R1 is as shown in figure or clockwise. V

Example 8 A solenoid has an inductance of 10 H and a resistance of 2 Ω. It is connected to a 10 V battery. How long will it take for the magnetic energy to reach 1/4 of its maximum value? (JEE 1996) Solution

U =

1 2 Li , 2

i.e. U ∝ i 2

1 U will reach th of its maximum value when current is reached half of its maximum value. In 4 L-R circuit, equation of current growth is written as i = i 0 (1 − e− t/ τL ) i 0 = Maximum value of current τL = Time constant = L /R 10 H τL = =5 s 2 Ω

Here,

i = i 0/2 = i 0 (1 − e− t/5 ) 1 1 = 1 − e− t/5 or e− t/5 = 2 2  1 or − t / 5 = ln   t / 5 = ln (2) = 0.693  2

Therefore, or or ∴ V

t = (5)(0.693) or

t = 3.465 s

Example 9 A circuit containing a two position switch S is shown in figure. C

R3

2Ω R1

2Ω 1 2

S

2 µ F 1Ω

E1 12 V A E2 3V

L

R2

B

2Ω

3Ω

R5

R4

10 mH

(a) The switch S is in position 1. Find the potential difference V A − V B and the rate of production of joule heat in R1. (b) If now the switch S is put in position 2 at t = 0. Find (i) steady current in R4 and (ii) the time when current in R4 is half the steady value. Also calculate the energy stored in the inductor L at that time.

510 — Electricity and Magnetism Solution (a) In steady state, no current will flow through capacitor. 2Ω

2 µF

i2 1

1Ω

i2

12 V

2Ω

i1 i2

i1 2

i1 A

i1 2Ω

3V

B 3Ω

10 mH

Applying Kirchhoff’s second law in loop 1, −2i 2 + 2 (i1 − i 2) + 12 = 0 ∴ 2i1 − 4i 2 = − 12 or i1 − 2i 2 = − 6 Applying Kirchhoff’s second law in loop 2, −12 − 2 (i1 − i 2) + 3 − 2i1 = 0 ∴ 4i1 − 2i 2 = − 9 Solving Eqs. (i) and (ii), we get i 2 = 2.5 A and i1 = − 1 A Now, V A + 3 − 2i1 = VB or V A − VB = 2i1 − 3 = 2 (−1) − 3 = − 5 V

(b) In position 2

PR1 = (i1 − i 2)2R1 = (−1 − 2.5)2 (2) = 24.5 W Circuit is as below 3V

2Ω 3Ω 10 mH

Steady current in R4, i0 =

3 = 0.6 A 3+2

Time when current in R4 is half the steady value,

i = i 0 (1 − e− t/ τL ) i = i 0 / 2 at t = t1/ 2, where L (10 × 10−3 ) t1/ 2 = τL (ln 2) = ln (2) = ln (2) R 5 = 1.386 × 10−3 s 1 1 U = Li 2 = (10 × 10−3 ) (0.3)2 2 2 = 4.5 × 10−4 J

…(i)

…(ii)

Chapter 27

Electromagnetic Induction — 511

Type 4. Based on L-C oscillations V

Example 10 In an L-C circuit, L = 3.3 H and C = 840 pF . At t = 0, charge on the capacitor is 105 µC and maximum. Compute the following quantities at t = 2.0 ms: (a) The energy stored in the capacitor. (b) The total energy in the circuit, (c) The energy stored in the inductor. Solution Given, L = 3.3 H , C = 840 × 10–12 F and q0 = 105 × 10–6 C The angular frequency of L-C oscillations is 1 1 ω= = LC 3.3 × 840 × 10–12 = 1.9 × 104 rad / s Charge stored in the capacitor at time t would be q = q0 cos ωt t = 2 × 10–3 s, q = (105 × 10–6 ) cos [ 1.9 × 104 ] [ 2 × 10–3 ] = 100.3 × 10–6 C ∴ Energy stored in the capacitor, 1 q2 UC = 2 C (100.3 × 10–6 )2 = 2 × 840 × 10–12

(a) At

= 6.0 J

Ans.

(b) Total energy in the circuit, U =

1 q02 (105 × 10–6 )2 = 2 C 2 × 840 × 10–12

= 6.56 J (c) Energy stored in inductor in the given time = total energy in circuit – energy stored in capacitor = (6.56 – 6.0) J = 0.56 J V

Ans.

Ans.

Example 11 An inductor of inductance 2.0 mH is connected across a charged capacitor of capacitance 5.0 µF and the resulting L-C circuit is set oscillating at its natural frequency. Let Q denotes the instantaneous charge on the capacitor and I the current in the circuit. It is found that the maximum value of Q is200 µC. (JEE 1998)

(a) When Q = 100 µC, what is the value of|dI / dt|? (b) When Q = 200 µC , what is the value of I? (c) Find the maximum value of I. (d) When I is equal to one-half of its maximum value, what is the value of|Q|?

512 — Electricity and Magnetism Solution This is a problem of L-C oscillations. Charge stored in the capacitor oscillates simple harmonically as Q = Q0 sin (ωt ± φ) Here, Q0= maximum value of Q = 200 µC = 2 × 10−4 C 1 1 = 104s −1 ω = = −3 −6 LC (2 × 10 )(5.0 × 10 ) Let at t = 0, Q = Q0, then Q (t ) = Q0 cos ωt dQ I (t ) = = − Q0ω sin ωt and dt dI (t ) = − Q0ω 2 cos ωt dt

…(i) …(ii) …(iii)

(a) Q = 100 µC Q0 1 at cos ωt = 2 2 π ωt = 3

or or 1 At cos ωt = , from Eq. (iii) : 2

dI = (2.0 × 10−4C)(104s −1 )2  1  2 dt dI = 104 A /s dt (b) Q = 200 µC or Q0 when cos ωt = 1, i. e. ωt = 0, 2π … At this time

I (t ) = − Q0ω sin ωt

or

I (t ) = 0

(c) I (t ) = − Q0ω sin ωt ∴ Maximum value of I is Q0 ω I max = Q0 ω = (2.0 × 10−4 )(104 ) I max = 2.0 A (d) From energy conservation, 1 2 1 1 Q2 LI max = LI 2 + 2 2 2 C or

2 Q = LC (I max − I 2)

I= ∴

I max = 1.0 A 2

Q = (2 .0 × 10−3 )(5.0 × 10−6 )(22 − 12) Q = 3 × 10−4 C

or

Q = 1.732 × 10−4 C

(sin 0° = sin 2π = 0)

Chapter 27

Electromagnetic Induction — 513

Type 5. Based on induced electric field V

Example 12 A uniform but time-varying magnetic field B(t) exists in a circular region of radius a and is directed into the plane of the paper as shown. The magnitude of the induced electric field at point P at a distance r from the centre (JEE 2000) of the circular region B(t) P

r

a

(a) is zero Solution

(b) decreases as 1/r  dφ 

dB or

(c) increases as r

(d) decreases as 1/r 2

dB E (2πr ) = πa 2   dt 

∫ E ⋅ d l = dt = S  dt 

For r ≥ a, E=

∴

a 2dB 2r  dt 

∴ Induced electric field ∝ 1 / r For r ≤ a,

dB E (2πr ) = πr 2  or  dt 

At r = a, E =

a 2

r dB E =   or 2 dt 

E∝r

dB  dt 

Therefore, variation of E with r (distance from centre) will be as follows E a dB 2 dt

E

∝

r

E ∝ 1r r=a

r

∴ The correct option is (b). V

Example 13 The magnetic field B at all points within a circular region of radius R is uniform in space and directed into the plane of the page in figure. If the magnetic field is increasing at a rate dB / dt, what are the magnitude and direction of the force on a stationary positive point charge q located at points a, b and c? (Point a is a distance r above the centre of the region, point b is a distance r to the right of the centre, and point c is at the centre of the region.)

x x x x

x B x x x x

a x x x x x x x xc x x x R x x x x x x x x x

x x x x x x x x

x x x x x x x x

x xx x b xx xx x

514 — Electricity and Magnetism Solution Inside the circular region at distance r, dφ  dB =S    dt  dt dB E (2πr ) = (πr 2) ⋅ ∴ dt r dB E= ∴ 2 dt qr dB F = qE = 2 dt At points a and b, distance from centre is r. qr dB ∴ F = 2 dt At point C, distance r = 0 El =

∴ F =0 ⊗ magnetic field is increasing. Hence, induced current in an imaginary loop passing through a and b should produce u magnetic field. Hence, induced current through an imaginary circular loop passing through a and b should be anti-clockwise. Force on positive charge is in the direction of induced current. Hence, force at a is towards left and force at b is upwards.

Type 6. Based on motion of a wire in uniform magnetic field with other element like resistance, capacitor or an inductor

Concept

P

X

⊗B

F

Q

A constant force F is applied on wire PQ of length l and mass m. There is an electrical element X in the box as shown in figure. There are the following three different cases : Case 1

If X is a resistance, then velocity of the wire increases exponentially.

Case 2

If X is a capacitor, then wire moves with a constant acceleration a (< F / m).

Case 3 If X is an inductor and instead of constant force F an initial velocity v 0 is given to the wire then the wire starts simple harmonic motion with v 0 as the maximum velocity ( = ωA) at mean position. V

Example 14 In the above case if X is a resistance R, then find velocity of wire as a function of time t. Solution At time t suppose velocity of wire is v, then due to motional emf a current i flows in the closed circuit in anti-clockwise direction. e Bvl i= = R R

Electromagnetic Induction — 515

Chapter 27

Due to this current magnetic force will act on the wire in the direction shown in figure, i

R

⊗B

F, v

Fm

 B2l2 Fm = ilB =  v  R   B2 l 2 dv Fnet = F − Fm = F −   v=m dt  R  dv

v

∫0

∴

B l  F − v  R  2 2

=

1 t dt m ∫0

Solving this equation, we get B 2l 2

− t FR (1 − e mR ) 22 Bl Thus, velocity of the wire increases exponentially. v - t graph is as shown below.

v=

v FR ——2 B2l

t

V

Example 15 If X is a capacitor C, then find the constant acceleration a of the wire. Solution At time t suppose velocity of wire is v. Then, due to motional emf e = Bvl capacitor gets charged . q = CV = C (Bvl)

q

+ –

F, v

Fm i

This charge is increasing as v will be increasing. Hence, there will be a current in the circuit as shown in figure. dq dv i= = (BlC ) dt dt

516 — Electricity and Magnetism  dv  = a as  dt 

i = (BlC ) a

or

Due to this current a magnetic force Fm will act in the direction shown in figure, Fm = ilB = (B2l2C ) a Now, Fnet = F − Fm ma = F − (B2l2C ) a F a= m + B2l2C

or ∴

Ans.

Now, we can see that this acceleration is constant but less than F /m. V

Example 16 A pair of parallel horizontal conducting rails of negligible resistance shorted at one end is fixed on a table. The distance between the rails is L. A conducting massless rod of resistance R can slide on the rails frictionlessly. The rod is tied to a massless string which passes over a pulley fixed to the edge of the table. A mass m tied to the other end of the string hangs vertically. A constant magnetic field B exists perpendicular to the table. If the system is released from (JEE 1997) rest, calculate B L

R

m

(a) the terminal velocity achieved by the rod and (b) the acceleration of the mass at the instant when the velocity of the rod is half the terminal velocity. Solution (a) Let v be the velocity of the wire (as well as block) at any instant of time t. Motional emf, e = BvL e BvL Motional current, i = = r R and magnetic force on the wire Fm = iLB =

vB2L2 R

Net force on the system at this moment will be Fnet = mg − Fm = mg − or

vB2L2 R vB2L2 a=g− mR

vB2L2 R

ma = mg −

…(i)

Chapter 27

Electromagnetic Induction — 517

Velocity will acquire its terminal value, i.e. v = vT when Fnet or acceleration a of the particle becomes zero. v B2L2 Thus, 0= g− T mR mgR or vT = 2 2 BL v mgR (b) When v= T = 2 2B2L2 Then from Eq. (i), acceleration of the block, 2 2 g  mgR   B L  a = g −  2 2   = g−  2B L   mR  2 a=

or V

g 2

Example 17 A loop is formed by two parallel conductors connected by a solenoid with inductance L and a conducting rod of mass m which can freely (without friction) slide over the conductors. The conductors are located in a horizontal plane in a uniform vertical magnetic field B. The distance between the conductors is l.

v0

x

m, l

At the moment t = 0, the rod is imparted an initial velocity v0 directed to the right. Find the law of its motion x (t) if the electric resistance of the loop is negligible. Solution Let at any instant of time, velocity of the rod is v towards right. The current in the circuit is i. In the figure, a

d

i

Fm

b

v

c

V a – Vb = V d – V c or i.e.

L

di dx = Bvl = Bl dt dt

Ldi = Bldx

dx  as v =   dt 

518 — Electricity and Magnetism Integrating on both sides, we get Li = Blx Bl or i= x L Magnetic force on the rod at this instant is B2l2 x L Since, this force is in opposite direction of v, so from Newton’s second law we can write  d 2x B2l2 m  2 = – x L  dt  Fm = ilB =

…(i)

…(ii)

 d 2x B2l2 x  2 = – mL  dt 

or Comparing this with equation of SHM,

d 2x = – ω 2x dt 2 Bl ω= mL

We have,

Bl . At time mL t = 0, rod was at x = 0 and it was moving towards positive x-axis. Hence, x-t equation of the rod is …(iii) x = A sin ωt

Therefore, the rod will oscillate simple harmonically with angular frequency ω =

dx has a value v0. Hence, dt dx = v = Aω cos ωt dt

To find A, we use the fact that at t = 0, v or

Aω = v0 v A= 0 ω

or or

(at t = 0)

Substituting in Eq. (iii), we have x=

v0 Bl sin ωt, where ω = ω mL

Ans.

Alternate method of finding A At x = A, v = 0, i.e. whole of its kinetic energy is converted into magnetic energy. Thus, 1 2 1 Li = mv02 2 2 Substituting value of i from Eq. (i), with x = A, we have 2

 Bl  L A = mv02 L  or as

mL v v0 = 0 Bl ω Bl ω= mL

A=

Ans.

Miscellaneous Examples V

Example 18 A sensitive electronic device of resistance 175 Ω is to be connected to a source of emf by a switch. The device is designed to operate with a current of 36 mA, but to avoid damage to the device, the current can rise to no more than 4.9 mA in the first 58 µs after the switch is closed. To protect the device it is connected in series with an inductor. (a) What emf must the source have? (b) What inductance is required? (c) What is the time constant? Solution (a) Given, R = 175 Ω and peak value current

i 0 = 36 × 10–3 A Applied voltage, V = i 0R = (175) (36 × 10–3 ) volt = 6.3 V (b) From the relation, We have, or ∴ or ∴ or

Ans.

i = i 0 (1 – e– t/ τL ) (4.9) = (36) [1 – e– t/ τL ] 4.9 e– t/ τL = 1 – = 0.864 36 t = – ln (0.864) = 0.146 τL t = 0.146 L /R Rt = 0.146 L Rt (175) (58 × 10–6 ) L= = 0.146 0.146 = 7.0 × 10–2 H

Ans.

(c) Time constant of the circuit, τL =

L 7.0 × 10–2 = R 175

= 4.0 × 10–4 V

Example 19 A conducting rod shown in figure of mass m and length l moves on two frictionless horizontal parallel rails in the presence of a uniform magnetic field directed into the page. The rod is given an initial velocity v0 to the right and is released at t = 0. Find as a function of time, (a) the velocity of the rod (b) the induced current and (c) the magnitude of the induced emf.

Ans. a l

v0

R

b

520 — Electricity and Magnetism The initial velocity will produce an induced emf and hence, an induced current in the circuit. The current carrying wire will now experience a magnetic force ( Fm ) in opposite direction of its velocity. The force will retard the motion of the conductor. Thus, HOW TO PROCEED

Initial velocity → motional emf → induced current → magnetic force → retardation. Solution (a) Let v be the velocity of the rod at time t. Current in the circuit at this moment is Bvl …(i) R From right hand rule, we can see that this current is in counterclockwise direction. The magnetic force is, B2l2 Fm = – ilB = – v R Here, negative sign denotes that the force is to the left and retards the motion. This is the only horizontal force acting on the bar, and hence, Newton’s second law applied to motion in horizontal direction gives dv B2l2 m = Fm = – v B a i dt R  B2l2 dv ∴ =–  dt v  mR  v R i=

Fm

Integrating this equation using the initial condition that, v = v0 at t = 0, we find that v dv B2l2 t ∫v0 v = – mR ∫0dt Solving this equation, we find that

b

…(ii) Ans. v = v0e– t/ τ mR where, τ= 22 Bl This expression indicates that the velocity of the rod decreases exponentially with time under the action of the magnetic retarding force. Bvl (b) i = R Substituting the value of v from Eq. (ii), we get Blv0 – t/ τ i= e R (c)

Ans.

Ans. e = iR = Blv0e– t/ τ i and e both decrease exponentially with time. v-t, i -t and e-t graphs are as shown in figure i

v

e

Blv0 R

v0

t

Blv0

t

t

Chapter 27

Electromagnetic Induction — 521

Alternate solution This problem can also be solved by energy conservation principle. Let at some instant velocity of the rod is v. As no external force is present. Energy is dissipated in the resistor at the cost of kinetic energy of the rod. Hence,  dK  –  = power dissipated in the resistor  dt  or

–

d 1 e2 2  mv  =  R dt  2 22 2  dv B l v – mv  =  dt  R

or ∴

v

∫v

∴

0

V

dv B2l2 =– dt v mR dv B2l2 t =– dt v mR ∫0

v = v0e– t/ τ ,

or

(as e = Bvl )

where τ =

mR B2l2

Example 20 A wire loop enclosing a semicircle of radius R is located on the boundary of a uniform magnetic field B. At the moment t = 0, the loop is set into rotation with a constant angular acceleration α about an axis O coinciding with a line of vector B on the boundary. Find the emf induced in the loop as a function of time. Draw the approximate plot of this function. The arrow in the figure shows the emf direction taken to be positive. Solution ∴

B θ O

1 2 αt 2 2θ t= = time taken to rotate an angle θ α

θ=

where, θ = 0 to π, 2π to 3π, 4π to 5π etc. ⊗ magnetic field passing through the loop is increasing. Hence, current in the loop is anti-clockwise or induced emf is negative. And for, θ = π to 2π, 3π to 4π, 5π to 6π etc. ⊗ magnetic field passing through the loop is decreasing. Hence, current in the loop is clockwise or emf is positive. So, 2π t1 = time taken to rotate an angle π = α t2 = time taken to rotate an angle 2π = …

…

…

…

…

…

tn = time taken to rotate an angle nπ = Now, from

0 to t1 emf is negative t1 to t2 emf is positive t2 to t3 emf is again negative

4π α 2 nπ α

522 — Electricity and Magnetism and

so on. θ=

Now, at time t, angle rotated is Area inside the field is

1 2 αt 2

 θ 1 S = (πR2)   = R2θ  2π  2 1 2 2 S = R αt 4 1 φ = BS = BR2αt 2 4 dφ 1 e= = BR2αt dt 2

or So, flux passing through the loop,

x

x

x

θ

x

x

x

x

x

x

x

x

x

x

x

e∝t i.e. e-t graph is a straight line passing through origin. e-t equation with sign can be written as 1  Ans. e = (–1)n  BR2αt 2  Here, n = 1, 2, 3 … is the number of half revolutions that the loop performs at the given moment t. The e-t graph is as shown in figure. e

t1

V

t2

t3

t

Example 21 A uniform wire of resistance per unit length λ is bent into a semicircle of radius a. The wire rotates with angular velocity ω in a vertical plane about a horizontal axis passing through C. A uniform magnetic field B exists in space in a direction perpendicular to paper inwards. ω

B

C θ < π/2 θ A

O

D

(a) Calculate potential difference between points A and D. Which point is at higher potential? (b) If points A and D are connected by a conducting wire of zero resistance, find the potential difference between A and C.

Chapter 27

Electromagnetic Induction — 523

θ Solution (a) Length of straight wire AC is l1 = 2a sin    2

ω

B

C

a π–θ

θ A

D

O

Therefore, the motional emf (or potential difference) between points C and A is 1  θ VCA = VC – VA = Bωl12 = 2a 2Bω sin 2   2 2

…(i)

From right hand rule, we can see that VC > V A Similarly, length of straight wire CD is  π θ  θ l2 = 2a sin  –  = 2a cos    2 2  2 Therefore, the PD between points C and D is 1  θ VCD = VC – VD = Bωl22 = 2a 2Bω cos 2   2 2

…(ii)

with VC > VD Eq. (ii) – Eq.(i) gives, θ θ  V A – VD = 2a 2Bω  cos 2 – sin 2   2 2 = 2a 2Bω cos θ

Ans.

A is at higher potential. (b) When A and D are connected from a wire current starts flowing in the circuit as shown in figure : Resistance between A and C is r1 = (length of arc AC) λ = aθλ r2 = (length of arc CD) λ = (π – θ ) aλ

and between C and D is

C E2

E1

r2

r1

i

A

In the figure,

 θ E1 = 2a 2Bω sin 2   2

D

 θ and E 2 = 2a 2Bω cos 2   2

with E 2 > E1 ∴ Current in the circuit is i=

E 2 – E1 2a 2Bω cos θ 2aBω cos θ = = r1 + r2 πaλ πλ

524 — Electricity and Magnetism and potential difference between points C and A is θ 2aBω cos θ  ′ = E1 + ir1 = 2a 2Bω sin 2  +  V CA  (aθλ )  2   πλ θ θ   = 2a 2Bω sin 2 + cos θ   2 π

Ans.

′ = E1 + ir1 when a current i flows in the circuit. Note VCA = E1 when no current flows through the circuit and VCA V

Example 22 A battery of emf E and of negligible internal resistance is connected in an L-R circuit as shown in figure. The inductor has a piece of soft iron inside it. When steady state is reached the piece of soft iron is abruptly pulled out suddenly so that the inductance of the inductor decreases to nL with n < 1 with battery remaining connected. Calculate L

R

E

(a) (b) (c) (d)

current as a function of time assuming t = 0 at the instant when piece is pulled. the work done to pull out the piece. thermal power generated in the circuit as a function of time. power supplied by the battery as a function of time.

When the inductance of an inductor is abruptly changed, the flux passing through it remains constant. HOW TO PROCEED

φ = constant ∴

φ  L =   i

Li = constant

Solution (a) At time t = 0, steady state current in the circuit is i 0 = E /R. Suddenly, L reduces to nL (n < 1), so current in the circuit at time t = 0 will increase to at time t. nL

R

i

E

Applying Kirchhoff's loop rule, we have  di  E – nL   – iR = 0  dt  di 1 ∴ = dt E – iR nL ∴

i

di

1

t

∫i / n E – iR = nL ∫0dt 0

i0 E . Let i be the current = n nR

Electromagnetic Induction — 525

Chapter 27

i   i = i 0 –  i 0 – 0  e– t/ τL  n E i i0 = i0 R n nL τL = R

Solving this equation, we get Here, and From the i-t equation, we get i =

i0 at t = 0 and i = i 0 at n

Ans.

i0

t=∞ The i-t graph is as shown in figure.

t

i n

Note At t = 0, current in the circuit is 0 . Current in the circuit in steady state will be again i0 . So, it will decrease exponentially from

i0 to i0 . From the i-t graph, the equation can be formed without doing any calculation. n i

i

i

i0 n i0 i n– 0

⇒ i0

+ i0 t

t

t

i  i = i 0 +  0 – i 0 e– t/ τL n 

∴ (b) Work done to pull out the piece,

W = Uf – Ui =

1 1 L f i f2 – Liii2 2 2 2

=

1 1  E  E (nL )   – (L )    nR  R 2 2

=

1  E  1  L    – 1  2  R  n

=

1 L 2

2

2

2

 E   1 – n      R  n 

Ans.

(c) Thermal power generated in the circuit as a function of time is P1 = i 2R Here, i is the current calculated in part (a). (d) Power supplied by the battery as a function of time is P2 = Ei

Ans.

Ans.

Exercises LEVEL 1 Assertion and Reason Directions : Choose the correct option. (a) If both Assertion and Reason are true and the Reason is correct explanation of the Assertion. (b) If both Assertion and Reason are true; but Reason is not the correct explanation of Assertion. (c) If Assertion is true, but the Reason is false. (d) If Assertion is false, but the Reason is true. 1. Assertion : A square loop is placed in x-y plane as shown in figure. Magnetic field in the $ . The induced current in the loop is anti-clockwise. region is B = − B0x k y

x

Reason : If inward magnetic field from such a loop increases, then current should be anti-clockwise.

2. Assertion : Magnetic field B (shown inwards) varies with time t as shown. At time t0 induced current in the loop is clockwise. B B t0 t

Reason : If rate of change of magnetic flux from a coil is constant, charge should flow in the coil at a constant rate.

3. Assertion : Electric field produced by a variable magnetic field can’t exert a force on a charged particle. Reason : This electric field is non-conservative in nature.

4. Assertion : Current flowing in the circuit is i = 2t − 8 2H a

At t = 1 s, V a − V b = + 4 V Reason : V a − V b is + 4 V all the time.

i

b

Chapter 27

Electromagnetic Induction — 527

5. Assertion : Angular frequency of L-C oscillations is 2 rad/s and maximum current in the circuit is 1 A. Then, maximum rate of change of current should be 2 A/ s.  dI  Reason :   = ( I max )ω.  dt  max

6. Assertion : A conducting equilateral loop abc is moved translationally with constant speed v in uniform inward magnetic field B as shown. Then : V a − V b = V b − V c. x

a

x b x

x

x

x

x

c x

x B

v

Reason :

Point a is at higher potential than point b.

7. Assertion : Motional induced emf e = Bvl can be derived from the relation e = − Reason :

dφ . dt

Lenz’s law is a consequence of law of conservation of energy.

8. Assertion : If some ferromagnetic substance is filled inside a solenoid, its coefficient of self induction L will increase. Reason : By increasing the current in a coil, its coefficient of self induction L can be increased.

9. Assertion : In the circuit shown in figure, current in wire ab will become zero as soon as switch is opened. a S

b

Reason :

A resistance does not oppose increase or decrease of current through it.

10. Assertion : In parallel, current distributes in inverse ratio of inductance 1 L In electrical circuits, an inductor can be treated as a resistor. i∝

Reason :

Objective Questions 1. The dimensions of self inductance are (a) [ MLT–2A –2] (c) [ML2 T−2A −2]

(b) [ML2 T−1A −2] (d) [ML2 T−2A −1 ]

2. When the number of turns in the two circular coils closely wound are doubled (in both), their mutual inductance becomes (a) four times (c) remains same

(b) two times (d) sixteen times

528 — Electricity and Magnetism 3. Two coils carrying current in opposite direction are placed co-axially with centres at some finite separation. If they are brought close to each other then, current flowing in them should (a) decrease (c) remain same

(b) increase (d) become zero

4. A current carrying ring is placed in a horizontal plane. A charged particle is dropped along the axis of the ring to fall under the influence of gravity (a) (b) (c) (d)

the current in the ring may increase the current in the ring may decrease the velocity of the particle will increase till it reaches the centre of the ring the acceleration of the particle will decrease continuously till it reaches the centre of the ring

5. Identify the incorrect statement. Induced electric field (a) (b) (c) (d)

is produced by varying magnetic field is non-conservative in nature cannot exist in a region not occupied by magnetic field None of the above

6. In the figure shown, V ab at t = 1 s is a

2Ω

4V

2H

2F

b

– + q = (4t 2)C

(a) 30 V (c) 20 V

(b) – 30 V (d) – 20 V

7. Two coils have a mutual inductance of 0.005 H. The current changes in the first coil according to equation I = I 0 sinωt, where I 0 = 10 A and ω = 100 π rad/ s. The maximum value of emf (in volt) in the second coil is

(a) 2π (c) π

(b) 5π (d) 4π

8. An inductance of 2 H carries a current of 2 A. To prevent sparking when the circuit is broken a

capacitor of 4 µF is connected across the inductance. The voltage rating of the capacitor is of the order of

(a) 103 V (c) 105 V

(b) 10 V (d) 106 V

9. A conducting rod is rotated about one end in a plane perpendicular to a uniform magnetic field with constant angular velocity. The correct graph between the induced emf (e) across the rod and time (t) is e

e t

(a)

e

(c)

t

(b)

e

t

(d)

t

Electromagnetic Induction — 529

Chapter 27

10. A magnet is taken towards a conducting ring in such a way that a constant current of 10 mA is induced in it. The total resistance of the ring is 0.5 Ω. In 5 s, the magnetic flux through the ring changes by (b) 25 mWb (d) 15 mWb

(a) 0.25 mWb (c) 50 mWb

11. A uniform but increasing with time magnetic field exists in a cylindrical

P

region. The direction of force on an electron at P is (a) (b) (c) (d)

towards right towards left into the plane of paper out of the plane of paper

12. A magnetic flux through a stationary loop with a resistance R varies during the time interval τ as φ = at ( τ − t ). Find the amount of heat generated in the loop during that time

aτ 2 2R 2a 2τ3 (c) 3R

a 2τ3 3R aτ (d) 3R

(a)

(b)

13. The current i in an induction coil varies with time t according to the graph shown in the figure. Which of the following graphs shows the induced emf ( ε ) in the coil with time? i

t

O ε

ε

ε

(a)

(b) O

t

ε

(c) O

(d) O

t

t

O

t

14. The network shown in the figure is a part of complete circuit. What is the potential difference V B − V A when the current I is 5 A and is decreasing at a rate of 103 A/s? 1Ω A

I

5 mH B

15 V

(a) 5 V (c) 15 V

(b) 10 V (d) 20 V

15. In the given branch AB of a circuit a current, I = (10t + 5) A is flowing, where t is time in second. At t = 0, the potential difference between points A and B (V A − V B ) is R = 3Ω

L = 1H B

(a) 15 V (c) – 15 V

10 V

I

(b) – 5 V (d) 5 V

A

530 — Electricity and Magnetism  dI  is   dt  max

16. In an LC circuit, the capacitor has maximum charge q0. The value of  C

L

q0 LC q (c) 0 − 1 LC

q0 LC q (d) 0 + 1 LC (b)

(a)

17. An alternating current I in an inductance coil varies with time t according to the graph as shown :

I

Which one of the following graphs gives the variation of voltage with time? t

(a)

V

V

V

(b)

(c)

(d)

V

t t

t

t

18. A loop of area 1 m 2 is placed in a magnetic field B = 2T, such that plane of the loop is parallel to the magnetic field. If the loop is rotated by 180°, the amount of net charge passing through any point of loop, if its resistance is 10 Ω, is (a) 0.4 C (c) 0.8 C

(b) 0.2 C (d) 0 C

19. A rectangular loop of sides a and b is placed in xy-plane. A uniform but time varying magnetic $ is present in the region. The magnitude of induced emf field of strength B = 20 t i$ + 10 t 2$j + 50k in the loop at time t is (a) 20 + 20 t (c) 20 t

(b) 20 (d) zero

20. The armature of a DC motor has 20 Ω resistance. It draws a current of 1.5 A when run by 200 V DC supply. The value of back emf induced in it will be (a) 150 V (c) 180 V

(b) 170 V (d) 190 V

21. In a transformer, the output current and voltage are respectively 4 A and 20 V. If the ratio of number of turns in the primary to secondary is 2 : 1, what is the input current and voltage? (a) 2 A and 40 V (c) 4 A and 10 V

(b) 8 A and 10 V (d) 8 A and 40 V

22. When a loop moves towards a stationary magnet with speed v, the induced emf in the loop is E. If the magnet also moves away from the loop with the same speed, then the emf induced in the loop is (a) E E (c) 2

(b) 2E (d) zero

Chapter 27

Electromagnetic Induction — 531

23. A short magnet is allowed to fall from rest along the axis of a horizontal conducting ring. The distance fallen by the magnet in one second may be (a) 5 m (c) 4 m

(b) 6 m (d) None of these

24. In figure, if the current i decreases at a rate α, then V A − V B is L A

B

i

(b) − αL (d) No relation exists

(a) zero (c) αL

25. A coil has an inductance of 50 mH and a resistance of 0.3 Ω. If a 12 V emf is applied across the coil, the energy stored in the magnetic field after the current has built up to its steady state value is (a) 40 J (c) 20 J

(b) 40 mJ (d) 20 mJ

26. A constant voltage is applied to a series R-L circuit by closing the switch. The voltage across inductor ( L = 2 H ) is 20 V at t = 0 and drops to 5 V at 20 ms. The value of R in Ω is

(a) 100 ln 2Ω (c) 100 ln 4Ω

(b) 100 (1 − ln 2) Ω (d) 100 (1 − ln 4)

27. A coil of area 10 cm 2 and 10 turns is in magnetic field directed perpendicular to the plane and changing at a rate of 108 gauss/s. The resistance of coil is 20 Ω. The current in the coil will be

(a) 0.5 A

(b) 5 × 10−3 A

(c) 0.05 A

(d) 5 A

28. In figure, final value of current in 10 Ω resistor, when plug of key K is inserted is 1H

10 Ω

3V

(a)

3 A 10

(b)

3 A 20

30 Ω K

(c)

3 A 11

(d) zero

29. A circuit consists of a circular loop of radius R kept in the plane of paper and an infinitely long current carrying wire kept perpendicular to the plane of paper and passing through the centre of loop. The mutual inductance of wire and loop will be (a)

µ 0 πR 2

R

(b) 0

(c) µ 0πR2

(d)

µ 0 R2 2

30. A flat circular coil of n turns, area A and resistance R is placed in a uniform magnetic field B. The plane of coil is initially perpendicular to B. When the coil is rotated through an angle of 180° about one of its diameter, a charge Q1 flows through the coil. When the same coil after being brought to its initial position, is rotated through an angle of 360° about the same axis a charge Q2 flows through it. Then, Q2 / Q1 is (a) 1

(b) 2

(c) 1/2

(d) 0

532 — Electricity and Magnetism 31. A small circular loop is suspended from an insulating thread. Another coaxial circular loop carrying a current I and having radius much larger than the first loop starts moving towards the smaller loop. The smaller l loop will (a) (b) (c) (d)

be attracted towards the bigger loop be repelled by the bigger loop experience no force All of the above

32. In the circuit shown in figure, L = 10H,R = 5 Ω , E = 15 V. The switch S is

L

R

closed at t = 0. At t = 2 s, the current in the circuit is

1 (a) 3 1 −  A  e 1  (c) 3   A  e

1 (b) 3 1 − 2 A  e 1 (d) 3  2 A e 

S

E

33. In the figure shown, a T-shaped conductor moves with constant angular

velocity ω in a plane perpendicular to uniform magnetic field B. The potential difference V A − V B is

(a) zero (c) 2Bωl

2

4l

1 (b) Bωl2 2 (d) Bωl2

o

a

uniform magnetic field B. The field vectors are inclined at an angle θ with the horizontal as shown in figure. If the instantaneous velocity of the rod is v, the induced emf in the rod ab is

v

l v

Blv Blv cos θ Blv sin θ zero

35. A semi-circular conducting ring acb of radius R moves with constant speed

ω

B

34. A conducting rod of length l falls vertically under gravity in a region of

(a) (b) (c) (d)

A l B

B

c

v in a plane perpendicular to uniform magnetic field B as shown in figure. Identify the correct statement. (a) V a − V c = BRv

(b) Vb − V c = BRv

(c) V a − Vb = 0

(d) None of these

v a

b

A

36. The ring B is coaxial with a solenoid A as shown in figure. As the switch S is

B

closed at t = 0, the ring B (a) (b) (c) (d)

is attracted towards A is repelled by A is initially repelled and then attracted is initially attracted and then repelled

S

37. If the instantaneous magnetic flux and induced emf produced in a coil is φ and E respectively, then according to Faraday’s law of electromagnetic induction (a) E must be zero if φ = 0 (c) E ≠ 0 but φ may or may not be zero

(b) E ≠ 0 if φ = 0 (d) E = 0 then φ must be zero

Electromagnetic Induction — 533

Chapter 27

38. The figure shows a conducting ring of radius R. A uniform steady magnetic field B lies

perpendicular to the plane of the ring in a circular region of radius r (< R ). If the resistance per unit length of the ring is λ, then the current induced in the ring when its radius gets doubled is

B

(a)

BR λ

2BR λ Br 2 (d) 4 Rλ (b)

(c) zero

39. A metallic rod of length l is hinged at the point M and is rotating about an axis perpendicular to the plane of paper with a constant angular velocity ω. A uniform magnetic field of intensity B is acting in the region (as shown in the figure) parallel to the plane of paper. The potential difference between the points M and N B

M

ω

(a) is always zero (c) is always

1 Bωl2 2

N

(b) varies between

1 Bωl2 to 0 2

(d) is always Bωl2

Subjective Questions Note You can take approximations in the answers.

1. An inductor is connected to a battery through a switch. The emf induced in the inductor is much larger when the switch is opened as compared to the emf induced when the switch is closed. Is this statement true or false?

2. A coil formed by wrapping 50 turns of wire in the shape of a square is positioned in a magnetic field so that the normal to the plane of the coil makes an angle of 30°, with the direction of the field. When the magnetic field is increased uniformly from 200 µT to 600 µT in 0.4 s, an emf of magnitude 80.0 mV is induced in the coil. What is the total length of the wire?

3. A loop of wire enclosing an area S is placed in a region where the magnetic field is perpendicular to the plane. The magnetic field B varies with time according to the expression B = B0 e– at where a is some constant. That is, at t = 0. The field is B0 and for t > 0, the field decreases exponentially. Find the induced emf in the loop as a function of time.

534 — Electricity and Magnetism 4. The long straight wire in figure (a) carries a constant current i. A metal bar of length l is moving at constant velocity v as shown in figure. Point a is a distance d from the wire. d

d

a v

l

a

d

l

v

b (a)

c

b (b)

(a) Calculate the emf induced in the bar. (b) Which point a or b is at higher potential? (c) If the bar is replaced by a rectangular wire loop of resistance R, what is the magnitude of current induced in the loop?

5. The switch in figure is closed at time t = 0. Find the current in the inductor and the current through the switch as functions of time thereafter. 8Ω

4Ω

4Ω

10 V

1H

S

6. A small coil is introduced between the poles of an electromagnet so that its axis coincides with the magnetic field direction. The cross-sectional area of the coil is equal to S = 3.0 mm 2, the number of turns is N = 60. When the coil turns through 180° about its diameter, a galvanometer connected to the coil indicates a charge q = 4.5 µC flowing through it. Find the magnetic induction magnitude between the poles, provided the total resistance of the electric circuit equals R = 40 Ω. 1.0

and of 8.5 Ω resistance, changes with time as shown in figure. Calculate the emf in the loop as a function of time. Consider the time intervals

(a) t = 0 to t = 2.0 s (b) t = 2.0 s to t = 4.0 s (c) t = 4.0 s to t = 6.0 s. The magnetic field is perpendicular to the plane of the loop.

B (T )

7. The magnetic field through a single loop of wire, 12 cm in radius

0.5 0

2.0 4.0 6.0 8.0 t (s)

8. A square loop of wire with resistance R is moved at constant speed v across a uniform magnetic field confined to a square region whose sides are twice the lengths of those of the square loop. 2L L

x

x

x

x

v

–2L

x

x

x

x

B x

x

x

x

x

x

x

x

–L

O

L

2L

Electromagnetic Induction — 535

Chapter 27

(a) Sketch a graph of the external force F needed to move the loop at constant speed, as a function of the coordinate x, from x = − 2L to x = + 2L. (The coordinate x is measured from the centre of the magnetic field region to the centre of the loop. It is negative when the centre of the loop is to the left of the centre of the magnetic field region. Take positive force to be to the right). (b) Sketch a graph of the induced current in the loop as a function of x. Take counterclockwise currents to be positive.

9. A square frame with side a and a long straight wire carrying a current i are located in the same plane as shown in figure. The frame translates to the right with a constant velocity v. Find the emf induced in the frame as a function of distance x.

a

v

i

x

a

10. In figure, a wire perpendicular to a long straight wire is moving parallel to the later with a speed v = 10 m/ s in the direction of the current flowing in the later. The current is 10 A. What is the magnitude of the potential difference between the ends of the moving wire? i = 10 A 1.0 cm

10.0 cm

v = 10 m/s

11. The potential difference across a 150 mH inductor as a function of time is shown in figure. Assume that the initial value of the current in the inductor is zero. What is the current when t = 2.0 ms? and t = 4.0 ms ? V (volt) 5.0 4.0 3.0 2.0 1.0 0

1.0

2.0

3.0

4.0

t (ms)

12. At the instant when the current in an inductor is increasing at a rate of 0.0640 A/ s, the magnitude of the self-induced emf is 0.0160 V. (a) What is the inductance of the inductor? (b) If the inductor is a solenoid with 400 turns, what is the average magnetic flux through each turn when the current is 0.720 A?

536 — Electricity and Magnetism 13. Two toroidal solenoids are wound around the same pipe so that the magnetic field of one passes through the turns of the other. Solenoid 1 has 700 turns and solenoid 2 has 400 turns. When the current in solenoid 1 is 6.52 A, the average flux through each turn of solenoid 2 is 0.0320 Wb. (a) What is the mutual inductance of the pair of solenoids? (b) When the current in solenoid 2 is 2.54 A, what is the average flux through each turn of solenoid 1?

14. A coil of inductance 1 H and resistance 10 Ω is connected to a resistanceless battery of emf 50 V

at time t = 0. Calculate the ratio of the rate at which magnetic energy is stored in the coil to the rate at which energy is supplied by the battery at t = 0.1 s.

15. A 3.56 H inductor is placed in series with a 12.8 Ω resistor. An emf of 3.24 V is then suddenly applied across the RL combination. (a) At 0.278 s after the emf is applied what is the rate at which energy is being delivered by the battery? (b) At 0.278 s, at what rate is energy appearing as thermal energy in the resistor? (c) At 0.278 s, at what rate is energy being stored in the magnetic field?

16. A 35.0 V battery with negligible internal resistance, a 50.0 Ω resistor, and a 1.25 mH inductor with negligible resistance are all connected in series with an open switch. The switch is suddenly closed (a) How long after closing the switch will the current through the inductor reach one-half of its maximum value? (b) How long after closing the switch will the energy stored in the inductor reach one-half of its maximum value?

17. A solenoid of inductance L with resistance r is connected in parallel to a

L, r

resistance R. A battery of emf E and of negligible internal resistance is connected across the parallel combination as shown in the figure. At time t = 0, switch S is opened, calculate (a) current through the solenoid after the switch is opened. (b) amount of heat generated in the solenoid

R

E

18. In the given circuit, find the current through the 5 mH inductor in steady state. 5 mH

10 mH 20 V 5Ω

19. In an oscillating L-C circuit in which C = 4.00 µF , the maximum potential difference across the capacitor during the oscillations is 1.50 V and the maximum current through the inductor is 50.0 mA. (a) What is the inductance L? (b) What is the frequency of the oscillations? (c) How much time does the charge on the capacitor take to rise from zero to its maximum value?

Electromagnetic Induction — 537

Chapter 27

20. In the L-C circuit shown, C = 1 µF. With capacitor charged to 100 V,

switch S is suddenly closed at time t = 0. The circuit then oscillates at 103 Hz.

(a) (b) (c) (d)

Calculate ω and T Express q as a function of time Calculate L Calculate the average current during the first quarter-cycle.

S q

C i

21. An L -C circuit consists of an inductor with L = 0.0900 H and a capacitor of C = 4 × 10−4 F. The initial charge on the capacitor is 5.00 µC, and the initial current in the inductor is zero. (a) (b) (c) (d)

What is the maximum voltage across the capacitor? What is the maximum current in the inductor? What is the maximum energy stored in the inductor? When the current in the inductor has half its maximum value, what is the charge on the capacitor and what is the energy stored in the inductor?

LEVEL 2 Single Correct Option 1. Two ends of an inductor of inductance L are connected to two parallel conducting wires. A rod of length l and mass m is given velocity v0 as shown. The whole system is placed in perpendicular magnetic field B. Find the maximum current in the inductor. (Neglect gravity and friction) mv0 L mv02 (c) L (a)

(b)

B l

v0

m v0 L

(d) None of these

2. A conducting rod is moving with a constant velocity v over the parallel conducting rails which are connected at the ends through a resistor R and capacitor C as shown in the figure. Magnetic field B is into the plane. Consider the following statements. A

E

R

H

L

v

C G

F B

(i) Current in loop AEFBA is anti-clockwise (iii) Current through the capacitor is zero Which of the following options is correct? (a) Statements (i) and (iii) are correct (c) Statements (i), (iii) and (iv) are correct

(ii) Current in loop AEFBA is clockwise 1 (iv) Energy stored in the capacitor is CB2L2v2 2 (b) Statements (ii) and (iv) are correct (d) None of these

538 — Electricity and Magnetism 3. A rod is rotating with a constant angular velocity ω about point O (its centre) in a magnetic field B as shown. Which of the following figure correctly shows the distribution of charge inside the rod? P ω

O

B Q

+P

−P

(a) − O

(b) + O

+Q

−Q

−P

+P

(c)

(d)

O

−Q

O

+Q

y

4. A straight conducting rod PQ is executing SHM in xy-plane from

x = − d to x = + d. Its mean position is x = 0 and its length is along y-axis. There exists a uniform magnetic field B from x = − d to x = 0 pointing inward normal to the paper and from x = 0 to = + d there exists another uniform magnetic field of same magnitude B but –d pointing outward normal to the plane of the paper. At the instant t = 0, the rod is at x = 0 and moving to the right. The induced emf ( ε ) across the rod PQ vs time ( t ) graph will be ε

ε

(b)

+d x Q

ε

ε t

t

(a)

P

t

t

(c)

(d)

5. Two parallel long straight conductors lie on a smooth plane surface. Two other parallel conductors rest on them at right angles so as to form a square of side a. A uniform magnetic field B exists at right angles to the plane containing the conductors. Now, conductors start moving outward with a constant velocity v0 at t = 0. Then, induced current in the loop at any time t is (λ is resistance per unit length of the conductors)

B v0

aBv0 λ (a + v0t ) Bv0 (c) λ (a)

aBv0 2λ Bv0 (d) 2λ (b)

Chapter 27

Electromagnetic Induction — 539

6. A conducting square loop is placed in a magnetic field B with its plane perpendicular to the field. Now the sides of the loop start shrinking at a constant rate α. The induced emf in the loop at an instant when its side is a, is

(a) 2aαB

(b) a 2αB

(c) 2a 2αB

(d) aαB

7. A conducting straight wire PQ of length l is fixed along a diameter of a non-conducting ring as shown in the figure. The ring is given a pure rolling motion on a horizontal surface such that its centre of mass has a velocity v. There exists a uniform horizontal magnetic field B in horizontal direction perpendicular to the plane of ring. The magnitude of induced emf in the wire PQ at the position shown in the figure will be P B v

Q

(a) Bvl

(b) 2Bvl

(c) 3Bvl/2

(d) zero

8. A conducting rod of length L = 0.1 m is moving with a uniform speed v = 0.2 m/ s on conducting

rails in a magnetic field B = 0.5 T as shown. On one side, the end of the rails is connected to a capacitor of capacitance C = 20 µF. Then, the charges on the capacitor’s plates are

A L B

(a) qA = 0 = qB (c) qA = + 0.2 µC and qB = − 0.2 µC

(b) qA = + 20 µC and qB = − 20 µC (d) qA = − 0.2 C and qB = − 0.2 µC

9. A wire is bent in the form of a V shape and placed in a horizontal plane. There exists a uniform magnetic field B perpendicular to the plane of the wire. A uniform conducting rod starts sliding over the V shaped wire with a constant speed v as shown in the figure. If the wire has no resistance, the current in rod will (a) increase with time (c) remain constant

v B

(b) decrease with time (d) always be zero

10. A square loop of side b is rotated in a constant magnetic field B at angular frequency ω as shown in the figure. What is the emf induced in it? ω

B

(a) b2Bω sin ωt (c) bB2ω cos ωt

(b) bBω sin 2 ωt (d) b2Bω

540 — Electricity and Magnetism 11. A uniform but time varying magnetic field exists in a cylindrical region as shown in the figure. The direction of magnetic field is into the plane of the paper and its magnitude is decreasing at a constant rate of 2 × 10−3 T/s. A particle of charge 1 µC is moved slowly along a circle of radius 1m by an external force as shown in figure. The plane of the circle lies in the plane of the paper and it is concentric with the cylindrical region. The work done by the external force in moving this charge along the circle will be

1m

(b) 2π × 10−9 J (d) 4π × 10−6 J

(a) zero (c) π × 10−9 J

12. Switch S is closed at t = 0, in the circuit shown. The change in flux in the inductor (L = 500 mH) from t = 0 to an instant when it reaches steady state is

5Ω 5Ω 20 V

50 µF 500 mH 10 V 5Ω

S t=0

(a) 2 Wb (c) 0 Wb

(b) 1.5 Wb (d) None of these

13. An L-R circuit is connected to a battery at time t = 0. The energy stored in the inductor reaches half its maximum value at time  2 − 1 L ln   R  2   2 − 1 R (d) ln   L  2 

 2  R ln   L  2 − 1  L 2  (c) ln   R  2 − 1

(a)

(b)

14. Electric charge q is distributed uniformly over a rod of length l. The rod is placed parallel to a long wire carrying a current i. The separation between the rod and the wire is a. The force needed to move the rod along its length with a uniform velocity v is

µ 0iqv 2 πa µ 0iqvl (c) 2 πa

µ 0iqv 4 πa µ 0iqvl (d) 4 πa

(a)

(b)

15. AB is an infinitely long wire placed in the plane of rectangular coil of dimensions as shown in the figure. Calculate the mutual inductance of wire AB and coil PQRS P

B

Q

c

A

(a)

µ 0b a ln 2π b

(b)

µ 0c b ln 2π a

a

S R b

(c)

µ 0abc 2π (b − a )2

(d) None of these

Electromagnetic Induction — 541

Chapter 27

16. PQ is an infinite current carrying conductor. AB and CD are smooth conducting rods on which a conductor EF moves with constant velocity v as shown. The force needed to maintain constant speed of EF is R

A

P

C

E

F

l v a Q

1 µ oIv (b)  ln vR  2π (a )  2 v µ oIv (b)  (c) ln R  2π (a ) 

b

D

B

2

(b)

(a)

v µ oIv (a )  ln R  2π (b) 

2

(d) None of these

17. The figure shows a circular region of radius R occupied by a time varying dB magnetic field B( t ) such that < 0. The magnitude of induced electric field at dt the point P at a distance r < R is

R P

(b) increasing with r (d) varying as r −2

(a) decreasing with r (c) not varying with r

18. Two circular loops P and Q are concentric and coplanar as shown in

l1

P

figure. The loop Q is smaller than P. If the current I1 flowing in loop P is decreasing with time, then the current I 2 in the loop Q (a) (b) (c) (d)

Q

flows in the same direction as that of P flows in the opposite direction as that of Q is zero None of the above

19. In the circuit shown in figure, the switch S is closed at t = 0. If V L is the

L

voltage induced across the inductor and i is the instantaneous current, the correct variation of V L versus i is given by VL

VL S

E

(a)

(b) i

i

O

O

VL

VL E

E

(c)

(d)

O

R

i

i O

E

542 — Electricity and Magnetism 20. In the figure shown, a uniform magnetic field |B | = 0.5 T is perpendicular to the plane of

circuit. The sliding rod of length l = 0.25 m moves uniformly with constant speed v = 4 ms−1. If the resistance of the slides is 2 Ω, then the current flowing through the sliding rod is B

4Ω

l

2Ω

12 Ω

v

(b) 0.17 A (d) 0.03 A

(a) 0.1 A (c) 0.08 A

21. The figure shows a non-conducting ring of radius R carrying a charge q. In a circular region of radius r, a uniform magnetic field B perpendicular to the plane of the ring varies at a constant dB rate = β. The torque acting on the ring is dt q r R

(a)

1 2 qr β 2

(b)

1 qR2β 2

(c) qr 2β

(d) zero

22. A conducting ring of radius 2R rolls on a smooth horizontal conducting surface as shown in figure. A uniform horizontal magnetic field B is perpendicular to the plane of the ring. The potential of A with respect to O is B A 2R v O

(a) 2 BvR (c) 8 BvR

1 BvR 2 (d) 4 BvR (b)

23. A uniformly wound long solenoid of inductance L and resistance R is cut into two parts in the ratio η : 1, which are then connected in parallel. The combination is then connected to a cell of emf E. The time constant of the circuit is

L R  η  L (c)    η + 1 R

(a)

L (η + 1) R  η + 1 L (d)    η  R (b)

Electromagnetic Induction — 543

Chapter 27

24. When a choke coil carrying a steady current is short-circuited, the current in it decreases to β (< 1) times its initial value in a time T . The time constant of the choke coil is (a)

T β

(b)

T  1 ln    β

(c)

T ln β

(d) T ln β

25. In the steady state condition, the rate of heat produced in a choke coil is P. The time constant of

the choke coil is τ. If now the choke coil is short-circuited, then the total heat dissipated in the coil is

(a) P τ (c)

(b)

Pτ ln 2

1 Pτ 2

(d) P τ ln 2

26. In the circuit shown in figure initially the switch is in position 1 for a long

L

R

time, then suddenly at t = 0, the switch is shifted to position 2. It is required that a constant current should flow in the circuit, the value of resistance R in the circuit

(a) (b) (c) (d)

should be decreased at a constant rate should be increased at a constant rate should be maintained constant Not possible

2

1

E

27. The figure shows an L -R circuit, the time constant for the circuit is R

L R

E

L (a) 2R

2L (b) R

(c)

2R L

(d)

R 2L

28. In figure, the switch is in the position 1 for a long time, then the switch is shifted to position 2 at t = 0. At this instant the value of i1 and i2 are 1

2 R L

E

i2 L

R

i1

E −E , R R

(a)

E ,0 R

(b)

(c)

E −E , 2R 2R

(d) None of these

544 — Electricity and Magnetism 29. In a decaying L-R circuit, the time after which energy stored in the inductor reduces to one-fourth of its initial value is (a) (ln 2) (c)

2

L R

L R  2  L (d)    2 − 1 R (b) 0.5

L R

30. Initially, the switch is in position 1 for a long time and then shifted to position 2 at t = 0 as shown in figure. Just after closing the switch, the magnitude of current through the capacitor is 1

2 R

E L R

C

(a) zero (c)

(b)

E R

E 2R

(d) None of these

31. When the switch S is closed at t = 0, identify the correct statement just after closing the switch as shown in figure C L

S

(a) (b) (c) (d)

R

E

The current in the circuit is maximum Equal and opposite voltages are dropped across inductor and resistor The entire voltage is dropped across inductor All of the above

32. Two metallic rings of radius R are rolling on a metallic rod. A magnetic field of magnitude B is applied in the region. The magnitude of potential difference between points A and C on the two rings (as shown), will be C ω

(a) 0 (c) 8 BωR2

A ω

(b) 4 BωR2 (d) 2 BωR2

Chapter 27

Electromagnetic Induction — 545

33. In the figure, magnetic field points into the plane of paper and the conducting rod of length l is moving in this field such that the lowest point has a velocity v1 and the topmost point has the velocity v2( v2 > v1 ). The emf induced is given by v2

v1

(b) Bv2l 1 (d) B(v2 − v1 ) l 2

(a) Bv1l 1 (c) B (v2 + v1 ) l 2

34. Find the current passing through battery immediately after key ( K ) is closed. It is given that initially all the capacitors are uncharged. (Given that R = 6 Ω and C = 4 µF) C

R K

R C

R E = 5V

C L C

R R

(b) 5 A (d) 2 A

(a) 1 A (c) 3 A

35. In the circuit shown, the key ( K ) is closed at t = 0, the current through the key at the instant t = 10−3 ln 2, is 4Ω

5Ω 20 V

L=10 mH

5Ω

K 6Ω

C = 0.1 mF

(a) 2 A

(b) 8 A

(c) 4 A

(d) zero

36. A loop shown in the figure is immersed in the varying magnetic field B = B0t, directed into the page. If the total resistance of the loop is R, then the direction and magnitude of induced current in the inner circle is B0(πa 2 − b2) R B0 (πa 2 + 4b2) (c) clockwise R

(a) clockwise

B0π (a 2 + b2) R B0 (4b2 − πa 2) (d) clockwise R

(b) anti-clockwise

b a

546 — Electricity and Magnetism 37. A square loop of side a and a straight long wire are placed in the same plane as shown in figure. The loop has a resistance R and inductance L. The frame is turned through 180° about the axis OO′. What is the electric charge that flows through the loop? O′ a b l O

(a)

µ 0Ia  2a + ln   b 2 πR

(c)

µ 0Ia  a + 2b ln    b  2 πR

b  

(b)

 b  µ 0Ia ln  2  2 πR  b − a 2

(d) None of these

More than One Correct Options 1. The loop shown moves with a velocity v in a uniform magnetic field

L

of magnitude B, directed into the paper. The potential difference between points P and Q is e. Then, 1 BLv 2 (b) e = BLv (c) P is positive with respect to Q (d) Q is positive with respect to P (a) e =

v P L/2

L B

Q

2. An infinitely long wire is placed near a square loop as shown in figure. Choose the correct options. I

a a

a

µ 0a ln (2) 2π µ a2 (b) The mutual inductance between the two is 0 ln (2) 2π (c) If a constant current is passed in the straight wire in upward direction and loop is brought close to the wire, then induced current in the loop is clockwise (d) In the above condition, induced current in the loop is anti-clockwise (a) The mutual inductance between the two is

3. Choose the correct options. (a) SI unit of magnetic flux is henry-ampere (b) SI unit of coefficient of self-inductance is J/A volt -second (c) SI unit of coefficient of self-inductance is ampere (d) SI unit of magnetic induction is weber

Electromagnetic Induction — 547

Chapter 27

4. In the circuit shown in figure, circuit is closed at time t = 0. At time

a

t = ln ( 2) second

(a) (b) (c) (d)

rate of energy supplied by the battery is 16 J/s rate of heat dissipated across resistance is 8 J/s rate of heat dissipated across resistance is 16 J/s V a − Vb = 4 V

b 2H

2Ω

8V

S

5. Two circular coils are placed adjacent to each other. Their planes are parallel

A

and currents through them i1 and i2 are in same direction. Choose the correct i1 options. (a) (b) (c) (d)

B i2

When A is brought near B, current i 2 will decrease In the above process, current i 2 will increase When current i1 is increased, current i 2 will decrease In the above process, current i 2 will increase

6. A coil of area 2 m 2 and resistance 4 Ω is placed perpendicular to a uniform magnetic field of 4 T. The loop is rotated by 90° in 0.1 second. Choose the correct options. (a) (b) (c) (d)

Average induced emf in the coil is 8 V Average induced current in the circuit is 20 A 2 C charge will flow in the coil in above period Heat produced in the coil in the above period can’t be determined from the given data

7. In L-C oscillations, 2π LC

(a) time period of oscillation is

(b) maximum current in circuit is

q0 LC

(c) maximum rate of change of current in circuit is

q0 LC

(d) maximum potential difference across the inductor is

q0 . Here, q0 is maximum charge on 2C

capacitor

8. Magnetic field in a cylindrical region of radius R in inward direction is as

y

shown in figure. (a) an electron will experience no force kept at (2R, 0, 0) if magnetic field increases with time (b) in the above situation, electron will experience the force in negative y-axis  R  (c) If a proton is kept at 0, , 0 and magnetic field is decreasing, then it will  2 

O

x

experience the force in positive x-direction (d) if a proton is kept at (−R, 0, 0) and magnetic field is increasing, then it will experience force in negative y-axis

9. In the figure shown, q is in coulomb and t in second. At time t = 1 s a

(a) V a − Vb = 4 V (c) V c − V d = 16 V

1H

b 2F c + – q = 2t 2

4Ω

d

(b) Vb − V c = 1 V (d) V a − V d = 20 V

548 — Electricity and Magnetism 10. An equilateral triangular conducting frame is rotated with angular velocity ω

a

in a uniform magnetic field B as shown. Side of triangle is l. Choose the correct options. Bωl2 2 Bωl2 (d) V c − Vb = − 2

(a) V a − V c = 0

B

(b) V a − V c =

Bωl2 (c) V a − Vb = 2

c

b ω

Comprehension Based Questions Passage I (Q. No. 1 to 3 ) A uniform but time varying magnetic field B = ( 2t3 + 24t ) T is present in a cylindrical region of radius R = 2.5 cm as shown in figure. R/2

1. The force on an electron at P at t = 2.0 s is (a) (b) (c) (d)

96 × 10−21 N 48 × 10−21 N 24 × 10−21 N zero

P R

2. The variation of electric field at any instant as a function of distance measured from the centre of cylinder in first problem is E

E

(a)

r

E

E

(b)

r

(c)

r

(d)

r

3. In the previous problem, the direction of circular electric lines at t = 1 s is (a) (b) (c) (d)

clockwise anti-clockwise no current is induced cannot be predicted

Passage II (Q. No. 4 to 7 ) A thin non-conducting ring of mass m, radius a carrying a charge q can rotate freely about its own axis which is vertical. At the initial moment, the ring was at rest in horizontal position and no magnetic field was present. At instant t = 0, a uniform magnetic field is switched on which is vertically downward and increases with time according to the law B = B0t. Neglecting magnetism induced due to rotational motion of ring.

4. The magnitude of induced emf on the closed surface of ring will be (a) πa 2B0 (c) zero

(b) 2a 2B0 1 (d) πa 2B0 2

5. The magnitude of an electric field on the circumference of the ring is (a) aB0 1 (c) aB0 2

(b) 2aB0 (d) zero

Chapter 27

Electromagnetic Induction — 549

6. Angular acceleration of ring is qB0 2m qB0 (c) m

qB0 4m 2qB0 (d) m

(a)

(b)

7. Find instantaneous power developed by electric force acting on the ring at t = 1 s. 2q2B02a 2 14m 3q2B02a 2 (c) m

q2B02a 2 8m q2B02a 2 (d) 4m

(a)

(b)

Passage III (Q. No. 8 to 10 ) Figure shows a conducting rod of negligible resistance that can slide on smooth U -shaped rail made of wire of resistance 1 Ω /m. Position of the conducting rod at t = 0 is shown. A time dependent magnetic field B = 2t tesla is switched on at t = 0.

20 cm

Conducting rod

40 cm

8. The current in the loop at t = 0 due to induced emf is (a) 0.16 A, clockwise (c) 0.16 A, anti-clockwise

(b) 0.08 A, clockwise (d) zero

9. At t = 0, when the magnetic field is switched on, the conducting rod is moved to the left at constant speed 5 cm/s by some external means. At t = 2 s, net induced emf has magnitude

(a) 0.12 V (c) 0.04 V

(b) 0.08 V (d) 0.02 V

10. The magnitude of the force required to move the conducting rod at constant speed 5 cm/s at the same instant t = 2s, is equal to

(b) 0.12 N (d) 0.064 N

(a) 0.096 N (c) 0.08 N

Passage IV (Q. No. 11 to 13) Two parallel vertical metallic rails AB and CD are separated by 1 m . They are connected at the two ends by resistances R1 and R2 as shown in the figure. A horizontal metallic bar L of mass 0.2 kg slides without friction, vertically down the rails under the action of gravity. There is a uniform horizontal magnetic field of 0.6 T perpendicular to the plane of the rails. It is observed that when the terminal velocity is attained, the powers dissipated in R1 and R2 are 0.76 W and 1.2 W respectively ( g = 9.8 m / s2 )

A

C R1 L 1m

B

11. The terminal velocity of the bar L will be (a) 2 m/s

(b) 3 m/s

D R2

(c) 1 m/s

(d) None of these

550 — Electricity and Magnetism 12. The value of R1 is

(b) 0.82 Ω (d) None of these

(a) 0.47 Ω (c) 0.12 Ω

13. The value of R2 is

(b) 0.5 Ω (d) 0.3 Ω

(a) 0.6 Ω (c) 0.4 Ω

Match the Columns 1. Match the following two columns. Column I

Column II

(a) Magnetic induction (b) Coefficient of self-induction (c) LC (d) Magnetic flux

(p) (q) (r) (s)

[MT −2A−1 ] [L2T −2] [ML2T −2A−2] None of these

2. In the circuit shown in figure, switch is closed at time t = 0. Match the following two columns. 2H VL

2Ω VR

10 V

Column I (a) VL at t = 0 (b) VR at t = 0

Column II (p) zero

(c) VL at t = 1 s (d) VR at t = 1 s

(q) 10 V (r) 10 V e (s) 1 − 1 10 V    e

1 F and maximum charge in the capacitor is 4 C. 4 Match the following two columns. Note that in Column II, all values are in SI units.

3. In an L-C oscillation circuit, L = 1 H, C =

Column I (a) Maximum current in the circuit (b) Maximum rate of change of current in the circuit (c) Potential difference across inductor when q = 2C (d) Potential difference across capacitor when rate of change of current is half its maximum value

Column II (p) 16 (q) 4 (r) 2 (s) 8

Chapter 27

Electromagnetic Induction — 551

4. In the circuit shown in figure, switch remains closed for long time. It is opened at time t = 0. Match the following two columns at t = (ln 2) second. a

b 9H

6Ω

9V

3Ω d

c

Column I

Column II

(a) Potential differences across inductor

(p) 9 V

(b) Potential difference across 3 Ω resistance

(q) 4.5 V

(c) Potential difference across 6 Ω resistance

(r) 6 V

(d) Potential difference between points b and c (s) None of these

5. Magnetic flux passing through a coil of resistance 2 Ω is as shown in

φ (Wb)

figure. Match the following two columns. In Column II all physical quantities are in SI units. 4

Column I

Column II

(a) Induced emf produced

(p) 4

(b) Induced current

(q) 1

(c) Charge flow in 2 s

(r) 8

(d) Heat generation in 2 s

(s) 2

2

t (s)

6. A square loop is placed near a long straight current carrying wire as shown. Match the following two columns.

i

Column I

Column II

(a) If current is increased

(p) induced current in loop is clockwise

(b) If current is decreased

(q) induced current in loop is anti-clockwise

(c) If loop is moved away from the wire (r) wire will attract the loop (d) If loop is moved towards the wire

(s) wire will repel the loop

552 — Electricity and Magnetism Subjective Questions 1. In the circuit diagram shown, initially there is no energy in the inductor and the capacitor. The switch is closed at t = 0. Find the current I as a function of time if R = L/ C .

I

R

L

R

C S

V

2. A rectangular loop with a sliding connector of length l is located in a uniform magnetic field perpendicular to the loop plane. The magnetic induction is equal to B. The connector has an electric resistance R, the sides ab and cd have resistances R1 and R2. Neglecting the self-inductance of the loop, find the current flowing in the connector during its motion with a constant velocity v. B

a

d

v

R

R1

R2

c

b

3. A rod of length 2a is free to rotate in a vertical plane, about a horizontal axis O passing through its mid-point. A long straight, horizontal wire is in the same plane and is carrying a constant current i as shown in figure. At initial moment of time, the rod is horizontal and starts to rotate with constant angular velocity ω, calculate emf induced in the rod as a function of time. i d O

ω

2a

4. In the circuit arrangement shown in figure, the switch S is closed at t = 0. Find the current in

the inductance as a function of time? Does the current through 10 Ω resistor vary with time or remains constant. 10 Ω

5Ω

S

36 V

1 mH

Electromagnetic Induction — 553

Chapter 27

5. In the circuit shown, switch S is closed at time t = 0. Find the current through the inductor as a function of time t. 2Ω

2V

1Ω

4V S

L = 1 mH

6. In the circuit shown in figure, E = 120 V, R1 = 30.0 Ω , R2 = 50.0 Ω , and L = 0.200 H. Switch S is closed at t = 0. Just after the switch is closed. + E

S a

b R1 R2

(a) (b) (c) (d)

c

d

L

What is the potential difference V ab across the inductor R1? Which point, a or b, is at higher potential? What is the potential difference V cd across the inductor L? Which point, c or d, is at a higher potential?

The switch is left closed for a long time and then is opened. Just after the switch is opened (e) (f) (g) (h)

What is the potential difference V ab across the resistor R1? Which point a or b, is at a higher potential? What is the potential difference V cd across the inductor L? Which point, c or d, is at a higher potential?

7. Two capacitors of capacitances 2C and C are connected in series with an inductor of inductance

L. Initially, capacitors have charge such that V B − V A = 4V 0 and VC − V D = V 0. Initial current in the circuit is zero. Find A

C

B 2C

D C

L

(a) maximum current that will flow in the circuit, (b) potential difference across each capacitor at that instant, (c) equation of current flowing towards left in the inductor.

8. A 1.00 mH inductor and a 1.00 µF capacitor are connected in series. The current in the circuit is

described by i = 20 t , where t is in second and i is in ampere. The capacitor initially has no charge. Determine

(a) the voltage across the inductor as a function of time, (b) the voltage across the capacitor as a function of time, (c) the time when the energy stored in the capacitor first exceeds that in the inductor.

554 — Electricity and Magnetism 9. In the circuit shown in the figure, E = 50.0 V, R = 250 Ω and C = 0.500 µF. The switch S is closed for a long time, and no voltage is measured across the capacitor. After the switch is opened, the voltage across the capacitor reaches a maximum value of 150 V. What is the inductance L? R

E

C

L

S

10. The conducting rod ab shown in figure makes contact with metal rails ca and db. The apparatus is in a uniform magnetic field 0.800 T, perpendicular to the plane of the figure. x x

c

x x

x

x

x

x

B

x

x

x

x

x

a

x x

v

x

50.0 cm

x

d x

b

x

(a) Find the magnitude of the emf induced in the rod when it is moving toward the right with a speed 7.50 m/s. (b) In what direction does the current flow in the rod? (c) If the resistance of the circuit abdc is 1.50 Ω (assumed to be constant), find the force (magnitude and direction) required to keep the rod moving to the right with a constant speed of 7.50 m/s. You can ignore friction. (d) Compare the rate at which mechanical work is done by the force (Fv) with the rate at which thermal energy is developed in the circuit (I 2R).

11. A non-conducting ring of mass m and radius R has a charge Q uniformly distributed over its circumference. The ring is placed on a rough horizontal surface such that plane of the ring is parallel to the surface. A vertical magnetic field B = B0t 2 tesla is switched on. After 2 s from switching on the magnetic field the ring is just about to rotate about vertical axis through its centre. (a) Find friction coefficient µ between the ring and the surface. (b) If magnetic field is switched off after 4 s, then find the angle rotated by the ring before coming to stop after switching off the magnetic field.

12. Two parallel long smooth conducting rails separated by a

x

distance l are connected by a movable conducting x connector of mass m. Terminals of the rails are connected R by the resistor R and the capacitor C as shown in figure. x A uniform magnetic field B perpendicular to the plane of the rail is switched on. The connector is dragged by a x constant force F. Find the speed of the connector as a function of time if the force F is applied at t = 0. Also find the terminal velocity of the connector.

x

x

x

x

x x

x

x

x

x

x

x

x

x

x

x

x

x

x

x C x

x

x

x

x

x

x

x

l

F

Electromagnetic Induction — 555

Chapter 27

13. A circuit containing capacitors C1 and C2, shown in the figure is in the steady state with key K 1 closed and K 2 opened. At the instant t = 0, K 1 is opened and K 2 is closed.

K1

20 V

R

C 1 = 2 µF

C 2 = 2 µF

K2

L = 0.2 mH

(a) Find the angular frequency of oscillations of L- C circuit. (b) Determine the first instant t, when energy in the inductor becomes one third of that in the capacitor. (c) Calculate the charge on the plates of the capacitor at that instant.

14. Initially, the capacitor is charged to a potential of 5 V and then connected to position 1 with the shown polarity for 1 s. After 1 s it is connected across the inductor at position 2. 100 Ω 1

–

+ E = 10 V

10 µF

–

L = 25 mH

+

2

(a) Find the potential across the capacitor after 1 s of its connection to position 1. (b) Find the maximum current flowing in the L- C circuit when capacitor is connected across the inductor. Also, find the frequency of LC oscillations.

15. A rod of mass m and resistance R slides on frictionless and resistanceless rails a distance l apart that include a source of emf E0. (see figure). The rod is initially at rest. Find the expression for the (a) velocity of the rod v (t). (b) current in the loop i(t).

×

×

×

×

× a ×

×

×

×

×

×

×

×

×

×

× E0 ×

×

×

×

×

×

×

×

×

×

×

×

× b ×

×

×

16. Two metal bars are fixed vertically and are connected on the top by a capacitor C. A sliding conductor of length l and mass m slides with its ends in contact with the bars. The arrangement is placed in a uniform horizontal magnetic field directed normal to the plane of the figure. The conductor is released from rest. Find the displacement x(t) of the conductor as a function of time t. ×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

l

556 — Electricity and Magnetism 17. A conducting light string is wound on the rim of a metal ring of radius r and mass m. The free end of the string is fixed to the ceiling. A vertical infinite smooth conducting plane is always tangent to the ring as shown in the figure. A uniform magnetic field B is applied perpendicular to the plane of the ring. The ring is always inside the magnetic field. The plane and the string are connected by a resistance R. When the ring is released, find × × × × × × × ×

× × × × × R× × × × × × × × × × ×

× × × × × × × ×

× × × × × × × ×

(a) the current in the resistance R as a function of time. (b) the terminal velocity of the ring.

18. A conducting frame abcd is kept in a vertical plane. A conducting rod ef of mass m and length l can slide smoothly on it remaining always horizontal. The resistance of the loop is negligible and inductance is constant having value L. The rod is left from rest and allowed to fall under gravity and inductor has no initial current. A magnetic field of constant magnitude B is present throughout the loop pointing inwards. Determine L

b

c

B e

f

a

d

(a) position of the rod as a function of time assuming initial position of the rod to be x = 0 and vertically downward as the positive x-axis. (b) the maximum current in the circuit. (c) maximum velocity of the rod

19. A rectangular loop with a sliding conductor of length l is located in a uniform magnetic field perpendicular to the plane of loop. The magnetic induction perpendicular to the plane of loop is equal to B. The part ad and bc has electric resistance R1 and R2, respectively. The conductor starts moving with constant acceleration a0 at time t = 0. Neglecting the self-inductance of the loop and resistance of conductor. Find a

b ⊗B

R1

l

d

R2

a0 c

(a) the current through the conductor during its motion. (b) the polarity of abcd terminal. (c) external force required to move the conductor with the given acceleration.

Chapter 27

Electromagnetic Induction — 557

20. A conducting circular loop of radius a and resistance per unit length R is moving with a constant velocity v0 ,parallel to an infinite conducting wire carrying current i0.A conducting rod v of length 2a is approaching the centre of the loop with a constant velocity 0 along the direction 2 of the current. At the instant t = 0,the rod comes in contact with the loop at A and starts sliding on the loop with the constant velocity. Neglecting the resistance of the rod and the self-inductance of the circuit, find the following when the rod slides on the loop. v0/2 C

P

Q

A

i0

v0 a√3 O a B

 a (a) The current through the rod when it is at a distance of   from the point A of the loop.  2 (b) Force required to maintain the velocity of the rod at that instant.

21. U-frame ABCD and a sliding rod PQ of resistance R, start moving with velocities v and 2v

respectively, parallel to a long wire carrying current i0. When the distance AP = l at t = 0, determine the current through the inductor of inductance L just before connecting rod PQ loses contact with the U-frame. 2v

A P

i0

B Q l

v D a

C a

Answers Introductory Exercise 27.1 1. Anti-clockwise 3. [ML2A–1 T –3 ]

2. No 4. Clockwise

5. Same direction, opposite direction. 7. 1600 µC

6. 6.74 V 8. 9.0 × 10 –7 Wb

Introductory Exercise 27.2 1. 4.4 V, north 3. (a)

B ωl 2

2

(b)

2. 0.00375 N −3Bωl 2

2

4. No

Introductory Exercise 27.3 1. 3 (t cos t + sin t )

2. − 80e −4t , − 40 e −4t

3. (a) 0.625 mH (b) 0.13 J, 0.21 J/s

4. (a) 4.5 × 10 –5 H (b) 4.5 × 10 –3 V

Introductory Exercise 27.4 1. 3.125 mH, 0.9375 V 3. (a) 0.27 V, Yes (b) 0.27 V

2. (a) 2 H

(b) 30 V

(c) 1 H

Introductory Exercise 27.5 2. (a) 0.2 s

(b) 10 A

3. No

(c) 9.93 A

Introductory Exercise 27.6 2. With KE as v ⇔ i and m ⇔ L. Therefore,

1 1 mv 2 = Li 2 2 2

3. (a) 45.9 µC (b) 23.3 V

4. 20.0 V

Introductory Exercise 27.7 1. (a) 3.1 × 10 –6 V (b) 2.0 × 10 –6 V / m 2. (a) 8.0 × 10 –21 N (downward and to the right perpendicular to r2 ) (b) 0.36 V / m (upwards and to the left perpendicular to r1 )

Exercises LEVEL 1 Assertion and Reason 1. (d)

2. (b)

3. (d)

4. (a,b)

5. (a)

6. (d)

7. (b)

8. (c)

9. (d)

10. (c)

Objective Questions 1. (c)

2. (a)

3. (b)

4. (c)

5. (c)

6. (b)

7. (b)

8. (a)

9. (c)

10. (b)

11. (a)

12. (b)

13. (c)

14. (c)

15. (a)

16. (a)

17. (c)

18. (a)

19. (d)

20. (b)

21. (a)

22. (d)

23. (c)

24. (b)

25. (a)

26. (c)

27. (d)

28. (d)

29. (b)

30. (d)

31. (b)

32. (a)

33. (a)

34. (d)

35. (c)

36. (b)

37. (c)

38. (c)

39. (a)

Electromagnetic Induction — 559

Chapter 27 Subjective Questions 1. True 3. e = aSB 0e

2. 272 m µ iv  l 4. (a) 0 ln  1 +   d 2π

– at

(b) a (c) zero

5. (a) (0.5) (1 – e –10t ) A (b) 1.50 A – (0.25 A) e –10t 6. 0.5 T

7. (a) 0.011 V

8.

F0 – 3L – L 2 2

X

O L 3L 2 2

L – 3L – 2 O 2 – I0

2

E e r

L 3L 2 2

I0

2 F0 = B L V R (a) 15. (a) 518 mW (b) 328 mW

17. (a)

(b) zero (c) – 0.011 V

I

F

R + r  – t  L 

X

E L 2r (R + r )

18.

µ 0 2ia2 v 4 π x (x + a)

10. (2 × 10 −5 ) ln (10) V 11. 12. 13. 14. 16.

i0 = BLV R (b) (c) 191 mW

2

(b)

9. e =

8 A 3

3.33 × 10 –2 (a) 0.25 H (a) 1.96 H 0.37 (a) 17.3 µs

A, 6.67 × 10 –2 A (b) 4.5 × 10 –4 Wb (b) 7.12 × 10 –3 Wb (b) 30.7 µs

19. (a) 3.6 mH (b) 1.33 kHz (c) 0.188 ms

20. (a) 6.28 × 103 rad/s , 10 −3 s (b) 10 −4 cos (6.28 × 103 t ) (c) 0.0253 H (d) 0.4 A 21. (a) 1.25 × 10 –2 V (b) 8.33 × 10 –4 A (c) 3.125 × 10 –8 J (d) 4.33 × 10 –6 C, 7.8 × 10 –7 J

LEVEL 2 Single Correct Option 1.(b)

2.(c)

3.(a)

4.(b)

5.(c)

6.(a)

7.(a)

8.(c)

9.(c)

10.(a)

11.(b)

12.(b)

13.(c)

14.(a)

15.(b)

16.(a)

17.(b)

18.(a)

19.(d)

20.(a)

21.(a)

22.(d)

23.(a)

24.(b)

25.(b)

26.(d)

27.(b)

28.(b)

29.(a)

30.(c)

31.(c)

32.(b)

33.(c)

34.(a)

35.(a)

36.(d)

37.(d)

More than One Correct Options 1. (a,c)

2. (a,c)

3. (a,c)

4. (a,b,d) 5. (a,c)

6. (b,c,d)

7. (b,c)

8. (b,c,d) 9. (a,b,c)

10. (a,c)

Comprehension Based Questions 1.(b)

2.(c)

3.(b)

11.(c)

12.(a)

13.(d)

4.(a)

5.(c)

Match the Columns 1. (a) → p

(b) → r

(c) → s

(d) → s

2. (a) → q

(b) → p

(c) → r

(d) → s

3. (a) → s

(b) → p

(c) → s

(d) → s

4. (a) → s

(b) → q

(c) → p

(d) → p

5. (a) → s

(b) → q

(c) → s

(d) → p

6. (a) → q, s

(b) → p, r

(c) → p, r

(d) → q, s

6.(a)

7.(d)

8.(a)

9.(b)

10.(c)

560 — Electricity and Magnetism Subjective Questions 1. I =

V R

2. i =

R1 R 2 Bvl where R ′ = R + R′ R1 + R 2

3. e =

µ 0 iω 2 π sin ωt

  d  d − a sin ωt  ln   + 2 a    sin ωt  d + a sin ωt 

4. 3.6 (1 − e – t / τ L ) A. Here, τL = 300 µs. Current through 10 Ω resistor varies with time. 5. i = 5(1 − e –2000t /3 ) A 6. (a) 120 V (b) a (c) 120 V (d) c

(e) −72 V

(f) b (g) −192 V

(c) i = q0 sinωt . Here, q0 = 2CV0 and ω =

7. (a) q0ω (b) 3V0 : 3V0

8. (a) 20 mV (b) 107 t 2 V (c) 63.2 µs 10. (a) 3 V 12. V =

(h) d 3 2LC 9. 0.28 H

(b) b to a (c) 0.8 N towards right

(d) both are 6 W

11.

(a)

2B 0QR mg

(b)

B 0Q m

FR B 2l 2 FR , vT = 2 2 (1 − e – αt ) Here, α = 2 2 B l mR + RB 2 l 2C B l

13. (a) 5 × 10 4 rad/s

(b) 1.05 × 10 –5 s

(c) 10 3 µC

14. (a) 8.16 V

2 2

15. (a) v =

E0 (1 – Bl

B l – t e mR )

(b) i =

E 0 – Blv R

(b) 5.16 A, 10 Hz 2

16. x =

mgt 2 (m + CB 2 l 2 )

−2B2r 2

− t mg mgR (1 − e mR ) (b) vT = 2Br 4B 2 r 2 v Bl 2mg g mL (b) imax = (c) vmax = v0 = 18. (a) x = 0 (1 – cos ωt ), ω = Bl Bl ω mL Bla0 t 19. (a) i = (R1 + R 2 ) (b) Polarity of a, b is positive and polarity of c, d is negative R1 R 2

17. (a) i =

  B 2l 2t (c) Fext = a0  m + (R1 + R 2 ) R1 R 2   9 v 0 µ 0 i0 9 µ 20 i02 v0 20. (a) i = ln (3) (b) (ln 3)2 32 aR π3 16 aR π 2

e 21. i =   [1 − e − lR /Lv ] , R 

where e =

µ 0 i0 v ln (2) 2π

Alternating Current Chapter Contents 28.1

Introduction

28.2

Alternating currents and phasors

28.3

Current and potential relations

28.4

Phasor algebra

28.5

Series L-R circuit

28.6

Series C-R circuit

28.7

Series L-C-R circuit

28.8

Power in an AC circuit

562 — Electricity and Magnetism

28.1 Introduction A century ago, one of the great technological debates was whether the electrical distribution system should be AC or DC. Thomas Edison favoured direct current (DC), that is, steady current that does not vary with time. George Westinghouse favoured alternating current (AC), with sinusoidally varying voltages and currents. He argued that transformers can be used to step the voltage up or down with AC but not with DC. Low voltages are safer for consumer use, but high voltages and correspondingly low currents are best for long distances power transmission to minimize i 2 R losses in the cables. Eventually, Westinghouse prevailed, and most present day household and industrial power distribution systems operate with alternating current. i or V

i or V

V0, i 0 +

+

t – –i 0 –V0

+

t

–

i = i 0 sin ωt or V = V 0 sin ωt

i = i 0 cos ωt or V = V 0 cos ωt

Fig. 28.1

A time varying current or voltage may be periodic and non-periodic. In case of periodic current or voltage, the current or voltage is said to be alternating if its amplitude is constant and alternate half cycle is positive and half negative. If the current or voltage varies periodically as sine or cosine function of time, the current or voltage is said to be sinusoidal and is what we usually mean by it.

28.2 Alternating Currents and Phasors The basic principle of the AC generator is a direct consequence of Faraday's law of induction. When a conducting loop is rotated in a magnetic field at constant angular frequency ω a sinusoidal voltage (emf) is induced in the loop. This instantaneous voltage is V = V0 sin ωt The usual circuit diagram symbol for an AC source is shown in Fig. 28.2. In Eq. (i),V0 is the maximum output voltage of the AC generator or the voltage amplitude and ω is the angular frequency equal to 2π times the frequency f.

…(i)

Fig. 28.2

ω = 2 πf The frequency of AC in India is 50 Hz, i.e. f = 50 Hz So,

ω = 2πf ≈ 314 rad/s

The time of one cycle is known as time period T, the number of cycles per second the frequency f. 1 2π or T= T= f ω

Chapter 28

Alternating Current — 563

A sinusoidal current might be described as i = i0 sin ωt If an alternating current is passed through an ordinary ammeter or voltmeter, it will record the mean value for the complete cycle, as the quantity to be measured varies with time. The average value of current for one cycle is 2π/ω

T

Thus,

i

One cycle

= ( iav )T

i

One cycle

=0

∫ idt = ∫0 (i0 sin ωt ) dt = 0 = 0T 2π / ω ∫0 dt ∫0 dt

Similarly, the average value of the voltage (or emf) for one cycle is zero. V

=0

One cycle

Since, these averages for the whole cycle are zero, the DC instrument will indicate zero deflection. In AC, the average value of current is defined as its average taken over half the cycle. Hence, π/ω

T /2

∫ idt ∫ (i0 sin ωt ) dt = 2 i i Half cycle = ( iav )T /2 = 0T /2 = 0 π/ω π 0 ∫0 dt ∫0 dt This is sometimes simply written as iav . Hence, iav = i Vav =

Similarly,

Half cycle

=

2 i ≈ 0.637 i0 π 0

2 V ≈ 0.637 V0 π 0

A DC meter can be used in an AC circuit if it is connected in the full wave rectifier circuit. The 2 average value of the rectified current is the same as the average current in any half cycle, i.e. times π the maximum current i0 . A more useful way to describe a quantity is the root mean square (rms) value. We square the instantaneous current, take the average (mean) value of i 2 and finally take the square root of that average. This procedure defines the root-mean-square current denoted as i rms . Even when i is negative, i 2 is always positive so i rms is never zero (unless i is zero at every instant). Hence, T 2

i

2

2π / ω

∫ i dt = ∫0 = 0T One cycle ∫0 dt

∴

i rms =

Thus,

i rms =

i2 i0 2

One cycle

≈ 0.707 i0

( i02 sin 2 ωt ) dt 2π / ω

∫0 =

i0 2

dt

≈ 0.707 i0

=

i02 2

564 — Electricity and Magnetism Similarly, we get V rms =

V0 2

≈ 0.707 V0

The square root of the mean square value is called the virtual value and is the value given by AC instruments. Thus, when we speak of our house hold power supply as 220 V AC, this means that the rms voltage is 220 V and its voltage amplitude is i V0 = 2 V rms = 311 V i0 i rms = 0.707 i 0 i av = 0.637 i 0

Form Factor The ratio, =

V / 2 rms value π = 0 = = 1.11 average value 2V0 /π 2 2

t

is known as form factor. The different values i0 , iav and i rms are shown in Fig. 28.3.

Fig. 28.3

Note (1) The average value of sin ωt, cos ωt, sin 2ωt, cos 2ωt, etc., is zero because it is positive in half of the time and negative in rest half of the time. Thus, sin ωt = cos ωt = sin 2 ωt = cos 2 ωt = 0 If i = i0 sin ωt then i = i0 sin ωt = i0 sin ωt = 0 1 2

(2) The average value of sin 2 ωt and cos 2 ωt is ⋅ or If

sin 2 ωt = cos 2 ωt =

1 2

i 2 = i02 sin 2 ωt

i02 2 (3) Like SHM, general expressions of current/voltage in an sinusoidal AC are i = i0 sin (ωt ± φ ) V = V0 sin (ωt ± φ ) or i = i0 cos (ωt ± φ ) and V = V0 cos (ωt ± φ ) (4) Average value of current or voltage over a half cycle can be zero also. This depends on the time interval (of course T /2) over which average value is to be found. Think why? then

i 2 = i02 sin 2 ωt = i02 sin 2 ωt =

Phasors If an AC generator is connected to a series circuit containing resistors, inductors and capacitors and we want to know the amplitude and time characteristics of the alternating current. To simplify our analysis of circuits containing two or more elements, we use graphical constructions called phasor diagrams. In these constructions, alternating (sinusoidal) quantities, such as current and voltage are rotating vectors called phasors. In these diagrams, the instantaneous value of a quantity that varies sinusoidally with time is represented by the projection onto a vertical axis (if it is a sine function) or onto a horizontal axis (if it

Chapter 28

Alternating Current — 565

is a cosine function) of a vector with a length equal to the amplitude ( i0 ) of the quantity. The vector rotates counterclockwise with constant angular velocity ω. ω

ω

i 0 sin ωt

i0

i0 or ωt

ωt O

i 0 cos ωt

O

Fig. 28.4

A phasor is not a real physical quantity with a direction in space, such as velocity, momentum or electric field. Rather, it is a geometric entity that helps us to describe and analyze physical quantities that vary sinusoidally with time. V

Example 28.1 Show that average heat produced during a cycle of AC is same as produced by DC with i = i rms . For an AC, i = i 0 sin ωt Therefore, instantaneous value of heat produced in time dt across a resistance R is

Solution

dH = i 2 Rdt = i 02 R sin 2 ωt dt ∴ Average value of heat produced during a cycle, T

2π /ω

∫ dH = ∫0 H av = 0T ∫0 dt =

( i 02 R sin 2 ωt ) dt 2 π/ω

∫0

dt

i 02  2π  R   = i 2rms RT 2 ω

i.e. AC produces same heating effects as DC of value i = i rms . V

Example 28.2 If the current in an AC circuit is represented by the equation, i = 5 sin ( 300t – π /4) Here, t is in second and i in ampere. Calculate (a) peak and rms value of current (b) frequency of AC (c) average current Solution

(a) As in case of AC,

∴ The peak value, and (b) Angular frequency, ∴ (c)

i = i 0 sin (ωt ± φ ) i0 = 5 A i0

5 = = 3.535 A 2 2 ω = 300 rad/s ω 300 f= = ≈ 47.75 Hz 2π 2π 2 2 i av =   i 0 =   ( 5) = 3.18 A  π  π

i rms =

Ans. Ans.

Ans. Ans.

566 — Electricity and Magnetism

28.3 Current and Potential Relations In this section, we will derive voltage current relations for individual circuit elements carrying a sinusoidal current. We will consider resistors, inductors and capacitors.

Resistor in an AC Circuit Consider a resistor with resistance R through which there is a sinusoidal current given by i = i0 sin ωt

…(i)

R a

b i Fig. 28.5

Here, i0 is the current amplitude (maximum current). From Ohm's law, the instantaneous PD between points a and b is VR = iR = ( i0 R ) sin ωt We can write as i0 R = V0 , the voltage amplitude …(ii) ∴ VR = V0 sin ωt From Eqs. (i) and (ii), we can see that current and voltage are in phase if only resistance is in the circuit. Fig. 28.6 shows graphs of i and VR as functions of time. i or VR i = i 0 sin ωt

i0 V0 t

i0

VR = V0 sin ωt

V0 ωt

O

Fig. 28.6

Fig. 28.7

The corresponding phasor diagram is shown in Fig. 28.7. Because i and VR are in phase and have the same frequency, the current and voltage phasors rotate together, they are parallel at each instant. Their projection on vertical axis represents the instantaneous current and voltage respectively. Note Direction of an alternating current is not shown in a circuit, as it keeps on changing. In the figure, the direction of instantaneous current is only shown.

Capacitor in an AC Circuit If a capacitor of capacitance C is connected across the alternating source, the instantaneous charge on the capacitor is q = CVC = CV0 sin ωt

Alternating Current — 567

Chapter 28

V = V0 sin ωt

and the instantaneous current i passing through it is given by dq i= = CV0ω cos ωt dt V = 0 sin (ωt + π/2) 1/ωC

a

i = i0 sin (ωt + π/2)

or

b i

Fig. 25.8

i V0 = 0 ωC

Here,

–q

q

1 is the effective AC resistance or the capacitive reactance of ωC the capacitor and is represented as X C . It has unit as ohm. Thus,

This relation shows that the quantity

XC =

1 ωC

It is clear that the current leads the voltage by 90° or the potential drop across the capacitor lags the current passing it by 90°. Fig. 28.9 shows V and i as functions of time t. i,V i0

i 0 sin ωt

i0 V0

i ωt

t VC

Fig. 28.9

90° V0 cos ωt

V0

Fig. 28.10

The phasor diagram 28.10 shows that voltage phasor is behind the current phasor by a quarter cycle or 90°.

Inductor in an AC Circuit Consider a pure inductor of self-inductance L and zero resistance connected to an alternating source. Again we assume that an instantaneous current i = i0 sin ωt flows through the inductor. Although, there is no resistance, there is a potential difference VL between the inductor terminals a and b because the current varies with time giving rise to a self-induced emf. di   VL = Vab = – ( induced emf ) = –  – L   dt  or

VL = L

di = Li0ω cos ωt dt

L a

b i

Fig. 28.11

568 — Electricity and Magnetism or

π  VL = V0 sin  ωt +   2

…(i)

Here,

V0 = i0 (ωL)

…(ii)

or

i0 =

∴

i=

V0 ωL V0 sin ωt ωL

…(iii)

Eq. (iii) shows that effective AC resistance, i.e. inductive reactance of inductor is X L = ωL and the maximum current, i0 =

V0 XL

The unit of X L is also ohm. From Eqs. (i) and (iii), we see that the voltage across the inductor leads the current passing through it by 90°. Fig. 28.12 shows VL and i as functions of time. i , VL

i 0 sin ωt

V0 i0 i

t

i0 V0 cos ωt

V0

90°

VL

Fig. 28.12

ωt

Fig. 28.13

Phasor diagram in Fig. 28.13 shows that VL leads the current i by 90°.

Extra Points to Remember ˜

˜

Circuit elements with AC

Circuit elements

Amplitude relation

Circuit quantity

Phase of V

Resistor

V0 = i0 R

R

in phase with i

Capacitor

V0 = i0 XC

XC =

1 ωC

lags i by 90°

Inductor

V0 = i0 X L

X L = ωL

leads i by 90°

In DC, ω = 0, therefore, X L = 0 and XC = ∞

Chapter 28 ˜

˜

˜

˜

Alternating Current — 569

The potential of point a with respect to point b is given by di VL = + L , the negative of the induced emf. This expression gives the dt correct sign of VL in all cases.

L a

i Fig. 28.14

b

If an oscillating voltage of a given amplitude V0 is applied across an inductor, the resulting current will have a smaller amplitude i0 for larger value of ω. Since, X L is proportional to frequency, a high frequency voltage applied to the inductor gives only a small current while a lower frequency voltage of the same amplitude gives rise to a larger current. Inductors are used in some circuit applications, such as power supplies and radio interference filters to block high frequencies while permitting lower frequencies to pass through. A circuit device that uses an inductor for this purpose is called a low pass filter. The capacitive reactance of a capacitor is inversely proportional to the capacitance C and the angular frequency ω. The greater the capacitance and the higher the frequency, the smaller is the capacitive reactance XC . Capacitors tend to pass high frequency current and to block low frequency current, just the opposite of inductors. A device that passes signals of high frequency is called a high pass filter. Figure shows the graphs of R, X L and XC as functions of angular frequency ω. R, X

XL R

XC

ω

Fig. 28.15 ˜

Remember that we can write, VR = iR, (V0 )R = i0 R, (V0 )L = i0 X L and (V0 ) C = i0 XC but can't write (for instantaneous voltages). VL = i X L

VC = i XC

or

This is because there is a phase difference between the voltage and current in both an inductor and a capacitor. V

Example 28.3 A 100 Ω resistance is connected in series with a 4 H inductor. The voltage across the resistor is V R = ( 2.0 V ) sin (103 rad / s) t : (a) Find the expression of circuit current (b) Find the inductive reactance (c) Derive an expression for the voltage across the inductor. Solution

(a)

i=

VR ( 2.0 V)sin (103 rad /s ) t = R 100 = ( 2.0 × 10–2 A ) sin (103 rad /s) t

(b)

Ans.

X L = ωL = (103 rad / s ) ( 4 H ) = 4.0 × 103 Ω

Ans.

(c) The amplitude of voltage across inductor, V0 = i 0 X L = ( 2.0 × 10–2 A ) ( 4.0 × 103 Ω ) = 80 V

Ans.

570 — Electricity and Magnetism In an AC, voltage across the inductor leads the current by 90° or π/2 rad. Hence, VL = V0 sin (ωt + π / 2) π = ( 80 V) sin  (103 rad /s ) t + rad 2   Note That the amplitude of voltage across the resistor ( = 2.0 V ) is not same as the amplitude of the voltage

across the inductor ( = 80 V ), even though the amplitude of the current through both devices is the same.

28.4 Phasor Algebra The complex quantities normally employed in AC circuit analysis, can be added and subtracted like coplanar vectors. Such coplanar vectors which represent sinusoidally time varying quantities are known as phasors. y In Cartesian form, a phasor A can be written as A = a + jb where, a is the x-component and b is the y-component of phasor A. The magnitude of A is

|A| = a + b 2

A

b

2

θ a

x

Fig. 28.16

and the angle between the direction of phasor A and the positive x-axis is  b θ = tan –1   a When a given phasor A, the direction of which is along the x-axis is multiplied by the operator j, a new phasor jA is obtained which will be 90° anti-clockwise from A, i.e. along y-axis. If the operator j is multiplied now to the phasor jA, a new phasor j 2 A is obtained which is along –x-axis and having same magnitude as of A. Thus, j2A = – A ∴

j 2 = – 1 or

j = –1

Now, using the j operator, let us discuss different circuits of an AC.

28.5 Series L-R Circuit As we know, potential difference across a resistance in AC is in phase with current and it leads in phase by 90° with current across the inductor.

VR

VL Fig. 28.17

Suppose in phasor diagram current is taken along positive x-direction. Then,VR is also along positive x-direction and VL along positive y-direction, so, we can write

Chapter 28

Alternating Current — 571

V = VR + jVL = iR + j( iX L ) = iR + j ( iω L) = iZ

(as X L = ωL)

VL

y

V φ

i

x

VR Fig. 28.18

Here, Z = R + j X L = R + j (ωL) is called as impedance of the circuit. Impedance plays the same role in AC circuits as the ohmic resistance does in DC circuits. The modulus of impedance is | Z | = R 2 + (ωL) 2 The potential difference leads the current by an angle, |V | X  φ = tan –1 L = tan –1  L   R  | VR |  ωL  φ = tan –1    R 

or

28.6 Series C-R Circuit Potential difference across a capacitor in AC lags in phase by 90° with the current in the circuit. Suppose in phasor diagram current is taken along positive x-direction. Then, VR is also along positive x-direction but VC is along negative y-direction. So, we can write VC

y

VR

Fig. 28.19 i

φ

VC

VR

x

V

Fig. 28.20

V = VR – jVC = iR – j( iX C )  i  = iR – j   = iZ  ωC  Here,

 1  impedance is Z = R – j    ωC 

572 — Electricity and Magnetism The modulus of impedance is  1  | Z| = R 2 +    ωC 

2

and the potential difference lags the current by an angle, V X /   1ωC φ = tan –1 C = tan –1 C = tan –1    VR R R   1  φ = tan –1    ωRC 

or

28.7 Series L-C-R Circuit Potential difference across an inductor leads the current by 90° in phase while that across a capacitor, it lags in phase by 90°. Suppose in a phasor diagram current is taken along positive x-direction. Then, VR is along positive x-direction, VL along positive y-direction and VC along negative y-direction.

VC

VL

VR

Fig. 28.21 2

V = VR + (VL – VC)2 y

VL – VC

VL VR i

φ

⇒

VR

VC

Fig. 28.22

Let us assume that X L > X C or VL > VC So, we can write V = VR + jVL – jVC = iR + j( iX L ) – j( iX C ) = iR + j [ i ( X L – X C )] = iZ 1   Here, impedance is Z = R + j ( X L – X C ) = R + j  ωL –   ωC  The modulus of impedance is

1   | Z | = R 2 +  ωL –   ωC 

2

and the potential difference leads the current by an angle, V − VC  X – XC  = tan –1  L φ = tan –1 L    R VR

or

1     ωL – ω C φ = tan    R   –1

x

Chapter 28

Alternating Current — 573

Note Let us take the most general case of a series L-C-R circuit in an AC. | Z | = R 2 + ( XL ~ XC ) 2 If XL = X C

or

ωL =

1 ωC

or

ω=

1 LC

or

f =

1 the modulus of impedance 2π LC

|Z | = R and the current is in phase with voltage, i.e. if V = V0 sin ωt, then i = i0 sin ωt V V where, i0 = 0 = 0 |Z| R Such a condition is known as resonance and frequency known as resonance frequency and is given by 1 f = 2π LC The current in such a case is maximum. If XL > XC , then the modulus of the impedance | Z | = R 2 + ( XL – XC ) 2 and the voltage leads the current by an angle given by  X – XC  φ = tan –1  L   R  i.e. if V = V0 sin ωt, then

i = i0 sin (ωt – φ ) V i0 = 0 |Z|

where,

If XC > XL, then the modulus of the impedance is | Z | = R 2 + ( XC – XL) 2 and the current leads the voltage by an angle given by  X – XL  φ = tan –1  C   R  i.e. if V = V0 sin ωt, then where,

i = i0 sin (ωt + φ ) V i0 = 0 |Z|

Extra Points to Remember ˜

˜

i0 =

V0 , |Z|

irms =

Vrms V . But in general i ≠ . |Z| |Z|

In L-C-R circuit, whenever voltage across various elements is asked, find rms values unless stated in the question for the peak or instantaneous value. The rms values are VR = i rmsR, VL = i rms X L

and

VC = i rms XC

The peak values can be obtained by multiplying the rms values by 2. The instantaneous values across different elements is rarely asked.

574 — Electricity and Magnetism ˜

Voltage magnification in series resonance circuit

1  At resonance  f =  , the PD across the  2 π LC 

inductor and the capacitor are equal and 180° out of phase and therefore, cancel out. Hence, the applied emf is merely to overcome the resistance opposition only. If an inductance or capacitance of very large reactance (X L or XC ) is connected with X L = XC (at resonance) then PD across them increases to a very high value. The ratio is known as voltage magnification and is given by, 1  i rms   PD across inductance (or capacitance) i rms(ω L) ω L  ωC  1 or = = = Applied emf i rms(R ) R i rms (R ) ωCR This ratio is greater than unity. ˜

Response curves of series circuit The impedance of an XL, XC, R, Z, i Z i L - C - R circuit depends on the frequency. The dependence is shown in figure. The frequency is taken on logarithmic scale because of its wide range. From the figure, we can see that at resonance, XL XC 1 (i) X L = XC or ω = Resonance LC (ii) Z = Z min = R and

R

log ω

XL – XC

(iii) i is maximum.

Fig. 28.23

Note Here, by Z we mean the modulus of Z and i means irms. ˜

V

Acceptor circuit If the frequency of the AC supply can be varied (e.g. in radio or television signal), then in series L-C-R circuit, at a frequency f = 1/ 2 π LC maximum current flows in the circuit and have a maximum PD across its inductance (or capacitance). This is the method by which a radio or television set is tuned at a particular frequency. The circuit is known as acceptor circuit.

Example 28.4 An alternating emf 200 virtual volts at 50 Hz is connected to a circuit of resistance 1 Ω and inductance 0.01 H. What is the phase difference between the current and the emf in the circuit? Also, find the virtual current in the circuit. Solution

In case of an L-R AC circuit, the voltage leads the current in phase by an angle, X  φ = tan −1  L   R 

Here,

X L = ωL = ( 2πfL ) = ( 2π ) ( 50) ( 0.01) = π Ω

and ∴ Further,

R = 1Ω φ = tan –1 ( π ) ≈ 72.3° V V rms i rms = rms = |Z | R2 + X 2

Ans.

L

Substituting the values, we have i rms =

200 (1) + ( π ) 2 2

= 60.67 A

Ans.

Chapter 28 V

Alternating Current — 575

Example 28.5 A resistance and inductance are connected in series across a voltage, V = 283 sin 314t The current is found to be 4 sin ( 314 t – π / 4) . Find the values of the inductance and resistance. Solution

In L-R series circuit, current lags the voltage by an angle, X  φ = tan −1  L   R  φ=

Here, ∴ ∴

π 4

X L = R or ωL = R 314 L = R

(ω = 314 rad/s) …(i)

V0 = i 0 | Z |

Further,

283 = 4 R 2 + X L2

∴

2

283 R 2 + (ωL ) 2 =   = 5005.56  4 

or

V

or

2R 2 = 5005.56

(as ωL = R)

∴

R ≈ 50 Ω

Ans.

and from Eq. (i),

L = 0.16 H

Ans.

Example 28.6 Find the voltage across the various elements, i.e. resistance, capacitance and inductance which are in series and having values 1000 Ω, 1 µF and 2.0 H, respectively. Given emf is V = 100 2 sin 1000 t volt Solution

The rms value of voltage across the source, V rms =

∴ ∴

i rms

100 2 2

= 100 V

ω = 1000 rad /s V rms V rms = = = |Z | R 2 + ( X L ~ X C )2

V rms 1  R 2 + ωL –   ωC 

100

=

  1 (1000) 2 + 1000 × 2 –  –6 1000 × 1 × 10   = 0.0707 A

2

2

576 — Electricity and Magnetism The current will be same every where in the circuit, therefore, PD across resistor, VR = i rms R = 0.0707 × 1000 = 70.7 V PD across inductor, VL = i rms X L = 0.0707 × 1000 × 2 = 141.4 V and 1 PD across capacitor, VC = i rms X C = 0.0707 × 1 × 1000 × 10–6 = 70.7 V

Ans.

Note The rms voltages do not add directly as `VR + VL + VC = 282.8 V which is not the source voltage 100 V. The reason is that these voltages are not in phase and can be added by vector or by phasor algebra. V = VR2 + (VL ~ VC ) 2

INTRODUCTORY EXERCISE

28.1

1. (a) What is the reactance of a 2.00 H inductor at a frequency of 50.0 Hz? (b) What is the inductance of an inductor whose reactance is 2.00 Ω at 50.0 Hz? (c) What is the reactance of a 2.00 µF capacitor at a frequency of 50.0 Hz? (d) What is the capacitance of a capacitor whose reactance is 2.00 Ω at 50.0 Hz?

2. An electric lamp which runs at 100 V DC and consumes 10 A current is connected to AC mains at 150 V, 50 Hz cycles with a choke coil in series. Calculate the inductance and drop of voltage across the choke. Neglect the resistance of choke. 360 3. A circuit operating at Hz contains a 1µF capacitor and a 20 Ω resistor. How large an inductor 2π must be added in series to make the phase angle for the circuit zero? Calculate the current in the circuit if the applied voltage is 120 V.

28.8 Power in an AC Circuit In case of a steady current, the rate of doing work is given by P = Vi In an alternating circuit, current and voltage both vary with time and also they differ in time. So, we cannot use P = Vi for the power generated. Suppose in an AC, the voltage is leading the current by an angle φ. Then, we can write V = V0 sin ωt and i = i0 sin (ωt – φ ) The instantaneous value of power in that case is P = Vi = V0 i0 sin ωt sin (ωt – φ ) 1   or …(i) P = V0 i0 sin 2 ωt cos φ – sin 2ωt sin φ  2   Now, the average rate of doing work (power) in one cycle will be T = 2π / ω

Pdt ∫ P One cycle = 0T = 2π /ω dt ∫0

…(ii)

Chapter 28

Alternating Current — 577

Substituting the value of P from Eq. (i) in Eq. (ii) and then integrating it with proper limits, we get V i 1 P One cycle = V0 i0 cos φ = 0 ⋅ 0 cos φ 2 2 2 or

P

One cycle

= V rms i rms cos φ

Here, the term cos φ is known as power factor. It is said to be leading if current leads voltage, lagging if current lags voltage. Thus, a power factor of 0.5 lagging means current lags the voltage by 60° (as cos –1 0.5 = 60°). The product of V rms and i rms gives the apparent power. While the true power is obtained by multiplying the apparent power by the power factor cos φ. Thus, Apparent power = V rms × i rms True power = apparent power × power factor

and

For φ = 0°, the current and voltage are in phase. The power is thus, maximum (= V rms × i rms ). For φ = 90°, the power is zero. The current is then stated wattless. Such a case will arise when resistance in the circuit is zero. The circuit is purely inductive or capacitive. The case is similar to that of a frictionless pendulum, where the total work done by gravity upon the pendulum in a cycle is zero.

Extra Points to Remember Let us consider a choke coil (used in tube lights) of large inductance L and low resistance R. The power factor for such a coil is given by R R R (as R X C , voltage leads by an angle φ given by R X ~ XL and tan φ = C cos φ = Z R (v) Instantaneous power = instantaneous current × instantaneous voltage (vi) Average power = Vrms irms cos φ , where R cos φ = = power factor. Z

Note Power is also equal to But this is not equal to

2 P = i rms R 2 Vrms P≠ R

R . If we substitute in P = Vrms i rms cos φ, then we get the first relation Z but not the second one. This implies that power is consumed only across resistance. V V (vii) i 0 = 0 or irms = rms Z Z (viii) (VC ) rms = (irms ) XC , (VL ) rms = (irms ) XL and (VR )rms = (irms ) R This is because Vrms = i rms Z and cos φ =

(ix) V = VR2 + (VC ~ VL )2 Here, V is the rms value of applied voltage VR is the rms value of voltage across resistance. VC across capacitor and VL across inductor etc.

580 — Electricity and Magnetism 10. ω = ωr =

1 is called resonance frequency. LC

11. At ω = ωr , (i) X L = X C (ii) Z = minimum value = R Vrms Vrms = Z min R V0 V0 (iv) i 0 = maximum value = = Z min R (iii) irms = maximum value =

(v) Power factor cos φ = 1

12. In one complete cycle, power is consumed only by resistance. No power is consumed by a capacitor or an inductor.

13. Z XL R

XC

ωr

XC =

1 ωC

⇒

XC ∝

X L = ωL

⇒

XL ∝ ω

ω

1 ω

R does not depend on ω. It is a constant. At ω = ωr : X C = X L and Z = Z min = R

14. For ω > ωr , X L > X C . Hence, voltage will lead the current or circuit is inductive. For ω < ωr , X C > X L. Hence, current will lead the voltage function or circuit is capacitive. At ω = ωr , X C = X L. Hence, current function and voltage function are in same phase.

15.

Conditions

Phase angle

Power factor

R=0

90°

0

XC = X L ≠ 0 R≠0

0°

1

XC = X L = 0 R≠0

0°

1

ω = ωr

0°

1

In all other cases, phase difference between current function and voltage function is

If If

0° < φ < 90°  X − XL  −1  R  X C > X L, φ = tan−1 C  or cos     Z R  X − XC  X L > X C , φ = tan−1 L    R

R  or cos −1   Z

Solved Examples TYPED PROBLEMS Type 1. Based on a real inductor

Concept An ideal inductor wire has zero resistance ( R = 0). So, in an AC, only X L is the impedance. But, a real inductor has some resistance also. In DC, total resistance is only R. In AC, total resistance called impedance is Z = R 2 + X L2 Therefore, V DC R V = AC = Z

I DC =

V AC

and

I AC

If

V AC = V DC , then

V

R 2 + X L2 I AC < I DC

Example 1 A current of 4 A flows in a coil when connected to a 12 V DC source. If the same coil is connected to a 12 V , 50 rad / s AC source, a current of 2.4 A flows in the circuit. Determine the inductance of the coil. Also, find the power developed in the circuit if a 2500 µF capacitor is connected in series with the coil. Solution (i) A coil consists of an inductance (L) and a resistance (R). In DC, only resistance is effective. Hence, V 12 = =3 Ω i 4 V V rms = rms = Z R2 + ω 2L2

R= In AC,

i rms

2  1  V rms  – R2     2 ω  i rms   

∴

L2 =

∴

L=

1  V rms  2   –R ω  i rms 

L=

1  12  2   – (3) 50  2.4

2

Substituting the values, we have 2

= 0.08 H

Ans.

582 — Electricity and Magnetism (ii) When capacitor is connected to the circuit, the impedance is Z = R2 + (XL – XC )2 R=3 Ω XL = ωL = (50) (0.08) = 4 Ω 1 1 XC = = =8 Ω ωC (50) (2500 × 10–6 )

Here, and ∴

Z = (3)2 + (4 – 8)2 = 5 Ω

Now,

P = V rms i rms cos φ V R = V rms × rms × Z Z 2

V  =  rms  × R  Z  Substituting the values, we have 2

 12 P =   ×3 5 = 17.28 W

Ans.

Type 2. Different time functions in AC

Concept In an L-C-R series circuit, there are total five functions of time, V , I , V R , VC and V L . Now, the following points are important in these functions. (i) V and I have a phase difference of φ where 0° ≤ φ ≤ 90° (ii) V R and I are in same phase (iii) VC lags behind I by 90° (iv) V L leads I by 90° (v) The functions, V = V L + VC + V L (all the time) V

VR

VC

VL

I

V

Example 2 30 Ω

1 mF

0.5 H

V = 200 sin (100 t + 30°)

In the diagram shown in figure, V function is given. Find other four functions of time I, VC , V R and V L . Also, find power consumed in the circuit, V is given in volts and ω in rad/s.

Chapter 28

Alternating Current — 583

Solution Given, ω = 100 rad /s XL = ωL = 50 Ω 1 1 XC = = = 10 Ω ωC 100 × 10−3 Z = R2 + (XL − XC )2 = (30)2 + (50 − 10)2 = 50 Ω Current function Maximum value of current, I0 =

V 0 200 = =4 A Z 50

XL > XC , therefore voltage leads the current by a phase difference φ where, R 30 3 cos φ = = = Z 50 5 φ = 53°

or ∴

I = 4 sin (100 t + 30° − 53° )

or

I = 4 sin (100 t − 23° )

Ans.

VR,VC and VL functions Maximum value of VR = I 0R = 4 × 30 = 120 volt, VR and I are in same phase. Therefore, VR = 120 sin (100 t − 23° )

Ans.

Maximum value of VC = I 0 XC = 4 × 10 = 40 volt Now, VC function lags the current function by 90°. Therefore, VC = 40 sin (100 t − 23° − 90° ) VC = 40 sin (100 t − 113° )

or

Ans.

Maximum value of VL = I 0XL = 4 × 50 = 200 volt, VL function leads the current function by 90°. Therefore, VL = 200 sin (100 t − 23° + 90° ) VL = 200 sin (100 t + 67° )

or

Ans.

Note We can check at any time that, V = VR + VL + VC Power

Power is consumed in an AC circuit only across a resistance and this power is given by P = V rms I rms cos φ 2 = I rms R

Let us use the first formula,  200  4  P=     2   2 = 240 watt

 3    5 Ans.

584 — Electricity and Magnetism Type 3. Parallel circuits

Concept Two or more than two sine or cosine functions of same ω can be added by vector method. Actually, their amplitudes are added by vectors method. V

Example 3 In the circuit shown in figure, I2

I1

R2

C

L

R1

I V = 200 sin (100 t + 30°)

R1 = 30 Ω , R2 = 40 Ω , L = 0.4 H and C =

1 mF. 3

Find seven functions of time I , I 1 , I 2 , V R1 , V L , V R2 and VC . Also, find total power consumed in the circuit. In the given potential function, V is in volts and ω in rad/s. Solution Circuit 1 (containing L and R1) I1 :

XL = ωL = 100 × 0.4 = 40 Ω R1 = 30 Ω

∴

Z 1 = R12 + XL2 = (30)2 + (40)2

= 50 Ω V 0 200 Maximum value of current, I1 = = =4 A Z1 50 Since, there is only XL , so voltage function will lead the current function by an angle φ1, where R 30 3 cos φ1 = 1 = = Z 1 50 5 ∴ ∴ or VR1 : VR1

φ1 = 53° I1 = 4 sin (100 t + 30° − 53° ) I1 = 4 sin (100 t − 23° ) function is in phase with I1 function. Maximum value of VR1 = (maximum value of I1 ) (R1)

Ans.

= (4) (30) = 120 volt ∴

VR1 = 120 sin (100 t − 23° )

VL : VL function is 90° ahead of I1 function. Maximum value of VL = (maximum value of I1 ) (XL ) = (4)(40) = 160 volt

Ans.

Chapter 28 ∴ or Power

Alternating Current — 585

VL = 160 sin (100 t − 23° + 90° ) VL = 160 sin (100 t + 67° ) In this circuit, power will be consumed only across R1. This power is given by PR1 = ( rms value of I1 )2 R1

Ans.

2

 4 =   (30)  2 = 240 watt Circuit 2 (containing C and R 2 ) XC =

I2 :

1 1 = = 30 Ω ωC 100 × 1 × 10−3 3 R2 = 40 Ω Z 2 = R22 + XC2

∴

= (40)2 + (30)2 = 50 Ω V 200 Maximum value of I 2 = 0 = =4 A Z 2 50 Since, there is only XC , so I 2 function will lead the V function by an angle φ2, where R 40 4 cos φ2 = 2 = = Z 2 50 5 ∴ ∴ VR2 : VR2

φ2 = 37° I 2 = 4 sin (100 t + 30° + 37° ) = 4 sin (100 t + 67° ) function is in phase with I 2 function. Maximum value of VR2 = ( maximum value of I 2) (R2)

= 4 × 40 = 160 volt ∴ VR2 = 160 sin (100 t + 67° ) VC : VC function lags I 2 function by 90° Maximum value of VC = (Maximum value of I 2)(XC )

Ans.

Ans.

= 4 × 30 ∴ or Power

= 120 volt VC = 120 sin (100 t + 67° − 90° ) Ans. VC = 120 sin (100 t − 23° ) In this circuit, power will be consumed only across R2 and this power is given by PR2 = (rms value of I 2)2 R2 2

 4 =   (40)  2 = 320 W ∴ Total power consumed in the circuit, P = PR1 + PR2 = (240 + 320) W = 560 W

Ans.

586 — Electricity and Magnetism I = I1 + I 2 I = 4 sin (100 t − 23° ) + 4 sin (100 t + 67° ) Now, the amplitudes can be added by vector method. I:

4A 4 √2 A 67°

22° 23°

100 t 4A

Resultant of 4 A and 4 A at 90° is 4 2 A at 45° from both currents or at 22° from 100 t line. Ans. ∴ I = 4 2 sin (100 t + 22° )

Miscellaneous Examples V

Example 4 An AC circuit consists of a 220 Ω resistance and a 0.7 H choke. Find the power absorbed from 220 V and 50 Hz source connected in this circuit if the resistance and choke are joined (a) in series (b) in parallel. Solution (a) In series, the impedance of the circuit is Z = R2 + ω 2L2 = R2 + (2πfL )2 = (220)2 + (2 × 3.14 × 50 × 0.7)2 ∴ and

= 311 Ω V 220 i rms = rms = = 0.707 A Z 311 R 220 cos φ = = = 0.707 Z 311

Z XL φ R

∴ The power absorbed in the circuit, P = V rms i rms cos φ = (220) (0.707) (0.707) Ans. = 110.08 W (b) When the resistance and choke are in parallel, the entire power is absorbed in resistance, as the choke (having zero resistance) absorbs no power. V2 (220)2 Ans. ∴ P = rms = = 220 W R 220 V

Example 5 A sinusoidal voltage of frequency 60 Hz and peak value 150 V is applied to a series L-R circuit, where R = 20 Ω and L = 40 mH . (a) Compute T , ω , X L , Z and φ (b) Compute the amplitudes of current, V R and V L

Chapter 28 T=

Solution (a)

Alternating Current — 587

1 1 = s f 60

Ans.

ω = 2πf = (2π )(60) = 377 rad /s

Ans.

XL = ωL = (377) (0.040) = 15.08 Ω Z =

XL2

Ans.

+R

2

= (15.08)2 + (20)2 = 25.05 Ω

Ans.

X   15.08 –1 φ = tan –1  L  = tan −1   = tan (0.754)  R  20  = 37° (b) Amplitudes (maximum value) are i0 =

Ans.

V0 150 = ≈6 A Z 25.05

Ans.

(V 0 )R = i 0R = (6)(20) = 120 V

Ans.

(V 0 )L = i 0XL = (6) (15.08) = 90.5 V

Ans.

Note V0 = (V0 ) 2R + (V0 ) L2 V

Example 6 For the circuit shown in figure, find the instantaneous current through each element. V = V0 sin ωt

R

C

L

Solution The three current equations are diL dt dV 1 = iC dt C

V = iR R, V = L and

V =

q C

⇒

The steady state solutions of Eq. (i) are V iR = 0 sin ωt ≡ (i 0 )R sin ωt R V V iL = – 0 cos ωt ≡ – 0 cos ωt ≡ – (i 0 )L cos ωt ωL XL and

iC = V 0ωC cos ωt ≡

V0 cos ωt ≡ (i 0 )C cos ωt XC

where, the reactances XL and XC are as defined.

…(i)

588 — Electricity and Magnetism V

Example 7 In the above problem find the total instantaneous current through the source, and find expressions for phase angle of this current and the impedance of the circuit. Solution For the total current, we have i = iR + iL + iC 1   1 1  = V 0  sin ωt +  –  cos ωt   XC XL  R  Using the trigonometric identity, A sin θ + B cos θ =

A 2 + B2 sin (θ + φ)

where, φ = tan –1 (B/A ) We can write,

V0 Z

Here,

i0 =

where,

2  1 1 1   1 =   + –   R  XC XL  Z

and

V

i ≡ i 0 sin (ωt + φ)

2

 1 1  –    XC XL  tan φ = (1 /R)

Example 8 An L-C-R series circuit with 100 Ω resistance is connected to an AC source of 200 V and angular frequency 300 rad / s. When only the capacitance is removed, the current lags behind the voltage by 60° . When only the inductance is removed, the current leads the voltage by 60° . Calculate the current and the power dissipated in the L-C-R circuit Solution When capacitance is removed, then

or ∴

tan φ =

XL R

tan 60° =

XL R

XL = 3 R

…(i)

When inductance is removed, then

or

tan φ =

XC R

tan 60° =

XC R

∴

XC = 3 R

From Eqs. (i) and (ii), we see that

XC = XL

So, the L-C-R circuit is in resonance. Hence,

Z =R

…(ii)

Chapter 28 ∴

i rms =

V rms 200 = =2 A Z 100

Alternating Current — 589 Ans.

P = V rms i rms cos φ At resonance current and voltage are in phase, φ = 0°

or ∴ V

P = (200) (2) (1) = 400 W

Ans.

Example 9 A series L-C-R circuit containing a resistance of 120 Ω has resonance frequency 4 × 105 rad / s. At resonance the voltages across resistance and inductance are 60 V and 40 V , respectively. Find the values of L and C. At what angular frequency the current in the circuit lags the voltage by π /4? Solution At resonance, XL – XC = 0 and ∴ Also, ∴

Z = R = 120 Ω (V ) 60 1 i rms = R rms = = A R 120 2 (V ) i rms = L rms ωL (VL )rms 40 L= = ωi rms  1 (4 × 105 )    2 = 2.0 × 10–4 H = 0.2 mH

Ans.

The resonance frequency is given by ω=

1 LC

or C =

1 ω 2L

Substituting the values, we have C=

1 (4 × 105 )2 (2.0 × 10–4 )

= 3.125 × 10–8 F

Ans.

Current lags the voltage by 45°, when tan 45° =

ωL –

1 ωC

R

Substituting the values of L, C, R and tan 45°, we get ω = 8 × 105 rad /s V

Ans.

Example 10 A choke coil is needed to operate an arc lamp at 160 V (rms) and 50 Hz. The lamp has an effective resistance of 5 Ω when running at 10 A (rms). Calculate the inductance of the choke coil. If the same arc lamp is to be operated on 160 V ( DC ), what additional resistance is required? Compare the power loses in both cases.

590 — Electricity and Magnetism Solution For lamp, (V rms )R = (i rms ) (R) = 10 × 5 = 50 V L

R

Choke

Lamp

VL

VR

V = V0 sin ωt

In series, (V rms )2 = (V rms )R2 + (V rms )L2 ∴

(V rms )L = (V rms )2 – (V rms )R2 = (160)2 – (50)2

As, ∴

= 152 V (V rms )L = (i rms )XL = (i rms ) (2πfL ) (V rms )L L= (2πf ) (i rms )

Substituting the values, we get L=

152 (2π ) (50) (10)

Ans. = 4.84 × 10–2 H Now, when the lamp is operated at 160 V, DC and instead of choke let an additional resistance R′ is put in series with it, then V = i (R + R ′ ) or 160 = 10 (5 + R ′ ) ∴ R′ = 11 Ω In case of AC, as the choke has no resistance, power loss in choke is zero. In case of DC, the loss in additional resistance R′ is P = i 2R ′ = (10)2(11) = 1100 W

Ans.

Ans.

Exercises LEVEL 1 Assertion and Reason Directions : Choose the correct option. (a) If both Assertion and Reason are true and the Reason is correct explanation of the Assertion. (b) If both Assertion and Reason are true but Reason is not the correct explanation of Assertion. (c) If Assertion is true, but the Reason is false. (d) If Assertion is false but the Reason is true. 1. Assertion : In an AC circuit, potential difference across the capacitor may be greater than the applied voltage. Reason : VC = IXC , whereas V = IZ and XC can be greater than Z also.

2. Assertion : In series L-C-R circuit, voltage will lead the current function for frequency greater than the resonance frequency. Reason : At resonance frequency, phase difference between current function and voltage function is zero.

3. Assertion : Resonance frequency will decrease in L-C-R series circuit if a dielectric slab is inserted in between the plates of the capacitor. Reason : By doing so, capacity of capacitor will increase.

4. Assertion : Average value of current in the given graph is 3 A. i (A)

4

2

3

4

5

6

t (s) -

Reason :

Average value can’t be greater than the peak value of any function.

5. Assertion : In series L-C-R circuit, if a ferromagnetic rod is inserted inside an inductor, current in the circuit may increase or decrease. Reason : By doing so X L will increase.

6. Assertion : Potential difference across, resistor, capacitor and inductor each is 10 V. Then, voltage function and current functions should be in phase. Reason : At this condition current in the circuit should be maximum.

7. Assertion : At some given instant I1 and I 2 both are 2 A each. Then, I at this instant should be zero.

592 — Electricity and Magnetism I1

I2

I

Reason :

There is a phase difference of π between I1 and I 2 functions.

8. Assertion : Peak value of current in AC through a resistance of 10 Ω is 2 A. Then, power consumed by the resistance should be 20 W. Reason : Power in AC is P = I 2rms R

9. Assertion : An inductor coil normally produces more current with DC source compared to an AC source of same value of rms voltage. Reason : In DC source, applied voltage remains constant with time.

10. Assertion : In an L-R series circuit in AC, current in the circuit will decrease with increase in frequency. Reason : Phase difference between current function and voltage function will increase with increase in frequency.

11. Assertion : In series L-C-R, AC circuit, current and voltage are in same phase at resonance. Reason : In series L-C-R, AC circuit, resonant frequency does not depend on the value of resistance. Hence, current at resonance does not depend on resistance.

Objective Questions 1. The term cos φ in an AC circuit is called (a) form factor (c) power factor

(b) phase factor (d) quality factor

2. A DC ammeter cannot measure alternating current because (a) (b) (c) (d)

AC changes its direction DC instruments will measure the average value AC can damage the DC instrument AC produces more heat

3. As the frequency of an alternating current increases, the impedance of the circuit (a) increases continuously (c) remains constant

(b) decreases continuously (d) None of these

4. Phasor diagram of a series AC circuit is shown in figure. Then, (a) (b) (c) (d)

The circuit must be containing resistor and capacitor only The circuit must be containing resistor and inductor only The circuit must be containing all three elements L , C and R The circuit cannot have only capacitor and inductor

5. The rms value of an alternating current (a) (b) (c) (d)

is equal to 0.707 times peak value is equal to 0.636 times peak value is equal to 2 times the peak value None of the above

V

30°

I

Chapter 28

Alternating Current — 593

6. In an AC circuit, the applied potential difference and the current flowing are given by π  V = 200 sin 100 t volt, I = 5 sin 100 t −  amp  2 The power consumption is equal to (a) 1000 W (c) 20 W

(b) 40 W (d) zero

7. The impedance of a series L-C-R circuit in an AC circuit is R + (XL − XC )

(a)

(b)

(c) R

R2 + (XL2 − XC2 )

(d) None of these

8. If V 0 and I 0 are the peak current and voltage across the resistor in a series L-C-R circuit, then the power dissipated in the circuit is (Power factor = cos θ )

(a)

V 0I 0 2

(b)

V 0I 0 2

(c) V 0I 0 cos θ

(d)

V 0I 0 cos θ 2

9. A generator produces a time varying voltage given by V = 240 sin 120 t, where t is in second. The rms voltage and frequency are (a) 170 V and 19 Hz (c) 170 V and 60 Hz

(b) 240 V and 60 Hz (d) 120 V and 19 Hz

10. An L-C-R series circuit has a maximum current of 5 A. If L = 0.5 H and C = 8 µF, then the angular frequency of AC voltage is (a) 500 rad / s (c) 400 rad / s

(b) 5000 rad / s (d) 250 rad / s

11. The current and voltage functions in an AC circuit are

π  i = 100 sin 100 t mA , V = 100 sin 100t +  V  3 The power dissipated in the circuit is

(a) 10 W

(b) 2.5 W

(c) 5 W

(d) 5 kW

(c) both (a) and (b)

(d) None of these

12. A capacitor becomes a perfect insulator for (a) alternating current (b) direct current

13. For an alternating voltage V = 10 cos 100 πt volt, the instantaneous voltage at t = (a) 1 V (c) 5 3 V

1 s is 600

(b) 5 V (d) 10 V

14. In a purely resistive AC circuit, (a) voltage leads current (c) voltage and current are in same phase

(b) voltage lags current (d) nothing can be said

15. Identify the graph which correctly represents the variation of capacitive reactance XC with frequency Xc

Xc

(a)

(b)

O

f

O

Xc

Xc

(c)

(d)

f O

f

O

f

594 — Electricity and Magnetism 16. In an AC circuit, the impedance is 3 times the reactance, then the phase angle is (a) 60°

(b) 30°

(c) zero

(d) None of these

17. Voltage applied to an AC circuit and current flowing in it is given by π π   V = 200 2 sin ωt +  and i = − 2 cos ωt +    4 4 Then, power consumed in the circuit will be (a) 200 W (c) 200 2 W

(b) 400 W (d) None of these

18. When 100 volt DC source is applied across a coil, a current of 1 A flows through it. When 100 V AC source of 50 Hz is applied to the same coil, only 0.5 A current flows. Calculate the inductance of the coil. (a) (π / 3 ) H (c) (2 /π ) H

(b) ( 3 /π ) H (d) None of these

19. In the circuit shown in figure, the reading of the AC ammeter is 1 µF A

V = 200 √2 sin 100t

(a) 20 2 mA

(b) 40 2 mA

(c) 20 mA

(d) 40 mA

20. An AC voltage is applied across a series combination of L and R. If the voltage drop across the resistor and inductor are 20 V and 15 V respectively, then applied peak voltage is (b) 35 V (d) 5 7 V

(a) 25 V (c) 25 2 V

21. For wattless power in an AC circuit, the phase angle between the current and voltage is (a) 0° (c) 45°

(b) 90° (d) Not possible

22. The correct variation of resistance R with frequency f is given by R

(a)

(b) f

R

R

R

(c) f

(d) f

f

23. If L and R be the inductance and resistance of the choke coil, then identify the correct statement. (a) L is very high compared to R (c) Both L and R are high

(b) R is very high compared to L (d) Both L and R are low

Chapter 28

Alternating Current — 595

24. When an AC signal of frequency 1 kHz is applied across a coil of resistance 100 Ω, then the applied voltage leads the current by 45°. The inductance of the coil is (a) 16 mH

(b) 12 mH

(c) 8 mH

(d) 4 mH

25. The frequency of an alternating current is 50 Hz. The minimum time taken by it in reaching from zero to peak value is (a) 5 ms

(b) 10 ms

(c) 20 ms

(d) 50 ms

26. An alternating voltage is applied across the R-L combination. V = 220 sin 120 t and the current I = 4 sin (120t − 60° ) develops. The power consumption is

(a) zero (c) 220 W

(b) 100 W (d) 440 W

27. In the AC network shown in figure, the rms current flowing through the

C

inductor and capacitor are 0.6 A and 0.8 A, respectively. Then, the current coming out of the source is

L

(a) (b) (c) (d)

1.0 A 1.4 A 0.2 A None of the above

28. The figure represents the voltage applied across a pure inductor. The diagram which correctly represents the variation of current i with time t is given by V

t

O

i

(a) O

i

i

t

(b) O

t

(c) O

i

t

(d) O

t

29. A steady current of magnitude I and an AC current of peak value I are allowed to pass through identical resistors for the same time. The ratio of heat produced in the two resistors will be (a) 2 : 1 (c) 1 : 1

(b) 1 : 2 (d) None of these

30. A 50 Hz AC source of 20 V is connected across R and C as shown in figure. The voltage across R is 12 V. The voltage across C is (a) (b) (c) (d)

8V 16 V 10 V Not possible to determine unless value of R and C are given

R

C

596 — Electricity and Magnetism Subjective Questions Note You can take approximations in the answers.

1. A 300 Ω resistor, a 0.250 H inductor, and a 8.00 µF capacitor are in series with an AC source with voltage amplitude 120 V and angular frequency 400 rad/ s. (a) What is the current amplitude? (b) What is the phase angle of the source voltage with respect to the current? Does the source voltage lag or lead the current? (c) What are the voltage amplitudes across the resistor, inductor, and capacitor?

2. A series circuit has an impedance of 60.0 Ω and a power factor of 0.720 at 50.0 Hz. The source voltage lags the current. (a) What circuit element, an inductor or a capacitor, should be placed in series with the circuit to raise its power factor? (b) What size element will raise the power factor to unity?

3. Voltage and current for a circuit with two elements in series are expressed as V ( t ) = 170 sin ( 6280t + π / 3) volt i ( t ) = 8.5 sin ( 6280t + π/ 2) amp (a) (b) (c) (d)

Plot the two waveforms. Determine the frequency in Hz. Determine the power factor stating its nature. What are the values of the elements?

4. A 5.00 H inductor with negligible resistance is connected across an AC source. Voltage amplitude is kept constant at 60.0 V but whose frequency can be varied. Find the current amplitude when the angular frequency is (a) 100 rad /s (b) 1000 rad /s (c) 10000 rad /s

5. A 300 Ω resistor is connected in series with a 0.800 H inductor. The voltage across the resistor as a function of time is V R = ( 2.50 V ) cos [( 950 rad/ s ) t ].

(a) Derive an expression for the circuit current. (b) Determine the inductive reactance of the inductor. (c) Derive an expression for the voltage VL across the inductor.

6. An L-C-R series circuit with L = 0.120 H, R = 240 Ω , and C = 7.30 µF carries an rms current of 0.450 A with a frequency of 400 Hz. (a) (b) (c) (d) (e)

What are the phase angle and power factor for this circuit? What is the impedance of the circuit? What is the rms voltage of the source? What average power is delivered by the source? What is the average rate at which electrical energy is converted to thermal energy in the resistor? (f) What is the average rate at which electrical energy is dissipated ( converted to other forms) in the capacitor? (g) In the inductor?

LEVEL 2 Single Correct Option 1. A capacitor and resistor are connected with an AC source as shown in figure. Reactance of

capacitor is XC = 3 Ω and resistance of resistor is 4 Ω. Phase difference between current    3 I and I1 is tan−1   = 37°  4   XC = 3Ω

I2

R = 4Ω

I1 I

V = V0 sin ωt

(a) 90°

(b) zero

(c) 53°

(d) 37°

2. A circuit contains resistance R and an inductance L in series. An alternating voltage V = V 0 sin ωt is applied across it. The currents in R and L respectively will be R

L

AC

(a) IR = I 0 cos ωt , IL = I 0 cos ωt (c) IR = I 0 sin ωt , IL = − I 0 cos ωt

(b) IR = − I 0 sin ωt , IL = I 0 cos ωt (d) None of the above

3. In the circuit shown in figure, the AC source gives a voltage V = 20 cos ( 2000t ). Neglecting source resistance, the voltmeter and ammeter readings will be 6Ω A 5 mH, 4Ω

50 µF I I1

V

(a) 0 V, 2.0 A

(b) 0 V, 1.4 A

(c) 5.6 V, 1.4 A

(d) 8 V, 2.0 A

4. A signal generator supplies a sine wave of 200 V, 5 kHz to the circuit shown in the figure. Then, choose the wrong statement. 1 µF π 100 Ω 200 V, 5 kHz

(a) (b) (c) (d)

The current in the resistive branch is 0.2 A The current in the capacitive branch is 0.126 A Total line current is ≈ 0.283 A Current in both the branches is same

598 — Electricity and Magnetism 5. A complex current wave is given by i = ( 5 + 5 sin 100 ωt ) A. Its average value over one time period is given as (a) 10 A (c) 50 A

(b) 5 A (d) 0

6. An AC voltage V = V 0 sin 100 t is applied to the circuit, the phase difference between current and voltage is found to be π/ 4, then V, l

V

I

t

π/4

(a) R = 100 Ω, C = 1 µF (c) R = 10 kΩ, L = 1 H

(b) R = 1 kΩ, C = 10 µF (d) R = 1 kΩ, L = 10 H

7. In series L-C-R circuit, voltage drop across resistance is 8 V, across inductor is 6 V and across capacitor is 12 V. Then, (a) (b) (c) (d)

voltage of the source will be leading in the circuit voltage drop across each element will be less than the applied voltage power factor of the circuit will be 3 /4 None of the above

8. Consider an L-C-R circuit as shown in figure with an AC source of peak value V 0

and angular frequency ω. Then, the peak value of current through the AC source is

(a)

V0

(c)

1 1   +  ωL −  2  ωC  R V0 1   R +  ωL −   ωC 

1  1  (b) V 0  2 + ωC −   ω L   R

2

(d) None of these

2

2

L C

2

R

V, ω

9. The adjoining figure shows an AC circuit with resistance R, inductance L and source voltage V s . Then, R

L

V 70 V

V 20 V Vs

(a) (b) (c) (d)

the source voltage V s = 72.8 V the phase angle between current and source voltage is tan −1 (7 /2) Both (a) and (b) are correct Both (a) and (b) are wrong

Chapter 28

Alternating Current — 599

10. When an alternating voltage of 220 V is applied across a device P, a current of 0.25 A flows through the circuit and it leads the applied voltage by an angle π/ 2 radian. When the same voltage source is connected across another device Q, the same current is observed in the circuit but in phase with the applied voltage. What is the current when the same source is connected across a series combination of P and Q? 1 A lagging in phase by π /4 with voltage 4 2 1 (c) A leading in phase by π/4 with voltage 2

1 A leading in phase by π /4 with voltage 4 2 1 (d) A leading in phase by π/2 with voltage 4 2

(b)

(a)

11. In a parallel L-C-R circuit as shown in figure if I R , I L , IC and I represent the rms values of current flowing through resistor, inductor, capacitor and the source, then choose the appropriate correct answer. IR

R

IL

L

IC

C

I

(a) I = IR + IL + IC

(b) I = IR + IL − IC

(c) IL or IC may be greater than I

(d) None of these

12. In a series L-C-R circuit, current in the circuit is 11 A when the applied voltage is 220 V. Voltage across the capacitor is 200 V. If the value of resistor is 20 Ω, then the voltage across the unknown inductor is (a) zero

(b) 200 V

(c) 20 V

(d) None of these

13. In the circuit shown in figure, the power consumed is R

L

V = V0 sin ωt

(a) zero

(b)

V 02 2R

(c)

V 02R 2(R2 + ω 2L2)

(d) None of these

14. In a series L - C circuit, the applied voltage is V 0. If ω is very low, then the voltage drop across the inductor V L and capacitor VC are (a) VL =

V0 V ; VC = 0 2 2

(c) VL = V 0 ; VC = 0

(b) VL = 0 ; VC = V 0 (d) VL = − VC =

V0 2

L

C

600 — Electricity and Magnetism 15. A coil, a capacitor and an AC source of rms voltage 24 V are connected in series. By varying the frequency of the source, a maximum rms current of 6 A is observed. If coil is connected to a DC battery of emf 12 volt and internal resistance 4 Ω, then current through it in steady state is (b) 1.8 A (d) 1.2 A

(a) 2.4 A (c) 1.5 A

16. In a series C - R circuit shown in figure, the applied voltage is 10 V and the voltage across capacitor is found to be 8 V. The voltage across R, and the phase difference between current and the applied voltage will respectively be 8V

VR

C

R

10 V

 4 (a) 6 V, tan −1    3 −1  3  (c) 6 V, tan    4

 3 (b) 3 V, tan −1    4 (d) None of these

17. An AC voltage source described by V = 10 cos ( π/ 2) t is connected to a 1 µF capacitor as shown in figure. The key K is closed at t = 0. The time ( t > 0) after which the magnitude of current I reaches its maximum value for the first time is

V = 10 cos

π t 2

C = 1 µF K

(b) 2 s (d) 4 s

(a) 1 s (c) 3 s

18. An AC voltage source V = V 0 sin ωt is connected across resistance R and capacitance C as shown in figure. It is given that R = 1/ωC. The peak current is I 0. If the angular frequency of the voltage source is changed to ω/ 3, then the new peak current in the circuit is R

V0 sin ωt

I0 2 I0 (c) 3 (a)

C

I0 2 I0 (d) 3 (b)

Chapter 28

Alternating Current — 601

More than One Correct Options 1. In a R-L-C series circuit shown, the readings of voltmeters V1 and V 2 are 100 V and 120 V. Choose the correct statement(s). V2

V1 V = 130 V

(a) Voltage across resistor, inductor and capacitor are 50 V, 86.6 V and 206.6 V respectively (b) Voltage across resistor, inductor and capacitor are 10 V, 90 V and 30 V respectively 5 (c) Power factor of the circuit is 13 (d) Circuit is capacitive in nature

2. Current in an AC circuit is given by i = 3 sin ωt + 4 cos ωt , then (a) rms value of current is 5 A 6 π (c) if voltage applied is V = Vm sin ωt, then the circuit may contain resistance and capacitance (b) mean value of this current in positive one-half period will be

(d) if voltage applied is V = Vm cos ωt, then the circuit may contain resistance and inductance only

3. A tube light of 60 V, 60 W rating is connected across an AC source of 100 V and 50 Hz frequency. Then, 2 H may be connected in series 5π 250 (b) a capacitor of µF may be connected in series to it π 4 (c) an inductor of H may be connected in series 5π (d) a resistance of 40 Ω may be connected in series (a) an inductance of

4. In an AC circuit, the power factor (a) (b) (c) (d)

is unity when the circuit contains an ideal resistance only is unity when the circuit contains an ideal inductance only is zero when the circuit contains an ideal resistance only is zero when the circuit contains an ideal inductance only

5. In an AC series circuit, R = 10 Ω, X L = 20 Ω and XC = 10 Ω. Then, choose the correct options (a) Voltage function will lead the current function (b) Total impedance of the circuit is 10 2 Ω (c) Phase angle between voltage function and current function is 45° 1 (d) Power factor of circuit is 2

602 — Electricity and Magnetism 6. In the above problem further choose the correct options. (a) (b) (c) (d)

The given values are at frequency less than the resonance frequency The given values are at frequency more than the resonance frequency If frequency is increased from the given value, impedance of the circuit will increase If frequency is decreased from the given value, current in the circuit may increase or decrease

7. In the circuit shown in figure, 40 Ω

2A

XL = 20 Ω 100 V

V = V0 sin ωt

(a) VR = 80 V (c) VL = 40 V

(b) XC = 50 Ω (d) V 0 = 100 V

8. In L-C-R series AC circuit, (a) If R is increased, then current will decrease (b) If L is increased, then current will decrease (c) If C is increased, then current will increase (d) If C is increased, then current will decrease

Comprehension Based Questions Passage I (Q. No. 1 to 3) A student in a lab took a coil and connected it to a 12 V DC source. He measures the steady state current in the circuit to be 4 A. He then replaced the 12 V DC source by a 12 V, (ω = 50 rad / s ) AC source and observes that the reading in the AC ammeter is 2.4 A. He then decides to connect a 2500 µF capacitor in series with the coil and calculate the average power developed in the circuit. Further he also decides to study the variation in current in the circuit (with the capacitor and the battery in series). Based on the readings taken by the student, answer the following questions.

1. The value of resistance of the coil calculated by the student is (a) 3 Ω (c) 5 Ω

(b) 4 Ω (d) 8 Ω

2. The power developed in the circuit when the capacitor of 2500 µF is connected in series with the coil is (a) 28.8 W (c) 17.28 W

(b) 23.04 W (d) 9.6 W

3. Which of the following graph roughly matches the variations of current in the circuit (with the coil and capacitor connected in the series) when the angular frequency is decreased from 50 rad/s to 25 rad/s? i

(a)

(b)

25

50

ω

i

i

i

(c)

25

50

ω

(d)

25

50

ω

25

50

ω

Chapter 28

Alternating Current — 603

Passage II (Q. No. 4 to 6) It is known to all of you that the impedance of a circuit is dependent on the frequency of source. In order to study the effect of frequency on the impedance, a student in a lab took 2 impedance boxes P and Q and connected them in series with an AC source of variable frequency. The emf of the source is constant at 10 V. Box P contains a capacitance of 1 µF in series with a resistance of 32 Ω. And the box Q has a coil of self-inductance 4.9 mH and a resistance of 68 Ω in series. He adjusted the frequency so that the maximum current flows in P and Q. Based on his experimental set up and the reading by him at various moment, answer the following questions.

4. The angular frequency for which he detects maximum current in the circuit is (a) 105 /7 rad /s (c) 105 rad /s

(b) 104 rad /s (d) 104 /7 rad /s

5. Impedance of box P at the above frequency is (a) 70 Ω (c) 90 Ω

(b) 77 Ω (d) 100 Ω

6. Power factor of the circuit at maximum current is (b) 1 (d) 1/ 2

(a) 1/2 (c) 0

Match the Columns 1. Match the following two columns for a series AC circuit. Column I

Column II

(a) Only C in the circuit

(p) current will lead

(b) Only L in the circuit (c) Only R in the circuit (d) R and C in the circuit

(q) voltage will lead (r) φ = 90° (s) φ = 0°

2. Applied AC voltage is given as V = V 0 sin ωt Corresponding to this voltage, match the following two columns. Column I (a) (b) (c) (d)

I I I I

= I 0 sin ωt = − I 0 cos ωt = I 0 sin (ωt + π /6) = I 0 sin (ωt − π /6)

Column II (p) (q) (r) (s)

only R circuit only L circuit may be C-R circuit may be L-C-R circuit

3. For an L-C-R series AC circuit, match the following two columns. Column I (a) (b) (c) (d)

If resistance is increased If capacitance is increased If inductance is increased If frequency is increased

Column II (p) (q) (r) (s)

current will increase current will decrease current may increase or decrease power may decrease or increase

604 — Electricity and Magnetism 4. In the circuit shown in figure, match the following two columns. In Column II, quantities are given in SI units. XC = 30 Ω, XL = 15 Ω

2A

40 V

Column I

Column II

(a) Value of resistance R

(p) 60

(b) Potential difference across capacitor

(q) 20

(c) Potential difference across inductor

(r) 30

(d) Applied potential difference

(s) None of the above

5. Corresponding to the figure shown, match the two columns. Column I

1 2

Column II

(a) Resistance

(p) 4

3

(b) Capacitive reactance

(q) 1

4

(c) Inductive reactance

(r) 2

(d) Impedance

(s) 3

ω

Subjective Questions Note Power factor leading means current is leading.

1. A coil is in series with a 20 µF capacitor across a 230 V, 50 Hz supply. The current taken by the circuit is 8 A and the power consumed is 200 W. Calculate the inductance of the coil if the current in the circuit is (a) leading

(b) lagging

2. The current in a certain circuit varies with time as shown in figure. Find the average current and the rms current in terms of I 0. I0 O

τ

2τ

t

–I0 .

3. Two impedances Z1 and Z 2 when connected separately across a 230 V , 50 Hz supply consume 100 W and 60 W at power factor of 0.5 lagging and 0.6 leading respectively. If these impedances are now connected in series across the same supply, find (a) total power absorbed and overall power factor (b) the value of reactance to be added in series so as to raise the overall power factor to unity.

Chapter 28

Alternating Current — 605

4. In the figure shown, the reading of voltmeters are V1 = 40 V , V 2 = 40 V and V3 = 10 V. Find V1

V2

V3

R=4Ω

L

C

E = E0 sin 100 πt +

(a) the peak value of current (c) the value of L and C

π 6

(b) the peak value of emf

5. In the circuit shown in figure power factor of box is 0.5 and power factor of circuit is 3/ 2 . Current leading the voltage. Find the effective resistance of the box. Box 10 Ω

6. A circuit element shown in the figure as a box is having either a capacitor or an inductor. The power factor of the circuit is 0.8, while current lags behind the voltage. Find C Box 1A

R = 80 Ω

VC = 100 V

V, 50 Hz

(a) the source voltage V, (b) the nature of the element in box and find its value.

7. The maximum values of the alternating voltages and current are 400 V and 20 A respectively in a circuit connected to 50 Hz supply and these quantities are sinusoidal. The instantaneous values of the voltage and current are 200 2 V and 10 A, respectively. At t = 0, both are increasing positively. (a) Write down the expression for voltage and current at time t. (b) Determine the power consumed in the circuit.

8. An L-C circuit consists of an inductor coil with L = 5.00 mH and a 20.0 µF capacitor. There is negligible resistance in the circuit. The circuit is driven by a voltage source with V = V 0 cosωt. If V 0 = 5.00 mV and the frequency is twice the resonance frequency, determine (a) the maximum charge on the capacitor (b) the maximum current in the circuit (c) the phase relationship between the voltages across the inductor, the capacitor and the source.

606 — Electricity and Magnetism 9. A coil having a resistance of 5 Ω and an inductance of 0.02 H is arranged in parallel with

another coil having a resistance of 1 Ω and an inductance of 0.08 H. Calculate the power absorbed when a voltage of 100 V at 50 Hz is applied. 5Ω

0.02 H

1Ω

0.08 H

i 100V 50 Hz

10. A circuit takes a current of 3 A at a power factor of 0.6 lagging when connected to a 115 V – 50 Hz supply. Another circuit takes a current of 5 A at a power factor of 0.707 leading when connected to the same supply. If the two circuits are connected in series across a 230 V, 50 Hz supply, then calculate (a) the current

(b) the power consumed and

(c) the power factor

Answers Introductory Exercise 28.1 1. (a) 628 Ω

(b) 6.37 mH (c) 1.59 kΩ

(d) 1.59 mF

2. 0.036 H, 111.8 V

3. 7.7 H, 6 A

Introductory Exercise 28.2 1. 650 Hz, 0

2. 0.2

Exercises LEVEL 1 Assertion and Reason 1. (a)

2. (b)

3. (a)

4. (b)

5. (a or b)

6. (b)

7. (a)

8. (a,b)

9. (b)

10. (b)

11. (c)

Objective Questions 1. (c)

2. (b)

3. (d)

4. (d)

5. (a)

6. (d)

7. (d)

8. (d)

9. (a)

10. (a)

11. (b)

12. (b)

13. (c)

14. (c)

15. (b)

16. (d)

17. (d)

18. (b)

19. (c)

20. (c)

21. (b)

22. (a)

23. (a)

24. (a)

25. (a)

26. (c)

27.

(c)

28.

(c)

29.

(a)

30.

(b)

Chapter 28

Alternating Current — 607

Subjective Questions 1. (a) 0.326 A (b) 35.3°, lagging (c) 97.8 V, 32.6 V, 102 V 2. (a) Inductor (b) 0.133 H 3 , leading (d) R = 17.32 Ω ,C = 15.92 µF 2 4. (a) 0.12 A (b) 1.2 × 10 –2 A (c) 1.2 × 10 –3 A

3. (b) 1000 Hz (c)

5. (a) (8.33 mA)cos (950 rad/s ) t

(b) 760 Ω

(c) – (6.33 V) sin (950 rad/s )t

6. (a) 45.8° , voltage leads the curren,t 0.697 (b) 343 Ω

(c) 155 V (d) 48.6 W (e) 48.6 W (f) 0 (g) 0

LEVEL 2 Single Correct Option 1.(c)

2.(d)

3.(c)

4.(b)

5.(b)

6.(b)

7.(d)

8.(b)

11.(c)

12.(b)

13.(c)

14.(b)

15.(c)

16.(a)

17.(a)

18.(b)

5.(a,b,c,d)

6.(b,c,d)

7.(a,b,c) 8.(a)

5.(b)

6.(b)

More than One Correct Options 1.(a,c,d)

2.(c,d)

3.(c,d) 4.(a,d)

Comprehension Based Questions 1.(a)

2.(c)

3.(b)

4.(a)

Match the Columns 1. (a) → p,r

(b) → q,r

(c) → s

(d) → p

2. (a) → p,s

(b) → q

(c) → r,s

(d) → s

3. (a) → q,s

(b) → r,s

(c) → r,s

(d) → r,s

4. (a) → q

(b) → p

(c) → r

(d) → s

5. (a) → s

(b) → p

(c) → r

(d) → q

Subjective Questions 1.

(a) 0.416 H (b) 0.597 H

2. zero,

I0 3

3. (a) 99 W, 0.92 leading (b) 194.2 Ω

1 1 H, (d) F 25 π 100 π 1.6 6. (a) 100 V (b) inductor, L = H π 4. (a) 10 2A (b) 50 2V (c)

5. 5 Ω

7. (a) V = 400 sin (100 πt + π /4), i = 20 sin (100 πt + π /6) (b) P = 3864 W 8. (a) 33.4 nC (b) 0.211 mA (c) Source and inductor voltages in phase. Capacitor voltage lags by 180°. 9. 797 W

10. (a) 5.5 A

(b) 1.188 kW (c) 0.939 lag

9.(a)

10.(b)

23. Current Electricity INTRODUCTORY EXERCISE

23.1

or

vd =

q ne = t t

1. Q i =

= 0.735 × 10−6 m/s

n=

∴

1 8.5 × 1028 × 16 . × 10−19 × 10−4

= 0.735 µ m/s

it (0.7) (1) = e 1.6 × 10− 19

t=

= 4.375 × 10

18

2. Q q = it = (3.6) (3 × 60 × 60) = 38880 C 3. (a) Q q = it = (7.5) (45) = 337.5 C q 337.5 (b) n = = e 1.6 × 10− 19

1. R = ρ

= 6.6 × 1015 Hz

= 1.06 mA

5. ∆q = ∫ idt = ∫ (10 + 4 t ) dt = 300 C

6. Current due to both is from left to right. So, the two currents are additive.

INTRODUCTORY EXERCISE

23.2

1. False. Only under electrostatic conditions (when i = 0 ) all points of a conductor are at same potential.

INTRODUCTORY EXERCISE

10 × 103 yr 0.735 × 10−6 × 60 × 60 × 24 × 365

23.4

−8

(1.72 × 10 ) (35) l = 0.18 Ω = A (π / 4 ) (2.05 × 10− 3 )2

1 ⇒ ρσ = constant σ ρl 3. R = A ρl ∴ A= R m = (Vd ), where V = volume and d = density ∴ m = ( Ald )

= 1.06 × 10− 3 A 10

=

2. ρ =

I = qf = (1.6 × 10 − 19 ) (6.6 × 1015 )

0

10 × 103 s 0.735 × 10−6

INTRODUCTORY EXERCISE

v 2.2 × 106 4. f = = 2πr (2π ) (5.3 × 10− 11 )

0

=

= 431.4 yr

= 2.1 × 1021

10

l vd

23.3

1. i = neAvd i.e. vd ∝ i When current has increased from i = 1.2 A to i = 6.0 A, i.e five times, then drift velocity will also increase to five times. 2. From i = neAvd i We have vd = neA

=

ρl 2 d R

(1.72 × 10− 8 ) (3.5)2 (8.9 × 103 ) 0.125 = 15 × 10− 3 kg =

= 15 g ρ (L) ρ L ρ 4. R = = = A tL t i.e. R is independent of L. Hence, the correct option is (c).

INTRODUCTORY EXERCISE

23.5

1. Copper is metal and germanium is semiconductor. Resistance of a metal decreases and that of a semiconductor increases with decrease in temperature. ∴ Correct option is (d).

Chapter 23 2. 4.1 [1 + 4.0 × 10− 3 (θ − 20)]

INTRODUCTORY EXERCISE

= 3.9 [1 + 5.0 × 10− 3 (θ − 20)] Solving we get,

INTRODUCTORY EXERCISE

23.6

1. PD across each resistance is 10 V. 10 = 5A 2 10 = = 2.5A 4

i2Ω = i4Ω

2. VA = 0 V ∴ ∴ ∴ i1 Ω i2 Ω

23.7

1. Applying loop law equation in upper loop, we have

θ ≈ 85° C

∴

Current Electricity — 611

(as it is earthed)

VC − VA = 5 V VC = 5 V VB − VA = 2 V VB = 2 V VD − VC = 10 V VD = 10 + VC = 15 V VC − VB = = 3A from C to B as VC > VB 1 V − VA = D = 7.5A from D to A as VD > VA 2

…(i) E + 12 − ir − 1 = 0 Applying loop law equation in lower loop, we have where i = 1 + 2 = 3A …(ii) E + 6 −1= 0 Solving these two equations, we get E = − 5 V and r = 2Ω 2. Power delivered by a battery = Ei = 12 × 3 = 36 W Power dissipated in resistance = i 2R = (3) (2)2 = 12 W

INTRODUCTORY EXERCISE

23.8

1. (a) Equivalent emf (V) of the battery PD across the terminals of the battery is equal to its emf when current drawn from the battery is zero. In the given circuit,

3. VA = VB ∴

VAB = 0

or

E − ir = 0  15 + E  E−  (2) = 0  8 

∴

Solving this equation, we get E = 5V 4. Net emf = (n − 2m) E = (10 − 2 × 2) (1) =6V Net emf i= Net resistance =

6 = 0.5 A 10 + 2

5. VR 1 = 0 ∴ ∴

iR 1 = 0

VR 2 = VR 3 = 10 V 10 iR 2 = iR 3 = =1A 10

r2

i=0

V2

i=0

i B

A r1

V1

Current in the internal circuit, Net emf V + V2 i= = 1 Total resistance r1 + r2 Therefore, potential difference between A and B would be VA − VB = V1 − ir1  V + V2  V1r2 − V2r1 ∴ VA − VB = V1 −  1  r1 = r1 + r2  r1 + r2  So, the equivalent emf of the battery is V r − V2 r1 V = 12 r1 + r2 Note that if V1 r2 = V2 r1 : V = 0

612 — Electricity and Magnetism If V1r2 > V2r1 : VA − VB = Positive i.e. A side of the equivalent battery will become the positive terminal and vice-versa. (b) Internal resistance (r) of the battery r1 and r2 are in parallel. Therefore, the internal resistance r will be given by 1/ r = 1/ r1 + 1/ r2 rr or r= 12 r1 + r2

2.

0.5 Ω

4V

E=

E

ig =

Now,

= 7.5V

r = 0.5 Ω

INTRODUCTORY EXERCISE

23.9

∴

V − G = series resistance connected with ig

galvanometer  5  =  − 1 = 999 Ω  5 × 10− 3  S

2.

i – ig ig

G

ig

S = i − ig G ∴

 ig  G S =  i − ig    (50 × 10− 6 ) (100) ≈ 1.0 Ω = −3 −6   (5 × 10 − 50 × 10 ) 

V (G + R ) G

R = (n − 1) G

INTRODUCTORY EXERCISE

23.10

l  r = R  1 − 1  l2   0.52  = 5 − 1  0.4 

= 1.5 Ω E 2. (a) VAJ = or emf of lower battery 2 E i RAJ = ∴ 2  E   15r  E or  (l ) =    2  15r + r  600 Solving this equation, we get l = 320 cm  15r  (b) Resistance of 560 cm =   (560)  600 = 14r Now the circuit is as under,

1. V = ig (G + R ) ∴ R=

V G

nV = ig (G + R ) =

Σ (E / r) (6/1) − (2/1) = = 2V Σ (1/ r) (1/1) + (1/1)

1 1 1 1 = + + r 1 2 2 ∴

∴

1.

r

Now, net emf of E and 4 V is 2V as they are oppositely connected. Σ (E / r) Eeq = Σ (1/ r) (10 / 1) + (4 / 2) + (6/ 2) = (1/1) + (1/ 2) + (1/ 2)

3.

3. V = ig G

r E i1 r

14r i1–i2

i2 E/2

r

G

Applying loop law in upper loop we have, E − 14 r (i1 − i2 ) − i1r − i1r = 0 Applying loop law in lower law loop we have, E − − i2r + (i1 − i2 ) (14 r) = 0 2 Solving these two equations 3E we get, i2 = 22r

…(i)

…(ii)

Chapter 23 INTRODUCTORY EXERCISE

10 × 53 = 10.6 Ω 50 ∴ Correct option is (b). 3. Slide wire bridge is most sensitive when the resistance of all the four arms of bridge is same. Hence, B is the most accurate answer. ∴

23.11

1. R > 2 Ω ⇒ 100 − x > x 2Ω

Current Electricity — 613

R

X =

G

INTRODUCTORY EXERCISE 100 – x

x

1.

G x + 20

Applying We have

80 – x

P R = Q S 2 x = R 100 − x R x + 20 = 2 80 − x

⇒

 Q  1 X =  R=  R  P  10

R lies between 142 Ω and 143 Ω. Therefore, the unknown resistance X lies between 14.2 Ω and 14.3 Ω. 2. Experiment can be done in similar manner but now K 2 should be pressed first then K 1. 3. BC, CD and BA are known resistances. The unknown resistance is connected between A and D.

2Ω

R

P R = Q X

23.12

INTRODUCTORY EXERCISE …(i) …(ii)

Solving Eqs. (i) and (ii), we get R = 3 Ω ∴ Correct option is (a). 2. Using the concept of balanced, Wheatstone bridge, we have, P R X 10 = ⇒ = Q S (52 + 1) (48 + 2)

23.13

1. Yellow → 4 Red → 2 Orange → 103 Gold → 5 ∴ R = (4.2 × 103 ± 5%) Ω

2. 2 → Red 4 → Yellow 106 → Blue 5% → Gold

Exercises LEVEL 1 Assertion and Reason 1. If PD between two terminals of a resistance is zero, then current through resistance is zero, this is confirmed. But PD between any two points of a circuit is zero, this does not mean current is zero. 2. In parallel, V = constant ∴ From the equation V2 1 P= ⇒ P∝ R R

ρl ρ 1 R or = resistance per unit length = ∝ A A A l Near A, area of cross-section is less. Therefore, resistance per unit length will be more. Hence from the equation, H = i 2Rt , heat generation near A will be more. i 1 Current density, J = (as i is same) or J ∝ A A 4. Since net resistance decreases, therefore main current increases. Hence, net potential difference across voltmeter also increases.

3. R =

614 — Electricity and Magnetism 5. Even if ammeter is non-ideal, its resistance should be small and net parallel resistance is less than the smallest individual resistance. ∴ Rnet < resistance of ammeter in the changed situation. Hence, net resistance of the circuit will decrease. So, main current will increase. But maximum percentage of main current will pass through ammeter (in parallel combination) as its resistance is less. Hence, reading of ammeter will increase. Initial voltmeter reading = emf of battery Final voltmeter reading = emf of battery − potential drop across shown resistance. Hence, voltmeter reading will decrease. 6. If current flows from a to b, then equation will become Va − ir − E = Vb or Va − Vb = E + ir So, Va − Vb is always positive. Hence, Va is always greater than Vb . 7. Current in the circuit will be maximum when R = 0. 8. Resistance will increase with temperature on heating. Hence current will decrease. 1 V2 or P ∝ R R Resistance is increasing. Hence, power consumed across R should decrease. V = IR is just an equation between P D across a resistance current passing through it and its resistance. This is not Ohm’s law. 9. Electrons get accelerated by the electric. Then, suddenly collision takes place. Then, again accelerated and so on. E / r + E2 / r2 11. Eeq = 1 1 (1/ r1 ) + (1/ r2 ) P=

Further

 (1/ r ) + (E2 / E1 ) (1/ r2 )  = E1  1  (1/ r1 ) + (1/ r2 )   So, Eeq may be greater than E1 also, if E2 / E1 > 1 r1 and r2 are in parallel. Hence, req is less than both r1 and r2 individually.

Objective Questions 3. H = I 2 Rt ∴

H   I 2t   ML2T − 2  2 −3 −2 = 2  = [ ML T I ]  IT 

[R ] =

l σA l m = = ohm − 1 -m − 1 ∴ σ= RA ohm - m 2 E 6. 0.5 = r + 3.75 E 0.4 = r + 4.75 R=

4.

…(i) …(ii)

Solving these two equations, we get E = 2V 7. In parallel current distributes in increase ratio of resistance IG S ∴ = IS G I  G =  S  (S )  IG 

∴

 50 − 20 =  (12)  20  = 18 Ω I S 8. G = IS G I  S =  G G  IS 

∴

G  2 = G=  98 49

9. P =

V2 R

or P ∝

1 R

P2 R1 l1 = = P1 R2 l2 ∴

l   l  P2 =  1  P1 =   P = 1.11 P1  0.9 l  1  l2 

So, power will increase by 11%. 10. By symmetry, VA = VB VAB = 0  l1  75 11. r = R  − 1 = 10  − 1  60   l2  or

= 2.5 Ω 12. Let V0 = V I AO + I BO = I OC 6−V 3−V V − 2 ∴ + = 6 3 2 Solving this equation, we get V = 3V Now,

(as R ∝ l )

Current Electricity — 615

Chapter 23 13. vd =

P + 15 2 …(ii) = Q 3 Solving these two equations, we get P = 9Ω 23. Total potential of 10V equally distributes between 50 Ω and other parallel combination of 100 Ω and voltmeter. Hence, their net resistance should be same. Or 100 × R = 50 100 + R

i i i ⇒ vd ∝ 2 = neA ne (πr2 ) r

or

14. For making voltmeter of higher range, more resistance is required.

15. V20Ω = VTotal ∴ ∴ ∴

(20) (0.3) = (RTotal ) (0.8) 30 RTotal = Ω 4 4 1 1 1 = + + 30 R1 20 15

Solving we get R1 = 60 Ω 16. Net resistance will decrease by increasing the parallel resistors. Therefore, main current i will increase, further, (PD)Voltmeter = VTotal − (PD)Ammeter (resistance of ammeter) = VTotal − i Since, i has increased. Hence, PD across voltmeter will decrease. V2 1 17. P = …(i) or P ∝ R R Resistor is cut in n equal parts. Therefore, each R resistance will become . Now, these are n connected in parallel. Therefore, net resistance will 1 R R become times . or 2 . n n n Now, from Eq. (i), power will become n2 times. 18. If A is fused, then complete circuit is broken.

19. E − ir = 0  3 + 15  ∴ 3−  (1) = 0  1 + 2 + R Solving this equation, we get R = 3Ω  (2500 R )  20. 100 = (5)    2500 + R  Solving this equation, we get R = 20Ω 21. Five parallel combination, each of value R R R + = 10 10 5 P 20 22. or Q = 4 P = Q 100 − 20 ∴ Now,

P E2 Therefore, net current is anti-clockwise or from B to A. (b) Current through E1 is normal. Hence, it is doing the positive work. (c) Current flows from B to A ∴ VB > VA

(anti-clockwise)

(0.49) (π / 4 ) (0.84 × 10− 3 )2 (2.75 × 10− 8 ) (1)

= 9.9 A (b) PD between two points, 12 m apart = ( 0.49 V/m) (12 m ) = 5.88 V V 5.88 (c) R = = = 0.6 Ω i 9.9 12. Radius at distance x from end P,  b − a r=a+   x  l  Q a

P

b

x dx l

Resistance of element of thickness dx is

618 — Electricity and Magnetism dR =

R=∫

∴

13. i =

ρ (dx ) πr 2

E R+r

X=l

X=0

(Using R =

ρl ) A

= 0.569 mm

dR

2

…(i)

For power to be maximum, dP =0 dR dP By putting = 0 we get, R = r dR Further, by putting R = r in Eq. (i) E2 We get, Pmax = 4r 14. As derived in the above question, E2 Pmax = 4r Here, E = net emf= 2 + 2 = 4 V and r = net internal resistance = 1 + 1 = 2Ω (4 )2 Pmax = = 2W ∴ (4 ) (2)

15. In series,

=

R01 α 1 + R02 α 2 R01 + R 02 (600) (0.001) + (300) (0.004 ) 600 + 300

= 0.002 per ° C Now, Rt = R0 [1 + α ∆θ ] = (600 + 300) [1 + 0.002 × 30 ] = 954 Ω 16. In parallel current distributes in inverse ratio of resistance 1→ Aluminium 2 → Copper R1 i2 = R2 i1 ρ 1l1 / A1 2 = ρ 2 l2 / A2 3 ∴ ∴

ρ 1l1d22 2 = ρ 2l2d12 3  2ρ 2l2  d2 =   d1  3ρ 1l1 

V 0.938 = = 1.25 V/m l 0.75 (b) E = Jρ E 1.25 ∴ ρ= = J 4.4 × 107

17. (a) E =

⇒ P = power across R = i 2R

 E  P=  R  R + r

α eq =

 2 × 0.017 × 6  =  (1 mm )  3 × 0.028 × 7.5 

= 2.84 × 10− 8 Ω-m i V V 18. (a) J = = = A RA  ρl    A  A J= or

V ρl J∝

…(i) 1 l

lmin = d. So, J is maximum. Hence, potential difference should be applied across the face (2d × 3d ) From Eq. (i), V J max = ρd V V VA (b) i = = = R (ρl / A ) ρl A or i∝ l Across face (2d × 3d ), area of cross-section is maximum and l is minimum. Hence, current is maximum. V (2d × 3d ) 6Vd imax = = ρ (d ) ρ

19. (a) ρ =

RA (0.104 ) (π / 4 ) (2.5 × 10− 3 )2 = l 14 = 3.65 × 10− 8 Ω-m

(b) V = El = 1.28 × 14 = 17.92 V V 17.92 i= = = 172.3 A ∴ R 0.104 i (c) vd = neA 172.3 =  π 28 (8.5 × 10 ) (1.6 × 10− 19 )   (2.5 × 10− 3 )2  4 = 2.58 × 10− 3 m/s

Current Electricity — 619

Chapter 23 20.

R1 + R2 = 20 R α + R2α 2 α eq = 1 1 R1 + R2

…(i) (in series)

R1 (− 0.5 × 10− 3 ) + R2 (5.0 × 10− 3 ) 20 …(ii) ∴ R1 = 10R2 Solving Eqs. (i) and (ii), we get 20 Ω R2 = RFe = 11 = 1.82Ω R1 = RCu = 10R2 = 18.18 Ω 8Ω and 12Ω resistors are in parallel. 8 × 12 ∴ Rnet = = 4.8 Ω 8 + 12 24 ∴ i= 4.8 = 5A All four resistors are in parallel 1 1 1 1 1 ∴ = + + + R 8 4 6 12 8 R= Ω 5 24 ∴ i= 8/ 5 = 15 A All these resistors are in parallel. The given network is as shown below. 0=

21.

22.

23. 24.

6

V1

6

V4 2

The simple circuit is as shown below. V3

3

6 4

(anti-clockwise)

V3 − VG = 25 × 5 = 125 V3 = 125 V as VG = 0 VG − V2 = 10 × 5 = 50 ∴ V2 = −50 V V2 − V1 = 5 × 5 = 25 V ∴ V1 = V2 − 25 = − 75 V Now, V3 − 2 = V3 − V2 ∴

4.3V

27. (a)

50Ω

1.0Ω 2.0Ω

Rnet = 1.0 + 2.0 +

3

V1

Rest procedure is same. 200 26. i = =5A 5 + 10 + 25

200Ω

12

6

VA − VG = 12 V VA = 12 V, as VG = 0 VA − VB = 1 × 3 = 3 V ∴ VB = VA − 3 = 9 V VB − VC = 2 × 3 = 6 V ∴ VC = VB − 6 = 3 V VG − VD = 6 V ∴ VD = − 6 V, as VG = 0 In the second case, 12 − 6 i= =1A 1+ 2 + 3

Now, ∴

V3

4

V2

Now, this is a balanced Wheatstone bridge in parallel with 12 Ω resistance. 25. First case 12 + 6 (clockwise) i= = 3A 1+ 2 + 3

V4

2

V2

= 43 Ω 4.3 ∴ i= = 0.1 A 43 = Readings of ammeter Readings of voltmeter = (i) net resistance of 50 Ω and 200 Ω  50 × 200  = (0.1)    50 + 200 =4V

12

50 × 200 50 + 200

620 — Electricity and Magnetism 30. (a) (i) When switch S is open, V1 and V2 are in

1.0Ω

4.3V i

(b)

2.0 Ω

50Ω i1 i2

200 Ω

52 × 200 = 42.27 Ω 52 + 200

Rnet = 1.0 +

Now

 200 i1 =   (0.1) = 0.08 A  252

∴ ∴

4.3 ≈ 0.1 A 42.27 i1 200 = i2 52 i=

∴

= Reading of ammeter Reading to voltmeter = Potential difference across 50 Ω and 2.0 Ω = 0.08 × 52 ≈ 4.2 V 5Ω

28. i1

4Ω

A

i2

i 1 –i2

42V (1)

6Ω

1Ω

C i3

3000 Ω

10V

(2) i2 – i3

E

series, connected to 200 V battery. Potential will drop in direct ratio of their resistors. ∴ V1 : V2 = RV 1 : RV 2 = 3000 : 2000 = 3: 2 3 V1 = × 200 = 120 V ∴ 5 2 V2 = × 200 = 80 V 5 (ii) When S is closed then V1 and R1 are in parallel. Similarly, V2 and R2 are also in parallel. Now, they are in series and they come out to be equal. So, 200 V will equally distribute between them. 200 ∴ V1 = V2 = = 100 V each 2 100 1 (b) i2 = = A 2000 20 100 1 i4 = = A 3000 30

8Ω (3) 16Ω

i1 B

2000 Ω

D

− 42 − 6 (i1 − i2 ) − 5i1 − i1 = 0

100V

…(i)

Loop 2 − 4 i2 − 10 − 8 (i2 − i3 ) + 6 (i1 − i2 ) = 0 …(ii) Loop 3 …(iii) 8 (i2 − i3 ) = − 16 i3 + 4 = 0 Solving these equations, we get i1 = 4 A, i2 = 1.0 A and i3 = 0.5 A 29. Net resistance of voltmeter (R = 400 Ω ) and 400 Ω will be 200 Ω. Now, we are getting a balanced Wheatstone bridge with 100 Ω and 200 Ω resistors on each side. Potential difference across each side will be 10 V which will distribute in direct ratio of resistors 100 Ω and 200 Ω. V100 Ω 100 1 ∴ = = V200 Ω 200 2 V200 Ω

V2

i4

Loop 1

or

i3

i2

4V

20  2 =   (10) = V  3 3

2000 Ω

V1

P

3000 Ω 100V

If we apply junction law at P, then current through switch 1 = i2 − i4 = A in upward direction. 60 31. Power absorbed by resistor is i 2R or 2W. Therefore, remaining 3W is absorbed by the battery (= Ei ). Hence, E is 3 V and current of 1 A enters from the position terminal as shown below. 2Ω

A

3V

B

1A

VAB = E + ir = E + iR = 3 + (1) (2) = 5 V 32. 8.4 = E − 1.5 r 9.4 = E + 3.5 r Solving these two equations, we get r = 0.2 Ω and E = 8.7 V

(Here, r = R) …(i) …(ii)

Chapter 23 33. During charging, V = E + ir = 2 + (5) (0.1) = 2.5 V 34. Simple circuit is as shown below i

2i

4Ω

i 4Ω

4Ω 2V 2V

2V

By symmetry, currents on two sides will be same (let i) Now if we apply loop law in any of the closed loop, we will get i = 0. 35. Net resistance should remain unchanged. GS R + G = R′ + ∴ G+S ∴

R′ − R = G −

GS G2 = G+S G+S

36. Current through voltmeter r 4.96A

Current Electricity — 621

where, R = resistance of voltmeter. Solving the above equation, we get R = 1200 Ω In the new situation, (300) (1200) Rnet = 400 + = 640 Ω 300 + 1200 60 ∴ i= = 0.09375 A 640 Now voltage drop across Voltmeter = 60 − potential drop across 400 Ω resistor = 60 − (400) i = 60 − (400) (0.09375) = 22.5 V (60) (120) 38. Rnet = 60 + = 100 Ω 60 + 120 120 = 1.2 A ∴ i= 100 Now, reading of voltmeter = 120 − potential drop across R1 = 120 − (60) (1.2) = 48 V 39. In parallel current distributes in inverse ratio of resistance. S

5A V 0.04A 2500Ω

i –i g

V 100 = = 0.04 A R 2500 In parallel current distribution in inverse ratio of resistors. Hence, 4.96 2500 = 0.04 r ∴ r = 20.16 Ω 37. Voltmeter reads 30 V, half of 60 V. Hence, resistance of 400 Ω and voltmeter is also equal to 300 Ω. =

60 V

ig

∴

G R

 i − ig  G + R   = S  ig 

∴

 i − ig   S −G R=  ig   20  =  − 3  (0.005) − 20 = 80 Ω  10 

Note In calculations, we have taken i − ig ≈ i. V

i 300 Ω

400 Ω

40.

100Ω A 2Ω

V 120 Ω

∴

 400 × R  300 =    400 + R 

3.4V 3Ω

622 — Electricity and Magnetism Reading of voltmeter = 3.4 − Voltage drop across ammeter and 3Ω resistance = (3.4 ) − 0.04 × 2 − 0.04 × 3 = 3.2 V (0.04 ) (100 R ) Now, 3.2 V = (100 + R ) where, R = resistance of voltmeter ∴ R = 400 Ω If voltmeter is ideal, then 3.4 i= = 0.03238 A 2 + 100 + 3

43. P = i 2R ∴

R/2

= (15) (1.5) (2.4 ) = 54 W

44. V = E − ir E − V 2.6 − 2 = i 1 = 0.6 Ω Now, power generated in the battery P = i 2r

∴

…(i)

E 100 Substituting in Eq. (i), we get RV = 4.5 × 10− 3 Ω

∴ Net power supplied for external circuit = 2.6 − 0.6 = 2.0 W 2Ω

1

45.

r RV

7V

If

I ′A =

(1)

i1 – i2 3Ω

(2)

1V

i2

ε = (R + RA + r) I A

ε R+r Substituting the value of ε, we get  RA  I ′ A = I A 1 +  RA + r  Now,

2Ω i1

If RV is increased from this value, V will increase. ε 42. (a) IA = R + RA+ r ∴

r=

= (1)2 (0.6) = 0.6 W Power supplied by the battery = Ei = 2.6 W

(b) V =

1+

R

 3R  Total maximum power = (imax )2    2

 E  =E−  r  r + RV 

(c) V = E

36 = 15 A 2.4

imax

Reading of voltmeter = 100i = 3.238 V 41. (a) V = E − ir

 ERV  V =   r + RV 

Pmax = R

imax =

…(i)

RA → 0, I A′ → I A

(b) In Eq. (i) substituting I A = 0.99 I ′A and the given values, we get RA = 0.0045 Ω I A′ I A′ (c) I A = = RA 1 1+ 1+ r RA + r 1+ RA If RA is decreased from this value, then I A will increase from 99% of I ′ A .

Loop equation in loop (1) + 7 − 2i1 − 3 (i1 − i2 ) = 0 Loop equation in loop (2) − 1 + 3 (i1 − i2 ) − 2i2 = 0 Solving Eqs. (i) and (ii), we get i1 = 2 A and i2 = 1 A Power supplied by E1 = E1i1 = 14 W But power consumed by E2 = E2i2 = 1W 12 46. (a) i= =2A 5+1 ⇒ P1 = Ei = 24 W (b) P2 = i 2r = (2)2 (1) = 4 W (c) P = P1 − P2 = 20 W V 47. (b) i = in each resistance R V2 in each resistance (e) P = R

…(i) …(ii)

Chapter 23 V2 R 1 P∝ R

Now let us find VAB across path ACB, VAB = 2i1 + 8i2 = 61.94 V Now, VAB = inet Rnet = (10) Rnet ∴ 61.94 = 10 Rnet or Rnet = 6.194 Ω (f) Two resistors in vertical middle wire can be removed. (g) Now balanced Wheatstone bridge, in parallel with 1Ω resistance between points A and B. The encircled resistance of 2Ω can be removed from the Wheatstone bridge.

P=

(f) or

48. (a) P =

(as V is same)

V2 R

∴

V = PR = 5 × 15 × 103 = 273.8 V

(b) P =

Current Electricity — 623

V2 (120)2 = = 1.6 W R 9 × 103 4Ω

A

49. (a)

8Ω

6Ω

2Ω

4Ω

1Ω

2Ω

B 2Ω

B

One resistor of 2Ω has already removed from the original circuit given the question. As its two ends will be the same potential (by symmetry). 50. Ri + 5R

4Ω

2Ω

3Ω 1Ω

A

2Ω

3Ω 1Ω

If connected by two equal resistors between B and D and between C and E, the combination is a balanced Wheatstone bridge and two resistors in series. ∴ R f = R + R + R = 3R = 0.6 Ri

B

10Ω 5Ω

5Ω

(d) 10Ω

51.

10Ω 5Ω

B 10Ω

10A

8Ω

C i1

(e) A

i 1 – i2 2Ω

i2 (2)

B

10 –i1 4Ω

15Ω

6Ω

40Ω

20Ω

30Ω

8Ω

B

52. (b) Three resistors are in parallel. Then, one resistor

2Ω (1)

15Ω 2Ω

A

5Ω

A

1Ω 2Ω

(b) A balanced Wheatstone bridge in parallel with R.

(c)

2Ω

1Ω

4Ω

A

2Ω

10Ω 10–i 2

Let us take a current of 10 A between A and B Loop equation in loop (1) …(i) − 2i1 − 2 (i1 − i2 ) + 4 (10 − i1 ) = 0 Loop equation in loop (2) …(ii) − 8i2 + 10 (10 − i2 ) + 2 (i1 − i2 ) = 0 Solving these two equations, we get i1 = 6.53 A and i2 = 6.11 A

in series with this combination. (c) Balanced Wheatstone bridge. Hence, two resistors in vertical wire can be removed. (d) All four resistors are in parallel (e) A balanced Wheatstone bridge. 53. (a) A balanced Wheatstone bridge with one resistance in parallel. (b) A

B

624 — Electricity and Magnetism LEVEL 2

Further

RA = R (1.5R ) (3R ) RBC = =R 1.5 R + 3R

or ∴ Because

RA = RBC VA = VBC iRA = iRBC

Single Correct Option 1. No current will flow through voltmeter. As it is ideal (infinite resistance). Current through two batteries 1.5 − 1.3 0.2 i= = r1 + r2 r1 + r2

2A

V = E2 − ir2

Now,

8.

 0.2  1.45 = 1.5 −  ∴  (r2 )  r1 + r2  Solving this equation, we get r1 = 3r2 2. In series, PD distributes in direct ratio of resistance. In first case, 198 900 = VAB − 198 R1 In second case,

180 900 = VAB − 180 2R1

…(i) …(ii)

R

Applying loop equation in closed loop we have, + 100 − 30 − 35 − 2R = 0 ∴

2R = 35 V = VR V5Ω = 7 × 5 = 35 V V5Ω =1 VR

∴  l1  − 1  l2 

9. r = R 

(R is same)

4 (R + RA ) = 20 V ∴ R = 5 − RA where, RA = resistance of ammeter  l1   70  − 1 ≈ 22.1 Ω − 1 = 132.4   60   l2 

5. r = R 

6. Initial current, i1 =

3A

100 V

P = i 2R

4.

5Ω 2A

3. Maximum current will pass through A. P ∝ i2

C

5A

Solving these two equations, we get VAB = 220 V

or

B 7A

E1 + E2 R + r1 + r2

Final current, when second battery is short circuited is E1 i2 = R + r1 E1 E1 + E2 i2 > i1 if > R + r1 R + r1 + r2 E1R + E1r1 + E1r2 > E1R + E1r1 + E2R + E2 r1 or E1r2 > E2 (R + r1 ) 7. B and C are in parallel ∴ VB = VC

 y   y − x = R  − 1 =   R x   x 

10. Let R = at + b At t = 10 s, R = 20 Ω ∴ 20 = 10 a + b At t = 30a + b Solving these two equations, we get a = 1.0 Ω/ s and b = 10 Ω ∴ R = (t + 10) E 10 i= = R t + 10

…(i) …(ii)

30

∆q = ∫ idt 10

30  10  =∫   dt 10  t + 10

= 10 loge (2)

or

11. Suppose n (< 1) fraction of length is stretched to m times. Then, or

(1 − n) l + (nl ) m = 1.5l nm − n = 0.5

…(i)

Chapter 23 R=

ρl ρl = A (V / l )

(V = volume)

ρl V or (if V = constant) R ∝ l2 Now, the second condition is (1 − n) R + (nR ) m3 = 4 R =

2

…(iii) ∴ nm2 − n = 3 Solving these two equations, we get 1 n= 8 l1 X 2 12. Initially, = = 100 − l1 R 3 2 ∴ l1 = × 100 5 = 40 cm l2 X 12 3 Finally, = = = 100 − l2 R′ 8 2 3 ∴ l2 = × 100 5 = 60 cm ∴ J is displaced by l2 − l1 = 20 cm 13. In parallel, current distributes in inverse ratio of resistance.

15. When K 1 and K 2 both are closed R1 is short-circuited, Rnet = (50 + r) Ω When K 1 is open and K 2 is closed, current remains half. Therefore, net resistance of the circuit becomes two times. or (50 + r) + R1 = 2 (50 + r) Of the given options, the above equation is satisfied if r = 0 and R1 = 50 Ω 16. 100 Ω , 25 Ω and 20 Ω are in parallel. Their, net resistance is 10 Ω ∴ Rnet = 4 Ω + 10 Ω + 6 Ω = 20 Ω V = i Rnet = 80 V 17. All these resistors are in parallel. R Rnet = + r = 4 Ω ∴ 3 Hence, the main current E i= =1A Rnet Current through either of the resistance i 1 or is A 3 3  1 V = iR =   (9) = 3 V ∴  3

9Ω 0.9Ω 0.01A

I A

S

18.

(I – 0.01)A 0.1Ω

Solving we get, I =1A 14. Equivalent emf of two batteries ε 1 and ε 2 is ε=

ε 1 / r1 + ε 2 / r2 (2/ 2) + (4 / 6) = 1/ r1 + 1/ r2 (1/ 2) + (1/ 6)

= 2.5 V Now, ∴ or

VAN = ε (I AN ) (RAN ) = ε   12   (4 ) (l ) = 2.5  4 + 4 × 4

Solving this equation, we get 25 l= m 24

0.03 – IG

B

0.01 0.1 = I − 0.01 9 + 0.9

∴

Current Electricity — 625

IG

G

In parallel, current distributers in inverse ratio of resistance. 0.03 − IG G r = = =4 IG S (r / 4 ) Solving this equation, we get IG = 0.006 A 8Ω 4 Ω 19. = 6Ω 3Ω ∴ or 20. V = iR ∴ ∴

VA = VB VAB = 0 V ∝R

(as i = constant)  πrB2 

VA  ρlA  =    VB  πrA2   ρlB 

626 — Electricity and Magnetism rB VA lB = × rA VB lA

or

3 1 1 × = 2 6 2 20 2 21. Current decreases times or times. Therefore, 30 3 3 net resistance should become times. 2 3 ∴ R + 50 = (2950 + 50) 2 Solving we get, R = 4450Ω =

between 2kΩ and 2kΩ. Hence, reading of voltmeter 10 = =5V 2 V2 27. R2 = R3 as P = and in parallel V is same. R i/2

R1

 E   10  =    × 0.2  10  1  E = 5 In second case,  E   10  E0 =    × 0.3  10 + x   1 Solving Eqs. (i) and (ii), we get x = 5Ω 23. V and V0 are oppositely connected.

…(i)

Hence, PR 2 = PR 3 If R2 = R3 Now current through R1 is double so R1 should be 1 th of R2 or R3 for same power. As P = i 2R. 4

More Than One Correct Options 1. H =

…(ii)

24. Balanced Wheatstone bridge. Hence, 1.5 Ω

1.4 A

20 Ω

V2 t1 R1

R1 =

4Ω

R2 =

In series,

 V2  H =  t  R1 + R2 

In parallel,

H =

10 Ω

Solving we get,

t=

t1t2 t1 + t2

6V

25. Resistance between A and B can be removed due

= 2.5 A 26. Net resistance of 3kΩ and voltmeter is also 2kΩ. Now, the applied 10 V is equally distributed

 1 1 V2 t = V 2t  +  Rnet  R1 R2 

 H H  = V 2t  2 + 2   V t1 V t2 

 2.5  i1 =   (1.4 ) = 1 A  2.5 + 1

to balanced Wheatstone bridge concept. Now, RDE and RGH are in series and they are connected in parallel with 10 V battery. 10 10 ∴ I DE = = RDE + RHG 2 + 2

V 2t2 H

Similarly,

i1 50 + 10 2.5 = = i2 20 + 4 1 ∴

V 2t1 H

H (R1 + R2 ) V2 Substituting the values of R1 and R2, we get t = t1 + t2

i2 50 Ω

⇒

t=

resistance can be removed from the circuit. i1

i/2 R3

E0 = VAC = (i )AC (R )AC

22.

R2

i

2Ω

2.

5V

i

3Ω

6−5 i= = 0.2 A 2+ 3 V1 = E1 − ir1 = 6 − 0.2 × 2 = 5.6 V

Chapter 23 3. (a) In series, current is same. ∴

IA = IB (b) VA + VB = VC ∴ I ARA + I B RB = IC RC (d) In parallel, current distributes in inverse ratio of resistance. IB IA RC = = ∴ IC IC RA + RB

4. Same as above. 5. (a) V = iR In series i is same. Hence, V is also same as R is given same. ρl (b) R = A R is same. Hence, A should be smaller in first i 1 wire. Secondly, vd = or vd ∝ ne A A A of first wire is less. Hence, its drift velocity should be more. V 1 (V → same) (c) E = or E ∝ l l 7. If switch S is open, i1λl = E2 where, i1 = current in upper circuit and λ is resistance per unit length of potentiometer wire. E Null point length, l = 2 ∴ i1λ (a) If jockey is shifted towards right, resistance in upper circuit will increase. So, current i1 will decrease. Hence, l will increase. (b) If E1 is increased, i1 will also increase. So, l will decrease. (c) l ∝ E2 (d) If switch is closed, then null point will be obtained corresponding to V2 = E2 − i2 r2 which is less than E2. Hence, null point length will decrease. 8. By closing S 1 , net external resistance will decrease. So, main current will increase. By closing S 2, net emf will remain unchanged but net internal resistance will decrease. Hence, main current will increase.

9.

10 V

2Ω a c

Vb + 10 − 2i = Va

i

b

Current Electricity — 627

Vb − Va = 2i − 10 = 2 V ∴ i = 6A Now, VC − Va = 2 × 6 = 12 V 10. Between a and c, balanced Wheatstone bridge is formed. Across all other points simple series and parallel grouping of resistors.

Comprehension Based Questions l  r = R  1 − 1  l2 

1 and 2.

 500  10 = R  − 1  490  Solving this equation, we get R = 490 Ω E  Further r = R  − 1 V  2   or 10 = 490  − 1 V 

∴

Solving, we get

V = 1.96 V

Match the Columns 1. Let potential of point e is V volts. Then, I ae + I be + I ce + I de = 0 2−V 4 −V 6−V 4 −V ∴   +  +  +  =0  1   2   1   2  or V =4V Now current through any wire can be obtained by the equation, PD I = R 2. i1 = i2 or i is same at both sections. A1 < A2 i 1 (a) Current density = ∝ A A Resistance ρ 1 (c) = ∝ length A A (d) and (b) E or potential difference per unit length = (i ) (Resistance per unit length) 1  ρ = (i )   ∝  A A

3. By introducing parallel resistance R3 in the circuit, total resistance of the circuit will decrease. Hence, main current i will increase. Now, VR 1 = E − VR 2 = E − iR2 Since, i is increasing, so VR 2 will increase. Hence, VR1 or current passing through R1 will decrease.

628 — Electricity and Magnetism H i 2t ∴ [ R ] = [ ML2T − 2 ] / [ A2 T ] = [ ML2T − 3A− 2 ]

4. (a) R =

(b) V = iR ∴ [ V] = [ A ][ ML2 T − 3A− 2 ] = [ ML2 T − 3A− 1 ] RA (c) ρ = l [ ML2T − 3A− 2 ][ L2 ] = [ ML3T − 3A− 2 ] ∴ [ρ ] = [L ]

4 −1 = 1A 1+ 1+ 1

distribution on either side of the plane will be identical and points E and F will be at same potential and no current will flow through it. i1

i

E i1

A

D

i2

i2 i3

i3

F

B

 1 (d) [ σ ] =   = [ M − 1L− 3T 3A2 ] ρ 

5. i =

2. (a) Due to symmetry about the shaded plane, current

C E

(anti-clockwise)

A D

(a) VA = E − ir = 4 − 1 × 1 = 3 V (b) VB = E + ir = 1 + 1 × 1 = 2 V (c) | PA | = Ei − i 2r = (4 × 1) − (1)2 (1) = 3W

B

F

(d) | PB | = Ei + i 2r = (1) (1) + (1)2 (1) = 2W

C

Subjective Questions 1. (a) Points D and E are symmetrically located with respect to points A and C. The circuit can be redrawn as shown in figure. B R/2

R R/2 A

D, E

E

R R/2 A

C

R

D

This is a combination of a balanced Wheatstone bridge in parallel with a resistance R. So, the resistance between B and D (or E) can be removed. 1 1 1 1 = + + RAC R R + R 2R 2 2 2R Ans. or RAC = 5 (b) With respect to D and E, points A, B and C all are symmetrically located. Hence, the simplified circuit can be drawn as shown in figure. R

R R

R D

E R

R A, B, C

∴

RDE =

R R 2R + = 3 3 3

2 8 r× r 3 = 8r Ans. ∴ RAD = 3 2 8 r + r 15 3 3 (b) Redrawing the given arrangement for resistance across AB. Potentials VD = VE

Ans.

F

B C

VC = VF ∴ No current flows through DE and CF. 3 r×r 3 Ans. RAB = 2 = r ∴ 3 r+ r 5 2 3. (a) Current flowing through resistance 5Ω is 11 A Power dissipated = i 2R = (121)5 = 605 W (b) VB + 8V + 3V + 12V − 12V − 5V = VC VB + 11V − 5V = VC 6V = VC − VB (c) Both batteries are being charged. Ans.

Chapter 23 4. (a) VA − VB = 6 V

For voltmeter, range V = I g (99 + 101) V = 200 I g Also resistance of the voltmeter = 99 + 101 = 200 Ω

Since, VB = 0 ∴ VA = 6 V VA − VC = 4 V ⇒ VC = VA − 4 = 2 V (b) VA − VD = VA − VC = 4 V From unitary method, we can find that,  100 AD =   (4 ) = 66.67 cm  6 

Ig

G

12 Ω

X A

J

B

C

D

(c) AJ = 60 cm ⇒ BJ = 40 cm If no deflection is taking place. Then, the Wheatstone bridge is said to be balanced. Hence, X RBJ X 40 2 or = = = 12 RAJ 12 60 3 X =8Ω

or

99 Ω

I = 100 I g I

Ig

101 Ω

V

In Fig. 2, resistance across the terminals of the battery 200 × 2 R1 = r + = 2.99 Ω 200 + 2 ∴ Current drawn from the battery, 12 I1 = = 4.01 A 2.99 ∴ Voltmeter reading 4 V = 12 – I 1 r = 12 – 4.01 × 1.01 5 5 V = 7.96 × = 9.95 V 4 9.95 Using Eq. (ii), I g = = 0.05 A 200 Using Eq.(i), range of the ammeter I = 100 I g = 5 A

Ans.

Ans.

7. Applying Kirchhoff’s laws in two loops we have, 10 V

R 1

i1 3Ω

6V

Ans.

6. For ammeter 99I g = (I – I g ) 1 or

…(ii)

G

(c) Since, they are at same potential, no current will flow through it. (d) VA − VB is still 6 V ∴ VA = 6 V Further, VA − VC = 7.5 V ∴ VC = − 1.5 V Since, EMF of the battery in lower circuit is more than the EMF of the battery in upper circuit. No such point will exist. 5. (a) There are no positive and negative terminals on the galvanometer because only zero deflection is needed.

(b)

Current Electricity — 629

i2

…(i)

2

i1 + i2

99 Ω

6Ω

G

1Ω

I g is the full scale deflection current of the galvanometer and I the range of ammeter. For the circuit in Fig.1, given in the question 12 V = 3 A ⇒ r = 1.01 Ω Ans. 99 × 1 2+ r+ 99 + 1

10 − i1R − 6 + 3i2 = 0 6 − 6(i1 + i2 ) − 3i2 = 0 Solving these two equations, we get 6 i1 = R+2

…(i) …(ii)

Power developed in R, P = i12R =

36 R (R + 2)2

…(iii)

630 — Electricity and Magnetism PD across switch = 10 + (1)i2 = 10 + 2 = 12 V Ans.

For power to be maximum, dP =0 dR ∴ (R + 2)2 (36) − (36R )(2)(R + 2) = 0

A6

(b)

R + 2 − 2R = 0 R=2Ω

or ∴

20 V A1

2Ω

A2

15 V

6V

i1

Ans.

For maximum power from Eq. (iii), we have Ans. Pmax = 4.5 W

4Ω

i2

A5 i3

2Ω

A4

A3 1Ω

10 V i4

8. Applying loop law in loops 1, 2 and 3, we have 2A

When switch is closed −2(i1 − i2 ) + 15 = 0 2i1 − 9i2 – 2i3 − 11 + i4 = 0 i2 + i3 + 3 = 0 10 − (i4 − i2 ) = 0 Solving these four equation, we get i1 = 12.5 A, i2 = 5.0 A , i3 = − 8.0 A i4 = 15 A

R 3

1A

E1

7A

E2 5A

4Ω

3Ω

1

3A

2

6Ω

8A

E1 − 12 − 24 = 0 E1 = 36 V − E2 + 24 + 30 = 0 or E2 = 54 V −2R − E1 + E2 = 0 E − E1 or R= 2 2 =9Ω 9. Using the loop current method,

Ammeter

∴

20 V

Reading (amp)

A2

15 V

A2

A3

A4

2.5

10

7

AB = 10 m RAB = 30 Ω

10. C

A

B

1Ω 6V

i2

2Ω

G

A5

1.5V

i3

Potential gradient across wire, 2 AB = = 0.2 V/m 10 Now, VAC = 1.5 V or (0.2)( AC ) = 1.5 Ans. ∴ AC = 7.5 m

A4

A3 1Ω

10V

S

…(i) (a) −2(i1 − i2 ) + 15 = 0 −4 i2 + 20 − 2(i2 − i1 ) − 15 − i2 − 10 − 2(i2 + i3 ) − 6 = 0 or …(ii) 2i1 − 9i2 − 2i3 − 11 = 0 − 2(i2 + i3 ) − 6 = 0 or …(iii) i2 + i3 + 3 = 0 Solving these equations, we get i1 = 9.5 A, i2 = 2 A and i3 = − 5 A

 RAB   30  12 (a) VAB =  V  ×2 =  ×2 =  RAB + 5  30 + 5 7 2V 5Ω C1

A 1Ω

Ammeter Reading (amp)

Ans.

2V

4Ω

2Ω

i1

A1 12.5

And the current through switch is 15 A.

A6 A1

…(i) …(ii) …(iii) …(iv)

A1 9.5

A2 9.5

A3 2

A4 5

G 1.5 V

B

Current Electricity — 631

Chapter 23 ∴ Potential gradient across 12 AB = V/m 70

∴ Power generated in Rx is P=

VAC = 1.5 V

Now,

For P to be constant, dP =0 dRx

 12    ( AC 1 ) = 1.5  70

∴ ∴

AC 1 = 8.75 m

Ans.

(50Rx + 600)2 (900 V 2 )

2V

or

C2

A

V12 900Rx V 2 = Rx (50Rx + 600)2

− 1800 × 50 × Rx V 2 (50Rx + 600) =0 (50Rx + 600)4

or 50Rx + 600 − 100Rx = 0 ∴ Rx = 12 Ω

B

12. The two batteries are in parallel. Thermal power

1.5 V 1Ω

generated in R will be maximum when, total internal resistance = total external resistance R1R2 or R= R1 + R2

i 5Ω

(b) VAC2 = V or

 5  (0.2)( AC 2 ) =   (1.5)  5 + 1

or

AC 2 = 6.25 m

Eeq Ans.

11. V = constant

 E1 E2  +    R1 R2  =  1 1 +    R1 R2   E R + E2R1  = 1 2   R1 + R2 

20 Ω

V

Ans.

30 Ω

Rx

∴

i=

Rnet =

2R1R2 R1 + R2

Eeq

E1R2 + E2R1 2R1R2

Rnet

=

Maximum power through R 20 Ω

⇒

V

Pmax = i 2R =

30 Rx 30+ Rx

30Rx     30 + R x V V1 =   30Rx + 20    30 + Rx   30Rx  = V  50Rx + 600

13. V1

(E1R2 + E2R1 )2 4 R1R2 (R1 + R2 )

V2  dT  = k (T − T0 ) + C    dt  R dT dt or = 2 C V − k (T − T0 ) R T t dt dT or =∫ ∫T0 V 2 0 C − k (T − T0 ) R (at t = 0, temperature of conductor T = T0) Solving this equation, we get V2 (1 − e− kt / C ) T = T0 + kR

Ans.

Ans.

24. Electrostatics INTRODUCTORY EXERCISE

F = Force between two point charges  1   q × q =     4 πε 0   a2 

24.1

1. Due to induction effect, a charged body can attract a neutral body as shown below. + + +

+ + + +

1

+ +

– 2

–

Fnet = F 2 + F 2 + 2FF cos 60°

+

–

= 3F

+

Body-1 is positively charged and body-2 is neutral. But we can see that due to distance factor attraction is more than the repulsion. 4. Number of atoms in 3 gram-mole of hydrogen atom = number of electrons in it = 3 N 0 = (3 × 6.02 × 1023 ) where, N 0 = Avogadro number ∴ Total charge = − (1.6 × 10− 19 ) (3 × 6.02 × 1023 ) = − 2.89 × 105 C

INTRODUCTORY EXERCISE

1. Fe =

24.2

1 q1q2 4 π ε 0 r2 m1 m2 r2 Fe (1/ 4 π ε 0 ) q1q2 = Fg G m1m2

∴ =

(9 × 10+ 9 ) (1.6 × 10− 19 )2 (6.67 × 10− 11 ) (9.11 × 10− 31 ) (1.67 × 10− 27 )

= 2.27 × 1039 1 q1q2 4 π ε 0 r2 qq ε0 = 1 2 2 4 π Fr F=

2. ∴

q A

q B r

4.

F O

Units and dimensions can be found by above equation. q

–q

q D

C

Net force on − q from the charges at B and D is zero. So, net force on − q is only due to the charge at A.  1  q×q F =   4 π ε 0  r2 where,

Fg = G

and

 3   q 2 =     4 πε 0   a

+

+

∴

2a a = 2 2  1  q2 F =   4 π ε 0  (a/ 2 )2 r=

 1   q 2 =     2π ε 0   a

5. The charged body attracts the natural body because attraction (due to the distance factor) is more than the repulsion. 1 qq 7. F = ⋅ 1 2 4 π ε 0 r2 ∴

(q1 )min = (q2 )min = e2 1 e2 Fmin = 4 π ε 0 r2

9. Two forces are equal and opposite. INTRODUCTORY EXERCISE

q

24.3

a

a

3.

60° a

F

q

√3F = Fnet F

1. Electric field lines are not parallel and equidistant. 2. Electric lines flow higher potential to lower potential. ∴

VA > VB

Chapter 24

1  1 = (9 × 109 ) (− 2 × 10− 12 )  −   1.0 2.0

3. If charged particle is positive, and at rest. Electric field lines are straight then only it will move in the direction of electric field. 4. See the hint of above question. 5. Electric field lines start from positive charge and terminate on negative charge.

6. In case of five charges at five vertices of regular pentagon net electric field at centre is zero. Because five vectors of equal magnitudes from a closed regular pentagon as shown in Fig. (i). D

C

E

D B

C

E

A (i)

= − 9 × 10− 3 J = − 9 mJ 3. Work done by electrostatic forces W = − ∆U = U i − U f ∴ U f =Ui − W = (− 6.4 × 10− 8 ) − (4.2 × 10− 8 ) = − 10.6 × 10− 8 J

4. U ∞ = 0 Ur =

B A (ii)

For

Where one charge is removed. Then, one vector (let AB) is deceased hence the net resultant is equal to magnitude of one vector 1 q = | BA | = 4 πε 0 a2  1   q    (rp − rq )  4 πε 0   r3 

7. E = 

Here,

r = (3)2 + (4 )2 = 5 m

∴

E=

qq 1 ⋅ 1 2 4 π ε0 r

Ur ≠U∞ r≠∞ 1  q1q2 q2q3 q3q1  Ur = + + 4 πε 0  r12 r23 r31 

INTRODUCTORY EXERCISE

W ab 12 = − 2 = 1200 V q 10 λ C/m C 2. (a) α = = = 2 x m m ∴ Vba =

(b)

24.4

x

A x = –d

x=0

dV =

dx

1 dq ⋅ 4 πε 0 x + d

 1   λdx  =     4 πε 0   x + d 

 1 1 2  1    (q1q2 )  −  m  4 πε 0   ri rf 

 1   αx dx  =     4 πε 0   x + d 

2 1  1 = (9 × 109 ) (− 2 × 10− 12 )  −   1.0 0.5 10− 4 = 18.97 m/s 2. Work done by electrostatic forces = − ∆U =Ui −U f  1 1  1  =  (q1q2 )  −   4 π ε0  ri rf 

dq

x +d

1 q1q2 1 1 q1q2 = mv 2 + 4 πε 0 ri 2 4 πε 0 rf

∴v =

24.5

W ab = ∆U a − b = U b − U a = q (Vb − Va ) = q Vba

1. K i + U i = K f + U f ∴

(For two charges)

Now,U r can be equal to U ∞ for finite value of r.

1.

(9 × 109 ) (− 2 × 10−6 ) $ (3i + 4 $j) (5)3 = − (4.32 $i + 5.76 $j) × 102 N/C

INTRODUCTORY EXERCISE

Electrostatics — 633

∴ V =∫

x=L x=0

dV =

α 4 πε 0

L    L − d ln 1 + d    

P

3.

r

d x = –l x = 0

dq x

dx

x=l

634 — Electricity and Magnetism  q dq =   dx  2l 

Substituting in Eq. (i), we have Qq W = 2πε 0L

At point P,  1   dq dV =      4 πε 0   r 

INTRODUCTORY EXERCISE

 q   1   2l ⋅ dx  =     4 π ε0  d 2 + x2    ∴

V =2∫

x=l x=0

 ∂V $ ∂V $  i+ j ∂y   ∂x = − a [(2x ) $i − (2 y) $j ] = − 2a [ x$i − y$j ]

1. (a) E = − 

(b) Again

dV

 ∂V $ ∂V E=− i+ ∂y  ∂x $ $ = − a [ yi + xj ]

x r

4.

dx

R

Vapex ∴

= − ∆U = U apex − U ∞ = qVapex =0 R = L  R r=  x  L

W W U∝ r x

Surface charge density, Q σ= πRl dq = (σ ) (dA )  Q  =  (2πr) dx  πRl   θ  R  =  (2π ) x dx  πRL L   2Q  =  2  x dx  L  Now,

 1   dq dV =      4 πε 0   x   2Q  =  dx  4 πε 0L2 

∴

$j  

dV = − Slope of V -x graph. dx From x = − 2m to x = 0, slope = + 5 V/m ∴ E = − 5 V/m From x = 0 to x = 2m, slope = 0, ∴ E=0 From x = 2m to x = 4m, slope = + 5V/m, ∴ E = − 5 V/m From x = 4 m to x = 8m, slope = − 5 V/ m ∴ E = + 5 V/m Corresponding E-x graph is as shown in answer. ∂V − 50 3. = = − 10 V/m ∂x 5

2. E = −

dq

∴ as

24.6

L

V = ∫ dV = 0

Q 2πε 0L

…(i)

2

Now,

2 2  ∂V   ∂V   ∂V  |E | =   +   +  ∂x   ∂z   ∂y 

No information is given about |E | ≥

∂V ∂x

or

∂V ∂V and . Hence, ∂y ∂z

| E | ≥ 10 V/m

4. VA = VD and VB = VC as the points A and D or B and C are lying on same equipotential surface (⊥ to electric field lines). Further, VA or VD > VB or VC as electric lines always flow from higher potential to lower potential VA − VB = VD − VC = Ed = (20) (1) = 20 V

INTRODUCTORY EXERCISE

24.7

1. (a) Given surface is a closed surface. Therefore, we can directly apply the result.

Chapter 24 φ=

qin =0 ε0

as

qin = 0

Electrostatics — 635

B

(b) Again given surface is a closed surface. Hence, we can directly apply the result. q q as qin = q φ = in = ε0 ε0 (c) Given surface is not closed surface. Hence, we cannot apply the direct result of Gauss’s theorem. If we draw a complete sphere, then q φ through complete sphere = ε0

3. (a) A

C

Net flux entering from AB = net flux entering from BC.

(b)

S

1  q ∴ φ through hemisphere =   2  ε0

φ = ES = E (πR 2 )

2. Net charge from any closed surface in uniform electric field = 0 ∴ Net charge inside any closed surface in uniform electric field = 0

4. Given electric field is uniform electric field. Net flux from any closed surface in uniform electric field = 0.

Exercises LEVEL 1

A

6. VA − VB = − ∫ E ⋅ dr B

Assertion and Reason 1. An independent negative charge moves from lower potential to higher potential. In this process, electrostatic potential energy decreases and kinetic energy increases. 2. Two unlike charges come together when left freely.  ∂V $ ∂V $ ∂V $  3. E=− i+ j+ k  ∂x ∂y ∂z  ∴ ∴ or

2

2  ∂V   ∂V   ∂V  |E | =   +   +  ∂x   ∂z   ∂y  2

∂V ∂x = 10 V/m

|E | ≥

kq or kq = VR R For inside points (r ≤ R ), kq E = 3r R or E∝r R At distance r = , 2 (VR )  R  V E= 3   = R  2  2R

4. V =

∴

(4,0)

=−∫

(0,4)

=−∫

(0, 4)

(4, 0)

(4 $i + 4 $j) ⋅ (dx$i + dy$j) (4 dx + 4 dy) = 0

VA = VB

7. At stable equilibrium position, potential energy is minimum.

8. In uniform electric field, net force on an electric dipole = 0 Therefore, no work is done in translational motion of the dipole. Electric lines also flow from higher potential to lower potential. Electrostatical force on positive charge acts in the direction of electric field. Therefore, work done is positive.

9. Charge on shell does not contribute in electric field just inside the shell. But it contributes in the electric field just outside it. So, there is sudden change in electric field just inside and just outside it. Hence, it is discontinuous. 1 q 10. | E | = 0 minimum at centre and |V | = ⋅ is 4 πε 0 R maximum at centre.

636 — Electricity and Magnetism Objective Questions

1 q 1 q ⋅ 1 + ⋅ 2 4 πε 0 3R 4 πε 0 3R qnet = (4 πε 0 ) (3R )

V =

8.

V 1. φ = ES →   (m 2 ) → volt-m   m

qE 2. ge = g −    m

 1   qnet  R=     4 πε 0   3V 

∴

or ge will decrease. Hence, T = 2π

l will ge

(9 × 109 ) (3 × 10− 6 ) (3) (9000) =1m

=

increase. 3. Electric lines terminate on negative charge. W = ∆U i

4.

=U f −Ui  1  1 qq  qq  ⋅  =3 ⋅ −3 4 πε 2 l 4 πε l     0 0

9.

R

r

2 3  3  1 q  = −   ⋅ 2  l = − Fl  2  4 πε 0 l  2

r = 3R kq k (3q) k q Vp = net = = r 3R R

2qV m 2eV m

v=

5.

vp = vd =

2e (2V ) 2m

vα =

2 (2e) (4V ) 4m

10.

7.

 1   q2  r=      4 πε 0   mg sin θ  =

(9 × 109 ) (2.0 × 10− 6 )2 (0.1) (9.8) sin 30°

= 27 × 10− 2 m = 27 cm Q1

 1   2q V V′ =     =  4 πε 0   4 r 2 q

1 q2 = mg sin θ 4 πε 0 r2 ∴

The ratio is 1 : 1 : 2. 1 q 6. V = ⋅ 4 πε 0 r

–q

P

2√2R

q

11. –q

a

q

F3 Q1

⇐

Q1

a

P Q1 F1

q

–q

–q

q

W = ∆U = U f − U i

1  − q2 q2 q2 q2 q2 q2  = − + + − −   4 πε 0  a a a 2a 2 a a  − q2 q2 q2 q2 q2 q2   − + − − + −  a 2a a 2 a a  a =

q2 [4 − 2 2 ] 4 πε 0a

F1 F2

Q1 = (2 2 − 1) Q F1 = Force between Q1 and Q1 at distance a F2 = Force between Q1 and Q1 at distance 2a a F3 = Force between Q1 and q at distance 2 For F3 to be in the shown direction, q and Q1 should have opposite signs. For net charge to be zero on Q1 placed at P. | F3 | = Resultant of F1 , F2 and F1

Chapter 24 ∴

k Q1 q = 2F1 + F2 (a/ 2 )2  k Q1 Q1  k Q1 Q1 +  a2  ( 2a)2 Q  1   2 2 + 1 |q | = 1 2 + =  Q1  2  2   4 

7 Q or 4 qE qσ | a| = = m 2ε 0 m

12.

t= =

2s = a

q=−

∴

7 Q 4

This is because, (q1 + q2 )2 = (q1 − q2 )2 + 4 q1q2

4 s ε0 m | q | |σ |

2

+q

–q

A

B

∴

F2 > F1

17. 4Q

Between A and B two forces on third charge will act in same direction. So, this charge cannot remain in equilibrium. To the right of B or left of A forces are in opposite directions but their magnitudes are different. Because charges have equals magnitudes but distances are different. 14. Between 2q and − q, two electric fields are in same direction. So their resultant can’t be zero. To the right of 2q left of − q they are in opposite directions. So, net field will be zero nearer to charge having small magnitude. q 3

2

 q1 + q2    > q1q2  2 

or

4 × 0.1 × 8.86 × 10− 12 × 1.67 × 10− 27 1.6 × 10− 19 × 2.21 × 10− 9

q

(q1 + q2 )2 > 4 q1q2

or

= 4 × 10− 6 s

13.

q1 + q2 2 k (q1 + q2 / 2)2 F2 = r2 2  q1 + q2    > q1q2  2  q′1 = q′2 =

where,

 2 2 + 1 =  (2 2 − 1)Q 4   =

 1  k =   4 πε 0 

k q1 q2 r2 k q′ q′ F2 = 12 2 r

16. F1 =

= 2

∴

Electrostatics — 637

–4Q +2Q –2Q +8Q

18. 1000  πr3 = πR 3 ⇒ R = 10 r   4 3

4 3

i.e. radius has become 10 times. Charge will become 1000 times. 1 (Charge) Charge or V ∝ V = ⋅ 4 πε 0 (Radius) Radius Hence, potential will become 100 times. q 1 1 19. PD = in  −  or PD ∝ qin 4 πε 0  R 2R 

E5

15.

q

E4 E1

1 E3

4

q

20. Q

2a E 2 5 q

E1 and E4 are cancelled. E2 and E5 are cancelled. 1 q q ∴ Enet = E3 = ⋅ = 4 πε 0 (2a)2 16πε 0a2

–Q 3Q –3Q +2Q

The desired ratio is −

3Q 3 =− 2Q 2

638 — Electricity and Magnetism 21. S = (1) $i ⇒ φ = B ⋅ S = 5 V-m q

θ

T

26.

Q

θ qE

22. F1 √2F1 Q

mg

F2

or

F2 = Force between q and q 1 q×q = 4 πε 0 ( 2 a)2

Similarly, T cos θ = mg σ 27. E = ε0

F1 = Force between Q and q. For net force on q to be zero . F2 = 2 F1 ∴

with sign,

∴

y

28.

+ q

kqA kqB + =0 rB rB qB = − qA = − q

–q q

From Gauss’s theorem, electric field at any point is given by kq E = 2in r qin inside A and outside B is zero. Therefore, E = 0 kq 3  kq =   2 r 2  R

or

r=

4 R 3

∴ Distance from surface = r = R R = 3 q in 25. φ = ε0 ∴

∴

q=−2 2 Q

Charge distribution is as shown below.

24.

V ∝R V1 R1 a = = V2 R2 b

or

| q | = 2 2Q

23. VB = 0 ∴

E1 = E2 σ1 = σ 2 σR V = ε0

∴

 1 qQ  1 q2 = 2 ⋅ 2 4 πε 0 2a2  4 πε 0 a 

∴

T sin θ = qE qE T = sin θ

q

F1

qin = 0 φ=0

(as σ → same)

E – q

x

E

Potential is zero at infinite and at origin. Therefore, PD = 0. Hence, the work done asked in part (c) is also zero. 29. According to principle of generator PD in this case only depends on the charge on inner shell. k | q| 30. …(i) 500 = 2 r k (− q) …(ii) − 3000 = r Solving these two equations, we get r=6m 500 r2 ∴ |q | = k (500) (6)2 = 9 × 109 = 2 × 10− 6 C = 2 µC qq 1 q1q2 1 31. ⋅ 2 = ⋅ 12 2 4 πε 0 r1 4 πε 0 k r2 ∴

r2 =

r1 50 = k 5

= 22.36 m

Chapter 24 32. W A → B = q (VB − VA )

38. E = 0, inside a hallow charged spherical

B A = q − ∫ Edr = q ∫ E dr B  A 

=q∫

2a

+λ

W = F⋅r

39.

= (QE) ⋅ r = Q (E ⋅ r ) = Q (E1 a + E2 b)

Subjective Questions –

33.

conducting shell.

λ qλ  1 ln   dr =  2 2πε 0r 2πε 0

a

Electrostatics — 639

1. F =

+ p

 1   where, k =   4 πε 0 

kq (Q − q) r2

For F to be maximum, Negative charge of dipole is near to positively charge line charge. Hence, attraction is more.

By putting

r = (4 − 1)2 + (2 − 2)2 + (0 − 4 )2

34.

=5m 1 q V = ⋅ 4 πε 0 r

3. E = ∴

(9 × 109 ) (2 × 10− 8 ) = = 36 V 5 Field is in the direction of r = rp − rq kq 35. V = R ∴

kq = VR Kq VR E= 2 = 2 r r

∴

W E= qs cos θ =

4 0.2 × 2 × cos 60°

= 20 N/C –Q

Q

37.

k= C1

C2

d

VC1 = VC2 = ∴

VC1 − VC2

kQ − R

2

1  −  R

−

σ = (2E ε 0 ) = 2 × 3.0 × 8.86 × 10− 12 = 5.31 × 10− 11 C/m 2

4.

C

R S

C E

P Q

(ii)

If loop is complete, then net electric field at centre C is zero. Because equal and opposite pair of electric field vectors are cancelled. If PQ portion is removed as shown in figure, then electric field due to portion RS is not cancelled. Hence, electric field is only due to the option RS. 1 qRS  1  (q/ 2πR ) x E= = ∴  4 πε 0 R 2  4 πε 0  R2 =

qx 8π 2 ε 0R 3

5. Let q1 and q2 are the initial charges. After they are

R2 + d 2

R +d

Q = 2πε 0

1 4πε0

kQ

kQ 2

dF = 0, we get dq Q q= 2

(i)

W = Fs cos θ = qEs cos θ

36.

σ 2ε 0

dF =0 dq

kQ R   R 2 + d 2 

connected by a conducting wire, final charge on them become.  q + q2  q′1 = q′2 =  1   2  Now, given that

1

0.108 =

(9 × 109 ) (q1 ) (q2 ) (0.5)2

…(i)

640 — Electricity and Magnetism  q + q2  (9 × 109 )  1   2  0.036 = (0.5)2

2

8. If we make a bigger cube comprising of eight small cubes of size given in the question with charge at centre (or at D). q Then, total flux through large closed cube = . ε0

…(ii)

Solving Eqs. (i) and (ii), we can find q1 and q2. 6. Since, net force on electric dipole in uniform electric field is zero. Hence, torque can be calculated about point. This comes out to be a constant quantity given by τ = p× E

7.

There are 24 symmetrical faces like EFGH on outermost surface of this bigger cube. q Total flux from these 24 faces is . Hence, flux ε0 1  q from anyone force =  . 24  ε 0 

E2 θ θ

E1

E1

Net = E

P r

r

y q

a

–2q

Electric lines are tangential to face AEHD. Hence, flux is zero.

P

θ θ

9. (a) a

q

q1

E1 = Electric field at P due to q

∴

E2 = Electric field at P due to − 2q  1  2q =   4 πε 0  y2

(b)

 1   q   y  1   2q =2      +     4 πε 0   r2   r   4 πε 0   y2 

or

E=−

3qa2 $ j 4 πε 0 y4

–a O

+a

+ 3a 2

q2

Q

Or magnitude wise q1 = 25 q2 with sign q1 = − 25q2

=

3qa2 4 πε 0 y4

q2

Therefore, q1 and q2 should be of opposite signs. Further, kq1Q kq2Q = (5a/ 2)2 (a / 2)2

E = 2E1 cos θ + E2

=−

+a

Q

q1 = 9q2

q1

∴ Net electric field at P,

− 3/ 2 2q  y  1 a2   3 + 1 + 2  = + 2 4 πε 0  y y y     2 a Applying binomial expression for 2 < < 1 y we get, 2q  3 a3  E= −  4 πε 0  2 y4 

O

+ a 2

q1 and q2 should be of same sign. Kq1Q Kq2Q Further, = 2 (3a/ 2) (a/ 2)2

 1  q =   4 πε 0  r2

2q  y   1  + 4 πε 0  r3   y2  2q  y 1 = + 2  2 2 3/ 2 4 πε 0  (a + y ) y 

–a

y

x

10.

+q

–Q

+4q

x + y=L Let force on − Q charge should be zero. kQ. q kQ (4 q) ∴ = x2 y2 x 1 = ∴ y 2 From Eqs. (i) and (ii), we get L 2L and y = x= 3 3 Net force on 4q should be zero. k | Q | (4 q) kq (4 q) = ∴ (2L / 3)2 L2

…(i)

…(ii)

Chapter 24 ∴

4 q 9

|Q | =

Now, applying Lami’s theorem for the equilibrium of ball we have mg 3F = sin (90° + 30° ) sin (90° + 60° )

With sign, 4 Q=− q 9 Similarly net force on q should be zero. 4 k | Q | q kq (4 q) or | Q | = q ∴ = 9 (L / 3)2 L2

F= ∴

=

mgr2 3 (1/ 4 πε 0 ) (0.1 × 10− 3 ) (9.8) (10 3 × 10− 2 )2 3 × 9 × 109

= 3.3 × 10− 8 C

N

11.

mg 1 (q) (q) mg = = 3 4 πε 0 r2 3

q=

If we, sightly displace − Q towards 4q, attraction between these charges will increase, hence − Q will move towards + 4q and it will not return back. Hence, equilibrium is unstable. 30°

60°

Electrostatics — 641

a

q

q

F

13.

a

mg

∴ 30°

12.

q= T

p

q

Using Lami’s theorem, we have mg F = sin (90° + 30° ) sin (90° + 60° ) mg 1 (q)(q) mg or F= = ∴ 4 πε 3 3 R2 0

E1 E2

E1

E1 = Electric field due to charge q at distance a  1   q =     4 πε 0   a2 

4 π ε 0 mg R 2 3

E2 = Electric field due to charge at distance

2a

 1  q E = 1 =   4 πε 0  ( 2 a)2 2

30° √3F

Net electric field at P, E2 (In the direction of E2) 2 E 2E1 + 1 2 1 2 +  E1 2

Enet = 2 E1 +

mg

F = Electrostatic force between two charged balls. 3F = Resultant of electrostatic force on any one ball from rest two balls.

=  = 

30° 60° a

(2 2 + 1)q 8πε 0a2 1 q E= (rp − rq ) 4 πε 0 r3 =

l

30°

r

a = l cos 60° = 10 cm r = 2 a cos 30°  3 = (2) (10)   = 10 3 cm  2

14.

r = (1.2)2 + (16 . )2 = 2 m ∴ E=

(9 × 109 ) (− 8 × 10− 9 ) (1.2 $i − 10.8 $i ) (2)3

= (14.4 $j − 10.8$j) N/ C

642 — Electricity and Magnetism y

dq

dq dE θ E net θ

+dq

(b)

15.

θ θ

O

dθ

dE

+dq

θ θ

θ θ dE

Enet = 2 ∫ =2∫ =2∫

90° 0

dE

Enet

–dq

Enet

dE

dE sin θ

dE

(c)

90° 

1   dq      sin θ  4 πε 0   R 2 

0

x

dq

90° 

1   λRdθ  λ  sin θ =     4 π ε0  R2  2πε 0R

0

18. Six vectors of equal magnitudes are as shown in figure.

Enet dE

–12q

θ

θ

θ

θ

–q

–11q

dE

–10q

16.

r

2

a dq

–2q 1

dx

x

3 4 5

–9q

dq

–3q 6

–8q

Enet = 2∫

x = L/ 2 x=0

=2∫

L/ 2 0

dE cos θs –7q

 1   dq  a    ⋅   4 πε 0   r2   r 

 Q  ⋅ dx  L / 2 1   L = 2∫     0  4 πε  a2 + x 2 0    

Now, resultant of two vectors of equal magnitudes (= E say ) at 120° is also E and passing through their bisector line. So, resultant of 1 and 5 is also E in the direction of 3. Similarly, resultant of 2 and 6 is also E in the direction of 4. Finally, resultant of 2E in the direction of 3 and 2E in the direction of 4 passes through the bisector line of 3 and 4 (or 9.30)

a a + x2 2

y Enet dE θ θ

17. (a) dq

–5q –6q

Solving this integration, we find the result.

−dq

–4q

2u sin θ 2 × 25 × sin 45° 5 = = s g 10 2 1 R = S x = uxT + axT 2 2 1  qE  = (u cos 45° ) (T ) +   T 2 2  m

19. T = x

Electrostatics — 643

Chapter 24 2 −6 7  1   5   2 × 10 × 2 × 10   5  = (25)    +     2  2  1   2

= 62.5 + 250 = 312.5 m qE 20. (a) ge = g + m (1.6 × 10− 19 ) (720) = 10 + 1.67 × 10− 27

R=

dx

dq

u2 sin 2θ ge

X 2 × 10− 2 4 = × 10− 7 s = vx 15 . × 105 3

∴

Total potential = 6∫

= − 6.0 × 10− 4 J (b) ∆V = Ed = E (∆x ) = (250) (0.2) = 50 V

25. k A + U A = k B + U B 1 2 1 mvA + qVA = mvB2 + qVB 2 2 2 vB = vA2 + q (VA − VB ) m

4   −  2.1 × 1013 × × 10− 7  $j   3

200 cm 100 cm q1 P x = –200 cm x = 0

= (5)2 +

2 × (− 5 × 10− 6 ) (200 − 800) 2 × 10− 4

= 7.42 m/s vB − vA as the negative charge is moving (freely) from lower potential at A to higher potential at B. So, its electrostatic PE will decrease and kinetic energy will increase.

r1 = distance from q1 and r2 = distance from q2

First point is at x = 40 cm where,

dV

= − (12 × 10− 6 ) (50)

∴

= (1.5i$ + 2$j) × 105 m/s kq k | q2 | 22. 1 = r1 r2 r1 q 2 = 1 = ∴ r2 | q2 | 3

x = a/ 2 x=0

 1  Q k =   4 πε 0 

∆U = − q (∆V )

24. (a)

∴

v = u + at = (1.5 × 105 $i + 3.0 × 10 6 $j)

60 cm 40 cm q1 P x = 0 x = 40 cm

x

dq = λdx Potential at C, due to charge dq is k (dq) dV = r

(9.55 × 103 )2 sin 2θ 6.9 × 1010 Solving this equation, we get θ = 37° and 53° 2u sin θ (b) Apply T = ge (1.6 × 10− 19 ) (120$j) qE 21. (a) a = =− m (91 . × 10− 31 ) = − (2.1 × 1013 $j) m/s2

where,

a 2√3

(a/2)

∴ (1.27 × 10− 3 ) =

(b) t =

C r

≈ 6.9 × 1010 m/s2 Now,

23.

26. (a) VC =

k qnet r where, r = distance of P from any point on circumference  1   − 5Q  =   4 πε 0   R 2 + z2 

(b) Vp =

q2

r1 2 = r2 3 q2

Second point is at x = − 200 cm, where

k . qnet  1   − 5Q  =     4 πε 0   R  R

27. W = − ∆U = U i − U f = q2 Vi − q2 V f r1 2 = . r2 3

 kq  = q2  1  − q2  ri 

 kq1     rf 

644 — Electricity and Magnetism 1 1 = kq1 q2  −   ri rf  = (9 × 109 ) (2.4 × 10− 6 ) (− 4.3 × 106 ) 1   1 −    0.15 0.25 2  = − 0.356 J  q1q2 q1 q3 q2q3  28. (a) U = k  + +   r12 r13 r23  where, k =

…(i)

1 4 πε 0

Q=−

q 2

B

30. Apply VB − VA = − ∫ E ⋅ dr A

…(i)

E is given in the question. and dr = dx$i + dy $j ∴ E ⋅ dr = (5dx − 3dy) ∴ − ∫ E ⋅ dr = (3 y − 5x ) With limits answer comes out to be VB − VA = 3 ( yf − yi ) − 5 (x f − xi ) 31. Procedure is same as work done in the above question. The only difference is, electric field is E = (400$j) V/m B

∴ VB − VA = − ∫ E ⋅ dr = − 400 ( yf − yi ) A

32. Similar to above two problems. But electric field here is E = (20$j) N/C

∴ | E | = (− 20)2 + (− 20)2 + (− 20)2 = 20 3 N/C

34. See the hints of Q No. 31 to Q No. 33.

(b) Suppose q3 is placed at coordinate x (> 0.2 m or 20 cm), then in Eq. (i) of part (a), put U = 0, r12 = 0.2 m, r13 = x and r23 = (x − 0.2) Now, solving Eq. (i) we get the desired value of x. 29. U = 0  q × q q × Q q × Q ∴ k + +  =0  a a a  or

(c) Substituting, A = 10, x = 1, y = 1 and z = 1in the expression of part (b) we have E = (− 20$i − 20$j − 20k$ ) N/C

V f − Vi = VB − VA = − 20 (x f − xi ) − 30 ( yf − yi ) or V f = − 20 (2 − 0) − 30 (2 − 0) = − 100 V (as Vi = 0)  ∂V $ ∂V $ ∂V $  35. (a) E=− i+ j+ k  ∂x ∂y ∂z  = − [( Ay − 2 Bx ) $i + ( Ax + C )$j ] ∴ Ex = (− Ay + 2 Bx ) and E y = (− Ax − C ) (b) E = 0 if Ex and E y are separately zero. …(i) Q − Ay + 2 Bx = 0 and …(ii) − Ax − C = 0 Solving these two equations We get C 2 BC and y = − x=− A A2 36. Sphere is a closed surface. Therefore, Gauss’s theorem can be applied directly on this q φ total = in ∴ ε0 or

qin = (φ total ) (ε 0 )

= (360) (8.86 × 10− 12 )

= 3.19 × 10−9 C

37. (a) φ total

= 3.19 nC qin = ε0

qin = (φ total ) (ε 0 ) 38. S = (0.2) $i (b)

∴

3  φ = B ⋅ S = 0.2  E0 5 

B

= 0.2 × 0.6 × 2 × 103

A

= 240 N-m 2 /C

∴ VB − VA = − ∫ E ⋅ dr = − 20 (x f − xi )

[V ] [ ML2 T −3A−1 ] 33. (a) [ A ] = = [ xy ] [ L ][ L ] = [ MT −3 A−1 ]

 ∂V $ ∂V $ ∂V $  (b) E = −  i+ j+ k ∂y ∂z   ∂x

39. Electric flux enters from the surface parallel to

y - z plane at x = 0. But E = 0 at x = 0. Hence, flux entering the cube = 0. Flux leaves the cube from the surface parallel to y - z plane at x = a.

Electrostatics — 645

Chapter 24 Flux leaving the cube = ES

42. We have q θ

l

2

(at x = a) +

 E x  E a =  0  (a2 ) =  0  (a2 )  l   l 

2

l

√R

Substituting the values, we get (5 × 103 ) (10− 2 )3 φ= (2 × 10− 2 )

R

= 2.5 × 10− 1

φ=

= 0.25 N-m 2 /C

Here, flux due to + q and − q are in same direction. q (1 − cosθ ) φ total = 2φ = ∴ ε0

At all other four surfaces, electric lines are tangential. Hence, flux is zero. q ∴ φ net = (0.25) N - m 2 /C = in ε0 ∴

y

= 2.2 × 10− 12 C

40. See the hint of Example-1 of section solved

43.

examples for miscellaneous examples. We have (1 − cos θ ) Q φ= 2ε 0 Q But, = φ total ε0  1− cos θ  φ = (φ total )    2 

Given that

O

x q

C

Flux passing through hemisphere = flux passing through circular surface of the hemisphere. For finding flux through circular surface of hemisphere we can again use the concept used in above problem.

...(i)

q 45°

φ=

Substituting in Eq. (i), we get, θ = 60° R = tan 60° = 3 b ∴ R = 3b S j = (L2 ) (− $j ) φ s1 = E ⋅ S j = − CL2

Similarly, we can find flux from other surfaces. Take area vector in outward direction of the cube. (b) Total flux from any closed surface in uniform electric field is zero.

Note

a

a

b

R

∴

  R 2 + l 2  l

C = (a, 0, 0)

1 φ = (φ total ) 4 60°

41. (a)

q  1 − ε0  

=

qin = (0.25) (ε 0 ) = (0.25) (8.86 × 10− 12 )

∴

q (1 − cos θ ) 2 ε0

q (1 − cos θ ) 2ε 0

=

q (1 − cos 45° ) 2ε 0

=

q  1 1 −  2ε 0  2 σ

σ

44.

C r

A

B

R

σ (4 πr2 + 4 πR 2 ) = Q

646 — Electricity and Magnetism ∴

σ=

Potential at centre = potential due to A + potential due to B σr σR = + ε0 ε0 =

 σa  a  σb  b σc VC =     −     +  ε 0   c  ε 0   c ε 0

Q 4 π (r2 + R 2 )

σ (r + R ) ε0

=

 σ  a2 b2 + c  − ε0  c c 

(c) W A → C = 0 ∴

 a2 − b2  (a − b + c) =   +c  c 

or

 Q   r+ R  =     4 πε 0   r2 + R 2 

VA = VC

a+ b=c

or

P

45.

Q C2

120°

48. (a)

C1

–Q

Sphere

+Q

Q Ring (Total charge = q0)

σi =

q Charge on arc PQ of ring = 0 3 This is also the charge lying inside the closed sphere. q ∴ φ through closed sphere = in ε0 (q0 / 3) q = = 0 ε0 3ε 0

–Q Q +q

σi =

2q –2q zero

If outermost shell is earthed. Then, charge on outer surface of outermost shell in this case is always zero. σa σb σc 47. (a) VA = − + ε0 ε0 ε0 σ (a − b + c) ε0

 σa  a σb σc VB =     − +  ε 0   b ε 0 ε0 =

σ  a2 −b+ ε 0  b

−Q 4 πa2

(Q + q) 4 πa2 (c) According to Gauss’s theorem, 1 qin E= 4 πε 0 r2

–q

=

Q 4 π a2

Q

and

+q A

σ0 =

and

(b)

46.

B

−Q 4 πa2

 c 

σ0 =

For x ≤ R, qin = Q and r = x both cases. 1 Q E= ∴ 4 πε 0 x 2

49. Let Q is the charge on shell B (which comes from earth) ∴ ∴

VB = 0 kq kQ kq + − =0 b b c b  Q =  − 1 q c 

Chapter 24 Charges appearing on different faces are as shown below.

q –q Q +q – (Q +q) Q

Total charge on A + C is 3q. Therefore, …(i) − q + q1 − q2 + q3 = 3q VB = 0  q − q2   q − q  q − q1  ∴ k 1  +k 2  +k 3  =0  2R   2R   3R  …(ii) VA = VC  q − q  q − q1   q − q2  ∴ k 1  +k 2  +k 3   R   2R   3R   q − q  q − q1   q − q2  =k  1  +k 2  +k 3   3R   3R   3R 

 b Q + q=  q  c

k=

A

=

Total charge on A + C is 3q ∴ q1 − q2 + q3 = 3q VB = 0 kq1  q2 − q1   q − q2  +k ∴  +k 3  =0  2R   3R  2R

…(i) …(ii)

VA = VC kq1  q − q1   q − q2  +k 2  +k 3     3R  R 2R

∴

kq1  q − q1   q − q2  +k 2  +k 3   3R   3R  3R 1 . In the above equations, k = 4 πε 0 =

1 4 πε 0

Solving above there equations, we can find q1 , q2 and q3. 52. (a) From Gauss’s theorem  1   qin   1   2Q  E=    =     4 πε 0   r2   4 πε 0   r2 

O q1 –q1 q2 –q2 q3

C

…(iii)

Solving these three equations, we can find the asked charges.

Q 2πε 0r2

(b) According to principle of generator, potential difference depends only on qin . 2θ  1 1 Q ∴ PD =  − = 4 πε 0  R 3R  3πε 0R (c) According to principle of generator, whole inner charge transfers to outer sphere. (d) Vin = 0 Q k qin kQ ∴ − = 0 ⇒ qin = 3 R 3R

53. (a) At r = R, V =

1  Q 2Q 3Q  − + 4 πε 0  R 2R 3R 

 1   Q =     4 πε 0   R  At r = 3R 1  Q 2Q 3Q  Q V = − + =   4 πε 0  3R 3R 3R  6πε 0R

51.

A B C

q –q q1

–q1

q2 –q2 q3

…(iii)

In the above equations,

50.

B

Electrostatics — 647

 1   qin  (b) E =      4 πε 0   r2   1   Q − 2Q  −Q = =   4 πε 0   (5R / 2)  25πε 0R 2

648 — Electricity and Magnetism Minus sign implies that this electric field is radially towards centre.

qA = q =

Now,

Q 2

7 Q 2 1 q 1  q − 2Q  − 3Q (e) E = ⋅ in = = 4 πε 0 r2 4 πε 0  (5R / 2)2  50πε 0R 2 qC = 4 Q − q =

(c) U3

U2

Minus sign indicates that electric field is radially inwards.

U1 Q –Q –Q +Q +2Q

LEVEL 2 Single Correct Option 1. Let us conserve angular momentum of + 2q about

UT = U 1 +U 2 + U 3 1Q 2 C1

where,

U1 =

where,

1 1 C 1 = 4 πε 0  −   R 2R  U2 =

the point at + Q. mv1r1 sin θ 1 = mv2 r2 sin θ 2 (m)(v )(R )sin 150°  v  = m  rmin sin 90°  3

2

1 Q2 2 C2

3 R 2 2. (vA ) y = v ⇒ (vB ) y = 2v sin 30° = v ∴

where ,

1  1 C 2 = 4 πε 0  −   2R 3R 

and

U3 =

where,

1  1 C 3 = 4 πε 0  −   3R ∞ 

Since, y -component of velocity remains unchanged. Hence electric field is along (− $i ) direction. Work done by electrostatic force in moving from A to B = change in its kinetic energy 1 ∴ (eE ) (2a − a) = m (4 v 2 − v 2 ) 2 mv 2 E= 2ea 3mv 2 $ or E=− i 2ea Rate of doing work done = power = Fv cos θ

1 (2Q )2 2 C3

C B

(d)

A

q –q –2Q + q 2Q – q 2Q

2

 3mv 2  =  (e) (2v ) cos 30°  2ea  =

Total charge on ( A + C ) is 3Q + Q = 4 Q. VA = VC kq k (2Q ) k (4 Q − q) − + ∴ R 2R 3R k (Q − 2Q + 3Q ) = 3R Solving this equation, we get Q q= 2

rmin =

3 3 mv 2 2 a

Now,

E

3. …(i)

a +q

b –q

x

Just to the right of a, electric field is along ab (∴positive) and tending to infinite. Similarly,

Chapter 24 electric field just to the left of b electric field is again along ab (∴positive) and tending to infinite. Q

4.

–Q E1 60° 60° E2

Q

–Q

E3 5 –Q

Q

 1   P E = E1 = E2 = E3 =      4πε 0   x 3   1  2Qa =   4 πε 0  x 3 Resultant of E1 and E3 is also equal to E along E2 (along E2) ∴ Enet = 2E Qa = πε 0 x 3

Electrostatics — 649

Let Q charge comes on shell-C from earth. Then, VC = 0 kq kQ kq ∴ + − =0 3a 3a 4 a Solving, we get q Q=− 4 kq kq/ 4 kq kq Now, VA = − − = 2a 3a 4 a 6a and VC = 0 kq ∴ VA − VC = 6a q 8. ρ = (4 / 3) πR 3 4 q = πρR 3 ∴ 3 3 1 1 q q ⋅ VC − VS =  ⋅  − 2  4 πε 0 R  4 πε 0 R =

5. From centre to the surface of inner shell, potential

q 8πε 0R

Substituting the value of q , we have ρR 3 VC − VS = 6 ε0

will remain constant = 10 V (given). 6. By closing the switch whole inner charge transfers to outer shell.

F2

9. U2 U1

–q q

⇐

U2

q

2

1 q 2 C 4 πε 0 C = = 8πε 0a (1/ a − 1/ 2)a =U1 =

∴

F1

q

B

Heat produced = U i − U f = (U 1 + U 2 ) − U 2

where,

A

Heat =

q2 kq2 = 16πε 0a 4 a –q

7. q A B C D

vA = 0 vC = 0

C

Hence, in between A and C there is a point B, where speed of the particle should be maximum. F1 = mg = constant F2 = electrostatic repulsion (which increases as the particle moves down) From A to B kinetic energy of the particle increases and potential energy decreases. Then, from B to C kinetic energy decreases and potential energy increases. 10. Over Q1, potential is + α. Hence, Q1 is positive. VA = 0 and A point is nearer to Q2. Therefore, Q2 should be negative and | Q1 | > | Q2 |. At A and B, potential is zero, not the force. Equilibrium at C will depend on the nature of charge which is kept at C.

650 — Electricity and Magnetism k min + qVC1 = 0 + qVC2 1 Q Q VC1 = −   4 πε 0  2R R 

11. V1 is positive and V2 is negative. Hence at all points, V1 > V2 12. Just to the right of q1, electric field is + α or in positive direction (away from q1). Hence, q1 is positive. Just to the left of q2, electric field is − α or towards left (or away from q2). Hence, q2 is also positive. Further E = 0 near q1. Hence, q1 < q2 13. Electric lines of forces of q will not penetrate the conductor. 14. E = 400 cos 45° $i + 400 sin 45° $j

VC2 =

…(i)

1 Q Q  −  4 πε 0  R 2R 

Substituting these values in Eq. (i), we can find K min . 20. E = Exi$ + E y$j Now we can use,

∫ dV

= − ∫ E ⋅ dr

two times and can find values of Ex and E y. 21. Let P = (x , y)

A

VA − VB = − ∫ E ⋅ dr

V=0

B

dr = dx i$ + dy $j

where,

C X=a

15. qA will remain unchanged. Hence, according to principle of generator potential difference will remain unchanged. VA′ − VB′ = VA − VB or (as VB′ = 0 ) VA′ = VA − VB 16. WT = 0 W Fe = (Fe ) (displacement in the direction of force) = Kinetic energy of the particle. 1 2 l l  ∴ mv = qE − cos 60° 2  2 2  qEl m

v=

∴

L∝t

18. U i + K i = U f + K f or

qVi +

1 2 mvmin = qV f + 0 2

 1   Q 1 2 3 Q  or q     + mvmin + q  ×   4 πε 0   R  2  2 4 πε 0R  From here, we can find vmin .

19.

Kmin C2

X = 9a

R = 4a

r`1 = (x + 3a)2 + y2 r2 = (x − 3a)2 + y2 Vp =

1 4 πε 0

 Q 2Q  r − r  = 0  2 1 

C1

–Q

2R +Q

kC1 + UC1 = kC2 + UC2

…(i)

Substituting values of r1 and r2 in Eq. (i), we can see that equation is of a circle of radius 4a and centre at 5a.

22. Fx = 0

17. L = mv r⊥ = m (at ) (x0 )  qE  = m  0  t (x0 ) or  m

X = 5a

∴

ax = 0 Fy = qE qE ∴ ay = m x = vt and 1 1  qE  y = ayt 2 =   t 2 2 2  m x Substituting t = in expression of y , we get v 1  qEx 2  y=   2  mv 2  1 m (vx2 + v 2y ) 2 qE where, vx = v and v y = ayt = t m KE =

Chapter 24 23. qE =

 1   1 q q ⋅   −2  4 πε 0 a 4 πε 0 b2 + a2    − 1 / 2   q  b2  2 − 2 1 + 2   = 4 πε 0a  a     Since, b < < a, we can apply binomial expansion  q  b2   Vp = ∴ 2 − 2 1 − 2   4 πε 0a  2a   

mv 2 r

30. Vp = 2   λ  mv 2 q  =  2πε 0r r

∴

gλ = 2πε 0m

v=

Now,

T =

2kqλ m

2πr v

=

Tmin

24.

π

C α

T

charge ± q. Then, 5E

T sin α = qE T cos α = mg α = tan − 1

5E

E1 = (5E )2 + (5E )2 = 5 2E 5E

1  q2   q2 q q2 q2 q2 q2   = − − − +    −  − 4 πε 0  a   a a a 2a a 2a

5E

E2 is again 5 2E.

26. On both sides of the positive charge V = + ∝ just (a = side of triangle)

Similarly, we can find E3 and E4 also.

32. U =

2

1   9 × 109 × × 10− 9  ε0 1  3 9 = ε0   = 2 J/m 2 2 (1)    

= 3U − U = 2U 28. U i + K i = U f + K f r∝

1 1  1 q ε 0E 2 = ε 0   2 2  4 πε 0 R 2 

2

 1 q2  W =U f −Ui = 3   −U  4 πε 0 a 

1 2 1 Qq mv = ⋅ + 0 or 2 4 πε 0 r

E2

(ii)

q2 = [ 2 + 1] 4 πε 0a over the charge. 1 q2 27. U = ⋅ 4 πε 0 a

E1

(i)  qE     mg 

Minimum tension will be obtained at α + π . 25. Energy required = ∆U = U f − U i

0+

qb2 4πε 0a3

31. Let E = magnitude of electric field at origin due to Tmax qE

mg

∴

Electrostatics — 651

1 v2

If v is doubled, the minimum distance r will 1 remain th. 4 29. See the hint of Sample Example 24.9 ρ 1.6 K = = =2 ρ − σ 1.6 − 0.8

B

33. A

They have a common potential in the beginning. This implies that only B has the charge in the beginning.

652 — Electricity and Magnetism kqB or kqB = Vb b Now, suppose qA charge is given to A. Then, kq kq VA = A + B = 0 a b  kqB  or kqA = − a   = − aV  b 

∴

V =

kqA kqB + b b a a  = − V + V = V 1 −   b b $ $ $ 34. Let E = Ex i + E y j + Ezk Now,

Apply

VB =

∫ dV

= − ∫ E ⋅ dv

three times and find values of Ex , E y and Ez. Then, again apply the same equation for given point.

Since | QB | > QA, electric field outside sphere B is inwards (say negative). From A to B enclosed charge is positive. Hence, electic field is radially outwards (positive).  ∂V $ ∂V $  38. E = −  i+ j = − [(− ky) $i + ( − kx ) $j ] ∂y   ∂x ∴

=k ∴

|E | ∝ r

kqA kqB + R 2R 3 kqA kqB VB = V = + 2 2R 2R Solving these two equations, we get qA 1 = qB 2

1. (a) VA = 2V =

35. q 2

B

Let charge on B is q′ ∴

VB = 0 (k (q/ 2)) kq′ + =0 d r

+q P

qA

– qA = q B

(b)

qr q′ = − 2d

∴

36.

x 2 + y2 = kr

More than One Correct Options

q′

A

| E | = (ky)2 + (kx )2

– – – –

+ + + –

q′A q = A = −1 q′B − qA

+

–

+

The induced charges on conducting sphere due to + q charge at P are as shown in figure. Now, net charge inside the closed dotted surface is negative. Hence, according to Gauss’s theorem net flux is zero.

(c) & (d) Potential difference between A and B will remain unchanged as by earthing B, charge on will not changed. ∴ VA′ − VB′ = VA − VB 3 V = 2V − V = 2 2 V ∴ VA′ = 2 as VB′ = 0

2. T =

37. E=0 +QA –QB

H = R=

2uy g u2y 2g

=

2 × 10 =2s 10

=

(10)2 =5m 20

1 1  qE  axT 2 =   T 2 2 2  m

Chapter 24 =

Electrostatics — 653

1  10− 3 × 104    (2)2 2 2 

qE

= 10 m 1 q 3. 100 = ⋅ 4 πε 0 (R + 0.05) 1 q 75 = ⋅ 4 πε 0 (R + 0.1)

…(i)

qE

…(ii)

Solving these equations, we get 5 q = × 10− 9 C 3 and R = 0.1 m 1 q (a) V = ⋅ 4 πε 0 R 5  (9 × 109 )  × 10− 9  3  = 0.1 = 150 V 1 q V 150 (c) E = ⋅ = = 4 πε 0 R 2 R 0.1

qin = 0, due to a dipole. kq 8. E = 2in r

9.

1. Vouter = 0 kQ kQ1 + =0 2r 2r ∴ Q1 = − Q = charge on outer shell

∴

 1   Here, k =   4πε 0 

EA = Ec = 0 EB ≠ 0 kq V = R kq V = r

(r ≤ R ) (r ≥ R )

Fe = qE

2. Vinner = 0 kQ2 kQ1 + =0 r 2r Q Q Q2 = − 1 = = charge on inner shell ∴ 2 2 Charge flown through S 2 = initial charge on inner shell − final charge on it Q = Q − Q2 = 2 Q 3. After two steps charge on inner shell remains or 2 half. So, after n-times Q qin = (2)n ∴

Now, according to the principle of generator, potential difference depends on the inner charge only. q 1 1  PD = in − ∴ 4 πε 0  r 2r 

Hinge force Fe = qE

Higher force = 2qE

If we displace the rod, τ 1 = τ 2 or τ net = 0 in displaced position too. Hence, equilibrium is neutral. 10. Along the line AB, charge q is at unstable equilibrium position at B (When displaced from B along AB, net force on it is away from B, whereas force at B is zero). Hence, potential energy at B is maximum. Along CD equilibrium of q is stable. Hence, potential energy at B is minimum along CD.

Comprehension Based Questions

= 1500 V/m (d) Vcentre = 15 . Vsurface 5. Electric field at any point depends on both charges Q1 and Q2. But electric flux passing from any closed surface depends on the charged enclosed by that closed surface only. q 6. Flux from any closed surface = in , ε0

∴ but,

T1 T2

(towards left)

=

1 2n + 1

 Q   4 πε r   0 

654 — Electricity and Magnetism 4. According to Gauss’s theorem,  1   qin  E=     4 πε 0   r2 

q2

–Q

…(i)

For r ≤ R r

qin = ∫ (4 πr2 ) ⋅ dr ⋅ ρ

(ii)

0

Vinner = 0 when solid sphere is earthed kq2 kQ Q − =0 a b  a Q q2 = Q    b

r r  = ∫ (4 πr2 ) (ρ 0 ) 1 −  dr 0  R  r3 r4  = 4 πρ 0  −   3 4 R

Substituting in Eq. (i), we get ρ  r r2  E= 0  −  ε  3 4r 

10. Whole inner charge transfers to shell. ∴ Total charge on shell = q2 − Q a  = Q  − 1 b 

5. For outside the ball, E= where,

1 qtotal 4 πε 0 r2

…(i)

R r  qtotal = ∫ (4 πr2 ) (ρ 0 ) 1 −  dr 0  R

Match the Columns 1. (a) EC and EF are cancelled. EE and ED at 60° (b) EB and EE are cancelled. EF and ED at 120°. (c) EB and EE are cancelled. Similar, EF and EC are cancelled. (d) EF and ED at 120°. So, their resultant is E in the direction of EE . Hence, net is 2E.

Substituting this value in Eq. (i), we get ρ R3 E= 0 2 12 εr

6. For outside the ball, electric field will continuously decrease. Hence, it will be maximum somewhere inside the ball. For maximum value, dE =0 dr d ρ 0  r r2   ∴  = 0   − dr  ε  3 4 R   2R Solving, we get r = 3 2R 7. Submitting r = in the same expression of 3 electric field, we get its maximum value. 8. Potential difference in such situation depends on inner charge only. So, potential difference will remain unchanged. Hence, ∆V = Va − Vb q1 = 0

Q

–Q

9. (i)

∫ dV

2. ∴

= − ∫ E ⋅ dr A

VA − VB = − ∫ E ⋅ dr B

kq ⇒ kq = VR R kq (a) V = 3 (1.5R 2 − 0.5r2 ) R 2 VR  3 1  R  = 3  R2 −    2  2  R 2 11 = V 8 kq VR V (b) V = = = r 2R 2 kq (c) E = 3 ⋅ r R V (V )  R  V = R3   = =   2 R 2 2 (R ) kq VR (d) E= 2 = r (2R )2 V V = = for R = 1 m 4R 4

3. V =

(if R = 1 m)

Electrostatics — 655

Chapter 24

q3 = − q2 = + 2 Q q4 = 2Q − q3 = 0 q5 = − q4 = 0

Subjective Questions 1. (a) By comparing this problem with spring-block system problem suspended vertically. Here, mg ≡ qE = 50 × 10−6 × 5 × 105 = 25 N X max = 2 mg/K X max = 2 qE / K =

Here,

= 50 cm

or

2. Total charge on ring = λ (2πa) = q (say) Electric field at distance x from the centre of ring. 1 qx λax = . 2 2 3/ 2 2 4 πε 0 (a + x ) 2ε 0 (a + x 2 )3/ 2

Restoring force on − Q charge in this position would be   λaQx F = − QE = −  2 2 3/ 2   2ε 0 (a + x )  For x (l + a), field intensity is positive. Therefore, Q1 is positive. (b) At x = l + a, field intensity is zero. 2

kQ1 kQ Q   l + a = 22 or  1 =  ∴  2 (l + a) a Q2  a  (c) Intensity at distance x from charge 2 would be kQ1 kQ − 22 E= (x + l )2 x dE For E to be maximum =0 dx 2kQ1 2kQ2 or − + =0 3 (x + l ) x3 or

 1 + 

3

l Q  l + a   = 1 =  x Q2  a 

2

In the second contact, S 1 again acquires the same charge Q. Therefore, total charge in S 1 and S 2 will be  R  Q + q1 = Q 1 +  R + r  This charge is again distributed in the same ratio. Therefore, charge on S 2 in second contact,  R   R  q2 = Q 1 +    R + r  R + r  2  R  R   =Q +    R + r  R + r  Similarly, 2 3  R  R   R   q3 = Q  +  +    R + r   R + r  R + r  and 2 n  R  R   R   qn = Q  + + … +      R + r   R + r  R + r 

qn = Q

or

2/ 3

or or

or

1+

l  l + a =  x  a  l x= 2/ 3  l + a −1    a  b=

l  l + a    a 

Ans.

2/ 3

−1

Electrostatics — 657

n R  R   1 −    r   R + r   

 a (1 − rn )  S n = (1 − r)    Therefore, electrostatic energy of S 2 after n such contacts a2 qn2 qn2 or U n = Un = n = 2C 2(4 πε 0R ) 8πε 0R where, qn can be written from Eq. (ii).

10. Capacities of conducting spheres are in the ratio of

n− 1   R  QR  R +…+…+  1 +   R+r R+r  R + r   as n → ∞

their radii. Let C 1 and C 2 be the capacities of S 1 and S 2 , then C2 R = C1 r

(b) qn =

(a) Charges are distributed in the ratio of their capacities. Let in the first contact, charge acquired by S 2 , is q1. Therefore, charge on S 1 will be Q − q1. Say it is q1′ . q1 q1 C R ∴ = = 2= q1′ Q − q1 C 1 r

q∞ =

QR R+r

1    R  1 −   R + r 

=

QR R+r

R  R + r =Q   r  r

It implies that Q charge is to be distributed in S 2 and S 1 in the ratio of R/r.  R  …(i) q1 = Q  ∴   R + r

…(ii)

∴

U∞ =

q∞2 Q 2R 2 / r2 = 2C 8πε 0R

or

U∞ =

Q 2R 8πε 0r2

 a  S ∞ = 1 − r   

Ans.

658 — Electricity and Magnetism 2qV  l 11. vx = v =  =l , t=  m 

vx

ay =

14. From energy conservation principle,

m 2qV

qE qat dv y = = m m dt

or

–

or y

+

v

E

x +

Integrating both sides, we get qat 2 vy = 2m 2  qa   m  al or v y =   l2   =  2m  2qV  4V Now, angle of deviation  al 2  vy θ = tan −1   = tan −1   vx   4V  al 2 = tan −1   4V

Ki + U i = K f + U f 1 2 mv + (+ q)Vi = 0 + (+ q)V f 2 2q v= (− Vi + V f ) m

/

=

2q  − Q 8Q Q 8Q  1 + + − m  10R 5R R 4 R  4 πε 0

=

Qq  3 10 − 5   2πε 0mR  5 10 

15. (a) E =

kQx , (R 2 + x 2 )3/ 2

3  2  (R + x 2 )3/ 2 − x. (R 2 + x 2 )1/ 2 (2x )   dE 2 = kQ   dx (R 2 + x 2 )3    

2qV   m 

m   2eV 

or

 R 2 + x 2 − 3x 2  dE = kQ  2 2 5/ 2  dx  (R + x ) 

or

dE =

Ans.

12. From energy conservation, UC + KC = U D + K D −5

∴ | ∆E | = −5

 9 × 10 × (5 × 10 )( − 5 × 10 )  or 2  + 4 5   9

Q 4 πε 0

∴ F = | q∆E | =

Ans.

=

U i + Ki = U f + K f qVi + K i = qV f + K f  Q  1 2 or q   + mv  4 πε 0r 2

 R 2 − 2x 2   2 2 5/ 2   (R + x ) 

aqQx πε 0 (R 2 + x 2 )3/ 2

Ans.

conservation of angular momentum about point O, we have v1

 Q   R  =q . R 2 − 0.5  + 0  15 4  4 πε 0R 3   2

1 2 11 Qq Qq mv = − 2 3πε 0R 4 πε 0r Qq  r − R 3 +   2πε 0mR  r 8

Qqa 2πε 0

16. From conservation of mechanical energy and

or

v=

Ans.

= − pE cos 180° + pE cos 0° = 2pE  1  Qx = 2(q)(2a)  2 2 3/ 2   4 πε 0 (R + x ) 

13. From conservation of energy,

or

 R 2 − 2x 2  ∆x  2 2 5/ 2   (R + x ) 

(b) W = U f − U i

OD = (9)2 − (3)2 = 72 m

Q  R 2 − 2x 2  dx 4 πε 0  (R 2 + x 2 )5/ 2 

Here, ∆x = 2a

 9 × 109 × (5 × 10−5 )(− 5 × 10−5 )  =2 + 0 AD   Solving we get AD = 9 m ∴ Maximum distance,

or

Ans.

B

r2

O r1 +Q

A

v2

Ans.

1 2 1 Qq 1 2 1 Q. q …(i) mv1 − . = mv2 − . 2 4 πε 0 r1 2 4 πε 0 r2

Electrostatics — 659

Chapter 24 mv1r1 sin 90° = mv2r2 sin 90°

and

v1r1 = v2r2

or

∴

E=

…(ii)

Solving these two equations, we have v1 =

Qqr2 2πε 0mr1 (r1 + r2 )

Qqr1 = v2 2πε 0mr2 (r1 + r2 )

and

=

1 q1 . 4 πε 0 r 2 9 × 109 × (200/9) × 10−12 (125 . × 10−2 )2

= 1.28 × 103 V/m

18. (a) Fnet = 2F cos θ Ans.

17. Let q1 : q2 and q3 be the respective charges. Then,

=

2kQ. q ( R + 2

x02 )2

.

x0 R + 2

+Q R

and

9 × 109 10−2

0=

9 × 10  q1 q2 q3  + + 2 4  10−2  2

40 =

9 × 109  q1 q2 q3  + + 4 4  10−2  4

Solving these equations, we get 200 q1 = + × 10−12 C, q2 = − 200 × 10−12 C and 9 3200 q3 = × 10−12 C 9 (a) At r = 1.25 cm  (200/9) × 10−12 200 × 10−12  −  9 × 10  1.25 2 V =   −2 10 (3200/9) × 10−12   +   4 Ans. =6V 9

(b) Potential at r = 2.5 cm  (200/ 9) × 10−12 200 × 10−12  −  9 × 10  2.5 2.5 V =   −2 − 12 10 (3200/9) × 10   +   4 Ans. = 16 V 9

(c) Electric field at r = 1.25 cm will be due to charge q1 only.

 1   Here, k =  4 πε 0  

√R 2 + x 02 F

x0 θ θ –q F

 q1 q2 q3  + +  1 2 4 

9

x02

2kQqx0 = 2 (R + x02 )3/ 2

q1 q 2 q3

10 =

Ans.

+Q

We can generalised the force by putting x0 = x, we have 2kQqx Ans. F =− 2 (R + x 2 )3/ 2 (b) Motion of bead will be periodic between Ans. x = ± x0 x (c) For τC2 Further, ∴

τC = CR τC ∝ R

q C 1 V ∝ C

V =

if q is same. 6. Charge (or current) will not flow in the circuit as they have already the same potential, which is a condition of parallel grouping. Further, q1 C 1V C 1 1 = = = q2 C 2V C 2 2

7. Capacitor and R2 are short-circuited. Hence, (as C = constant)

current through R2 is zero and capacitor is not charged.

662 — Electricity and Magnetism 8. Capacitor and resistance in its own wire are directly connected with the battery. Hence, time constant during charging is CR. 1 q2 1 or U ∝ as q is same in capacitors 9. U = 2C C (if initially they are uncharged) 10. By inserting dielectric slab, value of C 2 will increase. In series, potential difference distributes in inverse ratio of capacitance. If capacitance C 2 is increased PD across C 2 will decrease. If C 2 is increased, charge on capacitors will also increase. So, positive charge or current flows in clockwise direction.

Objective Questions  q   σ   q will not change.  =q  2 Aε 0   2ε 0 

9. At t = 0, when capacitor is under charged, equivalent resistance of capacitor = 0 In this case, 6 Ω and 3 Ω are parallel (equivalent = 2Ω) ∴ Rnet (1 + 2) Ω = 3 Ω 12 = 4A ∴ Current from battery = 3 = Current through 1Ω resistor 10. Final potential difference = E ∴ Final charge = EC a

i/4 i/4

11.

1. F = qE = q  ∴

F = constant

E

= (4 πε 0 a) + (4 πε 0 b) = 4 πε 0 (a + b) (in series)

=V + V +… = nV

∴

5. V32 = V5 = 6 V ∴

q5 = CV = 30 µC  3 × 2 q32 =   × 6 = 7.2 µC  3 + 2 q5 30 = q2 7.2

∴

∴

…(i)

 5  4 E      × R = 10  4   7R 

15 V

60 V

i Va − R − iR = Vb 4 5 Va − Vb = iR = 10 4 (3R ) (R ) Rnet = R + (3R + R ) 7 = R 4 E  4 E i= =  (7/ 4 )R  7R 

Substituting in Eq. (i), we have

A

6.

i

3i/4 i

3. C = C 1 + C 2

4. Vnet = V1 + V2 + …

b

30 V

∴

E = 14 V

12. All capacitors have equal capacitance. Hence,

B 15 V

qE = mg

7. Q

equal potential drop (= 2.5 V) will take place across all capacitors. VN − VB = 2.5 V

V   4  q   =  πr3ρ g  d  3  V ∝ +σ

0 − VB = 2.5 V ∴

3

r q

Further,

VB = − 2.5 V VA − VN = 3 (2.5) V = 7.5 V

–σ

8. E=0

E=0 E0

σ ⇒ E= = constant ε0

∴

VA = + 7.5 V

(as VN = 0)

13. q = CV = 200 µC In parallel, the common potential is given by

Chapter 25  σ  d  2ε 0 

Total charge Total capacity 200 µC = = 50 V (2 + 2) µF

V =

20. V = Ed = 

d=

∴

Heat loss = U i − U f 1 1 = (2 × 10− 6 ) (100)2 − (4 × 10− 6 ) (50)2 2 2 = 5 × 10− 3 J = 5 mJ

(8.86 × 10− 12 ) (5) 10− 7 = 0.88 × 10− 3 m =2

= 0.88 mm short-circuited).

= (i02R ) e− 2t / η

22. The equivalent circuit is as shown below.

= P0e− t / (η/ 2) η Hence, the time constant is . 2 15. Common potential in parallel grouping Total charge = Total capacity EC E = = 2C 2 9 16. VA − 6 − 3 × 2 + − 3 × 3 = VB 1 ∴ VA − VB = 12 V 17. In steady state condition, current flows from outermost loop. 12 i= = 1.5 A 6+ 2

1 µF

u2y 2g

=d

Dividing Eq. (ii) by Eq. (i), we have 1  uy  d   = 4  ux  l or

uy ux

=

u sin θ u cos θ

= tan θ =

4d l

Y

2 µF

23. C 1 = C RHS + C LHS K 2 ε 0 ( A / 2) K 1ε 0 ( A / 2) + d d ε0A 5ε 0 A = (K 1 + K 2 ) = 2d 2d ε0A C2 = d /2 d /2 + d − d /2 − d /2 + K1 K2 =

=

= 1.5 × 6 = 9 V

Maximum height, H =

1 µF

1 µF

X

Now, VC = V6Ω = iR ∴ q = CVC = 18 µC 19. Horizontal range, 2ux × uy R= =l g

2ε 0V σ

21. Three capacitors (consisting of two loops are

P = i 2R = (i0e− t / η )2 R

14.

Capacitors — 663

2ε 0 A  K 1K 2    d  K 1 + K 2

12 ε 0 A 5d C 1 25 = C 2 24 =

∴

…(i)

24. A balanced Wheatstone bridge is parallel with C. …(ii)

25. First three circuits are balanced Wheatstone bridge circuits.

26.

C = C LHS + C RHS =

=

K 1ε 0 ( A / 2) ε 0 ( A / 2) + d /2 d /2 d d − d /2 − d /2 + + K2 K3 ε0A  K 1 K 2K 3  + d  2 K 2 + K 3 

664 — Electricity and Magnetism C V2 C C C/2

V1

27. V1

B

V2

ε0A d 3. All three capacitors are in parallel with the battery. PD across each of them is 10 V. So, apply q = CV for all of them.

ε0A = 7 µF d The equivalent circuit is as shown in figure.

4. Capacitor and resistor both are in parallel with the

C =

V1 C V 3 V2

A V1

V1 V4 V4 V3 C

V3 C

B V2

C

V2

11 11 C = (7 µF) 7 7 = 11µF

C AB =

Subjective Questions

qnet = C net V 2  =  µF (1200 V) 3 

1. Charge on outermost surfaces

qtotal (10 − 4 ) µC = 2 2 = 3 µC Hence, charges are as shown below. =

3 µC

7 µC

–7 µC

3 µC

2. Charge on outermost surfaces

qtotal 2q − 3q q = =− 2 2 2 Hence, charge on different faces are as shown below. =

–

q 2

2.5q –2.5q

battery. PD across capacitor is 10 V. Now, apply q = CV . 5. In steady state, current flows in lower loop of the circuit. 30 i= = 3A 6+ 4 Now, potential difference across capacitor = potential difference across 4 Ω resistance. = iR = (3) (4 ) = 12 V ∴ q = CV = (2 µF) (12 V) = 24 µC C C 2 6. (a) C net = 1 2 = µF C1 + C2 3

C/2 V1

 σ  2.5 q d  =  d=    ε0  ε0A  Capacitance, C =

V4 V3

C V1

A

V3

–

q 2

Electric field and hence potential difference between the two plates is due to ± 2.5 q. PD = Ed

= 800 µC In series, q remains same. ∴ q1 = q2 = 800 µC q V1 = 1 = 800 V C1 q and V2 = 2 = 400 V C2 (b) Now, total charge will become 1600 µC. This will now distribute in direct ratio of capacity. q1 C 1 1 = = ∴ q2 C 2 2 1600  1 q1 =   (1600) = µC  3 3  2  3200 q2 =   (1600) =   µC  3  3  They will have a common potential (in parallel) given by Total charge V = Total capacity

Chapter 25 =

1600 µC 3 µF

q EC (EC – q0) q0

1600 V 3 7. Charge, q = CV = 104 µC =

In parallel, common potential is given by Total charge V = Total capacity 20 =

(10 µC) (C + 100) µC 4

Solving this equation, we get C = 400 µF 8. Charge supplied by the battery, q = CV Energy supplied by the battery, E = qV = CV 2 Energy stored in the capacitor, 1 U = CV 2 2 ∴ Energy dissipated across R in the form of heat 1 = E − U = CV 2 = U 2 − t/ τ C 9. i = i0 e i Putting i = 0 , we get 2 t = (ln 2) τC = (0.693) τC

10. Both capacitors have equal capacitance. Hence, half-half charge distribute over both the capacitors. q q1 = q2 = 0 2 q q1 decreases exponentially from q0 to 0 while q2 2 q0 increases exponentially from 0 to . 2 Corresponding graphs and equation are given in the answer. Time constant of two exponential equations will be τC = (C net ) CR C  R=  R=  2 2

Capacitors — 665

t

∴

q = q0 + (EC − q0 ) (1 − e− t / τ C ) = EC (1 − e− t / τ C ) + q0e− t / τ C

Here, τC = CR

12. (a) Immediately after the switch is closed whole current passes through C 1. ∴ i = E / R1 (b) Long after switch is closed no current will pass through C 1 and C 2. E ∴ i= R1 + R3

13. (a) At t = 0 equivalent resistance of capacitor is zero. R1 and R2 are in parallel across the battery PD across each is E. ∴ iR1 = E / R1 iR 2 = E / R2 (b) In steady state, no current flow through capacitor wire. PD across R1 is E. ∴ iR 1 = E / R1 and iR 2 = 0 (c) In steady state, potential difference across capacitor is E. 1 1 U = CV 2 = CE 2 ∴ 2 2 (d) When switch is opened, capacitor is discharged through resistors R1 and R2. τC = CRnet = C (R1 + R2 ) 14. (a) Simple circuit is as shown below.

A

B

(b) The simple circuit is as shown below.

11. qi = q0 qf = EC Now, charge on capacitor changes from qi to qf exponentially.

A

B

666 — Electricity and Magnetism (c) Let C AB = x. Then,

VA − VB =

or 2C

∴

A C

q across each capacitor. C 20. See the answer. 21. In series, potential difference distributes in inverse ratio of capacitance. VA C B C 2 60 3 = = = = ∴ VB C A C 1 40 2 Now, V =

x

B

(2C ) (x ) 2C + x 2Cx x =C+ 2C + x

C AB = C +

Now, or

∴

Solving this equation, we get x = 2C 15. (a) V = 660 V across each capacitor

Now,

Now, q = CV for both (b) qnet = q1 + q2 = (3.96 − 2.64) × 103 C

= 132 V Now, apply q = CV for both capacitors. 1 16. u = ε 0 E 2 2 2

=

or

Kε 0 A d dC (Vmax ) (C ) A= = K ε 0 (K ) (Emax ) (ε 0 )

C =

18. 0.1 =

1 (C 1 + C 2 ) (2)2 2 1 CC  1.6 × 10− 2 =  1 2  (2)2 2  C 1 + C 2

…(i) …(ii)

Solving these two equations, we can find C 1 and C 2.

19.

A

B +

– q

+ 10 V

q shown in figure is in µC. q q Now, VA − + 10 − = VB 1 2

– q

…(i)

or …(ii) C 1 + 2 = 9C 2 Solving Eqs. (i) and (ii), we get C 1 = 0.16 µF and C 2 = 0.24 µF 22. (a) q = CV ε A 1 (b) C = 0 or C ∝ d d If d is doubled, C will remain half. Hence, q will also remain half. ε (πR 3 )V  ε A (c) q = CV =  0  V = 0  d  d

Now, common potential Total charge 13.2 × 10− 3 V = = Total capacity 10 × 10− 6

∴

C 2 = 1.5 C 1 VA′ C B′ = VB′ C A′ 10 C2 = 90 (C 1 + 2)

or

= 1.32 × 10− 3 C

1 V  ε0   2  d V 17. d = max Emax

3q − 10 = 5 2 q = 10 µC

q ∝ R2

R is doubled. Hence, q will become four times or 480 µC. 1 23. Energy lost = energy stored = CV 2 2 ε0A 24. (a) C = d (b) q = CV V (c) E = d 1 1 1 1 25. (a) = + + C net 8.4 8.2 4.2 ∴ C net = 2.09 µF qnet = C net V = (2.09) (36) = 75.14 µC ≈ 76 µC In series, charge remains same in all capacitors.

Chapter 25 1 C net V 2 2 (c) qtotal = (3) (76) µ C = 288 µC Now common potential in parallel, q 228 µC V = total = C total (8.4 + 8.2 + 4.2) µF 1 (d) U total = C net V 2 2 (b) U total =

q1

26.

q1 + 5V q1 –

6 µF

– q2 +

2 µF

+ q – 3

4 µF

1 (5.5 × 10− 4 ) = 1.8 × 10− 4 C 3 For finding PD across any capacitor, use the equation q C = V 28. In series, potential difference distributes in inverse ratio of capacitance. VA C 2 130 C 2 or ∴ = = VB C 1 100 C 1 C or C1 = 2 = CA 1.3 K is made 2.5 times. Therefore, C 1 will also become 2.5 times. 2.5 C 2 C 1′ = 2.5 C 1 = 1.3 C ′1 25 or = C 2 13 VA′ C 2 13 Now, = = VB′ C 1′ 25 q34 =

5V

+ – 3 µF

– q4 +

– + q4– q3 10 V q4– q3

If we see the charge on positive plate of 6 µF capacitor, then …(i) q2 = − q1 − (q4 − q3 ) Now, applying three loop equations, we have q q …(ii) 5− 1 + 2 =0 3 6 q q …(iii) 10 − 2 − 3 = 0 6 2 q q …(iv) 5− 3 − 4 =0 2 4 Solving these four equations, we can find q1 , q2 , q3 and q4. 27. (a) Simple series and parallel grouping of capacitors. (b) qnet = C net V = (2.5 × 10− 6 ) (220) = 5.5 × 10− 4 C C 1 , C 5 and equivalent of other three capacitors are in series. Hence, charges across them are same.

 13  VA′ =   (230) = 78.68 V  13 + 25

or

29. C 23

VB′ = 230 − VB′ = 151.32 V 2×3 = = 1.2 µF 2+ 3

qtotal = C 1V = 110 µC Common potential in parallel is given by Total charge V = Total capacity 110 = = 50 V 1 + 1.2 q23 = (C 23 )V = 60 µC So, this much charge flows through the switch. 30. (a) Simple circuit is as shown below. 2 µF

4.2 µF

C2

Capacitors — 667

C34 2.1 µF

qtotal will distribute between C 2 and C 34 in direct ratio of capacitance. q2 4.2 2 = = ∴ q34 2.1 1 2 ∴ q2 = (5.5 × 10− 4 ) = 3.7 × 10− 4 C 3

4 µF C3

4 µF

C 2 = 2 µF

20 V C 1 = 3 µF

3 µF

6 µF, V1 ⇒

20 V 6 µF, V2

668 — Electricity and Magnetism (b) qnet = (C net )V = (34 µF) (20 V) = 60 µC (c) Upper network and lower network both have same capacitance = 6 µF 20 V1 − V2 = = 10 V 2 VC1 = 10 V ∴ qC1 = (C 1 )(VC1 ) = 30 µC (d) VC2 = 10 V, ∴ qC2 = (C 2 )(VC2 ) = 20 µC (e) VC3 = 5 V, ∴ qC3 = (C 3 )(VC3 ) = 20 µC C1

C3

C2

C4

32. qtotal = C 1V0 After switch is thrown towards right, C 23 and C 1 are in parallel. The common potential is V =

Total charge = Total capacity

Now, qC1 = C 1V =

qC2 = qC3 = C 23V C 1V0    CC  C 2C 3  = 2 3     C 2 + C 3  C 1 + C2 + C3    ε A q = Ci V =  0  V  d 

12 V

1× 3 3 = µF 1+ 3 4 2×4 4 = = µF 2+ 4 3

C 13 = C 24

VC1 C3 = VC2 C4 = 12 V

∴

3 q1 = q3 = (C 13 ) (VC1 C3 ) = × 12 = 9 µC 4 4 q2 = q4 = (C 24 ) (VC2 C4 ) = × 12 = 16 µC 3 C1

C3

C2

C4

V1

V2 12 V

C 12 = C 34

1× 2 2 = µF 1+ 2 3

3 × 4 12 = = µF 3+ 4 7

V1 C 34 12/ 7 18 = = = V2 C 12 2/ 3 7 ∴

C 12V0  C C  C1 +  2 3   C 2 + C 3

This is the same result as given in the answer.

31. (a)

(b)

C 1V0  C 2C 3  C1 +    C 2 + C 3

 18  V1 =   (12) = 8.64 V  25

V2 = 12 − 8.64 = 3.36 V Now, we can apply q = CV for finding charge on different capacitors.

33. (a)

Vf =

q (ε AV / d ) = 0 = 2V Cf (ε 0 A / 2d )

1 1  ε A U i = Ci V 2 =  0  V 2 2 2 d 

(b)

1 1  ε A U f = C f V f2 =  0  (2V )2 2 2  2d   ε A =  0 V 2  d  (c) W = U f − U i =

1  ε0 A 2  V 2 d 

34. (a) After long time, capacitor gets fully charged by E1. ∴ and

iC = 0 iR1 = iR 2 = =

E1 R1 + R2

20 20 × 103

= 10− 3 A = 1 mA (b) In steady state (with E1 ). VC = VR 2 = VR1 E = 1 = 10 V 2 Now, when the switch is shifted to position B, capacitor (at t = 0) behaves like a battery of 10 V.

Chapter 25 The circuit in that case is as shown below.

(d) Initially,

3 mA R2 1 mA

R1

Capacitors — 669

q

10 V

10 V

q

+ – p + r

18 V

–

2 mA

Now, with the help of Kirchhoff’s laws we can find different currents. Final currents are shown in the diagram. 35. (a) Va = 18 Vand Vb = 0 as no current flow through the resistors. ∴ Va − Vb = 10 V (b) Va − Vb = + ve . Hence, Va > Vb (c) Current flows through two resistors, 18 − 0 i= = 2A 6+3 ∴ Vb − 0 = iR = 2 × 3 or Vb = 6V (d) Initially, V3 µF = V6 µF = 18 V (q = CV ) ∴ q3 µF = 54 µC and q6 µF = 108 µC Finally, V6 µF = V6 Ω = iR = 2 × 6 = 12 V ∴ q6 µF = 72 µC V3 µF = V3 Ω = 6 V ∴ q3µF = 18 µC ∆q = qf − qi = − 36 µC on both capacitors. 36. (a) In resistors (in series) potential drops in direct ratio of resistance and in capacitors (in series) potential drops in inverse ratio of capacitance.  6  18 − Va =   (18)  6 + 3 ∴

(c) V3 µF ∴ ∴

q = C net V = (2 µF) (18) = 36 µC Finally,

q1

q2

Vb − 0 = 6 V Vb = 6 V

– p + r –

12 V

6V

q1 = 72 µC q2 = 18 µC Charge flow from S = (Final charge on plates p and r) − (Initial charges on plates p and r) = (− 72 + 18) − (− 36 + 36) = − 54 µC 37. (a) In steady state, V VC = 2 ∴ Steady state charge, CV q0 = CVC = 2 For equivalent value of τC : We short circuit the battery and find the value of Rnet across capacitors and then

R

Va = 6 V  3  18 − Vb =   (18)  6 + 3 Vb = 12 V  18 − 0 = V3 Ω = iR =   (3) = 6 V  6 + 3

+

R

R

Rnet = ∴ Now,

3R 2

τC = CRnet =

3RC 2

q = q0 (1 − e− t / τ C )

670 — Electricity and Magnetism (b) At t = 0, capacitor offers zero resistance. 3R Rnet = 2 V 2V i= = ∴ 3R / 2 3R 1 V iC = iAB = = 2 3R At t = ∞, capacitor offers infinite resistance. So, iC = 0. V ∴ ibattery = iAB = 2R Now, current through AB increases exponentially V V from to with same time constant. 3R 2R (i - t ) graph is as shown below. i V 6R

V 2R V 3R t

(i - t ) equation corresponding to this graph is V V i= + (1 − e− t / τ C ) 3R 6R

LEVEL 2

3. At t = 0 when capacitors are initially uncharged, their equivalent resistance is zero. Hence, whole current passes through these capacitors. 4. Changing current is given by i = i0 e− t / τ C V − t / CR e R If we have take log on both sides, we have i=

or

V   1  ln (i ) = ln   −   t  R   CR  Hence, ln (i ) versus t graph is a straight line with  1 V  slope  −  and intercept + ln   .  CR   R Intercepts are same, but | slope|1 > | slope|2.

5. During charging of a capacitor 50% of the energy supplied by the battery is lost and only 50% is stored. 1 q2 1 (EC / 2)2 E 2C ∴ Total energy lost = = = 2 C 2 8 C Now, this total loss is in direct ratio r : 2r or 1 : 2 1 E 2C ∴ Energy lost in battery is rd of . 3 8 6. Equal and opposite charges should transfer from two terminals of a battery. For charging of a capacitor, it should lie on a closed loop. R

E

Single Correct Option +σ

+3σ

7.

1.

C E

–ve

a

b i

+ve E1

E2

E3

σ 3σ 2σ E1 = + = = E3 2ε 0 2ε 0 ε 0 3σ σ σ E2 = − = 2ε 0 2ε 0 ε 0 E1 and E2 are in the negative direction and E3 in positive direction. 2. Let E be the external field (toward right). Then, σ …(i) E− =8 2ε 0 σ …(ii) E+ = 12 2ε 0 Solving these equations, we get σ = 4 ε 0

E0

R0

i=

[ E − E0 ] R + R0

Now, Va − E + E0 + iR0 = Vb ∴ Va − Vb = (E − E0 ) − iR0  R0  = (E − E0 ) 1 − R + R0   = ∴

R (E − E ) R + R0

q = C (Va − Vb ) =

CR (E − E0 ) R + R0

Chapter 25

1 1 × (100)2 2 10 ln 2 500 = J ln 2

8. Initially

=

(C ) (C 0 ) C 0 = 0 = C0 + C0 2

C net

= 0.5 C 0 Finally

12. Net capacitance between points A and P will be C net

(C 0 / 2) (2C 0 ) = (C 0 / 2) + 2C 0

= 0.4 C 0 9. The simple circuit is as shown below. R R

Rnet =

C

R

V

∴

Capacitors — 671

equal to the net capacitance between points P and B. 13. Total charge = (2C ) (4V ) − CV = 7CV Common potential after they are connected is Total charge VC = Total capacitance 7 CV 7 = = V 2C + C 3 Heat = U i − U f 1 1 = CV 2 + (2C ) (4V )2 2 2

R 3

τC = CRnet =

CR 3

−

q = q0 (1 − e− t / τ C ) , where, q0 = CV

10. Common potential in parallel grouping, Total charge Total capacity (2 × 100) + (4 × 50) = 2+ 4 200 = V 3 Loss = U i − U f V =

1  1  = 10− 6  × 2 × 100 × 100 + × 4 × 50 × 50   2 2  200 200  1 − ×6× ×  2 3 3   = 1.7 × 10− 3 J

11. V0 = i0R = (10) (10) = 100 V 1 After 2 s, current becomes th. Therefore, after 4 1 s, current will remain half also called half-life. t1/ 2 = (ln 2) τC = (ln 2) CR (t ) 1 ∴ C = 1/ 2 = F (ln 2)R 10 ln 2 1 Total heat = CV02 2

1 7  × 3C ×  V  3  2

2

25 CV 2 3 1 14. Total heat produced = CV 2 2 1 = (2 µF) (5)2 2 = 25 µJ Now, this should distribute in inverse ratio of resistors, as they are in parallel. H 5Ω R ∴ = HR 5 =

or or

 R  H 5Ω =    R + 5

(Total heat)

 R  10 =   (25)  R + 5

Solving this equation, we get  10 R =  Ω  3

15. In position-1, initial maximum current is V 10 = = 2A R 5 At the given time, given current is 1A or half of the above value. Hence, at this is instant capacitor is also charged to half of the final value of 5 V. i0 =

672 — Electricity and Magnetism Now, it is shifted to position-2 wherein steady state it is again charged to 5V but with opposite polarity. 1 (Q V = 5 V ) U i = U f = CV 2 2 ∴ Total energy supplied by the lower battery is converted into heat. But double charge transfer (from the normal) takes place from this battery. ∴ Heat produced = Energy supplied by the battery = (∆q)V = (2CV )(V ) = 2CV 2

22.

V1 1.5 = V1.5 1  1.5  V1 =   (30)  1.5 + 1

∴

= 18 V V2.5 0.5 1 = = V0.5 2.5 5  1  V2.5 =   (30)  1 + 5

∴

= 2 × 2 × 10− 6 × (5)2

Now,

=5V |Vab | = V1 − V2.5 = 13 V

= 100 × 10− 6 J = 100 µJ

23. In the figure, +6 V

16. Equivalent capacitance of 6 µF and 3 µF is also 2 µF and charge across it is also q or circuit is balanced. Hence, there is no flow of charge.

60 µF

17. Two capacitors are in parallel. 1 U = C net V 2 2 1 = (2C )V 2 = CV 2 2  ε0 A 2 = V  d 

V (let) q2 –

20 µF

∴

q1 + q2 + q3 = 0 60 (V − 6) + 20 (V − 2) + 30 (V − 3) = 0 Solving this equation, we get 49 V V = 11

V = 7A R =6V = CV = 12 µC

i1Ω = V2 µF q2 µF

20. During charging capacitor and resistance of its wire are independently connected with the battery. Hence, τC = CR During discharging capacitor is discharged through both resistors (in series). Hence, τC = C (2R ) = 2CR

+3 V

+2 V

18. Initially, the rate of charging is fast. 19. V1Ω = 5 + 2 = 7 V ∴

A

1 µF

3 µF B

= 50 V

2 µF

3 µF

24.

1 µF

10 V

21. Total charge = 3 × 100 − 1 × 100 = 200 µC Common potential (in parallel) after S is closed, is Total charge V = Total capacity 200 µC = 4 µF

30 µF + – q3

+

∴

–q 1 +

VAB (C )BC 3 1 = = = VBC (C )AB 6 2 ∴

 1  VAB =   (10) V  1 + 2 =

10 V 3

C

Chapter 25 25. Applying Kirchhoff’s loop law in outermost loop, we have 3 µF + – q 18 V

4Ω

5Ω

1Ω

15 V 2r

– 3A

q

33. Four capacitors are in parallel charge across each is q = CV . Two surfaces of plate C marks two capacitors, one with B and other with D and C is connected to positive terminal of the battery. Hence, qC = 2CV = + 40 µC

2

 1 =   (10 µC)  e

1

Vab =

C 1C 2 C2 + C2

Electric field between two plates and hence the potential difference is due to q2 and q3 only. q Q PD = 2 = V + C 2C

t

V  − i1 =   e 6 CR  2R 

More than One Correct Options

t

V  − i2 =   e CR  R

1

1.

5t

∴

i1 e 6 CR = i2 2

We can see that this ratio is increasing with time.

31. τC = CR  K ε0 A  d  =     d   Aσ 

4

q1 = q4 =

The only change is by increasing the resistance τC increase. Hence, process of redistribution of charge slows down. 29. Just after the switch is closed C 1 is short-circuited and current passes through R1 and C 1 only.

∴

3

qtotal CV − CV + Q Q = = 2 2 2 Q Q  q2 = (Q + CV ) − =  + CV   2 2 Q   q3 = − q2 = −  + CV  2 

q  E1 + E2  =  C1 C 2  C 1 + C 2

28. H 1 = H 2 = U i − U f

30.

2

34.

27. q = (E1 + E2 ) C net = (E1 + E2 )

5 × 8.86 × 10− 12 7.4 × 10− 12

∴ Resistance of ammeter = 2Ω

q0 = CV = 10 µC q = q0e− t / τ C = (10 µC) e− 12/ 6

= (0.37)2 (10 µC)

=

exponentially decreasing equation. ∴ t = t1/ 2 = (ln 2) τC = (ln 2) CRnet t Rnet = ∴ (ln 2) C 2 (ln 2) µs = = 4Ω (ln 2) (0.5 µF)

−

Now,

Kε 0 σ

=6s

2.5 A

q q + 15 − 2 × 2.5 − − 3 × 1 + 18 = 0 3 2 Solving this equation, we get q = 30 µC 26. τC = CR = 6 s

=

32. The given time is the half-life time of the

+

2 µF

Capacitors — 673

l   R =   σA 

2

Q 2

Q 2

3 –Q 2

qtotal Q = 2 2 Q q2 = Q − q1 = 2 q1 = q4 =

4 Q 2

674 — Electricity and Magnetism q3 = − q2 =

−Q 2

EA = E1 + E4 Q Q Q = + = 4 Aε 0 4 Aε 0 2 Aε 0 EC = − EA Q = 2 Aε 0

(towards left)

(towards right)

EB = E2 + E3 Q Q Q = + = 4 Aε 0 4 Aε 0 2 Aε 0

(towards right)

2. In steady state, qC = EC and q2C = 2EC τC = 2CR of both circuits At time t , qC = EC (1 − et / τ C ) q2C = 2EC (1 − e− t / τ C ) ∴

qC 1 = q2C 2

3. In steady state, current through capacitor wire is

zero. Current flows through 200 Ω , 900 Ω and A2. q 4 × 10− 3 VC = = C 100 × 10− 6 = 40 V This is also potential drop across 900 Ω resistance and 100 Ω ammeter A2 (Total resistance = 1000 Ω). Now, this 1000 Ω and 200 Ω are in series. Therefore, V1000 Ω V2 = V200 Ω = 5 40 = =8V 5 Emf = V1000 Ω + V200 Ω = 48 V

Emf i= Net resistance 48 1 = = A 1200 25 4. Current through A is the main current passing through the battery. So, this current is more than the current passing through B. Hence, during charging more heat is produced in A. In steady state, iC = 0 and iA = iB

Hence, heat is produced at the same rate in A and B. Further, in steady state ε VC = VB = 2 1 1 ∴ U = CVC2 = Cε 2 2 8 2  σ  q 5. F = qE = (q)   =  2ε 0  2ε 0 A q remains unchanged. Hence, F remains unchanged. σ q E= = ε 0 Aε 0 q remains unchanged. Hence, E also remains unchanged. q2 1 or U ∝ U = 2C C C will decrease. Hence, U will increase. V = Ed or V ∝ d d is increasing. Hence, V will increase. (C ) (2C ) 2 6. C i = = C C + 2C 3 2 qi = C i E = EC 3 C f = 2C ∴ qf = 2EC ∆q = qf − qi 4 = CE 3 7. Let (+ q) µC charge flows in the closed loop in clockwise direction. Then, final charges on different capacitors are as shown in figure. + q

– q

+

+

–

(300 – q )

+

–

(360 – q )

Now, applying Kirchhoff’s loop law 360 − q 300 − q q + = 3 2 1.5 Solving the above equation, we get q = 180 µC 8. If the battery is disconnected, then q = constant ε A 1 or C ∝ C = 0 d d

Chapter 25 d is decreased. Hence, C will increase. 1 1 q2 or U ∝ U = C 2C C is increasing. Hence, U will decrease. q 1 or V ∞ V = C C C is increasing. Hence, V will decrease. 9. (a) At t = 0, emf of the circuit = PD across the capacitor = 6 V. 6 ∴ i= = 2A 1+ 2 Half-life of the circuit = (ln 2) τC (ln 2) CR = (6 ln 2) s. In half-life time, all values get halved. For example 6 VC = = 3V 2 2 i = = 1A 1 ∴ V1Ω = iR = 1V V2Ω = iR = 2 V

10.

1 µF

4 µF

9 µF

V1

V2

V3

In series, ∴ or

∴ Now,

1 (as q = constant) C V2 1 = V1 4 V 10 V2 = 1 = = 2.5 V 4 4 V3 1 = V1 9 V 10 V V3 = 1 = 9 9 10  E = 10 + 2.5 +  V  9 V ∝

Comprehension Based Questions 1. Finally, the capacitors are in parallel and total

charge (= q0 ) distributes between them in direct ratio of capacity.  C2  ∴ qC2 =   q0 → in steady state.  C 1 + C 2 But this charge increases exponentially.

Capacitors — 675

Hence, charge on C 2 at any time t is  C q  qC2 =  2 0  (1 − e−t / τ C )  C 1 + C 2 Initially, C 2 is uncharged so, whatever is the charge on C 2, it is charge flown through switches.

2. Common potential in steady state when they finally come in parallel is Total charge q0 V = = Total capacity C 1 + C 2 Total heat dissipated = U i − U f  q0  q2 1 = 0 − (C 1 + C 2 )    C 1 + C 2 2C 1 2

2

 q2   C C  = 0   1 2   2C 1   C 1 + C 2  V d E0 V = = K Kd

3. Eair = E0 = 4. Edielectric

Match the Columns 1. (a) C i =

4×4 = 2 µF 4+4 C 1C 2 8×2 Cf = = C1 + C2 8 + 2

= 1.6 µF q = CV Since, total capacity is decreasing. Hence, charge on both capacitors will decrease. 1 q2 (b) U 2 = 2C 16 . or 0.8 times but C is halved. q has become 2 Hence,U 2 will increase. q (c) V2 = C q has become 0.8 times and C is halved. Hence, V2 will increase. V (d) E2 = 2 or E2 ∝ V2 d

2. (a) C i = 2 µF ∴

qi = 60 µC C f = 6 µF ∴ qf = 180 µC ∴ ∆q from the battery = qf − qi =120 µC.

676 — Electricity and Magnetism (b) Between 4 µF and 2 µF charge distributes indirect ratio of capacity. Hence, on 2 µF  2  qi =   (60 µC) = 20 µC  2 + 4

5. Let C 0 =

ε 0 ( A / 2) K ε 0 ( A / 2) + d d C0 3C 0 = + C0 = 2 2 ε0 A 2 ε0 A = C2 = d /2 1  d − d /2 + d 1 +   K K C1 =

 2  qf =   (180) 60 µC  2 + 4 ∴ ∆q = ∆f − qi = 40 µC (c) On 3 µF, initial charge is 60 µF and final charge is zero. ∴ ∆q = 60 µC (d) On 4 µF  4  qi =   (60 µC) = 40 µC  2 + 4

4 ε0A 4 = C0 3 d 3 C1 9 = ∴ C2 8 Capacitors are in series. Hence, q1 = q2 q1 or =1 q2 =

 4  qf =   (180 µC) = 120 µC  2 + 4 ∴

∆q = qf − qi = 80 µC

3. (a) In second figure, VC1 = V = maximum Hence, qC1 is maximum. V (b) In first figure, VC2 = 3 In second figure, VC2 = V  C  V In third figure, VC2 =  V =  2C + C  3 In fourth figure, VC2

1 q2 2 C 1 U ∝ C U 1 C2 8 = = U 2 C1 9 q q1 = q8 = total = 7Q 2 q2 = 4 Q − q1 = − 3Q q3 = − q2 = + 3Q q4 = Q − q3 = − 2Q q5 = − q4 = + 2Q q6 = 2Q − q5 = 0 q7 = − q6 = 0 U =

or ∴

6.

 2C  2V = V =  C + 2C  3

Now, q = CV Hence, qC2 is minimum in first and third figures. (c) In second figure, VC1 = V = maximum (d) Similar to option (b) 4. After closing the switch, the common potential is parallel. Total charge CV V V = = = Total capacity 3C 3

ε0A d

Subjective Questions 1.

V2 V1 3 2 V3

1 1 V  (2C )   = CV 2  3 2 9

+

1

V1

Loss of energy = U i − U f 1 1 V  = CV 2 − × 3C ×    3 2 2 1 = CV 2 3

– 4

2

U 2C =

5

V3

1 V  1 UC = C   = CV 2 2  3 18 2

(as q is same)

2

ε0A d (between two successive plates). The effective capacity has to be found between V1 and V2. C =

Capacitors — 677

Chapter 25 C V2

V1

 CV − q  4 CV − q  9CV − q   +  +   C   2C   3C 

V0 V1 + V1

V1

C C

V3

C

V2

– V2

V3 2V 0 3

V0 3

C net = C +

 16CV − q +  =0  4C 

V3

(2C )(C ) 5 5 ε0A = C = 2C + C 3 3 d

q3 = + [C (V1 − V2 ) + C (V1 − V3 )] ε0A d

V  4 ε AV  V + 0 = 0 0  0 3  3d  ε A   2V  q5 = C (V3 − V2 ) =  0   0   d   3  2ε AV = 0 0 3d qnet C 1V0 V0 2. (a) V = = = C net C 1 + C 2 1 + C 2/C 1 =

=

120 = 80 V 1 + 4/8

24 CV 5 q 24 19 Now, V1 = V − = V − V =− V C 5 5 q 24 2 V2 = 2V − = 2V − V =− V 2C 10 5 q 24 7 V3 = 3V − = 3V − V = V and 3C 15 5 q 6 14 V4 = 4V − = 4V − V = V 4C 5 5 4. At t = 5 ms,V = 10 V V 10 Ans. ∴ iR = = = 2.5 A R 4 Further, q = CV = (300 × 10−6 )(2000t ) = 0.6t dq Ans. iC = = 0.6 A = constant dt or

5. Potential energy stored in the capacitor, 1 1 U = CV 2 = × 5 × 10−6 × (200)2 = 0.1 J 2 2 During discharging this 0.1 J will distribute in direct ratio of resistance, 400 H 400 = × 0.1 ∴ 400 + 500

1 1 (b) U i = C 1V02 = × 8 × 10−6 × (120)2 2 2 = 5.76 × 10–2 J 1 (C 1 + C 2 )V 2 2 1 = × 12 × 10−6 × (80)2 2 = 3.84 × 10–2 J

Uf =

= 44.4 × 10−3 J = 44.4 mJ

6.

3. Let + q charge rotates in the loop in clockwise direction for achieving equilibrium state. In final steady state, charges on the capacitors will be as shown below –

q = 4.8 CV =

+

3Ω

Ans. 6Ω

+

E/3

– 4Ω E/2

4Ω

(CV – q) (16CV – q)

+

–

–

+ +

(4CV – q)

–

(9CV – q)

Now, applying Kirchhoff’s loop law we have

E

(a) Current in lower branch = E/8 = 3 A Current in upper branch = E/9 = 24 / 9 = 2.67 A (b) PD across the capacitor = E/2 − E/3 = E/6 E From q = CV , we have 16 = (4 ) 6 ∴ E = 24 V

678 — Electricity and Magnetism (c) After short-circuiting the battery, we will have to find net resistance across capacitor to calculate equivalent value of τC in discharging. 3Ω and 6Ω are in parallel. Similarly, 4 Ω and 4 Ω are in parallel. They are then in series. ∴ Rnet = 4Ω, τC = CRnet = (4 × 4 ) µs = 16 µs During discharging q = q0e− t / τ C Solving this equation, we get t = 11.1 µs

7. –

Ans.

C

2C –

+ q1

+ q2

– C + q1 – q 2

110 V

110 V

Applying loop law in two closed loops, we have q q − q2 110 − 2 + 1 = 0 or q2 = (110 C ) C C q q − q2 and − 110 + 1 + 1 =0 2C C  440 C  or q1 =    3  Potential difference between points M and N is q − q2 VN − VM = 1 C 110 Ans. = volt 3

8. Let us first find charges on both capacitors before and after closing the switch. C1

C1 – + q0 + q0 –

uncharged capacitor. 3 +

3 µF

2 µF 4 – q2

2+ 1 –

q3

q1

– 5 +6

2 µF

8 = 16e− t/ 16

or

C2

9. (i) Charge on capacitor A, before joining with an

q1 E

E

+ C2 – 1

E

qA = CV = (100)(3) µC = 300 µC Similarly, charge on capacitor B qB = (180)(2) µC = 360 µC Let q1 , q2 and q3 be the charges on the three capacitors after joining them as shown in figure. (q1 , q2 and q3 are in microcoulombs) From conservation of charge Net charge on plates 2 and 3 before joining = net charge after joining …(i) ∴ 300 = q1 + q2 Similarly, net charge on plates 4 and 5 before joining = net charge after joining − 360 = − q2 − q3 or …(ii) 360 = q2 + q3 Applying Kirchhoff’s second law in closed loop q1 q2 q3 − + =0 3 2 2 or …(iii) 2q1 − 3q2 + 3q3 = 0 Solving Eqs. (i), (ii) and (iii), we get q1 = 90 µC q2 = 210 µC and q3 = 150 µC (ii) (a) Electrostatic energy stored before completing the circuit, 1 1 U i = (3 × 10−6 )(100)2 + (2 × 10−6 )(180)2 2 2 1  2 U = CV    2 = 4.74 × 10−2 J or

q2 2

q2 = EC 2 and q1 = 0  C C  q0 = E  1 2   C 1 + C 2 From 2, −q0 charge will flow, so that charge on right hand side plate of C 1 becomes zero. From 1, q2 charge will flow.

U i = 47.4 mJ

(b) Electrostatic energy stored after completing the circuit, 1 (90 × 10−6 )2 1 (210 × 10−6 )2 + Uf = 2 (3 × 10−6 ) 2 (2 × 10−6 ) +

1 (150 × 10−6 )2 2 (2 × 10−6 )

= 1.8 × 10−2

 1 q2  U = 2 C    Ans. J or U f = 18 mJ

Chapter 25 10. (a) In steady state, capacitors will be in parallel.

13. (a) Let q be the charge on smaller sphere. Then,

Charge will distribute in direct ratio of their capacity.  C1  ∴ q1 =   q0  C 1 + C 2

+1 µC +1 µC –1 µC

 C2  q2 =   q0  C 1 + C 2

and

Initial emf in the circuit is potential difference across capacitor C 1 or q0 /C 1. Therefore, initial current would be q /C q i0 = 0 1 = 0 R C 1R

Vinner = 0 Kq K (2) ∴ + = 0 or q = − 1µC 2 4 K (2 − 1) × 10−6 Now, Vouter = 4 × 10−2

Current as function of time will be i = i0e−t / τ C  C C  τC =  1 2  R  C 1 + C 2

Here, (b) U i =

1 q02 2 C1

and

Uf =

Ans. 1 q02 2 C1 + C2

Heat lost in the resistor  q2  C2 =Ui −U f = 0  2 C 1 (C 1 + C 2 ) 

11. τC = CR =  

Kε 0 A   d 

= 5 × 8.86 ×

=

Ans.

 ρd    = Kε 0 ρ  A

10−12 ≈6s 7.4 × 10−12

12. (a) C =

Fig. (b) Capacitor 1µF is short-circuited. Therefore, q1 = 0.   20 V26 =   × 100 = 40 V  20 + 20 + 10

(b)

Ans.

This 40 V will distribute in inverse ratio of capacity. 2 ∴ V6 = × 40 = 10 V 8 6 V2 = × 40 = 30 V 8 Ans. ∴ q6 = 60 µC, q2 = 60 µC

ε0A Q Q x ,U = = x 2C 2ε 0 A 2

2

dU Q2 = dx 2ε 0 A ∴

 Q2  dU =   dx  2ε 0 A   Q2    dx = dW = Fdx  2ε 0 A 

(c) ∴

F=

Ans.

(b) Charge distribution is as shown in above figure. 90 14. (a) Fig. (a) V6 = = 30 V, q6 = 6 × 30 = 180 µC 3 90 V3 = = 30 V 3 Ans. q3 = 30 × 3 = 90 µC

Now, current as function of time i = i0e− t / τ C i = (1.425)e−12/ 6 = 0.193 µA

9 × 109 × 10−6 4 × 10−2

= 2.25 × 105 V

Initial current, q /C q q 8.55 i0 = 0 = 0 = 0 = = 1.425 µA R CR τC 6 or

Capacitors — 679

(b) Fig. (a) When S is open, 6 µF is short-circuited or and

2

Fig. (b) When S is open, V1 = 100 V

Q 2ε 0 A

(d) Because E between the plates is due to both the plates. While F = (Q ) (field due to other plate) Ans.

V6 = 0, q6 = 0 V3 = 90 V, q3 = 270 µC q1 = 100 µC

∴

V26 = 100 V 2 V6 = × 100 = 25 V 8

Ans.

680 — Electricity and Magnetism 6 × 100 = 75 V 8 ∴ q6 = 150 µC, q2 = 150 µC Further, VA − 0 = V2 Ans. ∴ VA = V2 = 75 V 10 15. Current in the circuit, i = =1A 4 + 1+ 2 + 3 V2 =

and

V5 µF = V1,2Ω = 3 V q5 = 15 µC V3µF = V2, 3Ω = 5 V q3 = 15 µC

Now, ∴ Further, ∴

∴

Rnet = 3 +

4×6 = 5.4 MΩ 4+6

∴

τC = CRnet = (10 × 10−6 )(5.4 × 106 )

∴

= 54 s V2 = 6 + (10.8 − 6)(1 − e−t / C ) = 6 + 4.8(1 − et/ 54 )

Here, V2 is in kV and t is second.

17. Circuit can be drawn as shown in figure. R2

Ans.

E

16. (a) At t = 0, when capacitor is uncharged, its equivalent resistance is zero. 6×3 Rnet = 4 + = 6 MΩ ∴ 6+ 3 i1 =

or

18 × 10 A = 3 mA 6 × 106 3

This will distribute in inverse ratio of resistances. 3 i1 = 1 mA and i3 = 2 mA ∴ i2 = 6+ 3 At t = ∞, when capacitor is completely charged, equivalent resistance of capacitor is infinite. 18 × 103 Ans. ∴ i3 = 0, i1 = i2 = = 1.8 mA (4 + 6) × 106 (b) At

At

R1

In charging of capacitor, R3 has no role. In steady state, potential difference across capacitor = potential difference across R2 = E/2 Therefore, steady state charge across capacitor CE q0 = 2 To find time constant of circuit we will have to short circuit the battery, then we will find net resistance across capacitor. R CR Rnet = ⇒ τC = CRnet = 2 2 R

t = 0, V2 = i2R2 = (1 × 10−3 )(6 × 106 ) V

R

= 6 kV t = ∞, V2 = i2R2 = (1.8 × 10−3 )(6 × 106 ) V = 10.8 kV

Ans.

(c) To find time constant of the circuit we will have to short-circuit the battery and find resistance across capacitor. In that case, R1 and R2 are in parallel and they are in series with R3. V2 (kV)

∴ Charge in the capacitor at time t would be 2t − CE q = q0 (1 − e− t / τ C ) = (1 − e CR ) Ans. 2 18. q2 = 20 µC ∴ q1 = 10 µC (as they are in parallel) Energy stored at this instant, 1 q12 1 q22 U = + 2 C1 2 C2 =

10.8

C

R3

1 (10−5 )2 1 (2 × 10−5 )2 × + × 2 2 10−6 2 × 10−6

= 1.5 × 10−4 J 6 t

= 0.15 mJ In charging of a capacitor 50% of the energy is stored and rest 50% is dissipated in the form of heat.

Capacitors — 681

Chapter 25 Therefore, 0.15 mJ will be dissipated in the form of heat across all the resistors. In series in direct ratio of resistance (H = i 2Rt ) and in parallel in inverse ratio of resistance. ∴ H 2 = 0.075 mJ, H 3 = 0.05 mJ and Ans. H6 = 0.025 mJ

19. (a) At t = 0, capacitor is equivalent to a battery of E emf . 2 Net emf of the circuit = E − E/2 = E/2 Total resistance is R. Therefore, current in the circuit at t = 0 would be E/2 E Ans. i= = R 2R (b) Let in steady state there is total q charge on C. Initial charge on C was CE/2 . Therefore, charge on 2C in steady state would be  CE  − q with polarities as shown. This is   2  because net charge on lower plate of C and of upper plate on 2C should remain constant. Applying loop law in the circuit in steady state, we have q CE/2 − q E− + =0 C 2C 5 q = CE ∴ 6 R

+ C

–

E

– 2C

+

q

CE –q 2

20. When S 1 is closed and S 2 open, capacitor will

discharge. At time t = R1C , one time constant,  1 charge will remain q1 =   times of CV or  e CV q1 = e When S 1 is open and S 2 closed, charge will CV increase (or may decrease also) from to CE e exponentially. Time constant for this would be (R1C + R2C ). Charge as function of time would be q = qi + (qf − qi )(1 − e− t / τ C ) q=

CV  CV  −t / τ + CE −  (1 − e C )  e e 

After total time 2R1C + R2C or t = R1C + R2C , one time constant in above equation, charge will remain q=

CV  CV   + CE −  1 −  e e  

 = EC 1 − 

1  e

1 VC  + 2 e e

21. At t = 0, capacitor C 0 is like a battery of emf =

Q0 =1V C0

Net emf of the circuit = 4 − 1 = 3 V Total resistance is R = 100 Ω 3 ∴ Initial current = = 0.03 A 100 This current will decrease exponentially to zero. ∴ i = 0.03 e− t / τ C τC = C net R = (1 × 10−6 )(100)

Here,

= 10−4 s Therefore, charge on C increases from qi =

CE 2

5CE exponentially. 6 Equivalent time constant would be  C × 2C  2 τC =   R = CR  C + 2C  3 to qf =

=

 CE CE  + 1− e 2 3 

 

t

Ans.

22. From O to A VC A

Therefore, charge as function of time would be q = qi + (qf − qi )(1 − e−t / τ C ) 3t  − 2C R 

4

i = 0.03 e−10

∴

Ans.

t

O

∴

VC = at qC = CVC = Cat

(a = constant)

682 — Electricity and Magnetism dqC = aC dt VR = iR = aCR = constant From A onwards VC = constant

∴

i=

∴

qC = constant dq i= C =0 ∴ dt or VR = 0 Therefore, VR versus t graph is as shown in figure.

From A onwards When V = constant V V0 = at or t = 0 a ∴ VED = aCR (1 – e–V0 / aCR )

(say V0)

After this VED will decrease exponentially. Hence, a rough graph is as shown in figure. VED

VR

t

24. qi = CVi = 100 µC, qf = CV f = − 50 µC Therefore, charge will vary from 100 µC to − 50 µC exponentially. ∴ q = − 50 + 150 e− t / τ C , Here q is in µC

t

O

23. From O to A

V = at

τC = CR = (10−6 )(5 × 103 ) = 5 × 10−3

Here,

a is a positive constant. q ∴ at = + iR C Differentiating w.r.t. time, we have 1  dq  di  a=   +   R C  dt   dt  or ∴

∴

q = (− 50 + 150e− t / τ C ) V C VC = 50 (3e−200t − 1) VC =

or dq   as i =   dt 

i  di    R=a–  dt  C i t dt di ∫0 a – i /C = ∫0 R

∴

q = − 50 + 150 e− t / τ C

i=− =

i = aC (1 – e– t / CR )

dq 150 × 10−6 −200t = e τC dt

150 × 10−6 −200 t e = 30 × 10−3 e−200 t 5 × 10−3

VR = iR = 150e−200t

Ans.

25. At t = 0, equivalent resistance of an uncharged

V A

capacitor is zero and a charged capacitor is like a battery of emf = potential difference across the capacitor. (a) VC1 = q1/C 1 = 2 V

B

t

O

i.e. current in the circuit increases exponentially ∴

VED = iR = aCR (1 – e– t / CR )

or

VED also increases exponentially. q +

V = at –

+

– i

R

∴ Net emf of the circuit = 9 − 2 = 7 V or 9 + 2 = 11 V 30 × 60 Net resistance = 30 + = 50 Ω 30 + 60 7 ∴ Current at t = 0 would be i0 = A 50 11 or Ans. A 50 (b) In steady state, no current will flow through the circuit. C 2 will therefore be short-circuited, while PD across C 1 will be 9 V. Ans. ∴ Q2 = 0 and Q1 = 9 µC

Chapter 25 For t ≥ 250 µs

26. Time constant of the circuit is τC = CR = (0.5 × 10–6 ) (500) = 2.5 × 10–4 s For t ≤ 250 µs i = i0e– t / τ C 20 Here, i0 = = 0.04 A 500 ∴ i = (0.04 e–4000t ) amp t = 250 µs = 2.5 × 10–4 s

At

i = – i0′ e– t / τ C = – 0.11 e–4000 t The (i - t ) graph is as shown in figure. i (A ) 0.04 0.015

i = 0.04 e–1 = 0.015 amp

2.5

At this moment PD across the capacitor, VC = 20 (1 – e–1 ) = 12.64 V So when the switch is shifted to position 2, the current in the circuit is 0.015 A (clockwise) and PD across capacitor is 12.64 V

t ( × 10–4 s)

–0.11

27. Capacitor C 1 will discharge according to the equation, q = q0e− t / τ C

i = 0.015 A 500 Ω 20 V

Capacitors — 683

+

40 V

…(i)

Here, τC = C 1R and discharging current : i=

12.64 V

–

− dq q0 − t / τ C q0e− t / τ C = .e = τC dt C 1R

…(ii)

At the given instant i = i0 Therefore, from Eq. (ii) q0e− t / τ C = i0C 1 R at this instant 500 Ω

⇒

40 V

+ 12.64 V –

As soon as the switch is shifted to position 2 current will reverse its direction with maximum current. 40 + 12.64 i0′ = 500 = 0.11 A Now, it will decrease exponentially to zero.

or charge C 1 at this instant will be [From Eq. (ii)] q = (i0C 1R ) Now, this charge q will later on distribute in C 1 and C 2 . q2 Ui = ∴ 2C 1 and

q2 =U f 2(C 1 + C 2 )

∴ Heat generated in resistance, H =Ui −U f Substituting values of q : U i andU f , we get (I R )2C 1C 2 H = 0 2(C 1 + C 2 )

Ans.

26. Magnetics INTRODUCTORY EXERCISE

26.1

qE = Bqv sin θ

1. Q

[ E / B ] = [ v ] = [LT −1 ]

∴

2. From the property of cross product, F is always perpendicular to both v and B . 3. May be possible that θ = 0° or 180° between v and B, so that Fm = 0

4. F = q ( v × B) Here, q has to be substituted with sign.

5. Apply Fleming’s left hand rule. 6. Q F = Bqv sinθ ∴

F v= Bq sin θ =

4.6 × 10−15 3.5 × 10 × 1.6 × 10−19 × sin 60° −3

= 9.47 × 106 m/s

7. Q

F = Bqv sin 90°

4. Magnetic force may be non-zero. Hence, acceleration due to magnetic force may be non-zero. Magnetic force is always perpendicular to velocity. Hence, its power is always zero or work done by magnetic force is always zero. Hence, it can be change the speed of charged particle. 5. F = q ( v × B) F is along position y -direction. q is negative and v is along positive x -direction. Therefore, B should be along positive z -direction. mv 6. (a) r = Bq or r ∝ m as other factors are same. Bq (b) f = 2πm 1 or f ∝ m 2qVm 7. Q r = Bq

= 0.8 × 2 × 1.6 × 10−19 × 105 = 2.56 × 10

−14

INTRODUCTORY EXERCISE

N

INTRODUCTORY EXERCISE

1. Q l = l $i Now,

26.2

1. Path C is undeviated. Therefore, it is of neutron’s path. From Fleming’s left hand rule magnetic force on positive charge will be leftwards and on negative charge is rightwards. Therefore, track D is of electron. Among A and B one is of proton and other of α-particle. mv m Further, or r ∝ r= Bq q Since,

 m  m   >   q α  q P

∴ rα > rP or track B is of α-particle. 2km or r ∝ m ( k,q and B are same) 2. r = Bq mp > me

⇒ rp > re

3. The path will be a helix. Path is circle when it enters normal to the magnetic field.

26.3

F = i(l × B) = i [(l $i ) × ( B0$j + B0k$ )] = ilB ($i − $j) = | F | = 2ilB 0

0

2. No, it will not change, as the new $i component of B is in the direction of l. i = 3.5 A l = (− 10−2 $j)

3. Q

Now, apply F = i (l × B) in all parts.

4.

C

A

FACD = FAD ∴ | Fnet | = 2 | FAD | = 2ilB = 2 × 2 × 4 × 2 = 32 N

D

Chapter 26 INTRODUCTORY EXERCISE

26.4

M q = L 2m  q  q M =   L =   (Iω )  2m  2m

1. Q ∴

qR ω  q  1  =    mR 2 (ω ) =   2m  2 4 M = iA = i (πR 2 ) = (0.2) (π ) (8 × 10−2 )2 Now,

= (4.0 × 10−3 ) A -m 2 $ = (4.0 × 10−3 ) (0.6 $i − 0.8 $j) M = MM

(a) τ = M × B (b) U = − M ⋅ B

3. Q

L 2π

L = 2πR ⇒ R = M = iA = i (πR 2 )

4. Q ∆U = U 0° − U 180° = − MB cos 0° + MB cos180° = − 2MB

26.5

1. (a) From screw law, we can see that direction of magnetic field at centre of the square is inwards as the current is clockwise.

45°

Magnetic field due to vertical is µ i B = 0 (sin 90° + sin 0° ) 4π x µ i (inwards) = 0 4π x 3. Both straight and circular wires will produce magnetic fields inwards. µ i µ i + 0 ∴ B= 0 2π R 2R −7 (2 × 10 ) (7) (4 π × 10−7 )(7) = + 0.1 2 × 0.1

4. Magnetic field at O due to two straight wires = 0

iBL2 = MB sin 90° = 4π

INTRODUCTORY EXERCISE

2. Magnetic field due to horizontal wire is zero.

= 5.8 × 10−5 T

2

iL2  L = (i ) (π )   =  2π  4π τ max

(b) 2πR = 4 (0.4) 1.6 2R = m π µ i (4 π × 10−7 ) (10) B= 0 = 2R (1.6 / π ) = 24.7 × 10−6 T

2

2. Q

Magnetics — 685

45°

i =10 A

0.2 m

Magnetic field due to circular wire, 1 B = (due to whole circle) 4 1  µ i (inwards) =  0 4  2R  =

1 (4 π × 10−7 ) (5) × 4 2 × 0.03

= 2.62 × 10−5 T

5. Magnetic field at P due to straight wires = 0 Due to circular wires one is outwards (of radius a) 1 and other is inwards. 60° means th of whole 6 circle. 1 µ 0i µ 0i  (outwards) ∴ B= − 6  2a 2b 

INTRODUCTORY EXERCISE

1. Applying Ampere’s circuital law,  µ i B = 4  0  (sin α + sin β )  4 π r 4 × 10−7 × 10 (sin 45° + sin 45° ) 0.2 = 2.83 × 10−5 T =

µ 0 iin 2π r (2 × 10−7 ) (1) = 10−3 = 2 × 10−4 T

BA =

26.6

686 — Electricity and Magnetism This is due to (⋅) current of 1 A. Hence, magnetic lines are circular and anti-clockwise. Hence, magnetic field is upwards. µ i Bb = 0 in 2π r (2 × 10−7 ) (3 − 1) = 3 × 10−3 −4

= 1.33 × 10

T

This is due to net ⊗ current. Hence, magnetic lines are clockwise. So, magnetic field at B is downwards.

2.

∫ B⋅ dl

3. Using Ampere’s circuital law over a circular loop of any radius less than the radius of the pipe, we can see that net current inside the loop is zero. Hence, magnetic field at every point inside the loop will be zero.

INTRODUCTORY EXERCISE  k  i= φ  NBA 

1. ∴

B=

= µ 0 (inet )

Along path (a), net current enclosed by this path is zero. Hence, line integral = 0 Along path (b), inet is ⊗. So, magnetic lines along this current is clockwise. But, we have to take line integral in counter clockwise direction. Hence, line integral will be negative.

26.7

kφ (10−8 ) (90) = NiA 100 × 10−6 × 10−4

= 90 T  k  i= φ  NBA 

2.

=

(0.125 × 10−7 ) (6) (π /180) 200 × 5 × 10−2 × 5 × 2 × 10−4

= 1.3 × 10−7 A

Exercises LEVEL 1

6. Q r =

Assertion and Reason 2. By changing the direction of velocity direction of magnetic force will change. So, it is not a constant force. 3. To balance the weight, force on upper wire should be upwards (repulsion). Further equilibrium can be checked by displacing the wire from equilibrium position. 4. τ = MB sin 90° ≠ 0

2qVm Bq

⇒ r∝

m q

m and q both are different. Ratio

m is not same for q

both.

7. Q ∴ ∴ 8. Q

Fe + Fm = 0 qE + q ( v × B) = 0 E = − ( v × B) = (B × v) Fm = q( v × B)

(always) Fm ⊥ v and P =F⋅ v ∴ Power of magnetic force is always zero. Fe = qE If Fe is also perpendicular to v, then its power is also zero.

5. Magnetic field is outwards and increasing with x. So, magnetic force will also increase with x. The force on different sections are as shown in figure.

9.

x

x

x

q+ Force will act in positive x-direction. But, no torque will act.

10. | v | = 2v0 = speed, which always remains constant.

Chapter 26 11. R ∝ v, by increasing the speed two times radius also becomes two times. Hence, acceleration (= v 2 / R ) will also become only two times.

Objective Questions 2πm , independent of v. Bq µ i 3. B = 0 , independent of diameter of wire. 2π r 8. In uniform B is force on any current carrying loop is always zero.

2. T =

9. M = NiA

1 Bc 8 µ 0NiR 2 1 µ 0Ni  ∴ = 2(R 2 + x 2 )3/ 2 8  2R  21. Electric field (acting along $j direction) will change the velocity component which is parallel to B (which is also along $j direction). B = B0$j and v = v0$j will rotate the particle in a circle. Hence, the net path is helical with variable pitch. 2Km K 22. r = ⇒ r∝ Bq B B

∴ B⋅α = 0 or 2x + 3 − 4 = 0 ∴ x = 0.5 12. In uniform magnetic field, force on any current carrying loop is always zero. P 1 or r ∝ (as P = constant) 13. r = Bq q

14. θ i = 180° , θ f = 180° − θ W = U f −Ui = − MB cos (180° − θ ) − (− MB cos 180° ) = MB cos θ − MB P

37°

15.

5 cm 4 cm 3 cm

Bx =

20.

10. B ⋅ F = 0 as B ⊥ F

37°

r = 5 cm

3 cm

BB + BD

We can see that all (a), (b) and (c) options are same. 18. F = Bqv or F ∝ v 2qV Now, v= m ∴ F∝ V

BA + BC Bnet

A

24. r =

D

2qVm  2Vm   1  =    Bq  q   B

26. In (c), two wires are producing u magnetic field and two wires are producing ⊗ magnetic field.

27. I 1 produces circular magnetic lines current I 2 is

each small circular element is parallel to (θ = 0° ) magnetic field. Hence, force is zero. 1 28. Arc of radius a  th of circle produces magnetic 4  $ field in k direction or outwards, while arc of radius b produces magnetic field in − k$ direction.

µ0 i (sin 37° + sin 37° ) 4π r µ NiR 2 µ Ni and Bc = 0 16. Bx = 02 c→ centre. 2R 2(R + x 2 )

17. F = I (l × B) = I ( ba × B)

C

23.

B=

∴

1  µ 0I  $ 1   k+ 4  2a  4 µ 0I  1 1 $ =  −  k 8  a b

B=

 µ 0I  2µ 0I Bnet = 2  = πR  2π (R / 2) 

 µ 0I    (− k$ )  2a 

29. Equivalent current, i = q f = ef µ i µ ef B= 0 = 0 2R 2R

30. C

19. Two fields are additive. ∴

Magnetics — 687

45° 45°

a 2

688 — Electricity and Magnetism µ I  BC = 4  0 (sin 45° + sin 45° ) 4 π a / 2   2 2 µ 0I = πa 31. At distance r from centre, µ (i ) (From Ampere’s circuital law) B = 0 in 2π r For path-1, iin ≠ 0 ∴ B1 ≠ 0 For path-2, iin = 0 ∴ B2 = 0 µ0 i 32. dB = (d l × r ) 4 π r3 ∴ dB is in the direction d l × r. Hence, d l is outwards and r is from d l towards P. 33. Magnetic field due the straight portions is zero. It is only due to arc of circle. φ µ0I  (Radius = x) B= ∴   2π  2x  µ0I φ 4 πx 34. From centre, r = (R − x ) µ I µ I B = 0 2 r = 0 2 (R − x ) 2π R 2π R 35. From Fleming’s left hand rule, we can see that magnetic force is outwards on the loop. =

F

∴

= (− 5.2 × 10−3 ) ∴ Bx = − 0.175 T Similarly, (7.8 × 10−6 ) (− 3.8 × 103 ) ( Bz ) = 7.6 × 10−3 or Bz = − 0.256 T (c) From the property of cross product. F is always perpendicular to B . Hence, F⋅B=0 4. Apply F = q ( v × B) For example, let us apply for charged particle at e.   v $ v $ Fe = q  j− k × ( B$i )   2 2   qvB $ $ = (− j − k ) 2

5. r =

2qVm Bq

∴

B= =

1 2Vm r q 1 0.18

2 × 2 × 103 × 9.1 × 10−31 1.6 × 10−19

= 8.38 × 10−4 T mv Bqr 6. (a) r = ⇒ v= Bq m T πm (b) t = = 2 Bq 2qVm (c) r = Bq ∴

So, it tends to expand.

(7.8 × 10−6 ) (3.8 × 10−3 ) ( Bx )

V =

r2 B 2q2 r2 B 2q = 2qm 2m

7. (a) Conservation of charge

Subjective Questions

B

1. F = q( v × B), where q = − 1.6 × 10− 19 C for an electron and q = + 1.6 × 10− 19 C for a proton

2. F = q( v × B) 3. (a) F = q ( v × B) [ (7.6 × 10−3 ) $i − (5.2 × 10−3 )k$ ] = (7.8 × 10−6 ) [(−3.8 × 103 )$j ( Bxi$ + B y$j + Bzk$ )] or

They will collide after time, T πm t= = 2 Bq

8. (a) At A, magnetic force should be towards right. From Fleming’s left hand rule, magnetic field should be inwards.

Chapter 26

13. W = Fm or mg = ilB sin 90°

mv Bq mv B= qr r=

Further, ∴

(r = 5 cm)

T πm = 2 Bq 9. Component of velocity parallel to B, i.e. vx will remain unchanged v y$j and B$j will rotate the particle in y z-plane (⊥ to B). (b) tAB =

Z

F

Y

vy t

θ vy

At the beginning direction to magnetic force is − k$ [from the relation, F = q (v y$j × B$i ) In the time t, particle rotates an angle  Bq θ = ω t =   t from its original path.  m In the figure, we can set that, y - component of velocity at time t is v y cos θ and z - component is − v y sin θ. 10. Fe + Fm = 0 or qE + q ( v × B) = 0 or E = − ( v × B) or E = (B × v) 11. Work is done only by electrostatic force. Hence, from work-energy theorem 1 = mv 2 = work done by electrostatic force only 2 2qE0 z = (qE0 )z or Speed v = m Particle rotates in a plane perpendicular to B, i.e. in xz-plane only. Hence, v y = 0 12. When they are moving rectilinearly, net force is zero. ∴ qE = Bqv sin (90° − θ ) E ∴ v= B cosθ When electric field is switched off,  2πm p=  v cos (90° − θ )  Bq  =

2πm E tan θ qB 2

Magnetics — 689

mg (13 × 10−3 ) (10) = Bl 0.44 × 0.62 = 0.47 A Magnetic force should be upwards to balance the weight. Hence, from Fleming’s left hand rule we can see that direction of current should be from left to right. 14. (a) ilB = mg V mgR or lB = mg or V = R lB (0.75) (9.8) (25) = 0.5 × 0.45 ∴

i=

≈ 817 V ilB − mg VlB (b) a = = −g m mR (817) (0.5) (0.45) = − 9.8 (0.75) (2.0)

V   as i =   R

≈ 112.8 m/s2 15. i = 5 A ⇒ B = (0.02$j) T Now, applying F = i (l × B) in all parts. Let us find l for anyone parts. l cd = rd − rc = (0.4 $j + 0.4 k$ ) − (0.4 $i + 0.4 k$ ) = (0.4 $j − 0.4 i$ )

16. Let surface charge density is σ. f r

dq

dr

dq = [(2πr)dr ]σ Equivalent current, i = (dq) f dM = iA = [(dq) f ][ πr2 ] ∴

R

M = ∫ dM 0

µ i µ (dq) f dB = 0 = 0 2r 2r ∴

R

B = ∫ dB 0

Now, we can find the ratio

M . B

690 — Electricity and Magnetism 17. (a) From energy conservation, U θ = K θ = U 0° + K 0° or (− MB cosθ ) + 0 = (− MB cos 0° ) + K 0° Substituting the given values, we can calculate. (b) To other side also it rotates upto the same angle. 2πr 18. (a) T = v  v  (b) I = qf = (e)    2πr evr  ev  2 (c) M = IA =   (πr ) =  2πr 2 and eh. 20. Assume equal and opposite currents in wires PQ and RS, then find M. z Q P y S R x

B (2$j) ⇒ τ = M × B

21. Bnet = B 2 + B 2 = 2 B µ0 i (sin 0° + sin 90° ) 4π r 22. Magnetic fields at O due to currents in wires ab and cd are zero. where, B =

a b

α β

c d r2

O

r1

Magnetic field due to current in wire da (say B2) is inwards due to current in wire bc (say B1) is outwards. µ i B1 = 0 (sin α + sin β ) 4 π r1 µ i B2 = 0 (sin α + sin β ) 4 π r2 ∴

B1 > B2 as Bnet = B1 − B2

outwards and due to I 2 is inwards. So, net magnetic field may be zero. Similarly, in third quadrant magnetic field due to I 1 is inwards and due to I 2 magnetic field is outwards. Hence, only in first and third quadrants magnetic field may be zero. Let magnetic field is zero at point P (xy), then BI1 = BI2 µ0 I1 µ0 I2 = ∴ 2π y 2π x ∴

19. Assume equal and opposite currents in wires cf

Now,

23. In first quadrant magnetic field due to I 1 is

r1 < r2 (outwards)

y=

I1 x I2

24. Two straight wires produces outward magnetic field by arc of circle produces inward magnetic field. Due to straight wires, µ i  B1 = 2 0 (sin θ + sin 90° )  4 π R  µ i (outwards) = 0 2πR θ  µ 0i  Due to circular arc, B2 = (inwards)   2π  2R  For net field to be zero, B1 = B2 or θ = 2 rad 25. (a) If currents are in the same direction, then above and below the wires magnetic fields are in the same direction. Hence, they can’t produce zero magnetic field. In between the wires, let B = 0 at a distance (r) cm from the wire carrying 75 A current. Then, µ 0  75  µ 0   25    =    2π  r   2π   40 − π  Solving, we get r = 30 cm. (b) If currents are in opposite direction, then in between the wire magnetic field are in the same direction. So, they cannot produce zero magnetic field. The points should be above or below the wires, nearer to wire having smaller current. Let it is at a distance r from the wire having 25 A current. Then, µ 0  25  µ 0   75    =    2π  r   2π   40 + r Solving this equation, we get r = 20 cm µ 0NiR 2 26. Apply B = 2(R 2 + x 2 )3/ 2

Chapter 26 27. I 2 produces inwards magnetic field at centre. Hence, I 1 should produce outward magnetic field. Or current should be towards right. Further, µ 0I 2  µ 0   I 1  =     2π   D  2R  πD  ∴ I1 =  I  R  2 µ 0Ni 2R µ 0NiR 2 (b) B2 = , 2(R 2 + x 2 )3/ 2

28. (a) B1 =

given B2 =

B1 2

29. 10 A and 8 A current produce inward magnetic field. While 20 A current produces outward magnetic field. Hence, current in fourth wire should be (20 − 10 − 8) A or 2 A and it should produce inward magnetic field. So, it should be downwards toward the bottom. 30. (a) B at origin BKLM = BKNM µ I = 0 (− $i + $j) 4R Now, we can apply F = q ( v × B) for finding force on it. (b) In uniform magnetic field, FKLM = FKNM = FKM = I (l × B) = I [{2R (− k$ )} × {B $j}]

Magnetics — 691

(b) The above calculated magnetic field can be written as 2

 sin π / n   µ i  π /n  B= 0 2r cos π / n π As, n→ ∝ , → 0 n  sin π / n Hence, lim  →1 π / n→ 0  π /n  lim (cot π / n) → 1 µ i or B→ 0 2r 32. Current per unit area, I σ= 2 πa − π (a/ 2)2 − (a/ 2)2 2I = 2 πa Total area is (πa2 ). Therefore, the total current is and

π / n→ 0

I 1 = (σ ) (πa2 ) = 2I Cavity area is π (a/ 2)2. Therefore, cavity current is I I 2 = (σ ) (πa2 / 4 ) = 2 Now, the given current system can be assumed as shown below.

0

= (2 B0 IR ) $i ∴

Net force is two times of the above value. π n

31. (a)

1

π n

(a) At P1 , a

a=

π π  µ i  B = n  0  sin + sin   n n   4π x   µ i (2 sin π / n) =n 0 4 π ( π r / n )cot π / n   =

 π  π µ 0in2 sin   tan    n  n 2π 2r

I/2

3

I/2

2I

x

1  2πr πr   = 2 n  n π  πr   π x = a cot =   cot    n n  n

2 +

and ∴

µ 0 2I µ 0I = 2π 5 πr µ0 I /2 B2 = 2π (r − a/ 2) µ I /2 B2 = 0 2π (r + a/ 2) B1 =

(towards left) (towards right) (towards right)

(towards left) Bnet = B1 − B2 − B2 1 1 1   r − 4 r − 2a − 4 r + 2a    µ 0I 16r3 − 4 a2 − 4 r2 − 2ar − 4 r2 + 2ar  =  π  r(16 r2 − 4 a2 )  µ I = 0 π

=

µ 0I πr

 2r2 − a2   2 2  4r − a 

(towards left)

692 — Electricity and Magnetism B1

√r 2+ a 2/4

a 2

(b) P2

r

θ θ B2

B3

θ

θ +

v

θ

L

 L Deviation, θ = sin −1   for L < r  r

P

B1

2 B1 B2

Q

At point P , B1 and B2 are in opposite directions. Hence, BP = 0 At point, Q , B1 and B2 are in same direction. µ λ Hence, BQ = 2  0  = µ 0λ  2  F µ 0 I 1I 2 = L 2π a µ I I (Repulsion or upwards) ∴ F= 0 1 2L 2π a M of the loop is inwards and magnetic field to I 1 on the plane of loop is outwards. Hence, τ = 0, as τ = M × B and angle between two is 180°.

35.

θ O

mv Bq

v

v C

1

B2

r=

where,

B

B in the above situation is given by µ λ B= 0 2

34.

θ

36.

(towards the top) Bnet = B1 − 2 B2 cos θ]   µ 2I µ I /2 r = 0 −2 0  2 2 2 2π R 2 π  r + a / 4  r + a2 / 4 µ I  2r2 + a2  (towards the top) = 0  2 πr  4 r + a2 

33.

v

r

For paraxial electron θ ≈ 0° and q = e, 2πmv ∴ d= Be

v

v

L=r

L>r

θ = π if L ≤ r qE0 (along negative z-direction) 37. aE = m Electric field will make z-component of velocity zero. At that time speed of the particle will be minimum and that minimum speed is the other component, i.e. v0. This is minimum when, vz = uz + az t qE or 0 = v0 − 0 t m mv0 or t= qE0 and

38. Path is helix and after one rotation only x-coordinate will change by a distance equal to pitch.  2πm ∴ x = p = (v0 cosθ )    Bq 

39. M = i(CO × OA) = i(CO × CB) = 4[(− 0.1 $i ) × (0.2 cos 30° $j + 0.2 sin 30° k$ ) = (0.04 $j − 0.07 k$ ) A-m 2

x- axis

Electrons touch the x-axis again after every pitch. Therefore, the asked distance is  2πm d = p = v11T = (v cosθ )    Bq 

C

y A

40.

B

τ O 30°

x C

z

Magnetics — 693

Chapter 26 M = Ni (OA × AB) = Ni (OA × OC)

LEVEL 2

= (100) (1.2) [(0.4 $j) × (0.03 cos 30° $i + 0.3 sin 30° k$ )]

Single Correct Option 1. τ mg about the left end (from where string is connected)

= (7.2 $i − 12.47 k$ ) A-m 2 τ=M×B = [(7.2 i$ − 12.47 k$ ) × (0.8 $i )]

or

= (− 9.98j$ ) N-m ∴ | τ | = 9.98 N-m Torque vector and expected direction of rotation is shown in figure. 41. BA = BB = BC = BD =B=

2. In uniform field, y P

µ 0 i (2 × 10−7 ) (5) = 2π r (0.2/ 2 )

A

a=

∴

C

F∆t = m 2gh or

or Net magnetic field

= 2 2B (towards bottom as shown)

2π br3 42. i = ∫ (2πr dr) j = ∫ (2πr dr) (br) = 0 0 3 (a) For r1 < R µ i B = 0 in 2π r1 r

=

∆q =

Hence,

= 2.0 × 10 T

µ 0  2πbr13 / 3   2π  r1 

4.

(ilB ) ∆t = m 2gh

i ∆t = ∆q (∆q) (lB ) = m 2gh

But, ∴

= ( BA + BC )2 + ( BB + BD )2 −5

x (m)

+2

F (0.16 $j) = = (16 . $j) m/s2 m 0.1 3. Linear impulse = mv

BB +BD Net

r

O

Magnetic force on POQ = magnetic force on straight wire PQ having the same current. Hence, F = i (l × B) = i (PQ × B) = 2 [(4 $i ) × (− 0.02 k$ )] = (0.16 $j)

D

B

Q

–2

= (5.0 2 ) × 10−6 T

BA+BC

= | M × B | = MB sin 90° (mgR ) = (NiA ) B0 = i (πR 2 ) B0 mg i= πRB0

or

m 2gh Bl

M q = L 2m ∴

 q  q M =   L =   (Iω )  2m  2m  q  2  1 =    mR 2ω = qR 2ω  2m  5  5

5.

µ br2 = 0 1 3

v

r

θ

θ d

(b) For r2 > R µ i B = 0 in 2π r2

+ x=0

 µ   2πbR / b µ 0bR =  0  =  2π   3r2 r2  3

3

sin θ =

v x=d

d d = ⇒ r (mv / Bq)

q v sin θ = m Bd

694 — Electricity and Magnetism 13. FMNPQ = FMQ and this force should be upwards to

6. Force

balance the weight. N

P

M

Variation of magnetic force on wire ACB is as shown in figure. Point of application of net force lies some where between A and C. 7. At a distance X from current I 2, µ I B= 0 2 2π X Magnetic force of small element dX of wire AB dF = I 2 (dX )B sin 90° F =∫

∴

x = 2a

x=a

dF

8. Apply screw law for finding magnetic field around a straight current carrying wire. µ 0IR 3 9. B1 = 2 (R 2 + x 2 )3/ 2 µ 0I (2R )2 B2 = 2 [(2R )2 + (2x )2 ] 3/ 2 B1 =2 B2

10. (b − a) = radius of circular part mv Bq Bq (b − a) + V = m b–a q = 2m 2 πqfl 2  q  q   ml  =   (Iω ) =     (2πf ) =  2m  2m  3  3 =

∴

11.

M L M

12. At x = 0, y P N M

y = ± 2m FMNP = FMP = i [ MP × B ] = 3 [ (4 $j) × (5k$ )] = 60 $i

x

Q

∴

ilB = mg, where a l = MQ = 2 2Mg ∴ i (a/ 2) B = mg or i = aB Force is upwards if current is clockwise or current in MQ is towards right. r 14. = radius of circular path 2 mv mv = = Bq (µ 0ni ) (q) µ 0qr ni v= ∴ 2m 15. (a) B ⊥ v, so it may along y - axis (b) F ⊥ v, ∴ a ⊥ v=0 or a⋅ v=0 or a1 b1 = a2 b2 (c) See the logic of option (a). (d) Magnetic force cannot change the kinetic energy of a particle. 16. Magnetic force is always perpendicular to velocity. So, it will always act in radial direction which will change tension at different points. But, time period and θ will remain unchanged. 17. Force on the wires parallel to x-axis will be obtained by integration (as B ∝ x and x coordinates vary along these wires). But on a loop there are two such wires. Force on them will be equal and opposite. Forces on two wires parallel to y-axis can be obtained directly (without integration) as value of B is same along these wires. But their values will be different (as x-coordinate and therefore B is different). (on two wires) Fnet = ∆F = Ia (∆B ) = Ia ( B0 ) (∆x ) = Ia B0 (a) = IB0a2 This is indecent of x ∴ F1 = F2 = IBla2 ≠ 0

Magnetics — 695

Chapter 26 18. Bx =

µ 0 I in 2π x

23. qE = Bqv

where,

=

v=

I (c2 − x 2 ) (c2 − b2 )

19. Fe = qE = (1.6 × 10− 19 ) (− 102.4 × 103 ) k$ = (− 1.6384 × 10− 14 k$ )N Fm = q ( v × B) = (1.6 × 10− 19 ) [ (1.28 × 106 $i ) × (8 × 10− 2 $j) ] = (1.6384 × 10− 14 k$ )Ν Now, we can see that Fe + Fm = 0 20. Due to $j component of B, magnetic force is zero. It is only due to $i component. l is towards − y direction, B is towards $i direction. Hence, l × B (the direction of magnetic force is + k$ ) 1

F = ∫ dF = ∫ (i )(dy) ( B ) 0

1

= ∫ (2 × 10− 3 ) (dy) (0.3 y) 0

= 3 × 10− 4 N

21. r =

E B mv m (E / B ) r= = Bq B (q) E (where, S = q/ m) = 2 BS

∴

  I I in = I −  [ π (x 2 − b2 )] 2 2  π ( c − b )  

mv Bq

F = weight per unit length l µ 0 i1i2 ∴ = 0.01 × 10 2π r (2 × 10− 7 ) (100 × 50) or = 0.01 × 10 r or r = 0.1 m When B wire is displaced downwards from equilibrium position, magnetic attraction from A wire will decrease (which is upwards). But, weight (which is downwards). So, net force is downwards, in the direction of displacement from the mean position or away from the mean position. Hence, equilibrium is unstable. 25. Magnetic field due to current in the wire along z-axis is zero. Magnetic field due to wire along x-axis is along $j direction and magnetic field due to wire along y - axis is along − $i direction. Both

24.

wires produce B=

(during circular path)

∴

v=

26. | FABC | + = | FAC | = ilB

Bqr m

qE = Bqv B 2qr E = Bv = ∴ m (0.1)2 (20 × 10− 6 ) (5 × 10− 2 ) E= ∴ (20 × 10− 9 ) = 0.5 V/m µ 0i1 (4 π × 10− 7 ) (5) 22. B1 = = 2R1  5  × 10− 2 (2)   2  Now,

= 2 2π × 10− 5 T B2 =

Bnet =

µ 0i2 (4 π × 10 ) (5 2 ) = 2R2 (2) (5 × 10− 2 ) B12 + B22

= (2) (5) (2) = 20 N 1 qx 27. E = 4 πε 0 (R 2 + x 2 )3/ 2 B=

µ 0 iR 2 2 (R 2 + x 2 )3/ 2

where,

 V  i = qf = q    2πR  c=

1 ε0µ 0

More than One Correct Options −7

= 2 2π × 10− 5 T

µ0 i 2π a

1. B1 =

µ 0N 1i1 2R`1 =

(4 π × 10− 7 ) (50) (2) 2 (5 × 10− 2 )

= 4 π × 10− 4 T

696 — Electricity and Magnetism B2 =

µ 0N 2i2 2R2

8. KE = qV

2qVm or r ∝ V Bq 2πm or T is independent of V . T = Bq

= 4 π × 10− 4 T

2. (a) v is parallel on anti-parallel to B. (c) qE + q ( v × B) = 0 or E = − ( v × B) = (B × v) mV (1) (10) 3. r = = =5m Bq (2) (1) 2πm (2) (π ) (1) T = = =π Bq (2) (1) = 3.14 s Plane of circle is perpendicular to B, i.e. xy-plane.

4. θ = 180° τ = MB sin θ = 0 U = − MB cos θ = + MB = maximum

5. If current flows in a conductor, then E≠0 E=0 µ i B= 0 2 r 2π R B = 0 at r = 0, i.e. at centre µ i B = 0 for outside points. 2π r 6. Fab = upwards

(for inside points) (for outside points) (for inside points)

Fabc = leftwards ∴ Net force on loop is neither purely leftwards or rightwards or upwards or downwards. 7. Fm = q ( v × B) Depending on sign of q, Fm may be along positive z-axis or along negative z-axis. Fe = qE Again, depending on the value of q it may be along positive z-axis or along negative z-axis. If q is positive, v × B and Fm comes along negative z-axis also. But, Fe comes along positive z-axis. So, it may also pass undeflected.

KE ∝ V

r=

(4 π × 10− 7 ) (100) (2) = (2) (10 × 10− 2 ) When currents are in the same direction, then Bnet = B1 + B2 When currents are in the opposite directions, then Bnet = B1 − B2

or

9. (a) Point a lies to the right hand side of ef and fg. Hence, both wires produce inward magnetic field. Hence, net magnetic field is inwards. Same logic can be applied for other points also. 10. See the hint of Q.No-3 of Assertion & Reason section for Level 1

Match the Columns 1. (a) F = q( v × B) q is negative, v is along + $i and B along + $j. Therefore, F is along negative z. (b) Same logic is given in (a). (c) B is parallel to v. So, magnetic force is zero. Charge is negative so, electric force is opposite to E. (d) Charge is negative. So, electrostate force is in opposite direction of E. 2. For direction of magnetic force apply Fleming’s left hand rule. According to that w and x are positively charged particles and y and z negatively charged particle. Secondly, 2Km r= Bq ∴

r∝

m q

(K → same)

3. Force on a current carrying loop is zero for all

angles. τ is maximum when Q, then angle between M and B is 90° minimum potential energy is at θ = 0°. Positive potential energy is for obtuse angle. Direction of M is obtained by screw law.

4. Two currents are lying in the plane of paper. Its point is lying to the right hand side of the current carrying wire, magnetic field is inward (− k$ direction). If point lies to left hand side, field is outward (k$ direction). µ i. i 5. Let us take F = 0 ⋅ 2π r Current in same direction means attraction and current in opposite direction means repulsion. Let

Chapter 26 us find force on wire-2 by other three wires 1,3 and 4.

= ∴

F2 = F F1 = F

 aµ = 2 ∫ 0  0 π  µ I 2 =  0 .  π 

Fnet is rightwards

=

6. FABC = FADC = FAC Floop = 2 FAC I  = (l × B)  2 

and

= I (AC × B) AC = (l i$ + l $j)

Hence,

µ0 I π 2a + 3x

x=a (dF cos 60° ) (− $i ) F231 = 2 ∫  x = 0 

F3 = F 2

∴

∴

Magnetics — 697

.

I 1 . I .(dx )  (− $i ) 2 2a + 3x

1 [ln (2a + 3

3x )]a0 (− $i )

µ 0I 2  2 + 3  ln   (− i$ ) 3π  2 

µ 0I 2 $ (i ) 2π µ I 2 1 1  2 + 3  $ = 0  − ln    (i ) Ans. π 2 3  2 

F12 = FTotal

2. Magnetic moment According to Bohr’s

B

hypothesis, angular momentum in nth orbit is  h L = n  .  2π  M q e or Further, = L 2m 2m ∴ Magnetic moment,  e  e   h   neh  M =   (L) =    n  =    2m  2m  2π   4 πm

C

A

D

Now, putting value of B we can find net force in different cases.

Subjective Questions µ0 I   (Ia) in positive  2π a

1. Force on wire 12 will be  x-direction. 2

x 60° 60° dF dF

r

3

1

Force on 2-3 and 3-1 : r = a + x sin 60° = a + Br =

µ0 I µ0 I = 2π r 2π a + 3 x 2

3

x 2

Magnetic field induction mv 2 1 (e)(e) = r 4 πε 0 r 2 From Bohr’s hypothesis : nh mvr = 2π Solving these two equations, we find e2 ε n2h2 and r = 0 2 v= 2ε 0nh πme Now, magnetic field induction at centre µ i B= 0 2r  v  Here, i = qf = (e)    2πr  µ   ev  µ 0e  v  ∴ B =  0   =    2r   2πr 4 π  r2 

…(i)

…(ii)

2 2 2 4  µ e  e   π m e  = 0    2 4 4  4 π   2ε 0nh  ε 0n h 

=

µ 0πm2e7 8ε 30h5n5

Ans.

698 — Electricity and Magnetism 3. r = 9 + 7 = 4 cm

−5

= 9.0 × 10

or

B √7 cm

r

(a) α

β

r r = 4 cm

(b)

Net field = 4 B sin θ = 4 × 9.0 × 10−5 × = 2.7 × 10−4 T

4. xy-plane Since,

y

rN

= 2.4 × 106 m/s

Ans.

2πr 2π (0.4 ) 8 = = π v 5 50 2πm m or T ∝ T = Bq q

E 120 × 103 = B 50 × 10−3

v=

∴

3 4

T =

t=0

Let n be the number of protons striking per second. Then, ne = 0.8 × 10−3 n=

or

p

r

0.8 × 10−3 1.6 × 10−19

= 5 × 1015 m/s

5 m/s

x

Force imparted = Rate of change of momentum = nmv = 5 × 1015 × 1.67 × 10−27 × 2.4 × 106 = 2.0 × 10−5 N

6. (a) Speed of particle at origin, v = 5 After collision mass has become times and 4 charge two times. 5 8 π  5 1 ∴ T′ =  ×  T = × π= s  4 2 8 50 10 T′ , i.e. combined mass will 4 complete one-quarter circle. Given time t =

Ans.

z-coordinate Mass of combined body has become 5 times of the colliding particle. Therefore, from conservation of linear momentum, velocity 1 component in z-direction will become times. Or 5 1 40 8 vz = × m/s = m/s 5 π π 8 π Ans. ∴ z = vz t = × = 0.2 m π 40 5. Fe = Fm or eE = eBv

3 cm

×

(as P = constant)

Since, charge has become two times r ∴ r′ = = 0.2 m 2 At t = (π /40) second, particle will be at P in xy -plane. ∴ x = r′ = 0.2 m y = r′ = 0.2 m

T θ

P Bq 1 r∝ q r=

Further

µ i B = 0 (2 sin α ) 4π r  30   3 = (10−7 )   2 ×  5  4 × 10−2  

Ans. 2qV = vx m

a m =a 2qV 2qV m 1 2 1  qE   2 m  a2E = y = ayt = a ×    2 2 m   2qV  4V t=

x = vx

(b) Component parallel to B is m   qE   v y = ayt =    a  = Ea  m   2qV 

q 2mV

Ans.

Chapter 26 Now, pitch = component parallel to B × time period  q   2πm = v y T =  Ea    2mV   Bq   πEa = B

Magnetics — 699

y S R x

2m qV

Ans. P

Q

7. To graze at C Using equation of trajectory of parabola, ax 2 2v cos2 θ

y = x tan θ − Her ,

a=

2

…(i)

qE 10−6 × 10−3 = m 10−10 = 10 m/s2

Substituting in Eq. (i), we have 10 × (0.17)2 0.05 = 0.17 tan 30° − 2 2v × ( 3 / 2)2 Solving this equation, we have v = 2 m /s In magnetic field, AC = 2r or 0.1 = 2r mv cos 30° or r = 0.05 m = Bq mv cos 30° B= ∴ (0.05)q (10−10 )(2)( 3 / 2) (0.05)(10−6 ) = 3.46 × 10−3 T =

= 3.46 mT

Ans.

8. Magnetic moment of the loop, M = (iA )$j = (I L2 )k$ 0

Magnetic field, B = ( B cos 45° ) i$ + ( B sin 45° )$j B $ $ = (i + j) 2 (a) Torque acting on the loop, τ = M × B  B $ $  = (I 0L2k$ ) ×  (i + j)  2  I 0L2 B $ $ ( j − i ) or | τ | = I 0L2 B Ans. 2 (b) Axis of rotation coincides with the torque and since torque is along $j − i$ direction or parallel to QS. Therefore, the loop will rotate about an axis passing through Q and S as shown in the figure. ∴

τ=

|τ | I where, I = moment of inertia of loop about QS. I QS = I PR = I ZZ (From the theorem of perpendicular axis) But, I QS = I PR 4 2I QS = I ZZ = ML2 ∴ 3 2 2 I QS = ML 3 |τ | I L2 B 3 I0B ∴ α= = 0 = I 2/ 3ML2 2 M Angular acceleration, α =

∴ Angle by which the frame rotates in time ∆t is 1 θ = α (∆t )2 2 3 I0B or Ans. . (∆t )2 θ= 4 M 9. In equilibrium, mg …(i) 2T0 = mg or T0 = 2 ω  Magnetic moment, M = iA =  Q (πR 2 )  2π  ωBQR 2 2 Let T1 and T2 be the tensions in the two strings when magnetic field is switched on (T1 > T2 ). For translational equilibrium of ring in vertical direction, …(ii) T1 + T2 = mg For rotational equilibrium, D ωBQR 2 (T1 − T2 ) = τ = 2 2 ωBQR 2 or …(iii) T1 − T2 = 2 Solving Eqs. (ii) and (iii), we have mg ωBQR 2 T1 = + 2 2D τ = MB sin 90° =

700 — Electricity and Magnetism As T1 > T2 and maximum values of T1 can be

3T0 , 2

We have

3T0 ω BQR 2 = T0 + max 2 2D DT0 Ans. ω max = ∴ BQR 2 µ (i/ω ). dx 10. dB = 0 2π x µ i d + ω dx ∴ B= 0 ∫ x 2πω d µ 0i  d + ω B= ln   (upwards) Ans.  d  2πω

 4 ∴ 0.0442 = (kx )(l sin 53° ) = (4.8)(x )(0.2)   5 x = 0.057 m

or

U =

1 2

kx 2 =

1 2

× (4.8)(0.057)2

= 7.8 × 10–3 J

Ans.

13. Let the direction of current in wire PQ is from P to Q and its magnitude be I. z

dB

Q y

P 1

x dx

x

 R  R −1  BqR  11. θ = tan −1   = tan −1    = tan   mv   r  mv / Bq

α

θ

×

0

R ×

The magnetic moment of the given loop is M = − Iabk$

Torque on weight of the loop about axis PQ is a  τ 2 = r × F =  $i  × (− mgk$ ) 2 

×

90°

r

×

O

×

C

×

× ×

×

r

=

12. (a) Yes, magnetic force for calculation of torque can be assumed at centre. Since, variation of torque about P from one end of the rod to the other end comes out to be linear. ∴

2  l  Il B τ = (IlB )   =  2 2

=

(6.5)(0.2)2 (0.34) 2

= 0.0442 N-m

R

Torque on the loop due to magnetic force is τ1 = M × B = (− Iab k$ ) × (3$i + 4 k$ ) B0$i = − 3IabB $j

×

×

 BqR  Deviation = 2θ = 2 tan −1    mv 

S

Ans.

(b) Magnetic torque on rod will come out to be clockwise. Therefore, torque of spring force should be anti-clockwise or spring should be stretched. (c) In equilibrium, Clockwise torque of magnetic force = anti-clockwise torque of spring force

mga $ j 2

We see that when the current in the wire PQ is from P to Q , τ 1 and τ 2 are in opposite directions, so they can cancel each other and the loop may remain in equilibrium. So, the direction of current I in wire PQ is from P to Q. Further for equilibrium of the loop |τ1 | = |τ2 | mga or 3IabB0 = 2 mg Ans. I = 6bB0 Magnetic force on wire RS is F = I (l × B) = I [(− b$j) × {(3i$ + 4 k$ ) B0}] F = IbB (3k$ − 4 $i ) 0

Ans.

27. Electromagnetic Induction INTRODUCTORY EXERCISE

27.1

1. ⊗ magnetic field passing through loop is increasing. Hence, induced current will produce magnetic field. So, induced current should be anti-clockwise. 2. It is true that magnetic flux passing through the loop is calculated by integration. But, it remains constant. dφ B 3. = [ Potential or EMF ] dt = [ ML2A− 1 T − 3 ]

4.

u is increasing. Hence, ⊗ is produced by the

induced current. So, it is clockwise. 5. By increasing the current in loop-1, magnetic field in ring-2 in downward direction will increase. Hence, induced current in ring-2 should produce upward magnetic field. Or current in ring should be in the same direction.

1

2

φ = | B ⋅ S | = 9 × 10− 7 Wb

INTRODUCTORY EXERCISE

27.2

e = Bvl = 1.1 × 5 × 0.8

1. Q

= 4.4 V Apply right hand rule for polarity of this emf. 2. Q e = Bvl e Bvl i= = R R B 2l 2v F = ilB = R (0.15)2 (0.5)2 (2) = 3 = 0.00375 N B ωl 2 3. VA − VC = 2 Bω (2l )2 VD − VC = 2 From these two equations, we find V A − V D = − 3 B ωl 2 / 2

4. Circuit is not closed. So, current is zero or

B

magnetic force is zero.

6. 2πR = 4 L ∴

8. S = [(5 × 10− 4 ) k$ ] m 2

πR (π ) (10) = 2 2 = (5π ) cm ∆S = S i − S f = (πR 2 ) − L2 L=

= π (0.1)2 − (5π )2 × 10− 4 = 0.0067 m ∆φ  ∆S  e= =B   ∆t  ∆t 100 × 0.0067 = 0.1 = 6.7 V 7. ∆φ = 2 (NBS ) ∆φ 2NBS ∆q = = R R 2 × 500 × 0.2 × 4 × 10− 4 = 50 = 1.6 × 10− 3 C 2

INTRODUCTORY EXERCISE

1. | e | = L

∆i ∆t

or

L

27.3

di dt

L = 1H di and = 3 [sin t + t cos t ] dt ∴ | e| = 3 (t cos t + sin t ) di d 2. VL = + L = (2) (10e−4t ) dt dt = − 80 e−4t di Further, Va − iR − L = Vb dt di Va − Vb = iR + L ∴ dt or Vab = (10e−4t ) (4 ) − 80e−4t Here,

= − 40e−4t

702 — Electricity and Magnetism 3. (a) dI / dt = 16A/s

INTRODUCTORY EXERCISE

e 10 × 10− 3 ∴ L= = dI / dt 16

1. Q

= 0.625 × 10− 3 H = 0.625 mH (b) At t = 1s, I = 21 A 1 1 U = LI 2 = × (0.625 × 10− 3 )(21)2 2 2 = 0.137 J P = Ei = (10 × 10− 3 ) (21)

E=

1 2 2 Li = i Rt 2

L has the units of time. R L 2 2. (a) τ L = = = 0.2 s R 10 E 100 (b) i0 = = = 10 A R 10 ∴

(c) i = i0 (1 − e− t / τ L )

= 0.21 J/s

4. (a)

= 10 (1 − e− 1/ 0.2 ) = 9.93 A

l

µ 0N 2S l l N = d µ 0lS (4 π × 10− 7 ) (0.4 ) (0.9 × 10− 4 ) L= 2 = (0.1 × 10− 2 )2 d L=

∴

= 4.5 × 10− 5 H (b) e = L

∆i (4.5 × 10− 5 ) (10) = = 4.5 × 10− 3 V ∆t 0.1

3. E = VR + VL INTRODUCTORY EXERCISE 1 q2 2 C q it LC = = = t i i 1 2

1. U = Li 2 = ∴

3. (a) VL = VC L

INTRODUCTORY EXERCISE

1. Q

27.5

M =

27.4

di dt = (0.75 × 18 × 10− 6 ) (3.4 )

q = (LC )

−3

e2 (50 × 10 ) = di1 / dt (8/ 0.5)

= 45.9 × 10− 6 C

= 3.125 × 10− 3 H = 3.125 mH −3  ∆i  (3.125 × 10 ) (6) e1 = M  2  =  ∆t  0.02

= 0.9375 V N φ (1000) (6 × 10− 23 ) 2. (a) M = 2 2 = =2H i1 3  ∆i  (2) (3) (b) | e2 | = M  1  = = 30 V  ∆t  0.2 N 1 φ 1 (600) (5 × 10− 3 ) = = 1H i1 3 di 3. (a) | e2 | = M  1   dt  (c) L1 =

= (3.24 × 10− 4 ) (830) = 0.27 V (b) Result will remain same.

di q = dt C

= 45.9 µC (b)

VL = VC = =

q C

4.2 × 10− 4 18 × 10− 6

= 23.3 V 1 2 1 2 4. Limax = CVmax 2 2  L ∴ Vmax =   imax  C  20 × 10− 3   (0.1) =  0.5 × 10− 6    = 20 V

27.6

Chapter 27 INTRODUCTORY EXERCISE

inductance between solenoid and coil, µ N N (πR12 ) M = 0 1 2 l1 µ 0N 1N 2S 1` = l1 di µ N N S di = 0 1 2 1 1 dt l1 dt

(4 π × 10− 7 ) (25) (10) (5 × 10− 4 ) (0.2) 10− 2 −6 = 3.14 × 10 V =

dφ =e dt e e E= = l 2πR2

El =

(b) ∴

=

F = qE =

27.7

1. In the theory we have already derived mutual

| e2 | = M

3.14 × 10− 6 (2π ) (0.25)

= 8.0 × 10− 21 N ⊗ magnetic field at the given instant is increasing. Hence, induced current in an imaginary circular loop passing through P2 should produced u magnetic field. Or, current in this should be anti-clockwise. Hence, electrons should move in clockwise direction. Or electron at P2 should experience force in downward direction (perpendicular to r2). (b) At P1 dφ El = dt dB  dB  ∴ E (2πr1 ) = s = (πr12 )    dt  dt ∴

2. (a) At P2 dφ dt

dB dt 2 R  dB  E=   2r2  dt 

E (2πr2 ) = πR 2 ⋅ ∴

eR 2 (6t 2 − 8t ) 2r2

Substituting the values, we have (1.6 × 10− 19 ) (2.5 × 10− 2 ) F= [ 6 (2)2 − 8 (2)] 2 × 5 × 10− 2

= 2 × 10− 6 V/m

El =

Electromagnetic Induction — 703

r1  dB    2  dt  r = 1 [ 6t 2 − 8t ] 2 0.02 = [ 6 (3)3 − 8 (3)] 2 = 0.3 V/m As discussed in the above part, direction of electric field is in the direction of induced current (anti-clockwise) in an imaginary circular conducting loop passing through P1. E=

Exercises LEVEL 1

2. At time t0, magnetic field is negative or and

Assertion and Reason

increasing. Hence, induced current will produce ⊗ magnetic field. Or induced current should be clockwise. dφ If = constant. Then, e = constant dt ∴ i or rate of flow of charge is constant. 3. It can exert a force on charged particle. di 4. = 2 A/s dt di Va − Vb = L dt = (2) (2) = 4 V

1. Due to non-uniform magnetic field (a function of x) magnetic flux passing through the loop obtained by integration. But that remains constant with time. dφ Hence, =0 dt or e=0 Magnetic field is along − k$ direction or in ⊗ magnetic is increasing. Hence, induced current should produce u magnetic field. Or induced current should be anti-clockwise.

704 — Electricity and Magnetism 5. Comparing with spring-block system  dI    is acceleration.  dt 

5.

amax = ω 2 A = ω (ωA )

If magnetic field in the shown cylindrical region is changing, then induced electric field exists even outside the cylindrical regions also where magnetic field does not exist. dq 6. i = = (8t )A dt di = 8 A/s dt At t = 1 s, q = 4 C, i = 8 A di and = 8 A/s dt

= ω (vmax )

 dI  = ω I max    dt  max

∴

6. Applying RHR, we can find that Va > Vb

8. Ferromagnetic substance will attack more number of magnetic lines through it. So, flux passing through it will increase. Hence, coefficient of self-inductance will increase. L depends on number of turns in the coil’s radius of coil etc. It does not depend on the current passing through it. 9. Current developed in the inductor wire will decrease exponentially through wire ab. 10. VL1 = VL2 di1 di = L2 2 dt dt L1di1 = L2di2 L1i1 = L2i2 1 i∝ L

∴

Charge on capacitor is increasing. So, charge on positive plate is also increasing. Hence, direction of current is towards left. a

or

∴ dI 7. = I 0 ω cos ωt dt

∴

1 2 Li 2 2 −2 U   ML T  [L ] =  2  =    i   A2 

∴

= [ML2T − 2A− 2 ]

2. M ∝ N 1 N 2 S

b

4 = Vb 2 Va − Vb = − 30 V

dI = MI 0 ω cos ωt dt = MI 0 ω = 0.005 × 10 × 100π = (5π ) V

8.

emax

1 2 1 Li = CV 2 2 2 L ∴ V = C

⇒ i=

2 (2) 4 × 10− 6

= 2 × 103 V

3. N

2F – + i

e=M

Objective Questions 1. U =

4V 2H

Now, Va + 2 × 8 − 4 + 2 × 8 +

L

or or

2Ω

S

N

When brought closer induced effects should produced repulsion. So, currents should increase, so that pole strength increases. Hence, repulsion increases. 4. Magnetic field of ring is also along its axis, or in the direction of velocity of charged particle. Hence, no magnetic force will act on charged particle. But, due to g velocity of charged particle will increase.

B ωl 2 = constant 2 ∆φ 10. ∆q = R ∆φ or i∆ t = R ∴ ∆φ = i (∆t )R = (10 × 10− 3 ) (5) (0.5)

9. e =

= 25 × 10− 3 Wb

11. ⊗ magnetic field is increasing. Therefore, induced electric lines are circular and anti-clockwise. Force on negative charge is opposite to electric field.

Chapter 27 12. e =

dφ = (aτ − 2at ) dt e aτ − 2at i= = R R

21.

τ

0

di 13. e = L = L (Slope of i -t graph) dt ⇒

e=0

Then in remaining two regions slopes are constants but of opposite signs. Hence, induced emfs are constants but of opposite signs. 14. VA − 1 × 5 + 15 + (5 × 10− 3 ) (103 ) = VB ∴

VB − VA = 15 V

 1 =   (4 ) = 2 A  2

22. Relative velocity = 0 ∴ Charge in flux = 0

23. In case of free fall,

q0  dI  =L   dt  max C q  dI  = 0    dt  max LC di 17. VL = L dt 18. φ i = BS cos 0° = 2 Wb

24.

25.

∴

26.

φ f = BS cos 180° = − 2 Wb | ∆φ | = 4Wb | ∆φ | | ∆q| = R 4 = = 0.4 C 10 19. S = (ab) k$ → perpendicular to x y -plane φ = B ⋅ S = (50) (ab) = constant dφ =0 dt ∴ e=0 20. Back emf = Applied voltage potential drop across armature coil = 200 − iR = 200 − 1. 5 × 20 = 170 V

1 2 gt 2 1 = (10) (1)2 = 5 m 2 Here due to repulsion from induced effects a E 3 (c) E 2 > E1 > E 3

Sphere-3

(b) E 3 > E1 > E 2 (d) E 3 > E 2 > E1

41. Four charges Q1 , Q2 , Q3 and Q4 of same magnitude are fixed along the x-axis at x = − 2a , − a , + a and +2a respectively. A positive charge q is placed on the positive y-axis at a distance b > 0. Four options of the signs of these charges are given in Column I. The direction of the forces on the charge q is given in Column II. Match Column I with Column II and select the correct answer using the code given below the lists. (Matching Type, 2014)

E2

+q (0,b)

Q2

E1 =1 E2 Q 3 (c) 1 = Q2 K

P

R

E1 1 = E2 K 2+K C (d) = C1 K

(b)

(a)

39. Let E1(r ), E2(r ) and E3 (r ) be the respective electric fields at a distance r from a point charge Q, an infinitely long wire with constant linear charge density λ, and an infinite plane with uniform surface charge density σ. If E1(r0 ) = E2(r0 ) = E3 (r0 ) at a given distance r0, then (More than One Correct Option, 2014)

(a) Q = 4 σπr02 r r (c) E1  0  = 2 E 2  0   2  2

λ 2 πσ r r (d) E 2  0  = 4E 3  0   2  2 (b) r0 =

Q2 (–a,0)

Q1 (–2a,0)

Q3 (+a,0)

Q4 (+2a,0)

Column I

Column II

1.

+x

Q. Q1, Q2 positive; Q3 , Q4 negative 2.

−x

R. Q1, Q4 positive; Q2 , Q3 negative 3.

+y

S. Q1, Q3 positive; Q2 , Q4 negative 4.

−y

P. Q1, Q2 , Q3 , Q4 all positive

Codes P (a) 3, (b) 4, (c) 3, (d) 4,

Q 1, 2, 1, 2,

R 4, 3, 2, 1,

S 2 1 4 3

Previous Years’ Questions (2018-13)

23

42. Two ideal batteries of emf V1 and V 2 and three resistances R1 , R2 and R3 are connected as shown in the figure. The current in resistance R2 would be zero if

45. A galvanometer gives full scale deflection with 0.006 A current. By connecting it to a 4990 Ω resistance, it can be converted into a voltmeter of range 0-30 V. If 2n connected to a Ω resistance, it 249 becomes an ammeter of range 0-1.5 A. The value of n is (Single Integer Type, 2014)

(More than One Correct Option, 2014)

V1

R1 R2

V2 R3

(a) V1 = V2 and R1 = R 2 = R 3 (b) V1 = V2 and R1 = 2R 2 = R 3 (c) V1 = 2V2 and 2R1 = 2R 2 = R 3 (d) 2V1 = V2 and 2R1 = R 2 = R 3

43. During an experiment with a metre bridge, the galvanometer shows a null point when the jockey is pressed at 40.0 cm using a standard resistance of 90 Ω, as shown in the figure. The least count of the scale used in the metre bridge is 1 mm. The unknown resistance is (Single Correct Option, 2014)

R

90 Ω

40.0cm

(a) 60 ± 0.15 Ω (c) 60 ± 0.25 Ω

(b) 135 ± 0.56 Ω (d) 135 ± 0.23 Ω

44. Two parallel wires in the plane of the paper are distance X 0 apart. A point charge is moving with speed u between the wires in the same plane at a distance X1 from one of the wires. When the wires carry current of magnitude I in the same direction, the radius of curvature of the path of the point charge is R1. In contrast, if the currents I in the two wires have directions opposite to each other, the radius of curvature of the path is R2. If X0 R = 3, and value of 1 is X1 R2 (Single Integer Type, 2014)

Passage (Q. Nos. 46-47) The figure shows a circular loop of radius a with two long parallel wires (numbered 1 and 2) all in the plane of the paper. The distance of each wire from the centre of the loop is d. The loop and the wires are carrying the same current I. The current in the loop is in the counter-clockwise direction (Passage Type, 2014) if seen from above.

46. When d ≈ a but wires are not touching the loop, it is found that the net magnetic field on the axis of the loop is zero at a height h above the loop. In that case (a) current in wire 1 and wire 2 is the direction PQ and RS, respectively and h ≈ a (b) current in wire 1 and wire 2 is the direction PQ and SR, respectively and h ≈ a (c) current in wire 1 and wire 2 is the direction PQ and SR, respectively and h ≈ 1.2 a (d) current in wire 1 and wire 2 is the direction PQ and RS, respectively and h ≈ 1.2 a

47. Consider d >> a, and the loop is rotated about its diameter parallel to the wires by 30° from the position shown in the figure. If the currents in the wires are in the opposite directions, the torque on the loop at its new position will be (assume that the net field due to the wires is constant over the loop) (a) (c)

µ 0I 2a 2 d 3 µ 0I2 a 2 d

(b) (d)

µ 0I 2a 2 2d 3 µ 0I2 a 2 2d

48. At time t = 0, terminal A in the circuit shown in the figure is connected to B by a key and an alternating current I ( t ) = I 0 cos (ωt ), with I 0 = 1 A and

24

Electricity & Magnetism ω = 500 rad s −1 starts flowing in it with the initial direction shown in the figure. 7π At t = , the key is switched from B to D. 6ω Now onwards only A and D are connected. A total charge Q flows from the battery to charge the capacitor fully. If C = 20 µF, R = 10 Ω and the battery is ideal with emf of 50 V, identify the correct statement(s). (More than One Correct Option, 2014)

B D A 50V

C=20µF R = 10 Ω

(a) Magnitude of the maximum charge on the 7π capacitor before t = is 1 × 10−3 C 6ω (b) The current in the left part of the circuit just 7π before t = is clockwise 6ω (c) Immediately after A is connected to D, the current in R is 10 A (d) Q = 2 × 10−3 C

49. Two non-conducting solid spheres of radii R and 2 R, having uniform volume charge densities ρ1 and ρ2 respectively, touch each other. The net electric field at a distance 2 R from the centre of the smaller sphere, along the line joining the ρ centre of the spheres, is zero. The ratio 1 ρ2 can be (More than One Correct Option, 2013) (a) − 4

(b) −

32 25

(d) 4

(c)

32 25

50. In the circuit shown in the figure, there are two parallel plate capacitors each of capacitance C. The switch S1 is pressed first to fully charge the capacitor C1 and then released. The switch S 2 is then pressed to charge the capacitor C2. After some time, S 2 is released and then S3 is pressed. After some time (More than One Correct Option, 2013)

S1

2V0

(a) (b) (c) (d)

S2 C1

S3 C2

V0

the charge on the upper plate of C1 is 2 CV 0 the charge on the upper plate of C1 is CV 0 the charge on the upper plate of C2 is 0 the charge on the upper plate of C2 is −CV0

51. Two non-conducting ρ –ρ spheres of radii R1 and R2 and carrying R2 uniform volume R1 charge densities + ρ and − ρ, respectively, are placed such that they partially overlap, as shown in the figure. At all points in the overlapping region (More than One Correct Option, 2013) (a) the electrostatic field is zero (b) the electrostatic potential is constant (c) the electrostatic field is constant in magnitude (d) the electrostatic field has same direction

52. A particle of mass M and positive charge Q, moving with a constant velocity u1 = 4ims−1 , enters a region of uniform static magnetic field normal to the x-y plane. The region of the magnetic field extends from x = 0 to x = L for all values of y. After passing through this region, the particle emerges on the other side after 10 milliseconds with a velocity u2 = 2( 3i + j) ms−1. The correct statement(s) is (are) (More than One Correct Option, 2013)

(a) the direction of the magnetic field is −z direction. (b) the direction of the magnetic field is + z direction 50 πM (c) the magnitude of the magnetic field is 3Q units. (d) the magnitude of the magnetic field is 100 πM units. 3Q

25

Previous Years’ Questions (2018-13) 53. A steady current I flows along an infinitely long hollow cylindrical conductor of radius R. This cylinder is placed coaxially inside an infinite solenoid of radius 2R. The solenoid has n turns per unit length and carries a steady current I. Consider a point P at a distance r from the common axis. The correct statement(s) is (are) (More than One Correct Option, 2013)

(a) In the region 0 < r < R, the magnetic field is non-zero (b) In the region R < r < 2R , the magnetic field is along the common axis (c) In the region R < r < 2R, the magnetic field is tangential to the circle of radius r, centered on the axis (d) In the region r > 2R, the magnetic field is non-zero

Passage (Q. Nos. 54-55) A point charge Q is moving in a circular orbit of radius R in the x-y plane with an angular velocity ω. This can be considered as equivalent Qω to a loop carrying a steady current ⋅ A 2π uniform magnetic field along the positive z-axis is now switched on, which increases at a constant rate from 0 to B in one second. Assume that the radius of the orbit remains constant. The applications of the magnetic field induces an emf in the orbit. The induced emf is defined as the work done by an induced electric field in moving a unit positive charge around a closed loop. It is known that, for an orbiting charge, the magnetic dipole moment is proportional to the angular momentum with a proportionality constant γ. (Passage Type, 2013)

54. The change in the magnetic dipole moment associated with the orbit, at the end of the time interval of the magnetic field change, is (a) γBQR 2 (c) γ

BQR 2 2

(b) −γ

BQR 2 2

(d) γBQR 2

55. The magnitude of the induced electric field in the orbit at any instant of time during the time interval of the magnetic field change is (a)

BR 4

(b)

− BR 2

(c) BR

(d) 2BR

Passage (Q. Nos. 56-57) A thermal power plant produces electric power of 600 kW at 4000 V, which is to be transported to a place 20 km away from the power plant for consumers’ usage. It can be transported either directly with a cable of large current carrying capacity or by using a combination of step-up and step-down transformers at the two ends. The drawback of the direct transmission is the large energy dissipation. In the method using transformers, the dissipation is much smaller. In this method, a step-up transformer is used at the plant side so that the current is reduced to a smaller value. At the consumers’ end a step-down transformer is used to supply power to the consumers at the specified lower voltage. It is reasonable to assume that the power cable is purely resistive and the transformers are ideal with a power factor unity. All the current and voltages mentioned are rms values. (Passage Type, 2013)

56. If the direct transmission method with a cable of resistance 0.4 Ω km −1 is used, the power dissipation (in %) during transmission is (a) 20

(b) 30

(c) 40

(d) 50

57. In the method using the transformers, assume that the ratio of the number of turns in the primary to that in the secondary in the step-up transformer is 1 : 10. If the power to the consumers has to be supplied at 200V, the ratio of the number of turns in the primary to that in the secondary in the step-down transformer is (a) 200 : 1 (c) 100 : 1

(b) 150 : 1 (d) 50 : 1

Answer with Explanations 1. (b, d) I1 =

tR  −   1− e L 

V R 

(d)

Z

 

R

B1 Bx

L

I1

I R

V

I2

I1

B2

2L

I2

V

X I2

I2 =

tR −  V  1 − e 2L   R  

At centre of ring, B due to wires is along X-axis. Hence Z-component is only because of ring which µ 0i $ ( − k ). 2R

B=

From principle of superposition, I = I1 − I2 ⇒

I=

V R

I1

3. (1.50)

tR tR −  −  e 2L  1 − e 2L 

 

I is maximum when

…(i)

 

dI = 0, which gives e dt

−

tR 2L

=

Before

1µF

+8µC –8µC

1 or 2

2L ln2 R Substituting this time in Eq. (i), we get V Imax = 4R

t=

3 µC +

C

–

1µF

2. (a,b,d) Z (0, 0, √3 R)

I

V0=8V

C

1µF

After 3µC –3µC

Applying loop rule, 5 3 3 − − =0 ⇒ 1 εr 1

3 =2 εr

+5µC –5µC

⇒ εr = 150 .

4. (2 m/s) If average speed is considered along X-axis, R1 =

Y X=–R O

X=R

X

mv 0 mv 0 mv 0 , R2 = = qB1 qB2 4qB1

R1 C1

–Y

(a) At origin, B = 0 due to two wires if I1 = I2, hence (B net ) at origin is equal to B due to ring. which is non-zero. (b) If I1 > 0 and I2 < 0, B at origin due to wires will be $ . Direction of B due to ring is along − k $ along + k direction and hence B can be zero at origin. (c) If I1 < 0 and I2 > 0, B at origin due to wires is $ and also along − k $ due to ring, hence along − k B cannot be zero.

⇒ R1 > R 2

C2 R2

Distance travelled along x-axis, ∆x = 2 ( R1 + R 2 ) = T1 T2 πm πm + = + 2 2 qB1 qB2 5 πm πm πm = + = 4qB1 4qB1 qB1 5mv 0 2 qB1 Magnitude of average speed = = 2 m/s 5 πm 4qB1

Total time =

5mv 0 2 qB1

27

Previous Years’ Questions (2018-13)

120° O

λ λ λ(2 l ) − = 2 πε0(d − l ) 2 πε0(d + l ) 2 πε0d 2 1 ⇒ E∝ 2 d σ (5) E = ⇒ E is independent of d 2 ε0

R

9. (b) When switch is closed for a very long time

5. (a, b) PQ = (2 ) R sin 60°

(4) E =

Z

l P R

Q

capacitor will get fully charged and charge on capacitor will be q = CV Energy stored in capacitor, 1 EC = CV 2 K(i) 2 R S

3 = ( 3R ) 2 q enclosed = λ ( 3R )  3 λR  q We have, φ = enclosed ⇒ φ =   ε0  ε0  = (2 R )

Also, electric field is perpendicular to wire, so Z-component will be zero. a=

6. (2)

F qE = = 10 3 sin(10 3t ) m m

dv = 10 3 sin(10 3t ) ⇒ dt ∴

v =

C

10 3

v

t

0

0

3 3 ∫ dv = ∫ 10 sin(10 t ) dt

[1 − cos (10 t )] 10 3 Velocity will be maximum when cos(10 3t ) = − 1 3

vmax = 2 m/s

7. (5.55) Given, N = 50, A = 2 × 10 −4 m2, C = 10 −4 , R = 50 Ω , . rad B = 0.02 T, θ = 02 Ni g AB = Cθ Cθ 10 −4 × 02 . ig = = = 0.1 A N AB 50 × 2 × 10 −4 × 0.02

∴ ⇒ ∴

Vab = i g × G = ( i − i g ) S 0.1 × 50 = (1 − 0.1) × S S i – ig

a

i

∴

ig

G

b

5 = 0.9 × S 50 S = Ω = 5.55 Ω 9

1 Q 4 πε0 d 2 1 2Q(2 l ) (2) E axis = 4 πε0 d 3 λ (3) E = 2 πε0d

Energy dissipated across resistance E D = (work done by eq. battery) − (energy stored ) 1 1 E D = CV 2 − CV 2 = CV 2 K(ii) 2 2 From Eqs. (i) and (ii), we get E D = EC

10. (a) For process (1) Charge on capacitor =

CV0 3

1 V02 CV02 C = 2 9 18 CV0 V CV02 Work done by battery = × = 3 3 9 CV02 CV02 CV02 Heat loss = − = 9 18 18 For process (2) 2CV0 Charge on capacitor = 3 CV0 Extra charge flow through battery = 3 CV0 2 V0 2CV02 Work done by battery = ⋅ = 3 3 9 Energy stored in capacitor =

2

Final energy stored in capacitor =

8. (b) List-II (1) E =

Work done by a battery, W = Vq = VCV = CV 2

⇒

E∝

⇒

E∝ ⇒

4CV02 CV02 3CV02 − = 18 18 18 Heat loss in process (2) = work done by battery in process (2) − energy store in capacitor process (2) 2CV02 3CV02 CV02 = − = 9 18 18 Energy stored in process 2 =

1 2

d 1

d3 E∝

1 d

1  2 V0  4CV02 C  = 2  3  18

28

Electricity & Magnetism α = 60 ° and β = − 30 ° µ I  3 1 = 0  −  4 πd  2 2 µ I  3 − 1 B12 = 0   4 πd  2  d =a B0 = 12 B12

For process (3) Charge on capacitor = CV0 Extra charge flown through battery 2CV0 CV0 = 3 3 Work done by battery in this process = CV0 −

CV CV02 =  0  ( V0 ) =  3  3 Final energy stored in capacitor = Energy stored in this process 1 2 4CV02 5CV02 CV0 − = 2 18 18 Heat loss in process (3)

µ 0I  3 − 1 µ 0I 6 [ 3 − 1]  = 4 πd  2  4 πa

= 12 ×

1 2 CV0 2

15. (b,c) (a)| ∆P | = 2 p

=

=

CV02

5CV02

p

p

CV02

− = 3 18 18 Now, total heat loss (E D ) =

d p

CV02 CV02 CV02 CV02 + + = 18 18 18 6

1 2 CV0 2 1 1 So we can say that E D =  CV02   3 2 Final energy stored in capacitor =

⇒ r sinθ =

(b) r (1 − cos θ) = R

θ θ cos 2 2 = 3 θ 2 2 sin2 2

sinθ 3 = 1 − cos θ 2

2 sin

θ

r

11. (c) For particle to move in negative y-direction, either its velocity must be in negative y-direction (if initial velocity ≠ 0) and force should be parallel to velocity or it must experience a net force in negative y-direction only (if initial velocity = 0)

3R 2

r

12. (a) Fnet = Fe + FB = qE + qv × B For particle to move in straight line with constant velocity, Fnet = 0 ∴

cot

qE + qv × B = 0

13. (c) For path to be helix with axis along positive z-direction, particle should experience a centripetal acceleration in xy-plane. For the given set of options only option (c) satisfy the condition. Path is helical with increasing pitch.

14. (a)

2 a

⇒ tan

1

µ 0I [sinα + sinβ ] 4 πd

θ 2 = 2 3

2 4 2    3 12 = 3 = tanθ = 4 5 5 1− 9 9

13

aα β d O

B12 =

⇒

θ 3 = 2 2

12

θ 5 12  12  3R 13R P 8P sinθ = r  = ; r= = ;B= 13  13  2 8 QB 13QR P 3R 2P (c) ,B> < QB 2 3QR mv 2 mv , d = 2r = (d) r = ⇒ d ∝m QB QB

29

Previous Years’ Questions (2018-13) 16. (a,b) At ω ≈ 0, X = 1 = ∞ . Therefore, current is C ωC

2π  − V0 sin ωt  3  2π = V0 sin ωt +  + V0 sin(ωt + π )  3  2π π φ= π− = 3 3 π  V0 ′ = 2 V0 cos   = 3 V0  6

19. (b,c) VXY = V0 sin  ωt +

nearly zero. Further at resonance frequency, current and voltage are in phase. This resonance frequency is given by, 1 1 = = 10 6 rad /s ωr = LC 10 −6 × 10 −6

⇒

We can see that this frequency is independent of R. 1 Further, X L = ωL, XC = ωC

⇒

VXY =

⇒

( VXY )rms = ( VYZ )rms =

At, ω = ω r = 10 6 rad /s, X L = XC .

17. (b,d) The net magnetic flux through the loops at time t is φ = B(2 A − A )cos ωt = BA cos ωt dφ  = BωA sinωt so,   dt  dφ  is maximum when φ = ωt = π /2 ∴  dt  The emf induced in the smaller loop, d εsmaller = − ( BA cos ωt ) = BωA sinωt dt ∴ Amplitude of maximum net emf induced in both the loops = Amplitude of maximum emf induced in the smaller loop alone. S V + –

R L2

Current through resistor at any time t is given by V (1 − R

V0 2

any instant is λ. Initial value of λ is suppose λ 0. Electric field s at a distance r at any instant is λ E= 2 πεr λ J = σE = σ 2 πεr dq i= = J( A ) = − Jσ 2 πrl dt d λl λ =− × σ 2 πrl dt 2 πεr λ

∫

λ0

t dλ σ = − ∫ dt λ ε 0

⇒ λ = λ 0e

λ

(q = λl ) σ − t ε

σ

σ

− t σ σλ 0 − ε t λ= e = J 0e ε 2 πεr 2 πεr σλ 0 Here, J 0 = 2 πεr ∴ J(t )decreases exponentially as shown in figure below. j(t)

Since inductors are connected in parallel dI dI VL1 = VL2 ; L1 1 = L2 2 dt dt I1 L2 = L1I1 = L2I2 ; I2 L1 I=

3

J=

L1

RT − e L

3V0 sin(ωt + φ)

20. (d) Suppose charger per unit length at

For ω > ω r , X L > XC . So, circuit is inductive.

18. (a,b,c)

⇒

), where L =

L1L2 L1 + L2

V R I1 + I2 = I L1I1 = L2I2 From Eqs. (i) and (ii), we get V L2 V L1 I1 = ⇒ I2 = R L1 + L2 R L1 + L2

t

21. (a,b,c,d) Just after pressing key,

After long time I =

(d) Value of current is zero at t = 0 Value of current is V / R at t = ∞ Hence option (d) is incorrect.

(0, 0)

K(i) K(ii)

5 − 25000i 1 = 0 5 − 50000i 2 = 0 (As charge in both capacitors = 0) ⇒ i 1 = 02 . mA ⇒ i 2 = 01 . mA and VB + 25000i 1 = VA ⇒ VB − VA = − 5 V After a long time, i 1 and i 2 = 0 (steady state) q 5 − 1 = 0 ⇒ q 1 = 200 µC ⇒ 40 q and 5 − 2 = 0 ⇒ q 2 = 100 µC 20 q2 VB − = VA ⇒ VB − VA = + 5 V 20

30

Electricity & Magnetism ⇒ (a) is correct. For capacitor 1, q 1 = 200 [1 − e −t /1 ] µC 1 i 1 = e −t /1 mA 5 For capacitor 2, q 2 = 100 [1 − e −t /1 ] µC 1 −t /1 i2 = e mA 10 q ⇒ VB − 2 + i 1 × 25 = VA 20 ⇒ VB − VA = 5 [1 − e −t ] − 5e −t = − 5 [1 − 2e −t ] At t = ln2, VB − VA = 5 [1 − 1] = 0 ⇒ (b) is correct. 1 1 −1 3 1 At t = 1, i = i 1 + i 2 = e −1 + e = ⋅ 5 10 10 e 1 1 3 At t = 0, i = i 1 + i 2 = + = 5 10 10 ⇒ (c) is correct. After a long time, i 1 = i 2 = 0 ⇒ (d) is correct.

24. (c,d) Because of non-uniform evaporation at different section, area of cross-section would be different at different sections. Region of highest evaporation rate would have rapidly reduced area and would become break up cross-section. Resistance of the wire as whole increases with time. Overall resistance increases hence power decreases.  V2 1 or P ∝ as V is constant  . At break up P =  R R  junction temperature would be highest, thus light of highest band frequency would be emitted at those cross-section.

25. (a,c) By reciprocity theorem of mutual induction, it can be assumed that current in infinite wire is varying at 10A/s and EMF is induced in triangular loop. y

dy

22. (d) Balls will gain positive charge and hence move towards negative plate. On reaching negative plate, balls will attain negative charge and come back to positive plate. and so on, balls will keep oscillating. But oscillation is not S.H.M., As force on balls is not ∝ x. ⇒ option (d) is correct.

23. (a) As the balls keep on carrying charge form one plate to another, current will keep on flowing even in steady state. When at bottom plate, if all balls attain charge q, V0 r  1  kq = V0  k =  ⇒ q = 4 πε0  k r  Inside cylinder, electric field E = [V0 − ( − V0 )] h = 2 V0h. ⇒ Acceleration of each ball, qE 2 hr 2 = ⋅ V0 a= m km ⇒ Time taken by balls to reach other plate, 2 h. k m 2h 1 km t = = = a V0 r 2 hrV02 If there are n balls, then Average current, nq Vr r i av = = n × 0 × V0 t k km

⇒ i av ∝ V02

i

2y

Flux of magnetic field through triangle loop, if current in infinite wire is φ, can be calculated as follows: µ i µ i dφ = 0 ⋅ 2 ydy ⇒ dφ = 0 dy 2 πy π µ 0i  l  ⇒ φ=   π  2 dφ µ 0  l  di =  ⋅ dt π  2  dt µ A µ = 0 (10 cm) 10  = 0 volt  π s π

⇒ EMF =

If we assume the current in the wire towards right then as the flux in the loop increases we know that the induced current in the wire is counter clockwise. Hence, the current in the wire is towards right. Field due to triangular loop at the location of infinite wire is into the paper. Hence, force on infinite wire is away from the loop. By cylindrical symmetry about infinite wire, rotation of triangular loop will not cause any additional EMF.

26. (a,c) For maximum range of voltage resistance should be maximum. So, all four should be connected in series. For maximum range of current, net resistance should be least. Therefore, all four should be connected in parallel.

31

Previous Years’ Questions (2018-13) 27. (8)

L1=1mH

r1=4Ω

L2=2mH

r2=4Ω

dφ = 0 e = 0, i = 0 dt F = 0 ⇒ x > 4L ⇒ e = Blv v

R=12 Ω

ε=5V ε 5 = A R 12  1 1 1 ε Imin = = ε + +  R eq  r1 r2 R 

Imax =

(Initially at t = 0) X

(finally in steady state)

Force also will be in left direction. BLv (clockwise) i = R 2 2 BLv dv a=− =v mR dx

1 1 1  10 = 5  + + A =  3 4 12  3 Imax =8 Imin

28. (b,c)

F =

v

R

x

When loop was entering (x < L) φ = BLx dφ dx = − BL e =− dt dt |e| = BLv e BLv (anticlockwise) i = = R R B2L2v (in left direction) F = ilB (Left direction) = R 2 2 F BLv dv a= =− ⇒ a=v ⇒ m mR dx x v dv B2L2v B2L2 v =− ⇒ ∫ dv = − dx ∫ dx mR mR 0 v

L

B2L2 dx = mR

vf

∫ dv

vi

B2L2v x mR (straight line of negative slope for x < L)

BL v ⇒ (I vs x will also be straight line of negative R slope for x < L) L ≤ x ≤ 3L

−

29. (6) ANBP is cross-section of a cylinder of length L. The line charge passes through the centre O and perpendicular to paper. P

O 30° 30° A

M

B

N a

a 3a , MO = 2 2 AM  −1  1  ∴ ∠AOM = tan−1   = tan   = 30°  OM   3 AM =

0

I=

x

∫−

B2L2 ( x − L) = v f − v i mR B2L2 vf = vi − ( x − L ) (straight line of negative slope) mR BLv I= → (Clockwise) (straight line of negative slope) R

⇒

⇒ v = v0 −

B2L2v R

Electric flux passing from the whole cylinder q λL φ1 = in = ε0 ε0 ∴ Electric flux passing through ABCD plane surface (shown only AB) = Electric flux passing through cylindrical surface ANB 60°  λL =   (φ ) =  360°  1 6 ε0 ∴

n=6

32

Electricity & Magnetism

30. (d) The sphere with cavity can be assumed as a

Here, q is total positive charge on whole sphere. It is in the direction of OP or a . Now, inside the cavity electric field comes out to be uniform at any point. This is a standard result.

2Ω 6Ω

10Ω

E2

2Ω

6Ω 2Ω 6.5V 12Ω 2Ω

S/ 2

6.5V

–

12Ω

S/ 2 C′′

4Ω 10Ω

2Ω

4Ω

2Ω 10Ω

4Ω

2Ω

d C

1Ω

6Ω

+ E1

4Ω

12Ω

C′ E1

8Ω 2Ω 4Ω

2Ω

6.5V

d/2

C

1Ω

⇒

31. (d)

33. (1)

⇒

complete sphere with positive charge of radius R1 + another complete sphere with negative charge and radius R 2. E + → E due to total positive charge E − → E due to total negative charge. E = E + + E− If we calculate it at P, then E − comes out to be zero. ∴ E =E+ 1 q and E + = (OP ), in the direction of OP. 4 πε0 R13

C′

6Ω

2Ω

12Ω

4Ω

6.5V –

+

s 2 ε0 ε 0s 2ε s 2 ,C = C1 = = 0 d d /2 d C′=

1 4.5Ω

6.5V

s 2 = 4 ε 0s d /2 d

4ε 0

s ε s 2 C ′′ = = 0 d d CC ′ 4 ε 0s ε s C2 = + C ′′ = + 0 C + C′ 3 d d 2 ε0

and

⇒

2Ω

C′′

7 ε 0s C 2 7 = 3 d C1 3  A Al A 1 1 1 1 32. (b) = + = + Fe  R R Al R Fe  ρAl ρFe  l =

 (7 2 − 2 2 ) 2 2  10 −6 1 = + ×  10  10 −8 50 × 10 −3  2.7 Solving we get, R =

1875 1875 × 10 − 6 Ω = µΩ 64 64

34. (c) At the shown position, net force on both charges is zero. Hence they are in equilibrium. But equilibrium of +q is stable equilibrium. So, it will start oscillations when displaced from this position. These small oscillations are simple harmonic in nature. While equilibrium of −q is unstable. So, it continues to move in the direction of its displacement.

35. (a,b,c) y P

I

2(L+R)

Q

x

Force on the complete wire = force on straight wire PQ carrying a current I. F = I(PQ × B ) = I[{2( L + R )$i } × B ]

33

Previous Years’ Questions (2018-13) This force is zero if B is along $i direction or x-direction. If magnetic field is along $j direction or k$ direction, |F| = F = ( I )(2 )( L + R )Bsin 90° or

F = 2 I( L + R )B

or F ∝ (L + R ) ∴ Options (a), (b) and (c) are correct.

36. (a,d) FB = Bev = Be Fe = eE eE = V = Ed =

BI nA

I BI = nAe nA

⇒ Fe = FB ⇒ E=

⇒ (d) is incorrect 1 4 πε0 k(2Q ) 2 kQ ⇒ E2 = E2 = R2 R2 k( 4Q ) R kQ ⇒ E3 = E3 = E 3 < E1 < E 2 (2 R )3 2R2

40. (c) E1 =

kQ R2

, where k =

41. (c) (P) Component of forces along x-axis will vanish. Net force along positive y-axis

B nAe

F4

F3

F2

F1

+q

BIw BI BI ⋅w = = nAe n( wd )e ned

V1 d 2 = V2 d 1 ⇒ if w1 = 2 w2 and d 1 = d 2 V1 = V2 ∴Correct answers are (a) and (d).

37. (a,c) V =

BI ned

⇒

+Q

+Q

+Q

+Q

(Q) Component of forces along y-axis will vanish. Net force along positive x-axis

V1 B n = 1 × 2 V2 B2 n1

F2 +q

If B1 = B2 and n1 = 2 n2, then V2 = 2 V1 If B1 = 2 B2 and n1 = n2, then V2 = 0.5V1 ∴Correct answers are (a) and (c).

F1 F4

F3

38. (a,d) C = C1 + C 2 Kε 0 A / 3 ε 2 A/3 , C2 = 0 d d ( K + 2 )ε0 A K +2 C ⇒ C = ⇒ = 3d C1 K V Also, E1 = E 2 = , where V is potential difference d between the plates. Q λ σ 39. (c, d) = = 4 πε0r02 2 πε0r0 2 ε0 C1 =

+Q

–Q

+Q

–Q

(R) Component of forces along x-axis will vanish. Net force along negative y-axis F1

F4 +q F2

F3

Q = 2 πσr02

λ πσ r (b) is incorrect, E1  0  = 4E1( r0 ) 2 1 As E1 ∝ 2 r 1 r E 2  0  = 2 E 2( r0 ) as E 2 ∝ 2 r

(a) is incorrect, r0 =

⇒ (c) is correct r E 3  0  = E 3( r0 ) = E 2( r0 ) 2 as E3 ∝ r 0

+Q

–Q

–Q

+Q

(S) Component of forces along y-axis will vanish. Net force along negative x-axis F1 +q F4

F3

F2

+Q

–Q

+Q –Q

34

Electricity & Magnetism

42. (a,b,d) Let us take VP = 0. Then potentials across

x0 3

Substituting x1 =

R1, R 2 and R 3 are as shown in figure (ii) In the same figure

B1 =

3µ 0 I 3µ 0 I 3µ 0 I − = 2 πx0 4 πx0 4 πx0

R1 =

mv qB1

and B2 =

R2 =

mv qB2

⇒

V1 V1 P

i1 R1

R1

R2

O

i2

V2

R2

R3

o

(i)

O R3

i3 –V2

45. (5)

V − ( − V2 ) V1 − V0 0 − V0 = 0 + R3 R1 R2

i g (G + 4990) = V ⇒

6 (G + 4990) = 30 1000

⇒ G + 4990 = ⇒

30,000 = 5000 6

G = 10 Ω Vab = Vcd

Current through R 2 will be zero if V1 R V0 = 0 ⇒ = 1 V2 R 3

⇒ ι γ Γ = (1.5 − ι γ ) Σ S

c

(where, R = 90 Ω )

ig

G

b

⇒

6 6  × 10 =  1.5 − S  1000 1000 

⇒

S =

60 2n = 1494 249

⇒ n=

249 × 30 2490 = =5 1494 498

46. (c) B R = B due to ring

X = 60 Ω

B1 = B due to wire-1 B 2 = B due to wire-2

l (100 − l )

X =R

a

1.5 A

43. (c) For balanced meter bridge X l = R (100 − l ) X 40 = 90 100 − 40

d

(1.5–ig)

In options (a), (b) and (d) this relation is satisfied.

∴

G

V

Solving this equation we get V1 V + 0− 2 R1 R3 V0 = 1 1 1 + + R1 R2 R3

∴

9µ 0I 4 πx0

R1 B 9 = 2 = =3 R2 B1 3

ig 4990 Ω

(ii)

i1 + i 2 = i 3 ∴

x0 = 3) x1

(as

∆X ∆l ∆l 0.1 0.1 = + = + X l 100 − l 40 60 ∆X = 0.25 X = ( 60 ± 0.25) Ω µ0I µ I 44. (3) B2 = 0 + 2 πx1 2 π ( x0 − x1 ) (when currents are in opposite directions) µ0I µ0I B1 = − 2 πx1 2 π ( x0 − x1 )

P θθ

So,

I

I

BR

B2 r

1

B1 h

In magnitudes B1 = B 2 =

x1

x0 x1

(when currents are in same direction)

2

a

µ 0I 2 πr

Resultant of B1 and B 2 µ I h µ Ih = 2B1 cos θ = 2  0    = 0 2  2 πr   r  πr

35

Previous Years’ Questions (2018-13) BR =

µ 0IR 2 2( R + x ) 2

2 3/ 2

=

2µ 0Iπa 2 4 πr 3

As, R = a, x = h and a 2 + h2 = r 2 For zero magnetic field at P, µ 0Ih 2µ 0Iπa 2 ⇒ πa 2 = 2 rh ⇒ η ≈ 1.2α = 4 πr 3 πr 2

47. (b) Magnetic field at mid-point of two wires

µ I = 2 (magnetic field due to one wire) = 2  0   2 π d  µ I = 0 ⊗ πd Magnetic moment of loop M = IA = I πa 2

µ I 2a 2 Torque on loop = MB sin 150° = 0 2d dQ 48. (c,d) = I ⇒ Q = ∫ I dt = ∫ ( I0 cos ωt ) dt dt I0 1 = = 2 × 10 −3 C ω 500 Just after switching

∴

Qmax =

49. (b,d) At point P

Q

C1 2R

P C2 2R

If resultant electric field is zero then KQ1 KQ 2 ρ = R ⇒ 1 =4 ρ2 4R 2 8R 3 At point Q If resultant electric field is zero then KQ1 KQ 2 + =0 4R 2 25R 2 ρ1 32 (ρ1 must be negative) =− ρ2 25

50. (b, d) After pressing S1 charge on upper plate of C1 is

–Q1 Q1=1mC + + + + Q1

50 V +

+ 2 CV0. After pressing S 2 this charge equally distributes in two capacitors. Therefore charge an upper plates of both capacitors will be + CV0 .

–

R=10Ω

When S 2 is released and S 3 is pressed, charge on upper plate of C1 remains unchanged ( = + CV0 ) but charge on upper plate of C 2 is according to new battery ( = − CV0 ).

In steady state Q2 + ++ +

–Q2

Substituting the values of Q1, C and R we get I = 10 A In steady state Q 2 = CV = 1 mC ∴ Net charge flown from battery = 2 mC

50 V

51. (c,d) For electrostatic field, Field at P E P = E1 + E 2 =

R=10Ω 7π 7π At or ωt = t = 6ω 6 Current comes out to be negative from the given expression. So, current is anti-clockwise. Charge supplied by source from t = 0 to 7π 7π t = ⇒ Q = ∫ 6 ω cos ( 500t ) dt 0 6ω 7π 7π sin sin 500t  6 ω  6 = − 1 mC = = 500  500  0

Apply Kirchhoff’s loop law just after changing the switch to position D Q 50 + 1 − IR = 0 C

( − ρ) ρ C1P + C2P 3ε 0 3ε 0

ρ (C1P + PC 2 ) 3ε 0 ρ EP = C1C 2 3ε 0 =

For electrostatic. Since, electric field is non-zero so it is not equipotential.

52. (a,c) u = 4i ; v = 2( 3i + j) v

u L

j

i

36

Electricity & Magnetism According to the figure, magnetic field should be in ⊗ direction, or along − z direction. v 1 2 Further, tanθ = y = = vx 2 3 3 π ∴ θ = 30° or 6 = angle of v with x-axis = angle rotated by the particle BQ  = Wt =  t  M  πM 50 πM units = (as t = 10 −3 second) ∴ B= 6Qt 3Q

53. (a,d)

where γ =

Q Q  =   ( Iω ) 2 m  2 m

Q  Qω R 2 2 =   ( mR ω ) =  2 m 2 Induced electric field is opposite. Therefore, ω ′ = ω − αt τ (QE )R = = I mR 2

α =

ω′ = ω −

∴

Q

Mf =

P

R

2R

P → Hollow cylindrical conductor Q → Solenoid

∴

M Q 54. (b) = L 2m

∆M = M f − M i = −

∫

E ⋅ dl = −

dφ dt

dB or El = − s    dt 

∴ E(2 πR ) = − ( πR 2 )( B) or

56. (b)

E=−

BR 2

P = Vi

P 600 × 10 3 = = 150 A V 4000 Total resistance of cables, R = 0.4 × 20 = 8Ω ∴Power loss in cables

∴

i =

= i 2R = (150)2 ( 8)

Np Ns or

⇒

QBR 2 2

M ∝ L,

 as γ = Q     2 m

55. (b) The induced electric field is given by,

x

Q  M =  L  2 m

QB 2m

Q 2BR 2 4m

57. (a) During step-up,

E

=

QB QB ⋅1 = ω − 2m 2m

= 180000 W = 180 kW This loss is 30% of 600 kW.

y Q

∴

mR 2

Qω ′ R 2 QB  R 2 = Q  ω −   2 2 m 2

M = −γ In the region, 0 < r < R BP = 0, BQ ≠ 0, along the axis ∴ Bnet ≠ 0 In the region, R < r < 2 R BP ≠ 0, tangential to the circle of radius r, centred on the axis. BQ ≠ 0, along the axis. ∴ Bnet ≠ 0 neither in the directions mentioned in options (b) or (c). In region, r > 2 R BP ≠ 0 BQ ≠ 0 ∴ Bnet ≠ 0

BR  (Q )  R  2 

=

Vp Vs

1 4000 = 10 Vs

or V s = 40,000 V In step, down transformer, Np V 40000 200 = p = = Ns Vs 200 1

Understanding Physics

JEE Main & Advanced

OPTICS AND MODERN PHYSICS

Understanding Physics

JEE Main & Advanced

OPTICS AND MODERN PHYSICS

DC PANDEY [B.Tech, M.Tech, Pantnagar, ID 15722]

ARIHANT PRAKASHAN (Series), MEERUT

Understanding Physics

JEE Main & Advanced

ARIHANT PRAKASHAN (Series), MEERUT All Rights Reserved

© SARITA PANDEY No part of this publication may be re-produced, stored in a retrieval system or by any means, electronic, mechanical, photocopying, recording, scanning, web or otherwise without the written permission of the publisher. Arihant has obtained all the information in this book from the sources believed to be reliable and true. However, Arihant or its editors or authors or illustrators don’t take any responsibility for the absolute accuracy of any information published and the damage or loss suffered thereupon. All disputes subject to Meerut (UP) jurisdiction only.

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Understanding Physics

JEE Main & Advanced

PREFACE The overwhelming response to the previous editions of this book gives me an immense feeling of satisfaction and I take this an opportunity to thank all the teachers and the whole student community who have found this book really beneficial. In the present scenario of ever-changing syllabus and the test pattern of JEE Main & Advanced, the NEW EDITION of this book is an effort to cater all the difficulties being faced by the students during their preparation of JEE Main & Advanced. The exercises in this book have been divided into two sections viz., JEE Main & Advanced. Almost all types and levels of questions are included in this book. My aim is to present the students a fully comprehensive textbook which will help and guide them for all types of examinations. An attempt has been made to remove all the printing errors that had crept in the previous editions. I am extremely thankful to (Dr.) Mrs. Sarita Pandey, Mr. Anoop Dhyani and Nisar Ahmad for their endless efforts during the project. Comments and criticism from readers will be highly appreciated and incorporated in the subsequent editions.

DC Pandey

Understanding Physics

JEE Main & Advanced

CONTENTS 29. ELECTROMAGNETIC WAVES 29.1 Introduction 29.2 Displacement Current

1-19 29.3 Electromagnetic Waves 29.4 Electromagnetic Spectrum

30. REFLECTION OF LIGHT 30.1 Introduction 30.2 General Concepts used in Geometrical Optics

21-70 30.3 Reflection of Light 30.4 Reflection from a Spherical Surface

31. REFRACTION OF LIGHT 31.1 Refraction of Light and Refractive Index of a Medium 31.2 Law of Refraction (Snell’s Law) 31.3 Single Refraction from Plane Surface 31.4 Shift due to a Glass Slab

71-190 31.5 Refraction from Spherical Surface 31.6 Lens Theory 31.7 Total Internal Reflection (TIR) 31.8 Refraction Through Prism 31.9 Deviation 31.10 Optical Instruments

32. INTERFERENCE AND DIFFRACTION OF LIGHT 32.1 Principle of Superposition 32.2 Resultant Amplitude and Intensity due to Coherent Sources 32.3 Interference

33. MODERN PHYSICS-I 33.1 Dual Nature of Electromagnetic Waves 33.2 Electromagnetic Spectrum

191-251

32.4 Young's Double Slit Experiment (YDSE) 32.5 Introduction to Diffraction 32.6 Diffraction from a Narrow Slit

253-323 33.3 Momentum and Radiation Pressure 33.4 de-Broglie Wavelength of Matter Wave

Understanding Physics

JEE Main & Advanced

33.5 Early Atomic Structure 33.6 The Bohr Hydrogen Atom 33.7 Hydrogen Like Atoms

33.8 X-Rays 33.9 Emission of Electrons 33.10 Photoelectric Effect

34. MODERN PHYSICS-II 34.1 Nuclear Stability and Radioactivity 34.2 The Radioactive Decay Law 34.3 Successive Disintegration 34.4 Equivalence of Mass and Energy

325-374 34.5 Binding Energy and Nuclear Stability 34.6 Nuclear Fission (Divide and Conquer) 34.7 Nuclear Fusion

35. SEMICONDUCTORS 35.1 Introduction 35.2 Energy Bands in Solids 35.3 Intrinsic and Extrinsic Semiconductors 35.4 p-n Junction Diode

375-414 35.5 35.6 35.7 35.8 35.9

Junction Diode as a Rectifier Applications of p-n Junction Diodes Junction Transistors Transistor as an Amplifier Digital Electronics and Logic Gates

36. COMMUNICATION SYSTEM 36.1 Introduction 36.2 Different Terms Used in Communication System 36.3 Bandwidth of Signals 36.4 Bandwidth of Transmission Medium 36.5 Propagation of Electromagnetic Waves or Communication Channels

415-431 36.6 Modulation 36.7 Amplitude Modulation 36.8 Production of Amplitude Modulated Wave 36.9 Detection of Amplitude Modulated Wave

Hints & Solutions JEE Main & Advanced Previous Years' Questions (2018-13)

433-518 1-28

Understanding Physics

JEE Main & Advanced

SYLLABUS JEE Main ELECTROMAGNETIC WAVES Electromagnetic waves and their characteristics. Transverse nature of electromagnetic waves. Electromagnetic spectrum (radio waves, microwaves, infrared, visible, ultraviolet, X-rays, g- rays). Applications of e.m. waves.

OPTICS Reflection and refraction of light at plane and spherical surfaces, Mirror formula, Total internal reflection and its applications, Deviation and Dispersion of light by a prism, Lens formula, Magnification, Power of a lens, Combination of thin lenses in contact, Microscope and astronomical telescope (reflecting and refracting) and their magnifying powers.

WAVE OPTICS Wave front and Huygens’ principle, Laws of reflection and refraction using Huygen’s principle. Interference, Young’s double slit experiment and expression for fringe width. Diffraction due to a single slit, width of central maximum. Resolving power of microscopes and astronomical telescopes, Polarisation, Plane polarized light; Brewster’s law, uses of plane polarized light and polaroids.

DUAL NATURE OF MATTER AND RADIATION Dual nature of radiation. Photoelectric effect, Hertz and Lenard’s observations; Einstein’s photoelectric equation; Particle nature of light. Matter waves-wave nature of particle, de-Broglie relation. Davisson-Germer experiment.

Understanding Physics

JEE Main & Advanced

ATOMS AND NUCLEI a-particle scattering experiment; Rutherford’s model of atom; Bohr model, Energy levels, Hydrogen spectrum. Composition and size of nucleus, Atomic masses, Isotopes, Isobars; Isotones. Radioactivity - a, b and g particles/rays and their properties; radioactive decay law. Mass-energy relation, Mass defect; Binding energy per nucleon and its variation with mass number, Nuclear fission and fusion.

ELECTRONIC DEVICES Semiconductors; semiconductor diode: I-V characteristics in forward and reverse bias; Diode as a rectifier; I-V characteristics of LED, photodiode, solar cell and Zener diode; Zener diode as a voltage regulator. Junction Transistor, transistor action, Characteristics of a transistor; transistor as an amplifier (common emitter configuration) and oscillator. Logic gates (OR, AND, NOT, NAND and NOR). Transistor as a switch.

COMMUNICATION SYSTEMS Propagation of electromagnetic waves in the atmosphere; Sky and space wave propagation, Need for modulation, Amplitude and frequency modulation, Bandwidth of signals, Bandwidth of transmission medium, Basic elements of a communication system (block diagram only).

Understanding Physics

JEE Main & Advanced

JEE Advanced GENERAL Focal length of a concave mirror and a convex lens using U V method.

OPTICS Rectilinear propagation of light, Reflection and refraction at plane and spherical surfaces, Total internal reflection, Deviation and dispersion of light by a prism, Thin lenses, Combinations of mirrors and thin lenses, Magnification.

WAVE NATURE OF LIGHT Huygens principle, Interference limited to Young’s double-slit experiment.

MODERN PHYSICS Atomic nucleus, a, b and g radiations, Law of radioactive decay, Decay constant, Half-life and mean life, Binding energy and its calculation, Fission and fusion processes, Energy calculation in these processes. Photoelectric effect, Bohr’s theory of hydrogen-like atoms, Characteristic and continuous X-rays, Moseley’s law, de-Broglie wavelength of matter waves.

Understanding Physics

JEE Main & Advanced

This book is dedicated to my honourable grandfather

(Late) Sh. Pitamber Pandey a Kumaoni poet and a resident of Village

Dhaura (Almora), Uttarakhand

29.1

Introduction

29.2

Displacement Current

29.3

Electromagnetic Waves

29.4

Electromagnetic Spectrum

2 — Optics and Modern Physics

29.1 Introduction Earlier we have learned that a time varying magnetic field produces an electric field. Is the converse also true? Does a time varying electric field can produce a magnetic field? James Clerk Maxwell argued that not only an electric current but also a time varying electric field generates magnetic field. Maxwell formulated a set of equations (known as Maxwell’s equations) involving electric and magnetic fields. Maxwell’s equations and Lorentz force formula make all the basic laws of electromagnetism. The most important outcome of Maxwell’s equations is the existence of electromagnetic waves. The changing electric and magnetic fields form the basis of electromagnetic waves. A combination of time varying electric and magnetic fields (referred as electromagnetic wave) propagate in space very close to the speed of light (= 3 ´ 10 8 m/s) obtained from optical measurements. We shall take a brief discussion of electromagnetic waves mainly developed by Maxwell around 1864.

29.2 Displacement Current An electric current produces magnetic field. Value of magnetic field (due to an electric current) at some point can be obtained by Biot-Savart law or Ampere’s circuital law. We have stated Ampere’s law as

ò B× d l = m 0 i

...(i)

where left hand side of this equation is the line integral of magnetic field over a closed path and i is the electric current crossing the surface bounded by that closed path. Ampere’s law in this form is not valid if the electric field at the surface varies with time. For an example if we place a magnetic needle in between the plates of a capacitor during its charging or discharging then it deflects. Although, there is no current between the plates, so magnetic field should be zero. Hence, the needle should not show any deflection. But deflection of needle shows that there is a magnetic field in the region between plates of capacitor during charging or discharging. So, there must be some other source (other than current) of magnetic field. This other source is nothing but the changing electric field. Because at the time of charging or discharging of capacitor electric field between the plates changes. The relation between the changing electric field and the magnetic field resulting from it is given by df E ...(ii) ò B × d l = m 0e 0 dt Here, f E is the flux of the electric field through the area bounded by the closed path along which line integral of B is calculated. Combining Eqs. (i) and (ii), we can make a general expression of Ampere’s circuital law and that is

ò B × d l = m 0 i + m 0e 0 or

df E df E ö æ =m0 ç i +e0 ÷ dt dt ø è

ò B × d l = m 0 (i + id )

...(iii)

Chapter 29 id = e 0

Here,

Electromagnetic Waves — 3

df E dt

...(iv)

is called the displacement current and which is produced by the change in electric field. The current due to the flow of charge is often called conduction current and is denoted by ic . Thus, Eq. (iii) can also be written as

ò B × d l = m 0 (ic + id )

...(v)

Example In the figure, a capacitor is charged by a battery through a resistance R. Charging of capacitor will be exponential. A time varying charging current i flows in the circuit (due to flow of charge) till charging continues. A time varying electric field is also produced between the plates. This causes a displacement current id between the plates. There is no current between the plates due to flow of charge, as a medium between the plates is insulator. Consider two closed paths a and b as shown in figure. Ampere’s circuital law in these two paths is Along path a

ò B × dl = m 0i

ò B × d l = m 0 ic

or

Along path b

ò B × d l = m 0 id E -

+ i a

id b

i = id

R

Fig. 29.1

Here, id = e 0

df E is in the direction shown in figure. In sample example 29.1, we have shown that dt ic = id

Extra Points to Remember Faraday’s law of electromagnetic induction says that changing magnetic field gives rise to an electric field and its line integral (= emf ) is given by the equation df ò E × d l = - dtB A changing electric field gives rise to a magnetic field is the symmetrical counterpart of Faraday’s law. This is a consequence of the displacement current and given by df ò B × d l = m 0 e0 dtE = m 0 id Thus, time dependent electric and magnetic fields give rise to each other.

4 — Optics and Modern Physics Maxwell’s Equations ˜

ò

E × d s = qin e0

˜

ò

B × ds = 0

˜

ò

E × dl = -

˜

ò

B × d l = m 0 ( i c + i d ) = m 0 i c + m 0 e0

V

(Gauss’s law for electricity) (Gauss’s law for magnetism)

dfB dt

(Faraday’s law) dfE dt

(Ampere-Maxwell’s law)

Example 29.1 During charging of a capacitor show that the displacement current between the plates is equal to the conduction current in the connecting wire. Let A is the area of plates, q is the charge on capacitor at some instant and d the separation between the plates. dq ...(i) Conduction current, i c = dt Electric field between the plates, q s q A E= = = e0 e0 Ae 0

q + -

Solution

ic

Fig. 29.2

The flux of the electric field through the given area is æ q fE = EA = çç è A e0 dfE 1 æ dq ö = ç ÷ dt e 0 è dt ø

\

ö q ÷A= ÷ e0 ø

Displacement current, id = e 0

dfE dt

é 1 dq ù = e0 ê × ú ë e 0 dt û dq id = dt

or

...(ii)

From Eqs. (i) and (ii), we can see that ic = id

INTRODUCTORY EXERCISE

Hence Proved.

29.1

1. A parallel-plate capacitor with plate area A and separation between the plates d , is charged by a constant current i . Consider a plane surface of area A/2 parallel to the plates and drawn symmetrically between the plates. Find the displacement current through this area.

Chapter 29

Electromagnetic Waves — 5

29.3 Electromagnetic Waves Stationary charges produce only electric field. Charges in uniform motion (or steady currents) produce both electric and magnetic fields. Accelerated charges radiate electromagnetic waves. It is an important result of Maxwell’s theory. Thus, an accelerated charge produces all three electric field, magnetic field and electromagnetic waves. Consider an oscillating charged particle. Let f is the frequency of its oscillations. This oscillating charged particle produces an oscillating electric field (of same frequency f ). Now, this oscillating electric field becomes a source of oscillating magnetic field (Ampere-Maxwell’s law). This oscillating magnetic field again becomes a source of oscillating electric field (Faraday’s law) and so on. The oscillating electric and magnetic fields regenerate each other and electromagnetic wave propagates through the space. The frequency of the electromagnetic wave is equal to the frequency of oscillation of the charge. Frequency of visible light is of the order of 1014 Hz, while the maximum frequency that we can get with modern electronic circuits is of the order of 1011 Hz. Therefore, it is difficult to experimentally demonstrate the production of visible light. Hertz’s experiment (in 1887) demonstrated the production of electromagnetic waves of low frequency (in radio wave region). Jagdish Chandra Bose succeeded in producing the electromagnetic waves of much higher frequency in the laboratory.

Extra Points to Remember ˜

˜

When electromagnetic waves propagate in space then electric and magnetic fields oscillate in mutually perpendicular directions. Further, they are perpendicular to the direction of propagation of electromagnetic wave also. Consider a plane electromagnetic wave propagating along the z-direction. The electric field E x is along the x-axis and varies sinusoidally. The magnetic field By is along the y-axis and again varies sinusoidally. We can write E x and By as E x = E0 sin (wt - kz) and

By = B0 sin (wt - kz)

Wavelength

Electric field Magnetic field Direction of wave

Fig. 29.3

Thus, electromagnetic wave travels in the direction of E ´ B. ˜

From Maxwell’s equations and the knowledge of waves we can write the following expressions, k = 2p l and w = 2 pf E w 1 c = = fl= 0 = k B0 e0m 0 where, f is the frequency of electromagnetic wave and l its wavelength.

Speed of light (in vacuum)

6 — Optics and Modern Physics ˜

˜

Unlike a mechanical wave (like sound wave) an electromagnetic wave does not require any material medium for the oscillations of electric and magnetic fields. They can travel in vacuum also. Oscillations of electric and magnetic fields are self sustaining in free space or vacuum. In a material medium (like glass, water etc.), electric and magnetic fields are different from the external fields. They are described in terms of permittivity e and magnetic permeabilitym. In Maxwell’s equations, e0 and m 0 are thus replaced by e and m and the velocity of light becomes, 1 v= em Thus, the velocity of light depends on electric and magnetic properties of the medium.

˜

Like other waves, electromagnetic waves also carry energy and momentum.In previous chapters, we have 1 B2 studied that, energy density in electric field = e0 E 2 and energy density in magnetic field = × 2 2 m0 An electromagnetic wave contains both electric and magnetic field. Therefore, energy density is associated with both the fields.

˜

Consider a plane perpendicular to the direction of propagation of the electromagnetic wave. If the total energy transferred to a surface in time t is E, then total momentum delivered to this surface for complete absorption is E (complete absorption) Dp= c If the wave is totally reflected, the momentum delivered is 2E Dp= c

˜

(completely reflected)

The energy transferred per unit area per unit time perpendicular to the direction of propagation of electromagnetic wave is called the intensity of wave. It is given by 1 I = e0 E 2c 2 Here, E is the rms value of electric field or Erms .

V

Example 29.2 A plane electromagnetic wave of frequency 25 MHz travels in free space along the x-direction. At a particular point in space and time, E = ( 63 . $j) V/m. What is B at this point? Solution

\

c=

E0 B0

=

or B=

E B

E c

Substituting the values in SI units, B=

6.3 3 ´ 108

= 2.1´ 10-8 T From the relation

c = E´ B

We can see that B is along positive z- direction. Because, E is along $j direction and c along $i direction.

Chapter 29 B = ( 2.1´ 10-8 T ) k$

\ V

Electromagnetic Waves — 7 Ans.

Example 29.3 The magnetic field in a plane electromagnetic wave is given by . ´ 103 x + 15 . ´ 1011 t) T B y = 2 ´ 10 -7 T sin ( 05 (a) What is the wavelength and frequency of the wave? (b) Write an expression for the electric field. (a) Wavelength From the given equation, we can see that

Solution

k = 0.5 ´ 103 m -1 2p l 2p 2p l= = k 0.5 ´ 103

k=

But, \

= 125 . ´ 10-2 m

Ans.

Frequency Angular frequency, w = 15 . ´ 1011 rad/s But,

w = 2pf

\

f=

. ´ 1011 w 15 = 2p 2p

\

= 2.39 ´ 1010 Hz (b) c =

Ans.

E0 B0

\

E 0 = cB 0 = ( 3.0 ´ 108 ) ( 2 ´ 10-7 ) = 60 V/m

From the relation, c = E´ B We can see that E is along z-direction. \ V

E z = ( 60 V/ m ) sin ( 0.5 ´ 103 x + 15 . ´ 1011 t ) V/m

Ans.

Example 29.4 Light with an energy flux of 18 W/cm 2 falls on a non-reflecting surface at normal incidence. If the surface has an area of 20 cm2 , find the average force exerted on the surface during a 30 minute time span. Solution

Total energy incident on the given surface in the given time interval is E = ( 18 ´ 104 W / m 2 ) ( 20 ´ 10-4 m 2 ) ( 30 ´ 60 s ) = 6.48 ´ 105 J

8 — Optics and Modern Physics Therefore, the total momentum transferred to the given surface for complete absorption is Dp =

E 6.48 ´ 105 = c 3.0 ´ 108

= 2.16 ´ 10-3 kg - m/s \

Fav =

Dp 2.16 ´ 10-3 = Dt 30 ´ 60

= 1.2 ´ 10-6 N V

Ans.

Example 29.5 In the above example what is the average force if surface is perfectly reflecting? Solution (a) If the surface is perfectly reflecting, then

Dp =

2E c

Therefore, average force is doubled or Fav = 2.4 ´ 10-6 N V

Ans.

Example 29.6 Calculate the electric and magnetic fields produced by the radiation coming from a 100 W bulb at a distance of 3 m. Assume that the efficiency of the bulb is 25 . % and it is a point source. Solution Intensity at a distance r from a point source of power P is given by

I=

P 4pr 2

So, intensity at a distance of 3 m from the bulb with 2.5% efficiency will be 2. 5 100 I= ´ 2 100 4p ( 3) = 0.022 W/ m 2 Half of the intensity is provided by electric field and half by magnetic field. I \ I E = = 0.011 W/ m 2 2 1 But, I E is given by e 0 E 2 c 2 \

IE =

2 IE 1 e 0 E 2 c or E = 2 e0 c

Substituting the values, we have E=

2 ´ 0.011 ( 8.85 ´ 10-12 ) ( 3 ´ 108 )

= 2. 9 V m

Ans.

Chapter 29

Electromagnetic Waves — 9

Note that this is actually the rms value of electric field.

c = E/ B E 2.9 B= = c 3.0 ´ 108

From the equation,

= 9.6 ´ 10-9 T

Ans.

This is again the rms value of magnetic field.

INTRODUCTORY EXERCISE 1. Show that the unit of

29.2

1 is m/s. e0 m 0

2. A capacitor is connected to an alternating current source. Is there a magnetic field between the plates?

3. The sunlight reaching the earth has maximum electric field of 810 Vm -1. What is the maximum magnetic field in this light?

4. The electric field in an electromagnetic wave is given by E = (50 NC-1) sin w (t - x /c ) . Find the energy contained in a cylinder of cross-section 10 cm 2 and length 50 cm along the x-axis.

29.4 Electromagnetic Spectrum The basic source of electromagnetic wave is an accelerated charge. This produces the changing electric and magnetic fields which constitute an electromagnetic wave. An electromagnetic wave may have its wavelength varying from zero to infinity. Not all of them are known till date. Today, we are familiar with electromagnetic waves having wavelengths as small as 30 fm (1 fm = 10 -15 m ) to as large as 30 km. The boundaries separating different regions of spectrum are not sharply defined, with visible light ranging from 4000 Å to 7000 Å. An approximate range of wavelengths associated with each colour are violet ( 4000 Å - 4500 Å), blue ( 4500 Å - 5200 Å), green (5200 Å - 5600 Å), yellow (5600 Å - 6000 Å), orange (6000 Å - 6250 Å) and red (6250 Å - 7000 Å). The classification of electromagnetic waves according to frequency or wavelength is called electromagnetic spectrum. Table below gives range of wavelengths and frequencies for different waves. Table 29.1 S.No.

Type

Wavelength range

Frequency range

1.

Radio waves

> 01 . m

< 3 ´ 109 Hz

2.

Micro waves

0.1 m to 1 mm

3 ´ 109 Hz to 3 ´ 1011 Hz

3.

Infrared

1 mm to 7000 Å

3 ´ 1011 Hz to 4.3 ´ 1014 Hz

10 — Optics and Modern Physics S.No.

Type

Wavelength range

Frequency range

4.

Visible light

7000 Å to 4000 Å

4.3 ´ 1014 Hz to 7.5 ´ 1014 Hz

5.

Ultraviolet

4000 Å to 10 Å

7.5 ´ 1014 Hz to 3 ´ 1017 Hz

6.

X-rays

10 Å to 0.01 Å

3 ´ 1017 Hz to 3 ´ 1020 Hz

7.

Gamma rays

< 0.01 Å

> 3 ´ 1020 Hz

Note In the above table, wavelength is decreasing from top to bottom. But, frequency is increasing. Now, let us discuss them in brief in the order of increasing wavelength.

These high frequency radiations are usually produced in nuclear reactions and also emitted by radioactive nuclei. They are used in medicines to destroy cancer cells. X-Rays X-rays were discovered in 1895 by W.Roentgen. These are produced by the rapid deceleration of electrons that bombard a heavy metal target. These are also produced by electronic transitions between the energy levels in an atom. X-rays are used as a diagnostic tool in medicine and as a treatment for certain forms of cancer. Ultraviolet Rays Ultraviolet radiation is produced by special lamps and very hot bodies. The sun is an important source of ultraviolet light. It plays an important role in the production of vitamin-D. But prolonged doses of UV radiation can induce skin cancers in human beings. Ozone layer in atmosphere at an altitude of about 40-50 km plays a protective role in this regarding. Depletion of this layer by chlorofluorocarbon (CFC) gas (such as Freon) is a matter of international concern now a days. Visible Light It is most familiar form of electromagnetic waves.Our eye is sensitive to visible light. Visible light emitted or reflected from objects around us provides information about world. Process of photosynthesis in plants needs visible light. Visible light is produced by the transition of electrons in an atom from one energy level to other. Infrared Radiation Infrared rays also sometimes referred as heat waves are produced by hot bodies. They are perceived by us as heat. In most of the materials, water molecules are present. These molecules readily absorb infrared rays. After absorption, their thermal motion increases, i.e. they heat up and heat their surroundings. Infrared rays are used for early detection of tumors. Infrared detectors are also used to observe growth of crops and for military purposes. Microwaves Microwaves may be generated by the oscillations of electrons in a device called klystron. Microwave ovens are used in kitchens. In microwave ovens frequency of the microwaves is selected to match the resonant frequency of water molecules so that energy from the waves is transferred to the kinetic energy of the molecules. This raises the temperature of any food containing water. Radio Waves Radio waves are generated when charges are accelerating through conducting wires. They are generated in L - C oscillators and are used in radio and television communication systems.

1. Gamma Rays 2.

3.

4.

5.

6.

7.

Chapter 29

Electromagnetic Waves — 11

Extra Points to Remember ˜

Our eyes are sensitive to visible light (lbetween 4000 Å to7000 Å). Similarly, different animals are sensitive to different ranges of wavelengths. For example, snakes can detect infrared waves.

˜

˜

˜

The basic difference between different types of electromagnetic waves lies in their frequencies and wavelengths. All of them travel with same speed in vacuum. Further they differ in their mode of interaction with matter. For example infrared waves vibrate the electrons, atoms and molecules of a substance. These vibrations increase the internal energy and temperature of the substance. This is why infrared waves are also called heat waves. Electromagnetic waves interact with matter via their electric and magnetic fields, which set in oscillation with the charges present in all matter. This interaction depends on the wavelength of the electromagnetic wave and the nature of atoms or molecules in the medium.

Microwave Oven In electromagnetic spectrum frequency and energy of microwaves is smaller than the visible light. Water content is required for cooking food in microwave oven. Almost all food items contain some water content. Microwaves interact with water molecules and atoms via their electric and magnetic fields. Temperature of water molecules increases by this. These water molecules share this energy with neighboring food molecules, heating up the food. Porcelain vessels are used for cooking food in microwave oven. Because its large molecules vibrate and rotate with much smaller frequencies and do not get heated up. We cannot use metal vessels. Metal vessels interact with microwaves. These vessels may melt due to heating.

Solved Examples V

Example 1 Long distance radio broadcasts use short wave bands. Explain why? Solution Short radio waves are reflected by ionosphere.

V

Example 2 It is necessary to use satellites for long distance TV transmission. Explain why? Solution TV waves (part of radio waves) range from 54 MHz to 890 MHz. Unlike short wave bands (used in radio broadcasts) which are reflected by ionosphere, TV waves are not properly reflected by ionosphere. This is why, satellites are used for long distance TV transmission.

V

Example 3 The ozone layer on the top of the stratosphere is crucial for human survival. Explain why? Solution Ozone layer protects ourselves from ultraviolet radiations. Over exposure to UV radiation can cause skin cancer in human beings. Ozone layer absorbs UV radiations. But unfortunately over use of Chlorofluoro Carbon Gases (CFCs) is depleting this ozone layer and it is a matter of international concern now a days.

V

Example 4 Optical and radio telescopes are built on ground but X-ray astronomy is possible only from satellites orbiting the earth. Explain why? Solution Visible and radio waves can penetrate the atmosphere, while X-rays are absorbed by the atmosphere. This is why X-ray telescopes are installed in satellites orbiting the earth.

V

Example 5 If the earth did not have an atmosphere, would its average surface temperature be higher or lower than what it is now? Solution Due to presence of atmosphere green house effect takes place. Heat radiated by earth is trapped due to green house effect. In the absence of atmosphere, temperature of the earth would be lower because the green house effect of the atmosphere would be absent.

V

Example 6 Some scientists have predicted the global nuclear war on the earth would be followed by a severe nuclear winter with a devastating effect on life on earth. What might be the basis of this prediction? Solution After nuclear war, clouds would perhaps cover the atmosphere of earth preventing solar light from reaching many parts of earth. This would cause a winter.

V

Example 7 Why is the orientation of the portable radio with respect to broadcasting station important? Solution Electromagnetic waves are plane polarised, so the receiving antenna should be parallel to electric and magnetic part of wave.

Chapter 29 V

Electromagnetic Waves — 13

Example 8 A plane electromagnetic wave propagating in the x-direction has a wavelength of 5.0 mm. The electric field is in the y-direction and its maximum magnitude is 30 Vm -1 . Write suitable equations for the electric and magnetic fields as a function of x and t. l = 5 mm = 5 ´ 10-3 m 2p 2p k= = m -1 l 5 ´ 10-3

Solution Given,

= 1257 m-1 w c= k

From the equation,

w = c k = (3 ´ 108 ) (1257) = 3.77 ´ 1011 rad/s E 0 = 30 V/m E c= 0 B0 B0 =

\

Now, and V

E0 30 = c 3 ´ 108

= 10-7 T E y = E 0 sin (wt - kx) = (30 V /m) sin [(3.77 ´ 1011 s -1 ) t - (1257 m-1 ) x] Bz = (10-7 T) sin [(3.77 ´ 1011 s -1 ) t - (1257 m-1 ) x]

Ans.

Example 9 A light beam travelling in the x-direction is described by the electric field E y = ( 300 Vm -1 ) sin w ( t - x c). An electron is constrained to move along the y-direction with a speed of 20 . ´ 10 7 ms -1 . Find the maximum electric force and the maximum magnetic force on the electron. Solution Maximum Electric Force Maximum electric field, E 0 = 300 V /m \ Maximum electric force F = qE 0 = (1.6 ´ 10-19 ) (300) = 4.8 ´ 10-17 N

Ans.

Maximum Magnetic Force From the equation, Maximum magnetic field, or \ Maximum magnetic force Substituting the values, we have Maximum magnetic force

c=

E0 B0

E0 c 300 = 10-6 T B0 = 3.0 ´ 108

B0 =

= B0qv sin 90o = B0qv = (10-6 )(1.6 ´ 10-19 )(2 . 0 ´ 107 ) = 3.2 ´ 10-18 N

Ans.

14 — Optics and Modern Physics V

Example 10 A parallel plate capacitor having plate area A and plate separation d is joined to a battery of emf V and internal resistance R at t = 0. Consider a plane surface of area A 2, parallel to the plates and situated symmetrically between them. Find the displacement current through this surface as a function of time. [The charge on the capacitor at time t is given by q = CV ( 1 - e - t t ), where t = CR] Solution

q = CV (1 - e- t t )

Given,

q CV = (1 - e- t t ) A A Electric field between the plates of capacitor, s CV E= = (1 - e- t t ) e0 e0 A

\

s=

Surface charge density,

Electric flux from the given area, EA CV = (1 - e- t t ) 2 2e0

fE =

i d = e0

Displacement current, or Substituting, We have, Again substituting,

i d = e0

dfE dt

d é CV (1 - e- t ê dt ë 2e0

ù CV - t )ú = e û 2t

t

t = CR V - t CR id = e 2R e A C= 0 d V id = e 2R

V

t

td e 0AR

Ans.

Example 11 About 5% of the power of a 100 W light bulb is converted to visible radiation. What is the average intensity of visible radiation (a) at a distance of 1 m from the bulb? (b) at a distance of 10 m? Assume that the radiation is emitted isotropically and neglect reflection. Solution Effective power (energy radiated per second) = 5% of 100 W P=5W

S r

Chapter 29

Electromagnetic Waves — 15

This energy will distribute on a sphere. At a distance r from the point source, area on which light is incident is S = 4 pr 2 \ Intensity at distance r from the point source, P 5 = Energy incident per unit area per unit time I= = S 4pr 2 (a) At r = 1m, I=

5 4p (1)2

= 0.4 W/m2

Ans.

(b) At r = 10 m, I=

5 4p (10)2

= 0.004 W/m2 V

Ans.

Example 12 Suppose that the electric field of an electromagnetic wave in vacuum is E = {( 3.0 N /C ) cos [( 1.8 rad / m) y + ( 5.4 ´ 10 6 rad / s) t]} $i. (a) What is the direction of propagation of wave? (b) What is the wavelength l? (c) What is the frequency f? (d) What is the amplitude of the magnetic field of the wave? (e) Write an expression for the magnetic field of the wave. Solution (a) From the knowledge of wave we can see that electromagnetic wave is travelling along negative y -direction, as wt and ky both are positive. (b) k = 1.8 rad /m 2p k= l 2p 2p l= = = 3.5 m \ k 1.8

Ans.

(c) w = 5.4 ´ 106 rad /s \

(d) From the relation, We have,

w = 2pf w 5.4 ´ 106 f = = 2p 2p = 8.6 ´ 105 Hz E 0 = 3.0 N/C E c= 0 B0 E0 3.0 B0 = = c 3.0 ´ 108 = 10-8 T

Ans.

Ans. $ (e) E is along i direction, wave is travelling along negative y-direction. Therefore, oscillations of B are along z-direction or $ Ans. B = (10-8 T) cos [(1.8 rad /m) y + (5.4 ´ 106 rad /s ) t ] k

16 — Optics and Modern Physics V

Example 13 A parallel plate capacitor made of circular plates each of radius R = 6.0 cm has a capacitance C = 100 pF . The capacitor is connected to a 230 V AC supply with a (angular) frequency of 300 rad / s.

(a) What is the rms value of the conduction current? (b) Is the conduction current equal to the displacement current? (c) Determine the amplitude of B at a point 3.0 cm from the axis between the plates. Solution (a) Capacitive reactance, XC = =

1 wC 1 300 ´ 100 ´ 10-12

108 W 3 There is only capacitance in the circuit. V \ i rms = rms XC =

=

230 (108 / 3)

Ans. = 6.9 ´ 10-6 A (b) Yes, the derivation in example 29.1 is true even if current is alternating. (c) Here, i d is displacement current and i the conduction current. Magnetic field at a distance r from the axis, m i B = 0 d2 r 2p R m 0 i rms \ Brms = r (i d = i = i rms ) 2 p R2 Substituting the values, we have Brms =

(2 ´ 10-7 ) (6.9 ´ 10-6 ) (3 ´ 10 (6 ´ 10-2)2

-2

)

= 1.15 ´ 10-11 T \

B0 = 2 Brms = ( 2 ) (1.15 ´ 10-11 ) T = 1.63 ´ 10-11 T

Ans.

Exercises Single Correct Option 1. One requires 11eV of energy to dissociate a carbon monoxide molecule into carbon and oxygen atoms. The minimum frequency of the appropriate electromagnetic radiation to achieve the dissociation lies in (a) visible region (c) ultraviolet region

(b) infrared region (d) microwave region

2. If E and B represent electric and magnetic field vectors of the electromagnetic wave, the direction of propagation of electromagnetic wave is along (a) E (c) B ´ E

(b) B (d) E ´ B

3. The ratio of contributions made by the electric field and magnetic field components to the intensity of an EM wave is (a) c : 1 (c) 1 : 1

(b) c2 : 1 (d) c : 1

4. Light with an energy flux of 20 W/cm 2 falls on a non-reflecting surface at normal incidence. If the surface has an area of 30 cm 2. The total momentum delivered (for complete absorption) during 30 minutes is (a) 36 ´ 10-5 kg -m/s (c) 1.08 ´ 104 kg -m/s

(b) 36 ´ 10-4 kg -m/s (d) 1.08 ´ 107 kg -m/s

More than One Correct Options 5. A plane electromagnetic wave propagating along x-direction can have the following pairs of E and B (a) E x , By (c) Bx , E y

(b) E y , Bz (d) E z , By

6. The source of electromagnetic waves can be a charge (a) moving with a constant velocity (c) at rest

(b) moving in a circular orbit (d) falling in an electric field

7. An electromagnetic wave of intensity I falls on a surface kept in vacuum and exerts radiation pressure p on it. Which of the following are true? (a) Radiation pressure is I /c if the wave is totally absorbed (b) Radiation pressure is I /c if the wave is totally reflected (c) Radiation pressure is 2 I /c if the wave is totally reflected (d) Radiation pressure is in the range I /c < p < 2 I /c for real surfaces

8. A charged particle oscillates about its mean equilibrium position with a frequency of 109 Hz. The electromagnetic waves produced (a) will have frequency of 109 Hz (c) will have a wavelength of 0.3 m

(b) will have frequency of 2 ´ 109 Hz (d) fall in the region of radio waves

18 — Optics and Modern Physics Subjective Questions 9. Can an electromagnetic wave be deflected by an electric field? By a magnetic field? 10. What physical quantity is the same for X-rays of wavelength 10-10m, red light of wavelength 6800 Å and radio waves of wavelength 500 m?

11. A plane electromagnetic wave travels in vacuum along z-direction. What can you say about the directions of its electric and magnetic field vectors. If the frequency of the wave is 30 MHz, what is its wavelength?

12. A radio can tune into any station in the 7.5 MHz to 12 MHz band. What is the corresponding wavelength of band?

13. The amplitude of the magnetic field part of a harmonic electromagnetic wave in vacuum is B0 = 510 nT. What is the amplitude of the electric field part of the wave?

14. Figure shows a capacitor made of two circular plates each of radius 12 cm, and separated by 5.0 cm. The capacitor is being charged by an external source (not shown in the figure). The charging current is constant and equal to 0.15 A.

(a) Calculate the capacitance and the rate of change of potential difference between the plates. (b) Obtain the displacement current across the plates. (c) Is Kirchhoff’s first rule (junction rule) valid at each plate of the capacitor? Explain.

15. Suppose that the electric field amplitude of an electromagnetic wave is E0 = 120 N C and that its frequency is 50.0 MHz. (a) Determine B0 , w, k and l, (b) Find expressions for E and B.

16. A variable frequency AC source is connected to a capacitor. How will the displacement current change with decrease in frequency?

17. A laser beam has intensity 2.5 ´ 1014 Wm -2. Find the amplitudes of electric and magnetic fields in the beam.

18. In a plane electromagnetic wave, the electric field oscillates sinusoidally at a frequency of 2 . 0 ´ 1010 Hz and amplitude 48 Vm -1.

(a) What is the wavelength of the wave? (b) What is the amplitude of the oscillating magnetic field? (c) Show that the average energy density of the field E equals the average energy density of the field B. [c = 3 ´ 108 ms -1 ].

19. The charge on a parallel plate capacitor varies as q = q0 cos 2pft. The plates are very large and

close together (area = A, separation = d ). Neglecting the edge effects, find the displacement current through the capacitor.

Answers Introductory Exercise 29.1 1. i/2

Introductory Exercise 29.2 2. Yes

4. 5.55 ´ 10 -12 J

3. 27 . mT

Exercises 1. 6. 10. 11. 12. 13. 14.

(c) 2. (d) 3. (c) 4. (b) 5. (b,d) (b,d ) 7. (a,c,d) 8. (a,c,d) 9. No, No The speed in vacuum is the same for all E and B lie in x-y plane and are mutually perpendicular, 10 m Wavelength band from 40 m to 25 m 153 N/C (a) 8.0 pF, 1.87 ´ 1010 Vs -1 (b) 0.15 A

(c) Yes, provided by current we mean the sum of conduction and displacement currents. 15. (a) 400 nT, 3.14 ´ 10 8 rad/s , 105 . rad/m, 6.00 m (b) E = (120 N/C) sin [(1.05 rad/m)] x - (3.14 ´ 10 8 rad/s)t] B = ( 400 nT) sin [(1.05 rad/m)] x - (3.14 ´ 10 8 rad/s) t] 16. Displacement current will decrease 17. 4.3 ´ 10 8 N/C , 1.44 T 18. (a) 15 . ´ 10 -2 m 19. -2pq 0 f sin 2p ft

(b) 16 . ´ 10 -7 T

30.1

Introduction

30.2

General concepts used in Geometrical Optics

30.3

Reflection of light

30.4

Reflection from a spherical surface

22 — Optics and Modern Physics

30.1 Introduction The branch of physics called optics deals with the behaviour of light and other electromagnetic waves. Light is the principal means by which we gain knowledge of the world. Consequently, the nature of light has been the source of one of the longest debates in the history of science. Electromagnetic radiation with wavelengths in the range of about 4000 Å to 7000 Å, to which eye is sensitive is called light. In the present and next two chapters we investigate the behaviour of a beam of light when it encounters simple optical devices like mirrors, lenses and apertures. Under many circumstances, the wavelength of light is negligible compared with the dimensions of the device as in the case of ordinary mirrors and lenses. A light beam can then be treated as a ray whose propagation is governed by simple geometric rules. The part of optics that deals with such phenomena is known as geometric optics. However, if the wavelength is not negligible compared with the dimensions of the device (for example a very narrow slit), the ray approximation becomes invalid and we have to examine the behaviour of light in terms of its wave properties. This study is known as physical optics.

30.2 General Concepts used in Geometrical Optics Some general concepts which are used in whole geometrical optics are given below. 1. Normal incidence means angle of incidence (with normal) is 0°. If angle of incidence is 90°, it is

called grazing incidence. 90°

Normal incidence Grazing incidence Ð i = 0° Ð i = 90° Fig. 30.1

2. An image is formed either by reflection or refraction. Minimum two (reflected or refracted) rays

are required for image formation. More the number of rays, more will be the intensity of image. 3. A light is reflected only from a silvered surface. Without any reflecting surface on the path of ray

of light it keeps on moving ahead. p

q (i)

(ii)

Fig. 30.2

In figure (i) For image formation, if reflected ray-p is required to the left of concave lens, then we will take it as dotted line.

Reflection of Light — 23

Chapter 30

In figure (ii) For image formation, if reflected ray-q is required to the right of concave mirror, then we will take it as dotted line. 4. Real object, virtual object, real image, virtual image The point where the rays meet (or appear to meet) before refraction or reflection is called object and the point where the rays meet (or appear to meet) after refraction or reflection is called image. Further, object (or image) is real if dark lines meet and virtual if dotted lines meet. In figure (a), object is real, while image is virtual. In figure (b), object is virtual while its image is real.

O

I

O

I

(a)

(b)

O

I2

I1

(c)

Fig. 30.3

In figure (c), the object O is real. Its image formed by the lens (i.e. I 1 ) is real. But, it acts as a virtual object for mirror which forms its real image I 2 . 5. The virtual images cannot be taken on screen. But, they can be seen by our eye. Because our eye lens forms their real image on our retina. Thus, if we put a screen at I in the above figure (a) no image will be formed on it. At the same time if we put the screen at I in figure (b), image will be formed. 6. Normally, the object is kept on the left hand side of the optical instrument (mirror, lens etc.), i.e. the ray of light travels from left to right. Sometimes, it may happen that the light is travelling in opposite direction. See the figure.

O

C

(a)

I

O ® Object

O

I

I ® Image Fig. 30.4

(b)

P

I

M

O

(c)

24 — Optics and Modern Physics In figures (a) and (b), light is travelling from left to right and in figure (c) it is travelling from right to left. 7. Sign convention The distances measured along the incident light are taken as positive while the distances against incident light are taken as negative. For example, in figures (a) and (b) the incident light travels from left to right. So, the distances measured in this direction are positive. While in figure (c) the incident light travels from right to left. So, in this case right to left direction will be positive. Distances are measured from pole of the mirror [point P in figure (b)], optical centre of the lens [point C in figure (a)] and the centre of the refracting surface [point M in figure (c)]. It may happen in some problem that sign convention does not remain same for the whole problem. For example, in the Fig. 30.5 shown, the ray of light starting A from O first undergoes refraction at A, then reflection at B B and then finally refraction at C. For refraction and reflection M P at A and B the incident light is travelling from left to right, so O C distances measured along this direction are positive. For final refraction at C the incident light travels from right to Fig. 30.5 left, so now the sign convention will change or right to left is positive. 8. Object distance (from P , C or M along the optic axis) in Fig. 30.4 is shown by u and image distance by v. 9. In front of mirror, object (or image) is always real, lines are always dark and u (or v ) are always negative. Behind the mirror object (or image) is always virtual, lines are always dotted and u (or v ) are always positive. This is because light always falls from front side of the mirror.

Real

Virtual

+ve –ve +ve

Fig. 30.6

10. In most of the cases, objects are real (whether refraction or reflection) and u for them is negative. +ve

O

+ve

O

u = –ve

u = –ve

Fig. 30.7

Chapter 30

Reflection of Light — 25

11. In case of mirror light always falls from front of the mirror. But in a lens (or some other refracting

surface like slab) light can fall from both sides.

or

Fig. 30.8 A

12. Total steps and reduced steps

1

2

In Fig. 30.9 total steps are five. Four of them are refraction. Only 3 D B third is reflection. But we have made a lens formula for steps 1 and 2 or 4 and 5. So, the reduced steps are three, 5 4 lens ® mirror ® lens. C 13. Image real or virtual In Fig.30.9, if we wish to find the nature Fig. 30.9 of I 2 (after 2nd refraction), then it is real if it is formed to the right of ABC because ray of light has moved to this side and it is virtual to the left of ABC. Similarly, I 5 is real to the left of ADC and virtual to the right of ADC. 14. Final image coincides with the object In most of these cases there will be one mirror, plane or spherical (convex or concave) and light will be falling normal ( Ði = 0° ) to this mirror. In case of spherical mirror it is normal if ray of light passes through centre of curvature. In case of normal incidence, ray of light retraces its path and final image coincides with the object.

C C

Fig. 30.10

In all above figures, Ð i = 0° (normal incidence). Ray of light retraces its path. 15. Image at infinity means rays after refraction or reflection have become parallel to the optic axis. If a screen is placed directly in between these parallel rays no image will be formed on the screen. But if a lens (or a mirror) is placed on the path of these parallel rays, then image is formed at focus. Sometimes, our eye plays the role of this lens and the image is formed on our retina. Screen Parallel rays

F

or

Parallel rays

Retina Eye lens

Fig. 30.11

26 — Optics and Modern Physics 16. Visual angle q

Angle subtended by an object on our eye is called the visual angle. The apparent size depends on the visual angle. As the object moves away from the eye, actual size remains the same but visual angle decreases. Therefore, apparent size decreases. h

h q2

q1 q 2 < q1

Fig. 30.12

30.3 Reflection of Light When waves of any type strike the interface between two different materials, new waves are generated which move away from the interface. Experimentally, it is found that the rays corresponding to the incident and reflected waves make equal angles with the normal to the interface and that the reflected ray lies in the plane of incidence formed by the incident ray and the normal. Thus, the two laws of reflection can be summarised as under: (i) Ð i = Ð r (ii) Incident ray, reflected ray and normal lie on the same plane. Normal Incident ray

Reflected ray

i r

Fig. 30.13

Two Important Points in Reflection Laws 1. The first law Ð i = Ð r can be applied for any type of surface. The main point is, normal at point of

incidence. In spherical surface (convex mirror or concave mirror) normal at any point passes through centre of curvature.

i

i

r

r

C

C

Concave mirror Convex mirror Fig. 30.14

2. Incident ray, reflected ray and normal are sometimes represented in the form of three vectors.

Then, these three vectors should be coplanar.

Chapter 30

Reflection of Light — 27

Reflection from a Plane Surface (or Plane Mirror) Almost everybody is familiar with the image formed by a plane mirror. If the object is real, the image formed by a plane mirror is virtual, erect, of same size and at the same distance from the mirror. The ray diagram of the image of a point object and of an extended object is as shown below.

O

B

B¢

A

A¢

I

Fig. 30.15

Important Points in Reflection from Plane Mirror 1. Relation between object distance ( u) and the image distance ( v ) in case of plane mirror is

v=-u Here, v and u are measured from the plane surface. Two conclusions can be drawn from this equation. (i) Negative sign implies that object and image are on opposite sides of the mirror. So, if object is real then image is virtual and vice-versa. (ii) | v | = | u | And this implies that perpendicular distance of the object from the mirror is equal to the perpendicular distance of image from the mirror. I

O

M O

I

M

Correct OM = MI

Wrong OM = MI

Fig. 30.16

2. Ray Diagram

Let us draw the ray diagram of a point object and an extended object. b

r i

O d

a

b¢

d a¢

I (i) O ® Real I ® Virtual

(ii) ab ® Real a¢b¢ ® Virtual

Fig. 30.17

28 — Optics and Modern Physics Just as we have drawn the ray diagram of point object O in figure (i), we can also draw the ray diagrams of points a and b in figure (ii). 3. Field of view of an object for a given mirror Suppose a point object O is placed in front of a small mirror as shown in Fig. 30.18 (a), then a question arises in mind whether this mirror will make the image of this object or not. Or suppose an elephant is standing in front of a small mirror, will the mirror form the image of the elephant or not. The answer is yes, it will form. A mirror whatever may be the size of it forms the images of all objects lying in front of it. But every object has its own field of view for the given mirror. The field of view is the region between the extreme reflected rays and depends on the location of the object in front of the mirror. If our eye lies in the field of view then only we can see the image of the object otherwise not. The field of view of an object placed at different locations in front of a plane mirror are shown in Fig. 30.18 (b) and (c). The region between extreme reflected rays (reflected from the end points of the mirror) is called the field of view. To see the image of object eye should lie in this region, as all reflected rays lie in this region.

O

I

O

I

O (a)

(b)

(c)

Fig. 30.18

4. Suppose a mirror is rotated by an angle q (say anti-clockwise), keeping the incident ray fixed then

the reflected ray rotates by 2q along the same direction, i.e. anti-clockwise. Y N¢

N

I

R¢ i – 2q

R

i–q

I

q i–q

i i

X

q (a)

(b)

Fig. 30.19

In figure (a), I is the incident ray, N the normal and R the reflected ray. In figure (b), I remains as it is N and R shift to N ¢ and R ¢. From the two figures, we can see that earlier the reflected ray makes an angle i with y-axis while after rotating the mirror it makes an angle i – 2q. Thus, we may conclude that the reflected ray has been rotated by an angle 2q.

Chapter 30

Reflection of Light — 29

H , where H is the height of 2 person. But, the mirror should be placed in a fixed position which is shown in Fig. 30.20.

5. The minimum length of a plane mirror to see one’s full height is

A x

F

B

x

C

(x + y)

y D

G

y E Person

Fig. 30.20

A ray starting from head (A) after reflecting from upper end of the mirror (F) reaches the eye at C. Similarly, the ray starting from the foot (E) after reflecting from the lower end (G) also reaches the eye at C. In two similar triangles ABF and BFC, AB = BC = x (say), Similarly in triangles CDG and DGE, CD = DE = y (say) Now, we can see that height of the person is 2 ( x + y) and that of mirror is (x + y), i.e. height of the mirror is half the height of the person. Note The mirror can be placed anywhere between the centre lines BF (of AC) and DG (of CE ). As the mirror is moved away on this line, image also moves away from the person. So, apparent size keeps on decreasing.

6. A person is standing exactly at midway between a wall and a mirror and he wants to see the full

height of the wall (behind him) in a plane mirror (in front of him). The minimum length of mirror H in this case should be , where H is the height of wall. The ray diagram in this case is shown in 3 Fig. 30.21 A

H

2x F

x

B

I x (x + y)

(x + y)

C y

K G

E

y

2y

J

D Wall

Person d

Mirror d

Fig. 30.21

In triangles HBI and IBC, HI = IC = x (say). Now, in triangles HBI and ABF, AF FB AF 2d or or AF = 2x = = HI BI x d Similarly, we can prove that DG = 2 y if, CK = KJ = y Now, we can see that height of the wall is 3 ( x + y) while that of the mirror is ( x + y).

30 — Optics and Modern Physics There are four important points related to object and image velocity. (i) Image speed is equal to the object speed. (ii) Image velocity and object velocity make same angles from the plane mirror on two opposite sides of the mirror. (iii) Components of velocities which are along the mirror are equal. (iv) Components of velocities which are perpendicular to the mirror are equal and opposite. The following four figures demonstrate the above four points.

7. Object and image velocity

v O v

v I

v

O

v O

I

q q

v

O

I v

v

I

q q

Fig. 30.22

Three Types of Problems in Reflection from Plane Mirror Type 1. Based on law of reflection Ð i = Ð r

Concept These problems are purely based on geometry. Proper normal at point of incidence is very important. V

Example 30.1 Two plane mirrors M 1 and M 2 are inclined at angle q as shown. A ray of light 1, which is parallel to M 1 strikes M 2 and after two reflections, the ray 2 becomes parallel to M 2 . Find the angle q. 2

M2

1 q M1

Fig. 30.23

Solution

Different angles are as shown in Fig. 30.24. In triangle ABC, B

q A

q

a q a aa q

a =90°– q q

C Fig. 30.24

q + q + q = 180° \

q = 60°

Ans.

Chapter 30 V

Reflection of Light — 31

Example 30.2 Prove that for any value of angle i, rays 1 and 2 are parallel. 1 i

2

90°

Fig. 30.25

PQ and MN are mutually parallel. Rays 1 and 2 are making equal angles ( = Ði ) from PQ and MN. So, they are mutually parallel.

Solution

1 i Q i 90° – i 90° – i

P

2

i M

N

Fig. 30.26

Important Result original path.

Two plane mirrors kept at 90° deviate each ray of light by 180° from its

Type 2. Based on field of view

Concept The region between extreme reflected rays is called field of view. To see the image of object our eye should lie in the field of view. V

Example 30.3 A point source of light S, placed at a distance L in front of the centre of a mirror of width d, hangs vertically on a wall. A man walks in front of the mirror along a line parallel to the mirror at a distance 2L from it as shown. The greatest distance over which he can see the image of the light source in the mirror is (a) d/2 (c) 2 d

(b) d (d) 3 d

S

d L

2L Fig. 30.27

(JEE 2000)

(d) The ray diagram will be as shown in Fig. 30.28. HI = AB = d d DS = CD = 2 Since, AH = 2AD d \ GH = 2 CD = 2 = d 2 Similarly, IJ = d \ GJ = GH + HI + IJ = d + d + d = 3d

G

Solution

C A

f f

D S E

B

H

I

F J

Fig. 30.28

32 — Optics and Modern Physics V

Example 30.4 A pole of height 4 m is kept in front of a vertical plane mirror of length 2 m. The lower end of the mirror is at a height of 6 m from the ground. The horizontal distance between the mirror and the pole is 2 m. Upto what minimum and maximum heights a man can see the image of top of the pole at a horizontal distance of 4 m (from the mirror) standing on the same horizontal line which is passing through the pole and the horizontal point below the mirror? E Solution PQ = Pole, MN = Image of pole HG BD = GN BN ( HG ) ( BN ) ( 2) ( 6) \ BD = = D GN 2 I =6m 2m C Minimum height required = AD = BD + AB = 10 m H 2m Q N IG BE B Further, = G GN BN 4m ( IG ) ( BN ) ( 4 ) ( 6) \ BE = = = 12 m P F A M GN 2 2m 2m 2m \ Maximum height required = AE Fig. 30.29 = BE + AB = 16 m

Type 3. Based on object and image velocity

Concept We have already discussed four points on this topic. V

Example 30.5 A plane mirror is lying in x-y plane. Object velocity is $ ) m/ s. Find the image velocity. v 0 = ( 2$i - 3$j + 4 k Components of object velocity parallel to plane mirror (or lying in x - y plane) remain unchanged. But, component perpendicular to plane mirror changes its direction but magnitude remains the same. Hence, the image velocity is Ans. v = ( 2$i - 3$j - 4k$ ) m/s Solution

I

V

Example 30.6 An object is falling vertically downwards with velocity 10 m/s. In terms of $i and $j , find the image velocity. O 10 m/s ^j ^ i

30°

Fig. 30.30

Chapter 30

Reflection of Light — 33

Image speed will also be 10 m/s. Further, image velocity will make same angle ( = 60° ) from the mirror on opposite side of it. In terms of i$ and $j ,

vI =10 m/s

Solution

vI = ( 5 3 i$ + 5 $j ) m/s

V

30°

60°

v0 =10 m/s Fig. 30.31

Ans.

Example 30.7 A point object is moving with a speed v before an arrangement of two mirrors as shown in figure. Find the magnitude of velocity of image in mirror M 1 with respect to image in mirror M 2 .

M1 v q

M2

Using the same concept used in above problem, we have to find magnitude of relative velocity between vI 1 and vI 2 . The angle between these two vectors is 2 q. Hence, vr = vI 1 - vI 2 \ | vr | = | vI 1 - vI 2 | M2 This is nothing but magnitude of subtraction of two velocity vectors of equal magnitudes v each and angle between them equal to 2q.

Fig. 30.32

Solution

Hence,

60°

30°

^i

vI = 10 cos 30° i$ + 10 sin 30° $j or

30°

^j

vI1 = v M1 q

q 2q

v0 = vI 2 = v

q

Fig. 30.33

| vr | = v 2 + v 2 - 2( v ) ( v ) cos 2q

Solving these two equations, we get | v2 | = 2 v sin q

INTRODUCTORY EXERCISE

Ans.

30.1

1. A man approaches a vertical plane mirror at speed of 2 m/s. At what rate does he approach his image?

2. An object M is placed at a distance of 3 m from a mirror with its lower 2m

end at 2 m from ground as shown in Fig. 30.34. There is a person at a distance of 4 m from object. Find minimum and maximum height of person to see the image of object.

3. In terms of q find the value of i , so that ray of light retraces its path after third reflection. i q

Fig. 30.35

2m Person 4 m M 3 m Fig. 30.34

34 — Optics and Modern Physics

30.4 Reflection from a Spherical Surface We shall mainly consider the spherical mirrors, i.e. those which are part of a spherical surface.

Terms and Definitions There are two types of spherical mirrors, concave and convex. A

A

Incident light

Incident light +ve

P

C

B

P

C

B (b) Convex mirror

(a) Concave mirror

Fig. 30.36

Centre C of the sphere of which the mirror is a part is called the centre of curvature of the mirror and P the centre of the mirror surface, is called the pole. The line CP produced is the principal axis and AB is the aperture of the mirror. The distance CP is called the radius of curvature (R). All distances are measured from point P. We can see from the two figures that R is positive for convex mirror and negative for concave mirror.

Principal Focus Observation shows that a narrow beam of rays, parallel and near to the principal axis, is reflected from a concave mirror so that all rays converge to a point F on the principal axis. F is called the principal focus of the mirror and it is a real focus, since, light actually passes through it. Concave mirrors are also known as converging mirrors because of their action on a parallel beam of light. They are used in car head-lights, search-lights and telescopes.

C

F

P

+ve

A converging mirror

P

F

C

A diverging mirror

Fig. 30.37

A narrow beam of rays, parallel and near to the principal axis, falling on a convex mirror is reflected to form a divergent beam which appears to come from a point F behind the mirror. A convex mirror thus has a virtual principal focus. It is also called a diverging mirror. The distance FP is called the focal length (f) of the mirror. Further, we can see that f is negative for a concave mirror and positive for convex mirror. Later, we will see that f = R /2 .

Chapter 30

Reflection of Light — 35

Paraxial rays Rays which are close to the principal axis and make small angles with it, i.e. they are nearly parallel to the axis, are called paraxial rays. Our treatment of spherical mirrors will be restricted to such rays which means we shall consider only mirrors of small aperture. In diagrams, however, they will be made larger for clarity.

Images formed by Spherical Mirrors In general position of image and its nature (i.e. whether it is real or virtual, erect or inverted, magnified or diminished) depend on the distance of object from the mirror. Case 1

Concave mirror

+µ

–µ C

F

P

Fig. 30.38

Table 30.1

Note

Object position

Image position

Image nature

Object speed and image speed

P

P

-

-

F

± µ

-

-

C

C

Real, inverted and same size

v I = vO

Between P and F

Between P and + µ

Virtual, erect and magnified

v I > vO

Between F and C

Between - µ and C

Real, inverted and magnified

v I > vO

Between C and - µ

Between C and F

Real, inverted and diminished

vO > v I

(i) The above table is only for real objects lying in front of the mirror for which u is negative. (ii) v I and vO are image and object speeds. (iii) From the above table we can see that image and object always travel in opposite directions as long as they move along the principal axis.

Case 2

Convex mirror

For real objects there is only one case

–µ

P

Fig. 30.39

F

C

+µ

36 — Optics and Modern Physics Object lies between P and - µ, then image lies between P and F . Image is virtual, erect and diminished. Object speed is greater than the image speed and they travel in opposite directions (along the principal axis). Ray diagrams We shall consider the small objects and mirrors of small aperture so that all rays are paraxial. To construct the image of a point object two of the following four rays are drawn passing through the object. To construct the image of an extended object, the image of two end points is only drawn. The image of a point object lying on principal axis is formed on the principal axis itself. The four rays are as under: 1 2

1 C

3 4

F

4

P

F

P

3

C

2

Fig. 30.40

Ray 1. A ray through the centre of curvature which strikes the mirror normally and is reflected back along the same path. Ray 2. A ray parallel to principal axis after reflection either actually passes through the principal focus F or appears to diverge from it. Ray 3. A ray passing through the principal focus F or a ray which appears to converge at F is reflected parallel to the principal axis. Ray 4. A ray striking at pole P is reflected symmetrically back in the opposite side.

Convex Mirror Image formed by convex mirror is always virtual, erect and diminished, no matter where the object is.

O

P

I

F

Fig. 30.41

Figure shows that convex mirror gives a wider field of view than a plane mirror, convex mirrors are therefore, used as rear view mirrors in cars or scooters. Although, they make the estimation of distances more difficult but still they are preferred because there is only a small movement of the image for a large movement of the object.

Chapter 30

Reflection of Light — 37

Concave Mirror In case of a concave mirror the image is erect and virtual when the object is placed between F and P. In all other positions of object the image is real. Object

Object

O

C

P

F

C

(a) Object beyond C Image between C and F, real, inverted, diminished

(b) Object at C Image at C, real, inverted, same size Object

Object

I C

O

P

F

F

P

C

(c) Object between C and F Image beyond C, real inverted, magnified

F

O

P

I

(d) Object between F and P Image behind mirror, virtual upright, magnified

Fig. 30.42

List of Formulae R 2 1 1 1 2 (ii) + = = v u f R (i) f =

(iii) Lateral, transverse or linear magnification, Image height I -v = = Object height O u 1 (iv) Power of a mirror (in dioptre) = Focal length (in metre) m=

(v) Image velocity Case 1

Along the principal axis, v I = - m 2 v O

Here, negative sign implies that object and image always travel in opposite directions. Case 2

Perpendicular to principal axis,

v I = m vO Here, m has to be substituted with sign. If m is positive, then v I and vO travel in same direction. If m is negative, then they travel in opposite directions.

38 — Optics and Modern Physics Important Points in Formulae R =¥

(i) For plane mirror,

1 1 2 2 R or f = ¥ and + = = = 0 or v = - u v u R ¥ 2 (ii) From the value of m, we can know nature of image, type of mirror and an approximate location of object. But, always remind that real (and inverted) image is formed only by a concave mirror (for real objects) but virtual image is formed by all three mirrors. The only difference is, in their sizes. Magnified image is obtained from concave mirror, same size from plane mirror and diminished size from convex mirror. Let us make a table : f =

\

Table 30.2 Value of m

Nature of image

Type of mirror

Object position

-4

Inverted, real and magnified

Concave

Between F and C

-1

Inverted, real and same size

Concave

At C

1 2

Inverted, real and diminished

Concave

Between C and - ¥

+3

Erect, virtual and magnified

Concave

Between P and F

+1

Erect, virtual and same size

Plane

For all positions

1 2

Erect, virtual and diminished

Convex

Between P and - ¥

+

(iii) Power Optical power means power of bending of light. By convention, converging nature is taken as the positive power and diverging nature as negative power. 1 Power of a lens (in dioptre) = + f ( in metre ) 1 Power of a mirror (in dioptre) = f ( in metre ) Now, let us make a table : Table 30.3 Lens/Mirror

f

P

Converging/Diverging

Convex lens

+ ve

+ ve

Converging

Concave lens

- ve

- ve

Diverging

Diagram

Chapter 30 Lens/Mirror

f

P

Converging/Diverging

Convex mirror

+ ve

- ve

Diverging

Concave mirror

- ve

+ ve

Converging

Reflection of Light — 39 Diagram

(iv) Image velocity Case 1 Along the principal axis

vI I

O vO

v

u

Fig. 30.43

By the motion of object and image v and u will change but focal length will remain unchanged. If we differentiate the mirror formula 1 1 1 + = v u f with respect to time, we get – v –2 . or

dv du – u –2 =0 dt dt æ v2 dv =–ç ç u2 dt è

(as f = constant) ö du ÷ ÷ dt ø

du is the rate by which u is changing. Or it is the object speed if mirror is stationary. dt dv Similarly, is the rate by which v (distance between image and mirror) is changing. Or it is dt image speed if mirror is stationary.

Here,

v2 u2

= m2

40 — Optics and Modern Physics So, the above relation becomes : image speed = m 2 ´ object speed As, object and image travel in opposite directions. So, in terms of velocity, the correct relation is v I = - m 2 vO Case 2

Perpendicular to axis vo O x y

I vI u v Fig. 30.44

This time v and u are constants. Therefore, m = -

v is also constant. u

But, x and y are variables y I = O x y = mx

m= Þ If we differentiate with respect to time, we get,

dy dx =m dt dt v I = m vO

Þ

Proofs of Different Formulae Discussed in Theory (i) Relation between f and R A ray AM parallel to the principal axis of a concave mirror of small aperture is reflected through the principal focus F. If C is the centre of curvature, CM is the normal to the mirror at M because the radius of a spherical surface is perpendicular to the surface. From first law of reflection, M

A

q q C

P

F

f R

Fig. 30.45

Chapter 30

Reflection of Light — 41

Ði = Ðr or

Ð AMC = Ð CMF = q

But,

Ð AMC = Ð MCF

\

Ð CMF = Ð MCF

(say) (alternate angles)

Therefore, DFCM is thus isosceles with FC = FM . The rays are paraxial and so M is very close to P. Therefore, FM » FP \

FC = FP

or

1 FP = CP 2 f =

or EXERCISE

R 2

Prove the above relation for convex mirror.

(ii) The mirror formula In Fig. 30.46 (a) and (b), a ray OM from a point object O on the principal axis is reflected at M so that the angle q, made by the incident and reflected rays with the normal CM are equal. A ray OP strikes the mirror normally and is reflected back along PO. The intersection I of the reflected rays MI and PO in figure (a) gives a real point image of O and in figure (b) gives a virtual point image of O. Let a , b and g be the angles as shown. As the rays are paraxial, these angles are small, we can take q

M q

O

g

a

P

I

C

q

q g

b

a

M

q

P

O

b I

v

R

R

v (a)

(b)

Fig. 30.46

a » tan a =

and

C

u

u

b=

MP CP

g=

MP IP

MP , OP

42 — Optics and Modern Physics Now, let us take the two figures simultaneously Table 30.4 Concave

Convex

In triangle CMO, b = a + q (the exterior angle) or …(i) q=b – a

In triangle CMO, q = a + b …(iv)

In D CMI, \

…(ii)

In DCMI g = q + b or q=g –b

…(v)

…(iii)

From Eqs. (iv) and (v), we get 2b = g – a

…(vi)

g =b + q q=g –b

From Eqs. (i) and (ii), we get 2b = g + a

(the exterior angle)

Substituting the values of a, b and g, we get 2 1 1 …(A) = + CP IP OP

Substituting the values of a,b and g, we get 2 1 1 …(B) = – CP IP OP

If we now substitute the values with sign, i.e. CP = – R, IP = – v and OP = – u 2 1 1 we get, = + R v u

If we now substitute the values with sign, i.e. CP = + R, IP = + v and OP = – u, we get 2 1 1 = + R v u

or

1 1 1 + = v u f

(iii) Magnification

Rö æ ç as f = ÷ 2ø è

1 1 1 + = v u f

or

Rö æ ç as f = ÷ 2ø è

The lateral, transverse or linear magnification m is defined as Height of image I ¢ I IP m= = = Height of object O¢ O OP

…(i)

(From similar triangles) O¢

q q

I O

P

I¢ v u

Fig. 30.47

Here, IP = – v and OP = – u, further object is erect and image is inverted so we can take I ¢ I as negative and O¢ O as positive and Eq. (i) will then become I ¢I v =– O¢O u or

m=–

v u

v 1 1 1 + = and m = – for special cases of the position of object but the same result u v u f can be derived for other cases also.

Note We have derived

Chapter 30

Reflection of Light — 43

Types of Problems in Spherical Mirror Type 1. To find image distance and its magnification corresponding to given object distance and focal length of mirror

How to Solve? l

In the mirror formula substitute signs of only u and f. Sign of v automatically comes after calculations. For real objects sign of u is always negative, sign of f is positive for convex mirror and negative for concave mirror.

V

Example 1 An object is placed at a distance of 30 cm from a concave mirror of focal length 20 cm. Find image distance and its magnification. Also, draw the ray diagram. Solution Substituting u = - 30 cm and f = - 20 cm in the mirror formula

1 1 1 + = , we have v u f

1 1 1 + = v -30 -20 v = - 60 cm

Solving, we get Magnification,

Ans.

-v (-60) m= =u (-30)

or m = -2 Magnification is -2, which implies that image is inverted, real and two times magnified. Ray diagram is as shown below.

Ans.

h

I

O

F

P

2h 20 cm 30 cm 60 cm

V

Example 2 An object is placed at a distance of 40 cm from a convex mirror of focal length 40 cm. Find image position and its magnification. Also, draw its ray diagram. Solution Substituting, u = - 40 cm and f = + 40 cm in the mirror formula 1 1 1 + = , we have v u f 1 1 1 + = v - 40 + 40 We get, v (+20) 1 Magnification, m = - = =+ u (-40) 2

v = + 20 cm

Ans.

44 — Optics and Modern Physics 1 Magnification is + , which implies that image is erect, virtual and half in size. 2 Ray diagram is as shown below.

h O

h/2

P

F

I

PO = PF = 40 cm PI = 20 cm Here, PO = PF = 40 cm or object is placed at a distance of its focal length, but object is not actually kept at F. Otherwise, image would be formed at infinity.

Note

Type 2. To find object/image distance corresponding to given magnification of image if focal length of mirror is also given

How to Solve? l l

l

In this type, substitute all three signs of u , v and f. Signs of u and f have been discussed in Type 1. Sign of v will be positive for virtual image and negative for real image. v m=Þ | v| = | mu | u

V

Example 3 Find the distance of object from a concave mirror of focal length 10 cm so that image size is four times the size of the object. Solution Concave mirror can form real as well as virtual image. Here, nature of image is not given in the question. So, we will consider two possible cases. Case 1 (When image is real) Real image is formed on the same side of the object, i.e. u , v and f all are negative. So let, u=–x then and Substituting in We have \

v v = – 4x as | | = |m| = 4 u f = – 10 cm 1 1 1 + = v u f 1 1 1 – = – 4x x –10

or

x = 12.5 cm

5 1 = 4x 10 Ans.

Note | f | < | x | < | 2 f | and we know that in case of a concave mirror, image is real and erect when object lies between F and C.

Chapter 30

Reflection of Light — 45

Case 2 (When image is virtual) In case of a mirror, image is virtual, when it is formed behind the mirror, i.e. u and f are negative, while v is positive. So let, u=–y and f = – 10 cm 1 1 = u f 1 1 = y –10 3 1 = 4 y 10

v = + 4y 1 + v 1 – 4y

then Substituting in We have or

y = 7.5 cm

or

Ans.

Note Here, | y | < | f | , as we know that image is virtual and erect when the object lies between F and P.

Type 3. Based on making some condition

How to Solve? l l l

Initially, substitute sign of f only. Make an equation of v. Now, for real image v should be negative and for virtual object v should be positive. With these concepts we can make the necessary condition.

V

Example 4 Find the condition under which a convex mirror can make a real image. Solution Substituting the sign of f only in the mirror formula, we have 1 1 1 + = v u +f

Þ

1 1 1 = v f u

For real image v should be negative and for this u should be positive and less than f. Object distance u is positive means object should be virtual and lying between P and F. The ray diagram is as shown below. i r I

P

O

F

C

Here, O = virtual between P and F I = real

Note Under normal conditions a concave mirror makes a real image. But, it makes a virtual image if a real object is kept between P and F. On the other hand a convex mirror makes a virtual image. But it makes a real image if a virtual object is kept between P and F.

46 — Optics and Modern Physics Type 4. To find nature of image and type of mirror corresponding to given optic axis of mirror a point object and a point image. With the help of ray diagram we have to find focus and pole of the mirror also. V

Example 5 An image I is formed of a point object O by a mirror whose principal axis is AB as shown in figure. O

A

B

I

(a) State whether it is a convex mirror or a concave mirror. (b) Draw a ray diagram to locate the mirror and its focus. Write down the steps of construction of the ray diagram. Consider the possible two cases: (1) When distance of I from AB is more than the distance of O from AB and (2) When distance of O from AB is more than the distance of I from AB

Solution D O M A

P

M

O F

C

B

A

C

F

P

B

I I Case (1)

D Case (2)

(a) As the image is on the opposite side of the principal axis, the mirror is concave. Because convex mirror always forms an erect image. (b) Two different cases are shown in figure. Steps are as under : (i) From I or O drop a perpendicular on principal axis, such that CI = CD or OC = CD. (ii) Draw a line joining D and O or D and I so that it meets the principal axis at P. The point P will be the pole of the mirror as a ray reflected from the pole is always symmetrical about principal axis. (iii) From O draw a line parallel to principal axis towards the mirror so that it meets the mirror at M. Join M to I, so that it intersects the principal axis at F. F is the focus of the mirror as any ray parallel to principal axis after reflection from the mirror intersects the principal axis at the focus.

Note In both figures, mirror should face towards the object.

Exercise In the above problem, find centre of curvature of the mirror with the help of only ray diagram.

Chapter 30

Reflection of Light — 47

Type 5. Based on velocity of image

How to Solve? l l l

Using the steps discussed in Type 1, find v and then m Along the axis v I = - m2 vO Perpendicular to axis Here, m has to be substituted with sign.

V

v I = m vO

Example 6 Focal length of the mirror shown in figure is 20 cm. Find the image position and its velocity. 5 mm/s 37°

P

O

30 cm

Solution Substituting the values, u = - 30 cm and f = - 20 cm in the mirror formula, we have 1 1 1 + = v - 30 - 20 Solving this equation, we get v = - 60 cm v (- 60) m=- == -2 u (- 30)

Further,

m2 = 4 Object velocity along the axis is 5 cos 37° = 4 mm/s (towards OP). Therefore, image velocity along the axis should be m2 times or 16 mm/s in the opposite direction of object velocity. Object velocity perpendicular to axis is 5 sin 37° = 3 mm/s (upwards). Therefore, image velocity will be m times or - 6 mm/s downwards. The position and velocity of image is shown below. and

16 mm/s q

I 6 mm/s

vI 60 cm

vI = (16)2 + (6)2 = 292 mm/s tan q =

6 3 = 16 8

or

æ3ö q = tan -1 ç ÷ è8ø

48 — Optics and Modern Physics Type 6. To find a rough image of a square or rectangular type of object kept along the axis

Concept (i) If object is towards P or C, then its image is also towards P or C. (ii) If object is towards F, then its image is towards infinity and it is more magnified. V

Example 7 A square mnqp is kept between F and C on the principal axis of a concave mirror as shown in figure. Find a rough image of this object. Solution Object is placed between F and C, therefore, image is real,

n

q P

C

m

p

F

inverted magnified and beyond C. Further pq is towards F. Therefore, its image p¢ q¢ is towards infinity and it is more magnified. A rough image is as shown alongside.

p¢

n

q

m

p

m¢ C

F

n¢ q¢

From the ray shown in figure, we can see that n ¢ q¢ will be a straight line.

Exercise In the given problem, focal length of mirror is 30 cm and side of square is 10 cm with pP = 40 cm. Find perimeter of the image Ans (90 + 15 10 ) cm

Type 7. Two mirror problems

Concept If an object is placed between two mirrors, then infinite reflections will take place. Therefore, infinite images are formed. But normally position of second image is asked. So, we have to apply mirror formula two times. Image from first mirror acts as an object for the second mirror. Sign convention for second reflection will change but sign of focal length will not change. V

Example 8 Focal length of convex mirror M 1 is 20 cm and that of concave mirror M 2 is 30 cm. Find position of second image I2 . Take first reflection from M 1 . M2

M1

20 cm

10 cm

Chapter 30 Solution

M1

M2

+ve

+ve P1 20 cm

I2 80 cm

For M1

Reflection of Light — 49

10 cm

I1 10 cm

P2

O

u1 = - 20 cm f1 = + 20 cm

Using the mirror formula, 1 1 1 + = , we have v u f 1 1 1 + = v1 - 20 + 20 Solving, we get For M 2

v1 = + 10 cm u2 = - 40 cm f2 = - 30 cm

(PI1 = 40 cm)

Using the mirror formula,

We have, \

1 1 1 + = v u f 1 1 1 + = v2 - 40 - 30 v2 = P2I 2 = - 120 cm

Note I1 is virtual from M1 point of view (behind M1 ). But it behaves like a real object for M2 (in front of M2 ).

Type 8. An extended object is kept perpendicular to principal axis and we have to make its image

How to Solve? With the help of type 1 first find v and then m. Now, suppose m = - 2 and object is 2 mm above the principal axis, then its image will be formed 4 mm below its principal axis. V

Example 9 c b a 30 cm

Focal length of the concave mirror shown in figure is 20 cm. ab = 1 mm and

bc = 2 mm

For the given situation, make its image

50 — Optics and Modern Physics u = - 30 cm and f = - 20 cm 1 1 1, Using the mirror formula, we have + = v u f 1 1 1 + = v -30 -20 Solving this equation, we get

Solution

v = - 60 cm v (-60) m=- == -2 u (-30)

Now,

Now c point is 2 mm above the principal axis and magnification is -2. Hence, image c¢will be formed 4 mm below the principal axis. Similarly a point is 1 mm below the principal axis and value of m is -2. Hence, image a¢ is formed 2 mm above the principal axis. Image with ray diagram of c is as shown below. a¢ b¢

c b a

a¢b¢ = 2 mm F

b¢c¢ = 4 mm

c¢

Type 9. An extended object is kept along the principal axis. Now, in this type further two cases are possible. Case 1

When object size is very small.

In this case image length, LI = m 2 ( Lo ) Here, Lo is the object length. So, using type 1, we have to find v and then m. Proof We have already proved that image speed. (Along the axis) v I = m 2 vO dv 2 æ du ö or =m ç ÷ dt è dt ø For small change in the values of v and u, we can write |Dv| =|m 2 ´ Du| Dv is nothing but difference in two values of v or image length LI . Similarly, Du is object length Lo . Case 2 When object size is large. If an extended object is lying along the principal axis, then we will get two values of u corresponding to its two ends. Now, apply mirror formula two times and find two values of v. The image length now becomes, LI =|v1 ~ v 2| Note If one end of the object is placed either at C or P, then its image will also be formed at C or P. So, we will have to apply the mirror formula only for the other end.

Reflection of Light — 51

Chapter 30 V

Example 10 a

b

A small object ab of size 1 mm is kept at a distance of 40 cm from a concave mirror of focal length 30 cm. Make image of this object. Solution Here, object size is very small. So, this is case 1. u = - 40 cm f = - 30 cm

and Applying the mirror formula

1 1 1 + = , we have v u f 1 1 1 + = v - 40 - 30 Solving this equation, we get v = - 120 cm v (-120) m=- == -3 u (-40)

Further,

m2 = 9 Now, LI = m L o = (9) (1 mm) = 9 mm Image diagram is as shown below. 2

b¢ –µ

a¢

C a

b

+µ

F

9 mm 120 cm

Note Point b is towards F, therefore its image b¢ should be towards - µ. V

f is lying along the principal axis of a concave 3 mirror of focal length f. Image is real, magnified and inverted and one of the end of rod coincides with its image itself. Find length of the image.

Example 11 A thin rod of length

Solution Image is real, magnified and inverted. So, the given rod lies between F and C. Further, one end of the rod is coinciding with its image itself. Therefore, it is lying at C. So, the thin rod CR is kept as shown below.

C

f 3 PR = 2f - f 3 5f = 3 CR =

R

P

F f 2f

52 — Optics and Modern Physics We have to apply mirror formula only for point R. 5f u = - , focal length = - f 3 1 1 1 Using the mirror formula + = , v u f 1 1 1 We have, + = 5 f v -f 3 Solving this equation, we get -5 f or -2.5 f v= 2 So, the image of rod CR is C ¢ R¢ as shown below. C C¢

R¢

R

2f

0.5f 2.5 f

f So, image length = C ¢ R¢ = 0.5 f or . 2

Note In this problem if magnification of rod is asked then we can write m=-

LI f /2 3 ==LO f /3 2

Negative sign has been used for inverted image.

Type 10. Based on u versus v graph or1/u versus1/v graph (only for real objects) V

Example 12 Draw u versus v graph or

1 1 versus graph for a concave mirror of u v

focal length f. Solution The mirror formula is 1 1 1 + = v u f If we take

1 1 along y-axis and along x-axis, then the above equation becomes v u æ 1 ö ças = constant÷ f è ø

y+ x=c Therefore,

1 1 versus graph will be a straight line. Let us take origin at pole. v u –µ

+µ C

F

O

Chapter 30

Reflection of Light — 53

Table 30.5 1 u

1 v

S.No.

u

v

1.

0 to - f

- 0 to + µ

- µ to -

2.

- f to - 2 f

- µ to - 2f

1 1 - to f 2f

3.

-2f to - µ

-2f to - f

u versus v and

-

1 to 0 2f

1 1 versus graphs are as shown below. u v v

1 –2f

O

–f

u

45° –f 3 P

Q

–2f

2

Note OP line cuts the u - v graph at Q ( -2 f , - 2 f ) 1 v 1 –1 f

–1 2f 2

O

–1 2f 3 –1 f

Exercise Draw above two graphs for convex mirror.

1 f

1 u

+ µ to 0

0 to -

1 2f

1 1 to 2f f

Miscellaneous Examples V

Example 13 An object is 30.0 cm from a spherical mirror, along the central axis. 1 The absolute value of lateral magnification is . The image produced is inverted. 2 What is the focal length of the mirror? Solution Image is inverted, so it is real. u and v both are negative. Magnification is therefore, v =

u . 2

1 , 2

Given, u = – 30 cm, v = – 15 cm Using the mirror formula, 1 1 1 + = v u f 1 1 1 1 We have, = – =– f –15 30 10 \ V

f = – 10 cm

Ans.

Example 14 A concave mirror has a radius of curvature of 24 cm. How far is an object from the mirror if an image is formed that is : (a) virtual and 3.0 times the size of the object, (b) real and 3.0 times the size of the object and (c) real and 1/ 3 the size of the object? Solution Given,

R = – 24 cm

(concave mirror)

R Hence, f = = – 12 cm 2 (a) Image is virtual and 3 times larger. Hence, u is negative and v is positive. Simultaneously, | v| = 3| u |. So let, u=–x then v = + 3x Substituting in the mirror formula, 1 1 1 + = v u f We have, 1 1 1 – = 3x x –12 \ x = 8 cm Therefore, object distance is 8 cm. (b) Image is real and three times larger. Hence, u and v both are negative and| v| = 3| u |. So let, u=–x

Ans.

Chapter 30

Reflection of Light — 55

then v = – 3x Substituting in mirror formula, we have 1 1 1 – =– –3 x x 12 or x = 16 cm Ans. \ Object distance should be 16 cm. 1 |u| . (c) Image is real and rd the size of object. Hence, both u and v are negative and| v| = 3 3 So let, u=–x x then v=– 3 Substituting in the mirror formula, we have 3 1 1 – – =– x x 12 \ \ Object distance should be 48 cm. V

x = 48 cm Ans.

Example 15 A ray of light is incident on a plane mirror along a vector $i + $j – k$ . The normal on incidence point is along $i + $j . Find a unit vector along the reflected ray. Solution Reflection of a ray of light is just like an elastic collision of a ball with a horizontal ground. Component of incident ray along the inside normal gets reversed while the component perpendicular to it remains unchanged. Thus, the component of incident ray vector A = $i + $j – k$ parallel to normal, i.e. $i + $j gets reversed while perpendicular to it, i.e. – k$ remains unchanged. Thus, the reflected ray can be written as R = – i$ – $j – k$ \ A unit vector along the reflected ray will be R – $i – $j – k$ = R 3 1 $ $ $ r$ = – (i + j + k ) 3 r$ =

or

Note V

Ans.

In this problem, given normal is inside the mirror surface. Think why?

Example 16 A gun of mass m1 fires a bullet of mass m2 with a horizontal speed v0 . The gun is fitted with a concave mirror of focal length f facing towards a receding bullet. Find the speed of separations of the bullet and the image just after the gun was fired. Solution Let v1 be the speed of gun (or mirror) just after the firing of bullet. From conservation of linear momentum, or

m2v0 = m1v1 mv v1 = 2 0 m1

…(i)

56 — Optics and Modern Physics du is the rate at which distance between mirror and bullet is increasing = v1 + v0 dt We have already read in extra points that : dv æ v2 ö du =ç ÷ \ dt çè u 2 ÷ø dt Now,

…(ii)

v2 = m2 = 1 (as at the time of firing, bullet is at pole). u2 dv du …(iii) = = v1 + v0 dt dt

Here, \

m1

v0

v1 m2

dv is the rate at which distance between image (of bullet) and mirror is increasing. So, if dt v2 is the absolute velocity of image (towards right), then dv v2 – v1 = dt

Here,

= v1 + v0 or v2 = 2v1 + v0 Therefore, speed of separation of bullet and image will be

…(iv)

vr = v2 + v0 = 2v1 + v0 + v0 vr = 2 (v1 + v0 ) Substituting value of v1 from Eq. (i), we have æ m ö vr = 2 çç1 + 2 ÷÷ v0 m1 ø è

or

Ans.

Exercises LEVEL 1 Assertion and Reason Directions : Choose the correct option. (a) If both Assertion and Reason are true and the Reason is correct explanation of the Assertion. (b) If both Assertion and Reason are true but Reason is not the correct explanation of Assertion. (c) If Assertion is true, but the Reason is false. (d) If Assertion is false but the Reason is true. 1. Assertion : A convex mirror can never make a real image. Reason :

For all real objects image formed by a convex mirror is virtual.

2. Assertion : Focal length of a convex mirror is 20 cm. If a real object is placed at distance 20 cm from the mirror, its virtual erect and diminished image will be formed. Reason :

If a virtual object is placed at 20 cm distance, its image is formed at infinity.

3. Assertion : In case of a concave mirror if a point object is moving towards the mirror along its principal axis, then its image will move away from the mirror. Reason : In case of reflection (along the principal axis of mirror) object and image always travel in opposite directions.

4. Assertion : Real view mirror of vehicles is a convex mirror. Reason :

It never makes real image of real objects.

5. Assertion : If magnification of a real object is – 2. Then, it is definitely a concave mirror. Reason :

Only concave mirror can make real images of real objects.

6. Assertion : Any ray of light suffers a deviation of (180°–2i) after one reflection. Reason :

For normal incidence of light deviation is zero.

7. Assertion : Two plane mirrors kept at right angles deviate any ray of light by 180° after two reflections. Reason :

The above condition is satisfied only for angle of incidence i = 45°.

8. Assertion : In reflection from a denser medium, any ray of light suffers a phase difference of p. Reason :

Denser medium is that medium in which speed of wave is less.

9. Assertion : For real objects, image formed by a convex mirror always lies between pole and focus. Reason :

When object moves from pole to infinity, its image will move from pole to focus.

10. Assertion : Light converges on a virtual object. Reason :

Virtual object is always behind a mirror.

58 — Optics and Modern Physics Objective Questions 1. A plane mirror reflects a beam of light to form a real image. The incident beam should be (b) convergent (d) not possible

(a) parallel (c) divergent

2. When an object lies at the focus of a concave mirror, then the position of the image formed and its magnification are (a) pole and unity (c) infinity and infinity

(b) infinity and unity (d) centre of curvature and unity

3. Two plane mirrors are inclined to each other at 90°. A ray of light is incident on one mirror. The ray will undergo a total deviation of (b) 90° (d) Data insufficient

(a) 180° (c) 45°

4. A concave mirror cannot form (a) (b) (c) (d)

virtual image of virtual object virtual image of real object real image of real object real image of virtual object

5. Which of the following is correct graph between u and v for a concave mirror for normal sign convention? v

v

(a)

(b) u

u

v

(c)

v

u

(d) u

6. Two plane mirrors are inclined at 70°. A ray incident on one mirror at incidence angle q, after reflection falls on the second mirror and is reflected from there parallel to the first mirror. The value of q is (a) 50° (c) 30°

(b) 45° (d) 25°

7. The radius of curvature of a convex mirror is 60 cm. When an object is placed at A, its image is formed at B. If the size of image is half that of the object, then the distance between A and B is (a) 30 cm (c) 45 cm

(b) 60 cm (d) 90 cm

Chapter 30

Reflection of Light — 59

8. A boy of height 1.5 m with his eye level at 1.4 m stands before a plane mirror of length 0.75 m fixed on the wall. The height of the lower edge of the mirror above the floor is 0.8 m. Then, (a) the boy will see his full image (c) the boy cannot see his feet

(b) the boy cannot see his hair (d) the boy can see neither his hair nor his feet

9. A spherical mirror forms an erect image three times the size of the object. If the distance between the object and the image is 80 cm, the nature and the focal length of the mirror are (a) concave, 30 cm (c) concave, 15 cm

(b) convex, 30 cm (d) convex, 15 cm

10. A convex mirror of focal length f produces an image (1/n) th of the size of the object. The distance of the object from the mirror is (a) nf (c) (n + 1) f

(b) f/n (d) (n - 1) f

11. An object is moving towards a concave mirror of focal length 24 cm. When it is at a distance of 60 cm from the mirror, its speed is 9 cm/s. The speed of its image at that instant, is (a) 4 cm/s towards the mirror (c) 4 cm/s away from the mirror

(b) 6 cm/s towards the mirror (d) 6 cm/s away from the mirror

12. All the following statements are correct except (for real objects) (a) (b) (c) (d)

the magnification produced by a convex mirror is always less than one a virtual, erect and same sized image can be obtained using a plane mirror a virtual, erect, magnified image can be formed using a concave mirror a real, inverted same sized image can be formed using a convex mirror

13. A particle moves perpendicularly towards a plane mirror with a constant speed of 4 cm/s. What is the speed of the image observed by an observer moving with 2 cm/s along the same direction? Mirror is also moving with a speed of 10 cm/s in the opposite direction. (All speeds are with respect to ground frame of reference) (a) 4 cm/s

(b) 12 cm/s

(c) 14 cm/s

(d) 26 cm/s

Subjective Questions Note You can take approximations in the answers.

1. Figure shows two rays P and Q being reflected by a mirror and going as P¢ and Q¢. State which type of mirror is this? P Q

P¢ Q¢

2. A candle 4.85 cm tall is 39.2 cm to the left of a plane mirror. Where does the mirror form the image, and what is the height of this image?

3. A plane mirror lies face up, making an angle of 15° with the horizontal. A ray of light shines down vertically on the mirror. What is the angle of incidence? What will be the angle between the reflected ray and the horizontal?

4. Two plane mirrors are placed parallel to each other and 40 cm apart. An object is placed 10 cm from one mirror. What is the distance from the object to the image for each of the five images that are closest to the object?

60 — Optics and Modern Physics 5. If an object is placed between two parallel mirrors, an infinite number of images result. Suppose that the mirrors are a distance 2b apart and the object is put at the mid-point between the mirrors. Find the distances of the images from the object.

6. Show that a ray of light reflected from a plane mirror rotates through an angle 2 q when the mirror is rotated through an angle q about its axis perpendicular to both the incident ray and the normal to the surface.

7. Two plane mirrors each 1.6 m long, are facing each other. The distance between the mirrors is 20 cm. A light incident on one end of one of the mirrors at an angle of incidence of 30°. How many times is the ray reflected before it reaches the other end?

8. Two plane mirrors are inclined to each other at an angle q. A ray of light is reflected first at one mirror and then at the other. Find the total deviation of the ray.

9. Assume that a certain spherical mirror has a focal length of – 10.0 cm. Locate and describe the image for object distances of (a) 25.0 cm (b) 10.0 cm (c) 5.0 cm.

10. A ball is dropped from rest 3.0 m directly above the vertex of a concave mirror that has a radius of 1.0 m and lies in a horizontal plane. (a) Describe the motion of ball’s image in the mirror. (b) At what time do the ball and its image coincide?

11. An object 6.0 mm is placed 16.5 cm to the left of the vertex of a concave spherical mirror having a radius of curvature of 22.0 cm. (a) Draw principal ray diagram showing formation of the image. (b) Determine the position, size, orientation, and nature (real or virtual) of the image.

12. An object 9.0 mm tall is placed 12.0 cm to the left of the vertex of a convex spherical mirror whose radius of curvature has a magnitude of 20.0 cm. (a) Draw a principal ray diagram showing formation of the image. (b) Determine the position, size, orientation, and nature (real or virtual) of the image.

13. How far should an object be from a concave spherical mirror of radius 36 cm to form a real image one-ninth its size?

14. As the position of an object in front of a concave spherical mirror of 0.25 m focal length is varied, the position of the image varies. Plot the image distance as a function of the object distance letting the later change from 0 to + ¥. Where is the image real? Where virtual?

15. An object is placed 42 cm, in front of a concave mirror of focal length 21 cm. Light from the concave mirror is reflected onto a small plane mirror 21 cm in front of the concave mirror. Where is the final image?

16. Prove that for spherical mirrors the product of the distance of the object and the image to the principal focus is always equal to the square of the principal focal length.

17. Convex and concave mirrors have the same radii of curvature R. The distance between the mirrors is 2R. At what point on the common optical axis of the mirrors should a point source of light A be placed for the rays to coverage at the point A after being reflected first on the convex and then on the concave mirror?

18. A spherical mirror is to be used to form on a screen 5.0 m from the object an image five times the size of the object. (a) Describe the type of mirror required.

(b) Where should the mirror be positioned relative to the object?

LEVEL 2 Single Correct Option 1. An insect of negligible mass is sitting on a block of mass M, tied with a spring of force constant k. The block performs simple harmonic motion with amplitude A in front of a plane mirror as shown. The maximum speed of insect relative to its image will be q = 60° M

(a) A

k M

(b)

A 3 2

k M

(c) A 3

k M

(d) 2 A

M k

2. A plane mirror is falling vertically as shown in the figure. If S is a point source of light, the rate of increase of the length AB is L Acceleration = g m/s2 Height = x m A

S

(a) directly proportional to x (c) inversely proportional to x

B

(b) constant but not zero (d) zero

3. A point object is placed at a distance of 10 cm and its real image is formed at a distance of 20 cm from a concave mirror. If the object is moved by 0.1 cm towards the mirror. The image will shift by about (a) (b) (c) (d)

0.4 cm away from the mirror 0.4 cm towards the mirror 0.8 cm away from the mirror 0.8 cm towards the mirror

4. Two plane mirrors L1 and L2 are parallel to each other and 3 m apart. A person standing x m from the right mirror L2 looks into this mirror and sees a series of images. The distance between the first and second image is 4 m. Then, the value of x is L1

L2

w x

(a) 2 m

(b) 1.5 m

(c) 1 m

(d) 2.5 m

62 — Optics and Modern Physics 5. A piece of wire bent into an L shape with upright and horizontal portion of equal lengths 10 cm each is placed with the horizontal portion along the axis of the concave mirror towards pole of mirror whose radius of curvature is 10 cm. If the bend is 20 cm from the pole of the mirror, then the ratio of the lengths of the images of the upright and horizontal portion of the wire is (a) 1 : 2

(b) 1 : 3

(c) 1 : 1

(d) 2 : 1

6. A point object at 15 cm from a concave mirror of radius of curvature 20 cm is made to oscillate along the principal axis with amplitude 2 mm. The amplitude of its image will be (a) 2 mm (c) 8 mm

(b) 4 mm (d) None of these

7. A ray of light falls on a plane mirror. When the mirror is turned, about an axis at right angles to the plane of mirror by 20° the angle between the incident ray and new reflected ray is 45°. The angle between the incident ray and original reflected ray was therefore (a) 35° or 50° (c) 45° or 5°

(b) 25° or 65° (d) None of these

8. A person AB of height 170 cm is standing in front of a plane mirror. His eyes are at

A

height 164 cm. At what distance from P should a hole be made in mirror so that he cannot see his hair? (a) 167 cm (c) 163 cm

(b) 161 cm (d) 165 cm B

P

9. Two blocks each of masses m lie on a smooth table. They are attached to two other masses as shown in figure. The pulleys and strings are light. An object O is kept at rest on the table. The sides AB and CD of the two blocks are made reflecting. The acceleration of two images formed in those two reflecting surfaces with respect to each other is m AOC m B

D

3m

(a)

5g 6

(b)

2m

5g 3

(c)

17 g 12

(d)

17 g 6 2 Ö3 m

in the figure. A light ray is incident at an angle of 30° at a point just inside one end of A. The number of times the ray undergoes reflections (including the first one) before it emerges out is (a) 29 (c) 31

(b) 30 (d) 32

0.2m

10. Two plane mirrors A and B are aligned parallel to each other as shown

B

30°

A

11. An object O is just about to strike a perfectly reflecting inclined plane of inclination 37°. Its velocity is 5 m/s. Find the velocity of its image. 5 m/s

37°

(a) 3$i + 4$j

(b) 4$i + 3$j

y

x

(c) 4.8 $i + 1.4$j

(d) 1.4 $i + 4.8$j

Chapter 30

Reflection of Light — 63

12. An elevator at rest which is at 10th floor of a building is having a plane mirror fixed to its floor. A particle is projected with a speed 2 m/s and at 45° with the horizontal as shown in the figure. At the very instant of projection, the cable of the elevator breaks and the elevator starts falling freely. What will be the separation between the particles and is image 0.5 s after the instant of projection?

u = Ö 2 m/s 45° Mirror

(b) 1 m (d) 1.5 m

(a) 0.5 m (c) 2 m

13. A plane mirror is moving with velocity 4$i + 4$j + 8k$ . A point object in front of the mirror moves with a velocity 3$i + 4$j + 5k$ . Here, k$ is along the normal to the plane mirror and facing towards the object. The velocity of the image is

(a) - 3$i - 4$j + 5k$ (c) -4$i + 5$j + 11k$

$ (b) 3$i + 4$j + 11 k $ $ $ (d) 7i + 9 j + 3k

14. Point A (0, 1 cm) and B (12 cm, 5 cm) are the coordinates of object and image. x-axis is the principal axis of the mirror. Then, this object image pair is (a) (b) (c) (d)

due to a convex mirror of focal length 2.5 cm due to a concave mirror having its pole at (2 cm, 0) due to a concave mirror having its pole at (- 2 cm, 0) Data is insufficient

15. Two plane mirrors AB and AC are inclined at an angle q = 20°. A ray of light starting from point P is incident at point Q on the mirror AB, then at R on mirror AC and again on S on AB. Finally, the ray ST goes parallel to mirror AC. The angle which the ray makes with the normal at point Q on mirror AB is S B

T

Q

A

(a) 20°

q i P

(b) 30°

C

R

(c) 40°

(d) 60°

16. A convex mirror of radius of curvature 20 cm is shown in figure. An object O is placed in front of this mirror. Its ray diagram is shown. How many mistakes are there in the ray diagram (AB is principal axis) O

C A 10 cm

(a) 3

(b) 2

B

20 cm

(c) 1

(d) 0

64 — Optics and Modern Physics More than One Correct Options 1. The image formed by a concave mirror is twice the size of the object. The focal length of the mirror is 20 cm. The distance of the object from the mirror is/are (b) 30 cm (d) 15 cm

(a) 10 cm (c) 25 cm

1 2

2. Magnitude of focal length of a spherical mirror is f and magnitude of linear magnification is . (a) (b) (c) (d)

If image is inverted, it is a concave mirror If image is erect, it is a convex mirror Object distance from the mirror may be 3 f Object distance from the mirror may be f

3. A point object is moving towards a plane mirror as shown in figure. Choose the correct options. (a) (b) (c) (d)

v

Speed of image is also v Image velocity will also make an angle q with mirror Relative velocity between object and image is 2v Relative velocity between object and image is 2v sin q

q

4. AB is the principal axis of a spherical mirror. I is the point image corresponding to a point object O. Choose the correct options. I B

A O

(a) (b) (c) (d)

Mirror is lying to the right hand side of O Focus of mirror is lying to the right hand side of O Centre of curvature of mirror is lying to the right hand side of O Centre of curvature of mirror is lying between I and O

5. A point object is placed on the principal axis of a concave mirror of focal length 20 cm. At this instant object is given a velocity v towards the axis O (event-1) or perpendicular to axis (event-2). Then, speed of image (a) (b) (c) (d)

In event-1 is 2v In event-1 is 4v In event-2 is 2v In event-2 is 4v

30 cm

6. A point object is placed at equal distance 3f in front of a concave mirror, a convex mirror and a plane mirror separately (event-1). Now, the distance is decreased to 1.5 f from all the three mirrors (event-2). Magnitude of focal length of convex mirror and concave mirror is f. Then, choose the correct options. (a) Maximum distance of object in event-1 from the mirror is from plane mirror (b) Minimum distance of object in event-1 from the mirror is from convex mirror (c) Maximum distance of object in event-2 from the mirror is from concave mirror (d) Minimum distance of object in event-2 from the mirror is from plane mirror

Chapter 30

Reflection of Light — 65

Comprehension Based Questions Passage : (Q. No. 1 to 4) A plane mirror ( M1 ) and a concave mirror ( M 2 ) of focal length 10 cm are arranged as shown in figure. An object is kept at origin. Answer the following questions. (consider image formed by single reflection in all cases) y

M2 20 cm 10 cm 45°

O

x

M1

1. The coordinates of image formed by plane mirror are (a) (- 20 cm, 0)

(b) (10 cm, - 60 cm)

(c) (10 cm, - 10 cm)

(d) (10 cm, 10 cm)

2. The coordinates of image formed by concave mirror are (a) (10 cm, – 40 cm)

(b) (10 cm, – 60 cm)

(c) (10 cm, 8 cm)

(d) None of these

3. If concave mirror is replaced by convex mirror of same focal length, then coordinates of image formed by M 2 will be (a) (10 cm, 12 cm)

(b) (10 cm, 22 cm)

(c) (10 cm, 8 cm)

(d) None of these

4. If concave mirror is replaced by another plane mirror parallel to x-axis, then coordinates of image formed by M 2 are (a) (40 cm, 20 cm)

(b) (20 cm, 40 cm)

(c) (– 20 cm, 20 cm)

(d) None of these

Match the Columns 1. For real objects, match the following two columns corresponding to linear magnification m given in Column I.

(a) (b) (c) (d)

Column I

Column II

m = -2 1 m=2 m=+2 1 m=+ 2

(p) convex mirror (q) concave mirror (r) real image (s) virtual image

2. For virtual objects, match the following two columns. Column I

Column II

(a) Plane mirror

(p) only real image

(b) Convex mirror

(q) only virtual image

(c) Concave mirror

(r) may be real or virtual image

66 — Optics and Modern Physics 3. Principal axis of a mirror ( AB), a point object O and its image I are shown in Column I, match it with Column II. Column I

Column II

O

(a)

(p) plane mirror

A

(b)

I

B

O A

B

(q) convex mirror

I

I

O

(c) A

(d)

(r) concave mirror B

I

O

(s) Not possible

4. Focal length of a concave mirror M1 is – 20 cm and focal length of a convex mirror M 2 is + 20 cm. A point object is placed at a distance X in front of M1 or M 2. Match the following two columns. Column I

Column II

(a) X = 20, mirror is M1

(p)

image is at infinity

(b) X = 20, mirror is M 2

(q)

image is real

(c) X = 30, mirror is M1

(r)

image is virtual

(d) X = 30, mirror is M 2

(s)

image is magnified

5. Focal length of a concave mirror is –20 cm. Match the object distance given in Column II corresponding to magnification (only magnitude) given is Column I. Column I

Column II

(a)

2

(p) 10 cm

(b)

1/2

(q) 30 cm

(c)

1

(r) 20 cm

(d)

1/4

(s) None of these

Subjective Questions 1. A point source of light S is placed at a distance 10 cm in front of the centre of a mirror of width 20 cm suspended vertically on a wall. An 20 cm insect walks with a speed 10 cm/ s in front of the mirror along a line parallel to the mirror at a distance 20 cm from it as shown in figure. Find the maximum time during which the insect can see the image of the source S in the mirror.

S 10 cm 20 cm

Chapter 30

Reflection of Light — 67

2. A concave mirror forms the real image of a point source lying on the optical axis at a distance of 50 cm from the mirror. The focal length of the mirror is 25 cm. The mirror is cut into and its halves are drawn a distance of 1 cm apart in a direction perpendicular to the optical axis. How will the image formed by the halves of the mirror be arranged?

1 cm

3. A point source of light S is placed on the major optical axis of the concave mirror at a distance of 60 cm. At what distance from the concave mirror should a flat mirror be placed for the rays to converge again at the point S having been reflected from the concave mirror and then from the flat one? Will the position of the point where the rays meet change if they are first reflected from the flat mirror? The radius of the concave mirror is 80 cm.

4. A balloon is moving upwards with a speed of 20 m/ s. When it is at a height of 14 m from ground in front of a plane mirror in situation as shown in figure, a boy drops himself from the balloon. Find the time duration for which he will see the image of source S placed symmetrically before plane mirror during free fall. 20 m/s

5m

2m

50 cm

S

12 m

Ground

5. A plane mirror and a concave mirror are arranged as shown in figure and O is a point object. Find the position of image formed by two reflections, first one taking place at concave mirror. R = 200 cm A

O

45°

B 110 cm

890 cm

6. Figure shows a torch producing a straight light beam falling on a plane mirror at an angle 60°. The reflected beam makes a spot P on the screen along y-axis. If at t = 0, mirror starts rotating about the hinge A with an angular velocity w = 1° per second clockwise. Find the speed of the spot on screen after time t = 15 s. d=3m

A 60° P

68 — Optics and Modern Physics 7. A thief is running away in a car with velocity of 20 m/ s. A police jeep is following him, which is sighted by thief in his rear view mirror, which is a convex mirror of focal length 10 m. He observes that the image of jeep is moving towards him with a velocity of 1 cm / s. If the 1 magnification of mirror for the jeep at that time is . Find 10 (a) the actual speed of jeep, (b) rate at which magnification is changing. Assume the police’s jeep is on the axis of the mirror.

8. A ball swings back and forth in front of a concave mirror. The motion of the ball is described approximately by the equation x = f cos wt, where f is the focal length of the mirror and x is measured along the axis of mirror. The origin is taken at the centre of curvature of the mirror.

C

x-axis

x=0

(a) Derive an expression for the distance from the mirror of the image of the swinging ball. (b) At what point does the ball appear to coincide with its image? T (c) What will be the lateral magnification of the image of the ball at time t = , where T is time 2 period of oscillation?

9. Show that a parallel bundle of light rays parallel to the x-axis and incident on a parabolic reflecting surface given by x = 2 by 2, will pass through a single point called focus of the reflecting surface. Also, find the focal length. y

O

F

x

Answers Introductory Exercise 30.1 1. 4 m/s

2.

20 40 m, m 3 3

3. 2 q

Exercises LEVEL 1 Assertion and Reason 1. (d)

2. (b)

3. (a)

4. (b)

5. (a or b)

6. (c)

7. (c)

8. (b)

9. (a,b)

6. (a)

7. (c)

8. (c)

9. (a)

10. (b)

Objective Questions 1. (b) 11. (c)

2. (c) 12. (d)

3. (a) 13. (d)

4. (a)

5. (b)

Subjective Questions 1. Plane mirror 2. 39.2 cm to the right of mirror, 4.85 cm 3. 15°, 60° 4. 5. 7. 8.

20 cm, 60 cm, 80 cm, 100 cm, 140 cm The images are at 2nb from the object with n as integer. 14 360° - 2 q

9. (a) – 16.7 cm, real (b) ¥

(c) + 10.0 cm, virtual

10. (a) A real image moves from – 0.6 m to – ¥, then a virtual image moves from + ¥ to 0. (b) 0.639 s and 0.782 s. 11. (b) 33.0 cm to the left of vertex 1.20 cm tall, inverted, real 12. (b) 5.46 cm to the right of vertex, 4.09 mm tall, erect, virtual 13. 180 cm V (m)

14.

0.5 0.25 0.25

0.5

u (m)

Image is virtual when object distance is from 0 to 0.25 m 15. 21 cm in front of plane mirror æ 3 + 1ö ÷ R from convex mirror 17. At a distance ç ç 2 ÷ø è 18. (a) A concave mirror with radius of curvature 2.08 m

(b) 1.25 m from the object

10. (d)

70 — Optics and Modern Physics

LEVEL 2 Single Correct Option 1. (c) 11. (c)

2. (d) 12. (b)

3. (a) 13. (b)

4. (c) 14. (b)

5. (c) 15. (b)

6. (c) 16. (b)

7. (d)

8. (a)

9. (c)

More than One Correct Options 1. (a,b)

2. (a,b,c,d)

3. (a,b,d)

4. (a,b,d)

5. (b,c)

6. (a,b,c)

Comprehension Based Questions 1.(c)

2.(d)

3.(d)

4.(d)

Match the Columns 1. (a) ® q,r

(b) ® q,r

(c) ® q,s

(d) ® p,s

2. (a) ® p

(b) ® r

(c) ® p

3. (a) ® r

(b) ® r

(c) ® p

(d) ® r

4. (a) ® p,s

(b) ® r

(c) ® q,s

(d) ® r

5. (a) ® p,q

(b) ® s

(c) ® s

(d) ® s

Subjective Questions 1. 6 s

2. At a distance of 50 cm from mirror and 2 cm from each other 2p 3. 90 cm, Yes 4. 1.7 s 5. 100 cm vertically below A 6. m /s 15 æ 2 + cos wt ö ÷ f (b) At x = 0 (c) m = ¥ 8. (a) Distance = çç 7. (a) 21 m /s (b) 10 –3 /s ÷ è 1 + cos wt ø 1 9. f = 8b

10. (c)

31.1

Refraction of light and Refractive Index of a medium

31.2

Law of refraction (Snell's law)

31.3

Single refraction from plane surface

31.4

Shift due to a glass slab

31.5

Refraction from spherical surface

31.6

Lens Theory

31.7

Total Internal Reflection (TIR)

31.8

Refraction through prism

31.9

Deviation

31.10 Optical Instruments

72 — Optics and Modern Physics

31.1 Refraction of Light and Refractive Index of a Medium When a ray of light travels from one medium to other medium with or without bending, the phenomenon is called refraction of light. Under following two conditions the ray of light does not bend in refraction. (i) For normal incidence ( Ð i = 0)

1 2

Fig. 31.1

(ii) If refractive index of both media is same, angle of incidence does not matter in this case.

1 2 m1 = m 2 Fig. 31.2

Here, m = refractive index of medium

Refractive Index (i) In general speed of light in any medium is less than its speed in vacuum. It is convenient to define refractive index m of a medium as, m=

Speed of light in vacuum c = Speed of light in medium v

(ii) As a ray of light travels from medium 1 to medium 2, its wavelength changes but its frequency remains unchanged. m 2 > m 1 , v1 > v 2 , l 1 > l 2 1

2

l1

l2

Fig. 31.3

Chapter 31 (iii) 1 m 2 =

m2 = Refractive index of 2 w.r.t. 1 m1

and

2m1

1m 2

\ (iv) 1 m 2 =

m2 , m1

2m 3

=

m3 m2

=

m1 = Refractive index of 1 w.r.t. 2 m2

=

1 2m1

and

3m1

\

=

1 m2

(v) v =

c m

Refraction of Light — 73

and l =

m1 m3 ´ 2 m 3 ´ 3 m1 = 1

l0 m

Here, l is the wavelength in a medium and l 0 the wavelength in vacuum. Thus, in travelling from vacuum to a medium speed and wavelength decrease m times but frequency remains unchanged. m c/ v 2 v1 f l1 l1 (vi) 1 m 2 = 2 = = = = m 1 c/ v1 v 2 f l 2 l 2 Here, f = frequency of light which remains same in both media. Thus, V

1m 2

m 2 v1 l 1 = = m1 v2 l 2

Example 31.1 (a) Find the speed of light of wavelength l = 780 nm (in air) in a medium of refractive index m = 1.55. (b) What is the wavelength of this light in the given medium? Solution

(a) v =

(b) l medium = V

=

c 3.0 ´ 108 = = 1.94 ´ 108 m/s m 1.55

Ans.

l air 780 = = 503 nm m 1.55

Ans.

Example 31.2 Refractive index of glass with respect to water is (9 /8) . Refractive index of glass with respect to air is ( 3 /2) . Find the refractive index of water with respect to air. Solution

As, \ \

Given,

wmg

= 9/ 8

and amg

amg

= 3/ 2

´ g m w ´ wm a = 1

1 amg = am w = am g ´ gm w = wma wmg amw

=

3/ 2 4 = 9/ 8 3

Ans.

74 — Optics and Modern Physics V

Example 31.3 A ray of light passes through two slabs of same thickness. In the first slab n 1 waves are formed and in the second slab n 2 . Find refractive index of second medium with respect to first. Solution One wave means one wavelength. So, if t is the thickness of slab, l the wavelength and n the number of waves, then t nl = t Þ l = n l1 n 2 1 or (as t is same) or lµ = n l 2 n1 Now, refractive index of second medium w.r.t. first medium is l1 n 2 = 1m 2 = l 2 n1

INTRODUCTORY EXERCISE 1. Given that 1 m 2 = 4 /3, 2 m

3

Ans.

31.1

= 3 /2. Find 1 m 3.

2. What happens to the frequency, wavelengths and speed of light that crosses from a medium with index of refraction m 1 to one with index of refraction m 2?

3. A monochromatic light beam of frequency 6.0 ´ 1014 Hz crosses from air into a transparent material where its wavelength is measured to be 300 nm. What is the index of refraction of the material?

31.2 Law of Refraction (Snell’s Law) 1 2

i1

m1 m2 i2

Fig. 31.4

If a ray of light passes through one medium to other medium, then according to Snell’s law, m sin i = constant For two media, or

…(i)

m 1 sin i1 = m 2 sin i2 m 2 sin i1 = = 1m 2 m 1 sin i2

…(ii)

From Eq. (i) we can see that i1 > i2 if m 2 > m 1 , i.e. if a ray of light passes from a rarer to a denser medium, it bends towards normal.

Refraction of Light — 75

Chapter 31 Eq. (ii) can be written as, 1m 2

=

sin i1 v1 l 1 m 2 = = = sin i2 v 2 l 2 m 1

…(iii)

Here, v1 is the speed of light in medium 1 and v 2 in medium 2. Similarly, l 1 and l 2 are the corresponding wavelengths.

i1

m1

Rarer

m2

Denser i2

Fig. 31.5

If m 2 > m 1 then v1 > v 2 and l 1 > l 2 , i.e. in a rarer medium speed and hence, wavelength of light is more.

i1

i1

1

Rarer

1

2

Denser

2

Denser Rarer

i2

i2

i1 < i 2 v2 > v1 m2 < m1 l2 > l1

i1 > i2 v2 < v1 m2 > m1 l2 < l1 Fig. 31.6

D

Experiments show that if the boundaries of the media are parallel, the emergent ray CD although laterally displaced, is parallel to the incident ray AB if m 1 = m 5 . We can also directly apply the Snell’s law (m sin i = constant) in medium 1 and 5, i.e.

i5 C

m 1 sin i1 = m 5 sin i5 So,

B

i1 = i5 if m 1 = m 5

If any of the boundary is not parallel we cannot use this law directly by jumping the intervening media.

i1

A 1

2

3 Fig. 31.7

4

5

76 — Optics and Modern Physics Extra Points to Remember ˜

m=

sin i is a special case of Snell’s law when one medium is air. sin r

In Fig. 31.8, if we apply the Snell’s law in original form then it is m air sin iair = m medium sin imedium or

˜

Medium

(1) sin i = (m ) sin r m=

\

i

Air

sin i sin r

r

Fig. 31.8

sin i In m = , angle i is not always the angle of incidence but it is the angle of sin r ray of light in air (with normal).

i

Air Medium r

Fig. 31.9

In the above figure, ray of light is travelling from medium to air. So, angle of incidence is actually r. But we sin i have to take i angle in air and now we can apply m = . sin r ˜

In m =

sin i sin i , if i is changed, then r angle also changes. But remains constant and this constant is sin r sin r

called refractive index of that medium. ˜

V

sin i can be applied for any pair of angles i and r except the normal incidence for which Ð i = Ð r = 0° sin r sin i is an indeterminant form. and m = sin r

m=

Example 31.4 A light beam passes from medium 1 to medium 2. Show that the emerging beam is parallel to the incident beam. Solution Applying Snell’s law at A, m 1 sin i1 = m 2 sin i 2 or Similarly at B, \ From Eqs. (i) and (ii), we have

m 1 sin i 2 = m 2 sin i1

i3

i2

…(i)

A i1

B

i2

m 2 sin i 2 = m 1 sin i 3 m 1 sin i 2 = m 2 sin i 3

…(ii)

m1

m2

m1

Fig. 31.10

i 3 = i1

i.e. the emergent ray is parallel to incident ray.

Proved.

Chapter 31

Problems Based on m = V

Refraction of Light — 77

sin i sin r

Example 31.5 A ray of light falls on a glass plate of refractive index m = 1.5. What is the angle of incidence of the ray if the angle between the reflected and refracted rays is 90°? Solution

In the figure, r = 90° – i Reflected

Incident

i

i

90° - i i r

Refracted Fig. 31.11

From Snell’s law,

1.5 =

i = tan –1 (1.5) = 56.3°

\ V

sin i sin i = = tan i sin r sin ( 90° – i ) Ans.

Example 31.6 A pile 4 m high driven into the bottom of a lake is 1 m above the water. Determine the length of the shadow of the pile on the bottom of the lake if the sun rays make an angle of 45° with the water surface. The refractive index of water is 4 /3. 4 sin 45° Solution From Snell’s law, = 3 sin r A 45 45

°

1m

°

D

1m

B

r 3m

C

Solving this equation, we get Further, \ Total length of shadow,

E Fig. 31.12

F

r = 32° EF = ( DE ) tan r = ( 3) tan 32° = 1.88 m L = CF or L = (1 + 1.88) m = 2.88 m

Ans.

78 — Optics and Modern Physics V

Example 31.7 An observer can see through a pin-hole the top end of a thin rod of height h, placed as shown in the figure. The beaker height is 3h and its radius h. When the beaker is filled with a liquid up to a height 2h, he can see the lower end of the rod. Then, the refractive index of the liquid is (JEE 2002)

3h

h 2h Fig. 31.13

(a)

5 2

Solution

(b)

5 2

3 2

(c)

(d)

3 2 P

PQ = QR = 2h Ði = 45° ST = RT = h = KM = MN

\ \

2

So, \

h

i

sin r =

=

h 5 \

2h R

1 5

K

sin i sin 45° 5 m= = = sin r 1/ 5 2

\ The correct answer is (b).

INTRODUCTORY EXERCISE

31.2

1. In the figure shown, find 1m 2. 60° 1 2 30° Fig. 31.15

l 2. If 1m 2 is 1.5, then find the value of 1 . l2

2h

S

2

KS = h + ( 2h ) = h 5 h

i

i

r T M 2h

Fig. 31.14

Q N

Chapter 31

Refraction of Light — 79

31.3 Single Refraction from Plane Surface (i) If ray of light travels from first medium (refractive index m 1 ) to second medium (refractive index m 2 ), then image distance (or v ) from the plane surface is given by m v= 2 u m1 (ii) From this equation, we can see that v and u are of same sign. This implies that object and image lie on same side of the plane surface. If one is real, then the other is virtual. (iii) If ray of light travels from denser medium to rarer medium (or m 1 > m 2 ), then we can see that v < u) or the image distance is less than the object distance. If the light travels from rarer to denser medium (m 1 < m 2 ), then v > u or the image distance is greater than the object distance. Further, if rarer medium is air (or vacuum), then this decrease or increase in image distance will be m times. (iv) In all the four figures, single refraction is taking place through a plane surface. Refractive index of medium (may be glass, water etc.) is m. In figures (a) and (d), the ray of light is travelling from denser to rarer medium and hence, it bends away from the normal. In figures (b) and (c), the ray of light is travelling from a rarer to a denser medium and hence, it bends towards the normal. Now, let us take the four figures individually. 1

D i

Air Medium

x m

A

Air

B

i

A

Medium

r

x

I

x

2

C

mx

O r

O

I

(a)

(b)

I O

mx

O I

x Air A

x

x m

Air A

Medium 1

(c)

Medium 2

(d) Fig. 31.16

Refer figure (a) Object O is placed at a distance x from A. Ray OA, which falls normally on the plane surface, passes undeviated as AD. Ray OB, which falls at angle r (with the normal) on the

80 — Optics and Modern Physics plane surface, bends away from the normal and passes as BC in air. Rays AD and BC meet at I after extending these two rays backwards. Let BC makes an angle i (> r) with normal. In the figure, Ð AOB will be r and Ð AIB is i. For normal incidence (i.e. small angles of i and r) AB …(i) sin i » tan i = AI AB …(ii) and sin r » tan r = AO Dividing Eq. (i) by Eq. (ii), we have sin i AO AO sin i æ ö or m = = =m ÷ ç as sin r AI AI sin r è ø AO x \ AI = = m m If point O is at a depth of d from a water surface, then the above result is also sometimes written as, d d apparent = actual m or the apparent depth is m times less than the actual depth. Refer figure (b) In the absence of the plane refracting surface, the two rays 1 and 2 would have met at O. Proceeding in the similar manner we can prove that after refraction from the plane surface, they will now meet at a point I, where (if AO = x) AI = mx Refer figure (c) In this case object is at O, a distance AO = x from the plane surface. When seen from inside the medium, it will appear at I, where AI = mx If point O is at a height of h from the water surface, then the above relation is also written as happ = mh Refer figure (d) The two rays 1 and 2 meeting at O will now meet at I after refraction from the AO x plane surface, where AI = = . m m Note In all the four cases, the change in the value of x is m times whether it is increasing or decreasing. All the relations can be derived for small angles of incidence as done in part (a).

Three immiscible liquids of refractive indices m 1 , m 2 and m 3 (with m 3 > m 2 > m 1 ) are filled in a vessel. Their depths are d1 , d 2 and d 3 respectively. Prove that the apparent depth (for almost normal incidence) when seen from top of the first liquid will be

EXERCISE

d app =

d1 d 2 d 3 + + m1 m 2 m 3

m1

d1

m2

d2

m3

d3

Fig. 31.17

Chapter 31 V

Refraction of Light — 81

Example 31.8 In Fig. 31.16, find position of second image I2 formed after two times refraction from two plane surfaces AB and CD. A C m1 =1 m2 =1.5 m3 =2 O

E 10 cm

F 10 cm

B D Fig. 31.18

Solution

We will apply v =

m2 u, two times with using the fact that object and image are on m1

same side of the surface. Refraction from AB m1 =1 m 2 = 1.5 u = EO = 10cm \

æm v = çç 2 è m1

or

v = EI 1 = 15cm

(towards left of AB)

ö 1.5 ÷÷ u = (10) 1 ø (towards left of AB)

Refraction from CD I 1 will act as an object for refraction from CD m 1 = 1.5 and m 2 = 2 u = FI 1 = FE + EI 1 = (10 + 15) cm = 25cm æm ö æ 2 ö v = çç 2 ÷÷ ( u ) = ç ÷ ( 25) m è 1.5 ø è 1ø 100 = cm 3

\

(towards left of CD)

(towards left of CD)

= FI 2 The correct figure is as shown below A

C

10 cm

I2

I1

O

E

F

15 cm 100 —— cm B 3 Fig. 31.19

D

Note I1 and I2 both are virtual as the light has moved towards right of AB and CD (because it is a refraction) but I1 and I2 are towards left of AB or CD.

82 — Optics and Modern Physics V

Example 31.9 Refractive index of the glass slab is 1.5. There is a point object O inside the slab as shown. To eye E1 object appears at a distance of 6 cm (from the top surface) and to eye E2 it appears at a distance of 8 cm (from the bottom surface). Find thickness of the glass slab. E1 d1

O

d2

E2 Fig. 31.20

Solution

Applying d app =

d m d = (m ) d app d1 = (1.5)( 6) = 9 cm d 2 = (1.5)( 8) = 12 cm

\

Therefore, actual thickness of the glass slab is d1 + d 2 = 21cm.

INTRODUCTORY EXERCISE

Ans.

31.3

1. In the figure shown, at what distance E1 E2

10 cm 10 cm m = 1.5

Fig. 31.21

(a) E2 will appear to E1 (b) E1 will appear to E2

31.4 Shift due to a Glass Slab (i) It is a case of double refraction from two plane surfaces. So, we are talking about the second (or final) image and let us call it I. (ii) If O is real then second image I is virtual and vice-versa. æ 1ö (iii) I is shifted (w.r.t.) O by a distance OI = shift = çç 1 - ÷÷ t in the direction of ray of light. m è ø (iv) If E1 observes E 2 , then E 2 is object. So, light travels from E 2 towards E1 . So, shift is also in the same direction or E1 will observe second image of E 2 at a distance d1 = d - shift

Refraction of Light — 83

Chapter 31 Same is the case when E 2 observes E1 . E1

E2

d Fig. 31.22

(v) If two or more than two slabs are kept jointly or separately, then total shift is added. æ 1 S Total = S 1 + S 2 = çç 1 è m1

\ (vi)

C

E

ö æ 1 ÷÷ t1 + çç 1 ø è m2

M

C

ö ÷÷ t 2 ø E

N N

M

I1

O

I

A

P

B

A

m D

B

O

I

m D

F t

F t

(a)

(b) Fig. 31.23

Refer figure (a) An object is placed at O. Plane surface CD forms its image (virtual) at I 1 . This image acts as an object for EF which finally forms the image (virtual) at I. Distance OI is called the normal shift and its value is æ 1ö OI = çç 1 – ÷÷ t è mø This can be proved as under Let then

\

OA = x AI 1 = mx BI 1 = mx + t BI t BI = 1 = x + m m OI = ( AB + OA ) – BI æ t ö æ 1ö = ( t + x ) – çç x + ÷÷ = çç 1 – ÷÷ t m m è ø è ø

Note For two refractions (at CD and EF) we have used, æm ö v = çç 2 ÷÷ m èm1 ø

(Refraction from CD) (Refraction from EF)

Hence Proved.

84 — Optics and Modern Physics Refer figure (b) The ray of light which would have met line AB at O will now meet this line at I after two times refraction from the slab. Here, æ 1ö OI = çç 1 – ÷÷ t è mø V

Example 31.10 Refractive index of glass slab shown in figure is 1.5. Focal length of mirror is 20 cm. Find (a) total number of refractions and reflections before final image is formed. (b) reduced steps. (c) value of x, so that final image coincides with the object. Solution (a) There are total four refractions and one reflection. (b) Reduced steps are three, first slab, then mirror and then again slab. (c) Shift due to the slab,

9 cm O

x

10 cm

Fig. 31.24

æ 1 ö æ 2ö s = çç 1 - ÷÷ t = ç 1 - ÷ ( 9) = 3cm è m ø è 3ø Actual distance from mirror to object is (19 + x ) cm. Slab will reduce this distance by 3 cm. So, apparent distance will be (16 + x ) cm. Now, if 16 + x = R = 2 f = 40 cm or x = 24 cm then ray of light will fall normal to the concave mirror. It will retrace its path and final image will coincide with the object x = 24 cm

\

Ans.

Note If ray of light falls normal to a mirror, then there is no need of applying the slab formula in return journey of ray of light. Path is retracing means, slab formula is automatically applied in return journey. But if it is not normal, then we will have to apply the slab formula in return journey too. V

Example 31.11 A point object O is placed in front of a concave mirror of focal length 10 cm. A glass slab of refractive index m = 3 /2 and thickness 6 cm is inserted between object and mirror. Find the position of final image when the distance x shown in figure is 6 cm

O

x 32 cm Fig. 31.25

(a) 5 cm

(b) 20 cm

Chapter 31 Solution

Refraction of Light — 85

As we have read in the above article, the normal shift produced by a glass slab is æ 1ö æ 2ö Dx = çç 1 – ÷÷ t = ç 1 – ÷ ( 6) = 2 cm 3ø è mø è

i.e. for the mirror the object is placed at a distance ( 32 – Dx ) = 30 cm from it. Applying mirror formula 1 1 1 1 1 1 or or v = – 15cm + = – =– v u f v 30 10 (a) When x = 5 cm The light falls on the slab on its return journey as shown. But the slab will again shift it by a distance Dx = 2cm. Hence, the final real image is formed at a distance (15 + 2) = 17 cm from the mirror.

I Dx

15 cm Fig. 31.26

(b) When x = 20 cm This time also the final image is at a distance 17 cm from the mirror but it is virtual as shown. 15 cm

I Dx

Fig. 31.27

INTRODUCTORY EXERCISE

31.4

1. At what distance eye E will observe the fourth image (after four refractions from plane surfaces) of object O from itself. 10 cm

E

m1 =1.5

10 cm 10 cm

m2 = 2

10 cm 10 cm O Fig. 31.28

86 — Optics and Modern Physics

31.5 Refraction from Spherical Surface (i) There are two types of spherical surfaces, concave and convex. (ii) +ve 1 2

R = – ve u = – ve

1

O

2

R = +ve u = –ve Fig. 31.29

If the ray of light is travelling from first medium to second medium, then for image distance v, we have the formula m 2 m1 m 2 - m1 = v u R (iii) For plane surface R = µ. Putting this value in the above formula, we get æm ö v = çç 2 ÷÷ u è m1 ø and this formula, we have already discussed in article 31.3. (iv) Ray of light has moved in medium-2, so image formed in medium-2 will be real and v in this medium will be positive.

Proof +ve Consider two transparent media having indices of refraction m 1 and m 2 , where the boundary between the q1 P two media is a spherical surface of radius R. We d q2 b a assume that m 1 < m 2 . Let us consider a single ray M C O leaving point O and focusing at point I. Snell’s law applied to this refracted ray gives, m1 m2 m 1 sin q 1 = m 2 sin q 2 u R Because q 1 and q 2 are assumed to be small, we can use the small angle approximation Fig. 31.30 (angles in radians) sin q » q and say that m 1q 1 = m 2 q 2 From the geometry shown in the figure, q1 = a + b and b = q2 + g The above three equations can be rearranged as, m b = 1 (a + b) + g m2 So, m 1a + m 2 g = (m 2 – m 1 ) b

g

I

v

…(i) …(ii) …(iii)

…(iv)

Refraction of Light — 87

Chapter 31

Since, the arc PM (of length s) subtends an angle b at the centre of curvature, s b= R s s Also in the paraxial approximation, a = and g = u v Using these expressions in Eq. (iv) with proper signs, we are left with, m1 m 2 m 2 – m1 m 2 m1 m 2 – m1 or + = – = –u v R v u R

…(v)

Although the formula (v) is derived for a particular situation, it is valid for all other situations of refraction at a single spherical surface. V

Example 31.12 A glass sphere of radius R = 10 cm is kept inside water. A point object O is placed at 20 cm from A as shown in figure. Find the position and nature of the image when seen from other side of the sphere. Also draw the ray diagram. Given, m g = 3 /2 and m w = 4 /3.

C A

O

B 10 cm

20 cm Fig. 31.31

A ray of light starting from O gets refracted twice. The ray of light is travelling in a direction from left to right. Hence, the distances measured in this direction are taken positive. m m m – m1 , twice with proper signs. We have, Applying 2 – 1 = 2 v u R 3/ 2 4 / 3 3/ 2 – 4 / 3 or AI 1 = – 30 cm – = AI 1 –20 10

Solution

Now, the first image I 1 acts as an object for the second surface, where BI 1 = u = – ( 30 + 20) = – 50 cm 4 / 3 3/ 2 4 / 3 – 3/ 2 – = BI 2 –50 – 10

\

N

M P

I2

I1

A

O

C

B

20 cm 30 cm 100 cm Fig. 31.32

\ BI 2 = – 100 cm, i.e. the final image I 2 is virtual and is formed at a distance 100 cm (towards left) from B. The ray diagram is as shown in Fig. 31.32

88 — Optics and Modern Physics Following points should be noted while drawing the ray diagram. (i) At P the ray travels from rarer to a denser medium. Hence, it will bend towards normal PC. At M, it travels from a denser to a rarer medium, hence, it moves away from the normal MC. (ii) PM ray when extended backwards meets at I 1 and MN ray when extended meets at I 2 .

INTRODUCTORY EXERCISE

31.5

1. If an object is placed at the centre of a glass sphere and it is seen from outside, then prove that its virtual image is also formed at centre.

2. A glass sphere (m = 1.5) with a radius of 15.0 cm has a tiny air bubble 5 cm above its centre. The sphere is viewed looking down along the extended radius containing the bubble. What is the apparent depth of the bubble below the surface of the sphere?

3. One end of a long glass rod (m = 1.5) is formed into a convex surface of radius 6.0 cm. An object is positioned in air along the axis of the rod. Find the image positions corresponding to object distances of (a) 20.0 cm, (b) 10.0 cm, (c) 3.0 cm from the end of the rod. 4 4. A dust particle is inside a sphere of refractive index . If the dust particle is 10.0 cm from the wall 3 of the 15.0 cm radius bowl, where does it appear to an observer outside the bowl.

5. A parallel beam of light enters a clear plastic bead 2.50 cm in diameter and index 1.44. At what point beyond the bead are these rays brought to a focus?

31.6 Lens Theory (i) A lens is one of the most familiar optical devices for a human being. A lens is an optical system with two refracting surfaces. The simplest lens has two spherical surfaces close enough together that we can neglect the distance between them (the thickness of the lens). We call this a thin lens.

Biconvex

Biconcave

Plano-convex

Convex meniscus

Plano-concave

Concave meniscus

Fig. 31.33 Types of lens.

Lenses are of two basic types convex (converging) which are thicker in the middle than at the edges and concave (diverging) for which the reverse holds.

Refraction of Light — 89

Chapter 31

Figure shows examples of both types bounded by spherical or plane surfaces. As there are two spherical surfaces, there are two centres of curvature C1 and C 2 correspondingly two radii of curvature R1 and R 2 . The line joining C1 and C 2 is called the principal axis of the lens. The centre P of the thin lens which lies on the principal axis, is called the optical centre. R1 > 0 R2 < 0

Incident light R2

C2

C1

P R1

(a) R1 < 0 R2 > 0

Incident light P PP

R1

C1

C2 R2

(b)

Fig. 31.34 (a) A converging thin lens and (b) a diverging thin lens

(ii) All lens formulae (which we will use in this chapter) can be applied directly under following two conditions. Condition 1 Lens should be thin or its thickness should be negligible. On both sides of the lens, medium should be same (not necessarily air) (iii) If either of the above two conditions are not satisfied, then apply refraction formulae m æ m 2 m1 m 2 - m1 for spherical surface or v = 2 u for plane surface) two times. = ç m1 u R è v Condition 2

(iv) In a biconvex (or equiconvex) or biconcave (or equiconcave) lens, | R1 | = | R 2 | (v) Use of thin lens If the lens is thin, then the first image distance v1 is exactly equal to the second object distance u2 . t I1

O u2 v1 Fig. 31.35

In the figure, we can see that But,

u2 = v1 - t u2 = v1 if

t =0

90 — Optics and Modern Physics (vi) Unlike a mirror, a lens has two foci : First focus ( F1 ) It is defined as a point at which if an object (real in case of a convex lens and virtual for concave) is placed, the image of this object is formed at infinity. Or we can say that rays passing through F1 become parallel to the principal axis after refraction from the lens. The distance PF1 is the first focal length f 1 . Incident light

P

F1

P

F1

+ve f1

f1

First focus Fig. 31.36

Second focus or principal focus ( F2 ) A narrow beam of light travelling parallel to the principal axis either converge (in case of a convex lens) or diverge (in case of a concave lens) at a point F2 after refraction from the lens. This point F2 is called the second or principal focus. If the rays converge at F2 , the lens is said a converging lens and if they diverge, they are called diverging lens. Distance PF2 is the second focal length f 2 . f2 P F2

Second focus

P F2 f2 Principal focus

Fig. 31.37

From the figure, we can see that f 1 is negative for a convex lens and positive for a concave lens. But, f 2 is positive for convex lens and negative for concave lens. (vii) We are mainly concerned with the second focus f 2 . Thus, wherever we write the focal length f , it means the second or principal focal length. Thus, f = f 2 and hence, f is positive for a convex lens and negative for a concave lens. (viii) If the two conditions mentioned in point number (ii) are satisfied, then | f1 | = | f 2 | although their signs are different.

Chapter 31

Refraction of Light — 91

(ix) If those two conditions are satisfied, then object can be placed on either side of the lens or light can fall from both sides of the lens. Following figures will help you to clear this concept. +ve O

30 cm

60 cm

I

+ve

60 cm

I

30 cm

O

+ve

F1

F2

+ve

F1

F2

Fig. 31.38

Image Position, its Nature and Speed Case 1

Convex lens +¥

–¥ 2F1

F1

O

F2

2F2

Fig. 31.39

Table 31.1 Object

Image

Nature of image

Speed

At F1

± ¥

-

-

At 2 F1

At 2 F2

Real, Inverted and same size

v I = vO

At - ¥

At F2

-

-

Between O and F1

Between O and - µ

Virtual, Erect and magnified

vO > v I

Between F1 and 2 F1

Between + ¥ and 2 F2

Real, Inverted and magnified

v I > vO

Between 2 F1 and - ¥

Between 2 F2 and F2

Real, Inverted and diminished

vO > v I

92 — Optics and Modern Physics Note

(i) The above table has been made only for real objects (lying between O and - ¥ ), object distance u for them is negative. (ii) Since | f1 | = | f2 | (when two conditions discussed earlier are satisfied). Therefore, F1 and F2 are sometimes denoted by F and 2 F1 (and 2 F2 ) by 2F. (iii) If object is travelling along the principal axis, then image also travels along the principal axis in the same direction.

Case 2 Concave lens In case of concave lens there is only one case for real objects. Object lies between O and - ¥, then image lies between O and F2 . Nature of image is virtual, erect and diminished. Object speed is always greater than image speed ( vO > v I ). Both travel in the same direction.

+¥

–¥ 2F2

F2

O

F1

2F1

Fig. 31.40

Three Standard Rays for making Ray Diagrams 1. A ray parallel to the principal axis after refraction passes through the principal focus or appears to

diverge from it.

F2

F2

Fig. 31.41

2. A ray through the optical centre P passes undeviated because the middle of the lens acts like a thin

parallel-sided slab.

P

Fig. 31.42

3. A ray passing through the first focus F1 becomes parallel to the principal axis after refraction.

F1

F1

Fig. 31.43

Chapter 31 Ray Diagrams

Refraction of Light — 93 Nature of image

(a)

2F

F

F

Real Inverted Diminished

2F

F

Real Inverted Same size

F

2F

Real Inverted Magnified

2F

Real Inverted Magnified

2F

(b)

2F

(c)

F

2F

(d)

F

2F

F

F Im

ag e

at inf in

ity

(e)

(f)

2F

F

2F

F

F

F

2F

Virtual Erect Magnified

2F

Virtual Erect Diminished

Fig. 31.44 Ray diagrams for a convex lens (a–e) and a concave lens (f).

94 — Optics and Modern Physics List of Formulae (i)

1 1 1 - = v u f m=

(ii) Linear magnification,

Image height I v = = Object height O u

(iii) Lens maker’s formula m1

m1

R1

m2 R2

Fig. 31.45

öæ 1 1 æ m2 1 ö ÷÷ - 1 ÷÷ çç = çç f è m1 ø è R1 R 2 ø æ 1 1 1 In air, m 1 = 1 and m 2 = m. Therefore, = (m - 1) çç f R R è 1 2 (iv) Power of a lens (in dioptre) =

ö ÷÷ ø

1 Focal length (in metre)

(v) Two or more than two thin lenses in contact.

Fig. 31.46

1 1 1 + = F f1 f 2

or

P = P1 + P2

(vi) Two or more than two thin lenses at some distance

d Fig. 31.47

1 1 1 d = + F f1 f 2 f1 f 2 or

P = P1 + P2 - dP1 P2

Note In the above two equations if d = 0, then 1 1 1 = + F f1 f2

and P = P1 + P2

Chapter 31

Refraction of Light — 95

(vii) Image velocity (a) Along the principal axis, v I = m2 v 0 v I and v 0 are in the same direction. (b) Perpendicular to principal axis, v I = mv 0 If m is positive then v I and v 0 are in the same direction and if m is negative, then v I and v 0 are in opposite directions. Note The above two formulae have been derived in the previous chapter of reflection. So, same method can be applied here.

Important points in formulae (i) On linear magnification m From the value of m, we can determine nature of image, type of lens and an approximate position of object. Following table illustrates this point. Table 31.2 Value of m

Nature of image

Type of lens

Object position

-3

Inverted, real and magnified

convex

Between F1 and 2 F1

-1

Inverted, real and same size

convex

at 2 F1

1 2

Inverted, real and diminished

convex

Between 2 F1 and - ¥

+2

Erect, virtual and magnified

convex

Between O and F1

1 4

Erect, virtual and diminished

concave

Between O and - ¥

+

Note For real objects, real image is formed only by convex lens. But virtual image is formed by both types of lenses. Their sizes are different. Magnified virtual image is formed by convex lens. Diminished virtual image is formed by concave lens.

(ii) On lens maker’s formula öæ 1 1 æ m2 1 = çç - 1 ÷÷ çç f è m1 ø è R1 R 2

ö ÷÷ ø

æ 1 1 ö ÷÷ in air = (m - 1) çç R R è 1 2 ø

K (i) K (ii)

æ 1 1 ö ÷÷ in Eq. (ii) comes For a converging lens, R1 is positive and R 2 is negative. Therefore, çç – è R1 R 2 ø out a positive quantity and if the lens is placed in air, ( m – 1) is also a positive quantity. Hence, the focal length f of a converging lens turns out to be positive. For a diverging lens however, R1 is negative and R 2 is positive and the focal length f becomes negative.

96 — Optics and Modern Physics Incident light C2

R1 > 0 R2 < 0 C1

R2 R1 (a)

Incident light C1

R1

C2 R2 (b)

Fig. 31.48

Rö æ Focal length of a mirror ç f M = ÷ depends only upon the radius of curvature R while that of a 2ø è lens [Eq. (i)] depends on m 1 , m 2 , R1 and R 2 . Thus, if a lens and a mirror are immersed in some liquid, the focal length of lens would change while that of the mirror will remain unchanged.

Fig. 31.49 Air bubble in water diverges the parallel beam of light incident on it.

Suppose m 2 < m 1 in Eq. (i), i.e. refractive index of the medium (in which lens is placed) is more æm ö than the refractive index of the material of the lens, then çç 2 – 1 ÷÷ becomes a negative quantity, è m1 ø i.e. the lens changes its behaviour. A converging lens behaves as a diverging lens and vice-versa. An air bubble in water seems as a convex lens but behaves as a concave (diverging) lens. (iii) Power of lens By optical power of an instrument (whether it is a lens, mirror or a refractive surface) we mean the ability of the instrument to deviate the path of rays passing through it. If the instrument converges the rays parallel to the principal axis its power is said to be positive and if it diverges the rays it is said a negative power.

f1

f2

Fig. 31.50

Chapter 31

Refraction of Light — 97

The shorter the focal length of a lens (or a mirror) the more it converges or diverges the light. As shown in the figure, f1 < f 2 and hence the power P1 > P2 , as bending of light in case 1 is more than that of case 2. For a lens, 1 P (in dioptre) = f (in metre) P (in dioptre) =

and for a mirror,

–1 f (in metre)

Following table gives the sign of P and f for different types of lens and mirror. Table 31.3 Power 1 1 PL = , PM = – f f

Converging/ diverging

+ ve

+ ve

converging

Concave mirror

– ve

+ ve

converging

Concave lens

– ve

– ve

diverging

Convex mirror

+ ve

– ve

diverging

Nature of lens/mirror

Focal length

Convex lens

(f )

Ray diagram

Thus, convex lens and concave mirror have positive power or they are converging in nature. Concave lens and convex mirror have negative power or they are diverging in nature. (iv) Based on two or more than two thin lenses in contact (or at some distance) If the lenses are kept in contact, then after finding the equivalent focal length F from the equation 1 1 1 = + F f1 f 2 We can directly apply the formula 1 1 1 - = v u F

98 — Optics and Modern Physics for finding the image distance v, but we will have to apply lens formula,

1 1 1 - = two times if v u f

the lenses are kept at some distance.

Proofs of Different Formulae (i) Consider an object O placed at a distance u from a convex lens as shown in figure. Let its image I after two refractions from spherical surfaces of radii R1 (positive) and R 2 (negative) be formed at a distance v from the lens. Let v1 be the distance of image formed by refraction from the refracting surface of radius R1 . This image acts as an object for the second surface. Incident light R1

R2 O

C2

m1 m2 m1

C1

I +ve

u

v

Fig. 31.51

Using, We have, and

m 2 m1 m 2 – m1 – = v u R m 2 m1 m 2 – m1 – = v1 u R1 m1 m 2 m1 – m 2 – = v v1 – R2

…(i) …(ii)

Adding Eqs. (i) and (ii) and then simplifying, we get öæ 1 1 1 æ m2 1 ö ÷÷ …(iii) – = çç – 1 ÷÷ çç – v u è m1 ø è R1 R 2 ø This expression relates the image distance v of the image formed by a thin lens to the object distance u and to the thin lens properties (index of refraction and radii of curvature). It is valid only for paraxial rays and only when the lens thickness is much less than R1 and R 2 . The focal length f of a thin lens is the image distance that corresponds to an object at infinity. So, putting u = ¥ and v = f in the above equation, we have öæ 1 1 æ m2 1 = çç – 1 ÷÷ çç – f è m1 ø è R1 R 2

ö ÷÷ ø

…(iv)

If the refractive index of the material of the lens is m and it is placed in air, m 2 = m and m 1 = 1 so that Eq. (iv) becomes æ 1 1 1 = (m – 1) çç – f è R1 R 2

ö ÷÷ ø

…(v)

This is called the lens maker’s formula because it can be used to determine the values of R1 and R 2 that are needed for a given refractive index and a desired focal length f.

Chapter 31

Refraction of Light — 99

Combining Eqs. (iii) and (v), we get 1 1 1 – = v u f

…(vi)

Which is known as the lens formula. (ii) Magnification The lateral, transverse or linear magnification m produced by a lens is defined by Height of image I m= = Height of object O O¢

I O

P

u

I¢

v

Fig. 31.52

A real image II ¢ of an object OO¢ formed by a convex lens is shown in figure. Height of image II ¢ v = = Height of object OO¢ u Substituting v and u with proper sign, II ¢ – I v = = OO¢ O – u or

I v =m= O u

Thus,

m=

v u

(iii) Focal length of two or more than two thin lenses in contact Combinations of lenses in contact are used in many optical instruments to improve their performance. f1 f2

I

O

u

v

Fig. 31.53

100 — Optics and Modern Physics Suppose two lenses of focal lengths f 1 and f 2 are kept in contact and a point object O is placed at a distance u from the combination. The first image (say I 1 ) after refraction from the first lens is formed at a distance v1 (whatever may be the sign of v1 ) from the combination. This image I 1 acts as an object for the second lens and let v be the distance of the final image from the combination. Applying the lens formula, 1 1 1 – = v u f 1 1 1 For the two lenses, we have …(i) – = v1 u f 1 and Adding Eqs. (i) and (ii), we have

1 1 1 – = v v1 f 2 1 1 1 1 1 – = + = v u f1 f 2 F

…(ii) (say)

Here, F is the equivalent focal length of the combination. Thus, 1 1 1 = + F f1 f 2 Similarly for more than two lenses in contact, the equivalent focal length is given by the formula, n

1 1 =å F i =1 f i Note Here, f1 , f2 etc., are to be substituted with sign.

Types of Problems in Lens Type 1. Based on two or more than two thin lenses in contact How to Solve

Apply,

1 1 1 = + F f1 f 2 P = P1 + P2

and V

Example 31.13 A convex lens of power 2 D and a concave lens of focal length 40 cm are kept in contact, find (a) Power of combination (b) Equivalent focal length Solution (a) Applying P = P1 + P2 = Pconvex + Pconcave é 1 1 ù =2+ P ( in D ) = ê f ( in m ) úû ( -0.4 ) ë = 2 - 2.5 = - 0.5 D

Ans.

Chapter 31 F=

(b)

Refraction of Light — 101

1 P

=

1 = -2 m - 0.5

= - 200 cm

Ans.

Note F and P are negative so, the system behaves like a concave lens. V

Example 31.14 A converging lens of focal length 5.0 cm is placed in contact with a diverging lens of focal length 10.0 cm. Find the combined focal length of the system. Solution

Here,

f1 = + 5.0 cm

f 2 = – 10.0 cm

and

Therefore, the combined focal length F is given by 1 1 1 1 1 1 = + = – =+ F f1 f 2 5.0 10.0 10.0 \

F = + 10.0cm

Ans.

i.e. the combination behaves as a converging lens of focal length 10.0 cm.

Type 2. Based on lens maker’s formula How to Solve Apply,

öæ 1 1 æ m2 1 ö ÷÷ or = çç - 1 ÷÷ çç f è m1 ø è R1 R 2 ø

æ 1 1 1 ö ÷÷ = (m - 1) çç f è R1 R 2 ø

Note If initial two conditions are satisfied (thin lens and same medium on both sides,) then we can find focal length of the lens (f or f2 ) from either side of the lens. Result comes out to be same. V

Example 31.15

m1 = 2

R

m1 = 2 m2 = 1.5

R = 40 cm

Fig. 31.54

Find focal length of the system shown in figure from left hand side. Solution m 1 = 2, m 2 = 1.5, R1 = + 40 and R 2 = ¥ Using the equation öæ 1 1 æm2 1 ö ÷÷ , we have = çç - 1÷÷ çç f è m1 ø è R1 R 2 ø 1 æ 1.5 ö æ 1 1ö =ç - 1÷ çç - ÷÷ Þ f è 2 + 40 ¥ øè ø

R1 R2

Fig. 31.55

f = - 160 cm

102 — Optics and Modern Physics Therefore, from left hand side it behaves like a concave lens of focal length 160 cm.

F2

160 cm Fig. 31.56

Exercise

Find focal length of the above system from right hand side and prove that it is also

-160 cm. V

Example 31.16 Focal length of a convex lens in air is 10 cm. Find its focal length in water. Given that m g = 3 /2 and m w = 4 /3. 1

Solution

and

f air 1 f water

æ 1 1 ö ÷÷ = (m g – 1) çç – è R1 R 2 ø

…(i)

æmg öæ 1 1 ö ÷÷ = çç – 1÷÷ çç – m R R 2 ø è w øè 1

…(ii)

Dividing Eq. (i) by Eq. (ii), we get (m g – 1) f water = f air (m g / m w – 1) Substituting the values, we have f water =

( 3/ 2 – 1) f air æ 3/ 2 ö çç – 1÷÷ è 4/ 3 ø

= 4 f air = 4 ´ 10 = 40 cm Note

Ans.

(i) Students can remember the result fwater = 4 fair , if m g = 3 / 2 and m w = 4 / 3.

1 th. This is 4 because, difference in refractive index between glass and water has been decreased (compared to glass and air). So, there will be less bending of light.

(ii) In water focal length has become four times and power (power of bending of light) remains

V

Example 31.17 A biconvex lens (m = 1.5) has radius of curvature 20 cm ( both ). Find its focal length. m

R1

Fig. 31.57

R2

Chapter 31

Refraction of Light — 103

R1 = + 20 cm, R 2 = - 20 cm, m = 1.5 Substituting the values in the equation Solution

æ 1 1 1 = (m - 1) çç f è R1 R 2

ö ÷÷, we have ø

1 1 ö æ 1 = (1.5 - 1) ç ÷ f è 20 -20 ø f = + 20 cm

or

Ans.

Note If a biconvex or biconcave lens has refractive index m = 1.5, then | R1 | = | R2 | = | f |

Type 3. To find image distance and its magnification corresponding to given object distance

How to Solve? l

Substitute signs of u and f. Sign of v automatically comes after applying the lens formula. Sign of u is negative for real objects. Sign of f is positive for convex lens and negative for concave lens. V

Example 31.18 Find distance of image from a convex lens of focal length 20 cm if object is placed at a distance of 30 cm from the lens. Also find its magnification. Solution u = - 30 cm, f = + 20 cm Applying the lens formula 1 1 1 - = v u f We have, 1 1 1 = v - 30 + 20 Solving, we get v = + 60cm m=

Ans.

v + 60 = = -2 u - 30

Ans.

m is - 2, it implies that image is real, inverted and two times magnified. The ray diagram is as shown below. h O

I F2

F1

30 cm

60 cm

Fig. 31.58

2h

104 — Optics and Modern Physics Type 4. To find object/image distance corresponding to given magnification of image and focal length of lens

How to Solve? l

l

Substitute all three signs of u , v and f. Signs of u and f have been discussed in the above type. Sign of v is positive for real image (see the above example) and it is negative for virtual image. v m= Þ | v| = | m ´ u | u V

Example 31.19 Find the distance of an object from a convex lens if image is two times magnified. Focal length of the lens is 10 cm. Convex lens forms both types of images real as well as virtual. Since, nature of the image is not mentioned in the question, we will have to consider both the cases. Solution

When image is real

Means v is positive and u is negative with | v | = 2 | u |. Thus, if u = – x,

Substituting in

then v = 2x and f = 10 cm 1 1 1 – = v u f 1 1 1 or + = 2x x 10

We have \

3 1 = 2x 10

x = 15cm

Ans.

x = 15 cm, means object lies between F and 2 F. When image is virtual

Means v and u both are negative. So let, u = – y,

Substituting in,

then v = – 2 y and f = 10 cm 1 1 1 – = v u f 1 1 1 or + = –2 y y 10

We have, \

1 1 = 2 y 10

y = 5 cm

Ans.

y = 5 cm, means object lies between F and O.

Type 5. To make some conditions

How to Solve? l

Initially, substitute sign of only f, then make equation of v. From this equation of v, find the asked condition. V

Example 31.20 Under what condition, a concave lens can make a real image. Solution

Substituting sign of f in the lens formula, we have 1 1 1 1 1 1 or - = = v u -f v u f

K (i)

Refraction of Light — 105

Chapter 31

For real image v should be positive. Therefore, from Eq. (i) we can see that u should be positive and less than f. Further, u is positive and less than f means a virtual object should lie between O and F1 . + ve

F1

O

Fig. 31.59

Important Result Under normal conditions, a convex lens makes a real image. But the image is virtual (and magnified) if a real object is placed between O and F1 . Opposite is the case with concave lens. Under normal conditions it makes a virtual image (for all real objects). But the image is real if a virtual object is placed between O and F1 .

Type 6. To find image nature, type of lens, its optical centre and focus for given principal axis, point object and its point image. V

Example 31.21 An image I is formed of point object O by a lens whose optic axis is AB as shown in figure. O A

B I

Fig. 31.60

(a) State whether it is a convex lens or concave? (b) Draw a ray diagram to locate the lens and its focus. Solution (a) (i) Concave lens always forms an erect image. The given image I is on the other side of the optic axis. Hence, the lens is convex. (ii) Join O with I. Line OI cuts the optic axis AB at optical centre (P) of the lens. The dotted line shows the position of lens.

M

O A

F

B

P

I Fig. 31.61

From point O, draw a line parallel to AB. Let it cuts the dotted line at M. Join M with I. Line MI cuts the optic axis at focus (F) of the lens.

106 — Optics and Modern Physics Type 7. Two lens problems

How to Solve? l

We have to apply lens formula two times. The first image behaves like an object for the second lens. V

Example 31.22 Focal length of convex lens is 20 cm and of concave lens 40 cm. Find the position of final image.

O 30 cm

40 cm

Fig. 31.62

Solution

Using the lens formula for convex lens, u = -30 cm, f = + 20 cm 1 1 1 = v - 30 + 20

\

Solving this equation, we get v = + 60 cm. Therefore, the first image is 60 cm to the right of convex lens or 20 cm to the right of concave lens. Again applying lens formula for concave lens, u = + 20 cm, f = - 40 cm, we have 1 1 1 = v + 20 - 40 v = + 40 cm So, the final image I 2 is formed at 40 cm to the right of concave lens as shown below.

O 30 cm

40 cm

20 cm 20 cm I1 I2

Fig. 31.63

Type 8. Based on image velocity

How to Solve? l

Using the methods discussed in Type 3, first find v and then m. Now,

(i) Along the axis, v I = m 2 v 0 (ii) Perpendicular to axis, v I = mv 0

Chapter 31 V

Refraction of Light — 107

Example 31.23 Focal length of the convex lens shown in figure is 20 cm. Find the image position and image velocity. 5 mm/s 37° O 30 cm

Fig. 31.64

Solution

For the given condition, u = - 30 cm, f = + 20 cm

Using the lens formula, we have 1 1 1 = v - 30 + 20 Solving this equation, we get v = + 60cm and v + 60 m= = = -2 u - 30 m2 = 4 Component of velocity of object along the axis = 5 cos 37° = 4 mm/s (towards the lens) component of velocity of image along the axis = m2 ( 4 mm/s ) = 4 ´ 4 = 16 mm/s. This component is away from the lens (in the same direction of object velocity component) Component of velocity of object perpendicular to the axis = 5 sin 37° = 3 mm/s (upwards). \ Component of velocity of image perpendicular to axis = m ( 3 mm/s ) or ( -2)( 3 mm/s) = - 6 mm/s or this component is 6 mm/s downwards. These all points are shown in the figure given below. 3 mm/s I O 4 mm/s

q

16 mm/s

6 mm/s vI

30 cm

60 cm

Fig. 31.65

v I = (16) 2 + ( 6) 2 = 292 mm/s tan q = or

6 3 = 16 8

q = tan -1 ( 3/ 8)

108 — Optics and Modern Physics Type 9. An extended object kept perpendicular to principal axis

How to Solve? l

Using the methods discussed in Type 3, first find v and then m. Now, suppose m is -2 and object is 1 mm above the principal axis, then image will be 2 mm below the principal axis. V

Example 31.24 Focal length of concave lens shown in figure is 60 cm. Find image position and its magnification. b c a

2 mm 1 mm

30 cm Fig. 31.66

Solution

For the given situation, u = - 30 cm, f = - 60 cm

Using the lens formula, we have 1 1 1 = v - 30 - 60 Solving this equation, we get v = - 20 cm v - 20 2 Further, m = = = + . Point b is 2 mm above the principal axis. Therefore, its image b¢ will u - 30 3 æ 2ö 4 be ( 2) ç ÷ or mm, above the principal axis. è 3ø 3 æ 2ö Similarly, point a is 1 mm below the principal axis. Therefore, its image a¢ will be (1) ç ÷ or è 3ø 2 mm below the principal axis. The final image is as shown in Fig. 31.67 3 b

b¢ c¢

a

a¢

20 cm 30 cm

Fig. 31.67

c¢b¢ =

4 mm 3

a ¢ c¢ =

2 mm 3

Refraction of Light — 109

Chapter 31

Type 10. To plot u versus v or 1 versus 1 graph u

v

Concept 1 1 versus graph will be a straight line. Further for real v u 1 objects, u is always negative. So, u varies from 0 to - ¥. Therefore, will vary from - ¥ to 0. u

In the previous chapter, we have seen that

V

Example 31.25 Plot u versus v and

1 1 versus graph for convex lens (only for u v

real objects) Solution +¥

–¥ 2F

0

F

2F

F

Fig. 31.68

Table 31.4 S.No.

u

v

1.

0 to - f

0 to - ¥

2.

- f to - 2 f

+ ¥ to + 2 f

3.

-2 f to - ¥

+ 2 f to + f

1 u

1 v

1 f 1 1 - to f 2f

- ¥ to 0

- ¥ to -

-

0 to +

1 to 0 2f

+

u versus graph v P

2

P = (- 2f, - 2f )

+ 2f

3

+f 45° – 2f

–f

u O

1

Fig. 31.69

1 2f

1 1 to + 2f f

110 — Optics and Modern Physics 1 1 versus graph u v 1 — v –1 — f –1 — 2f

2 1

1 — u

–1 — 2f

–1 — f

Fig. 31.70

Type 11. Problems of inclined lenses V

Example 31.26

+ ve

P O

O

1

Fig. 31.71

O1 P is the principal axis and O is the point object. Given, O1 P = 30 cm, f = 20 cm and OP = 2 mm. Find the image distance and its position. Solution For the given situation, P

O1Q = 60 cm

O

I O1

QI = 4 mm

Q

Fig. 31.72

and Using the lens formula, we have

u = -30 cm f = + 20 cm 1 1 1 = v - 30 + 20

Solving this equation, we get v = + 60cm

Chapter 31 m=

Further,

Refraction of Light — 111

v + 60 = = -2 u - 30

Therefore, image is at a distance of + 60 cm from the lens at a distance of (2 mm) (-2) or 4 mm from the principal axis on other side of the object. The image is as shown above in Fig. 31.72. Note Image will always lie on the line joining O and O1 . This is because the ray OO1 passes undeviated.

Type 12. To find focal length of an optical system for which either of the two conditions (thin lens and same medium on both sides) is not satisfied

Concept If focal length is asked then we have to find the second focal length f 2 . The definition of F2 is, if object is at infinity ( u1 = ¥ ) then final image after two refractions will be at F2 ( v 2 = f 2 or f ). The use of thin lens is v1 is exactly equal to u2 . V

Example 31.27 In the figure, light is incident on a thin lens as shown. The radius of curvature for both the surfaces is R. Determine the focal length of this (JEE 2003) system. m1 m 2

m3

Fig. 31.73

Solution

For refraction at first surface, m 2 m1 m 2 -m1 = v1 -¥ +R

For refraction at second surface,

Adding Eqs. (i) and (ii), we get

…(i)

m 3 m 2 m 3 -m 2 = v2 v1 +R m 3 m 3 -m1 = v2 R

v2 =

or m3

m1

m2

v2 Fig. 31.74

v1

… (ii) m 3R m 3 -m1

112 — Optics and Modern Physics Therefore, focal length of the given lens system is m 3R f= m 3 -m1 Note If we find the focal length of the above system from right hand side, then it will be different because medium on both sides is not same.

Important Points in Lens Theory 1. or

2f P/2

Biconvex lens f:P First

2f P/2

Second

Fig. 31.75

In the first figure, 1 1 ö æ2ö æ1 = (m - 1) ç ÷ = (m - 1) ç ÷ f èRø è R -R ø R f = 2 (m - 1)

\ In the second figure,

1 æ1 1 ö = (m - 1) ç - ÷ f¢ èR ¥ø R m -1

\

f¢=

or

f ¢ =2f

2. or f, P

f, P 2P f/2

f, P

Fig. 31.76

3. The system shown in Fig. 31.77 has single value of u but two different parts will have two focal

lengths. Therefore, we get two images I 1 and I 2 , horizontally separated from each other. The two focal lengths are æ 1 1 1 ö ÷÷ = (m 1 - 1) çç f1 è R1 R 2 ø

Chapter 31

Refraction of Light — 113

æ 1 1 1 ö ÷÷ = (m 2 - 1) çç f2 R R 2 ø è 1

and

v1 m1 R1

R2

R1

I1

m2 R2

u

I2

v2

Fig. 31.77

4. This system shown in Fig. 31.78 has single values of u and f .

Therefore, we will get single value of v. Still, we will get two 1 f P Q images, vertically separated from each other. Let us take an O N example in support of this. M 2 f If u = 30 cm, f is 20 cm, principal axis PQ of upper part 1 is 1 mm u above the object and principal axis MN of lower part 2 is 1 mm Fig. 31.78 below the object O. Then, after applying lens formula, we get v = + 60 cm and m = - 2 Now, O is 1 mm below PQ and m = - 2. Therefore, upper part will make I 1 , 2 mm above PQ. Similarly, O is 1 mm above MN , therefore lower part will make I 2 , 2 mm below MN . Ray diagram from the two parts is as shown in Fig. 31.79 below d = 1 mm 1 I1

F1

2 mm

d d

O F1

2

I2

20 cm 30 cm

2 mm

60 cm

Fig. 31.79

5. If a liquid is filled between two thin convex glass lenses, then it is a group of three lenses as shown

in figure. R2 R1

R3

R2 +

1

+

2

Fig. 31.80

\

1 1 1 1 = + + F f1 f 2 f 3

R3

R4

3

114 — Optics and Modern Physics æ 1 1 1 ö ÷÷ = (m g - 1) çç f1 R R 2 ø è 1

where,

æ 1 1 1 ö ÷÷ and = (m l - 1) çç f2 è R2 R3 ø æ 1 1 1 = (m g - 1) çç f3 è R3 R4

ö ÷÷ ø 6. The system shown in figure behaves like a lens of zero power (or f = ¥). This is because Air 1 2

Glass

3 4

A

Thin

Fig. 31.81

R1 » R 2 and R 3 » R 4 Now if we find focal length or power of part A, then 1 f

or

æ 1 1 P = (m g - 1) çç è R1 R 2

ö ÷÷ = 0 ø

as R1 » R 2 Similarly, we can prove that power of other part is also zero. 7. Minimum distance between real object and its real image from a convex lens is 4 f . Exercise Prove the above result. 8. Silvered lens A point object O is placed in front of a silvered lens as shown in figure. R1

O

I

m1

R2

m2 +ve

u

Fig. 31.82

Ray of light is first refracted, then reflected and then again refracted. In first two steps, light is travelling from left to right and in the last one direction of light is reversed. But we will take one sign convention, i.e. left to right as positive and in the last step will take v, u and R as negative. m 2 m1 m 2 – m1 …(i) – = v1 u R1 1 1 1 2 + = = v 2 v1 f mirror R 2

…(ii)

Chapter 31

Refraction of Light — 115

m1 m 2 m1 – m 2 – = – v – v2 – R1 Solving Eqs. (i), (ii) and (iii), we get 1 1 2 (m 2 /m 1 ) 2 (m 2 / m 1 – 1) + = – v u R2 R1 This is the desired formula for finding position of image for the given situation.

…(iii)

…(iv)

Note The given system behaves as a mirror because the ray of light finally reflects in the same medium. Whose focal length can be found by comparing Eq. (iv) with mirror formula 1/v + 1/u = 1/ f .

1 2 (m 2 /m 1 ) 2 (m 2 /m 1 – 1) = – f R2 R1

…(v)

Let us take one example in support of this. V

Example 31.28

m = 1.5 Air O 20 cm

R = 40 cm R

Fig. 31.83

(a) Find focal length of the system as shown in figure. (b) Find image position. Solution (a) m 1 = 1, m 2 = 1.5, R1 = R = + 40 cm and R 2 = ¥ Using the formula, 1 2 (m 2 / m 1 ) 2 (m 2 / m 1 - 1) = f R2 R1 =

2 (1.5) 2 (1.5 - 1) ¥ 40

or f = - 40 cm Thus, the given system behaves like a concave mirror of focal length 40 cm. (b) Using the mirror formula, we have 1 1 1 + = v u f \

1 1 1 + = v - 20 - 40

\ v = + 40 cm Therefore, image will be formed at a distance of 40 cm to the right hand side of the given system.

If the distance d between an object and screen is greater than 4 times the focal length of a convex lens, then there are two

9. Displacement method of finding focal length of a convex lens

116 — Optics and Modern Physics positions of the lens between the object and the screen at which a sharp image of the object is formed on the screen. This method is called displacement method and is used in laboratory to determine the focal length of convex lens. x

Screen

Object u

d–u d

Fig. 31.84

To prove this, let us take an object placed at a distance u from a convex lens of focal length f. The distance of image from the lens v = ( d – u). From the lens formula, 1 1 1 – = v u f 1 1 1 – = d – u –u f

We have,

u 2 – du + df = 0

or \

u=

d ± d (d – 4 f )

2 Now, there are following possibilities: (i) If d < 4 f , then u is imaginary. So, physically no position of the lens is possible. d (ii) If d = 4 f , then u = = 2 f . So, only one position is possible. From here we can see that the 2 minimum distance between an object and its real image in case of a convex lens is 4 f. (iii) If d > 4 f , there are two positions of lens at distances d + d (d – 4 f )

and

d – d (d – 4 f )

2 2 for which real image is formed on the screen. (iv) Suppose I 1 is the image length in one position of the object and I 2 the image length in second position, then object length O is given by O = I1 I 2 This can be proved as under | u1 | = \

d + d (d – 4 f ) 2

| v1 | = d – | u1 | =

d – d (d – 4 f ) 2

Chapter 31 d – d (d – 4 f )

| u2 | = \ Now,

Refraction of Light — 117

2

| v 2 | = d – | u2 | =

d + d (d – 4 f )

2 I I |v | |v | | m1 m2 | = 1 ´ 2 = 1 ´ 2 O O | u1 | | u2 | I1 I 2

Substituting the values, we get

O2

=1

O = I1 I 2

or

Hence Proved.

(v) Focal length of the lens is given by f = Proof

or

d2 - x2 4d

In the figure, we can see that difference of two values of u is x. Thus, | u1 | - | u2 | = x é d + d (d - 4 f ) ù é d - d (d - 4 f ) ù ê ú-ê ú=x 2 2 êë úû êë úû

Solving this equation, we can find that f = V

d2 - x2 4d

Example 31.29 A thin plano-convex lens of focal length f is split into two halves. One of the halves is shifted along the optical axis as shown in figure. The separation between object and image planes is 1.8 m. The magnification of the image formed by one of the half lens is 2. Find the focal length of the lens and separation between the two halves. Draw the ray diagram for image formation. (JEE 1996)

O

1.8 m

Fig. 31.85

For both the halves, position of object and image is same. Only difference is of magnification. Magnification for one of the halves is given as 2 ( > 1). This can be for the first one, because for this, | v | > | u |. Therefore, magnification, | m | = | v / u | > 1. Solution

118 — Optics and Modern Physics So, for the first half | v/ u | = 2 | v | = 2| u |

or Let

u =– x,

then

v = + 2x

and

| u | + | v| = 1.8 m

i.e. or

3x = 1.8 m x = 0.6 m u = – 0.6 m

Hence,

v = + 1.2 m 1 1 1 1 1 1 = – = – = f v u 1.2 –0.6 0.4

and Using \

f = 0.4 m

For the second half,

1 1 1 = – f 1.2 – d – ( 0.6 + d )

Ans.

1 1 1 = + 0.4 1.2 – d ( 0.6 + d )

or

d = 0.6 m

Solving this, we get

Ans.

Magnification for the second half will be m2 =

v 0.6 1 = =– u – (1.2) 2

m1 =

v 1.2 = =– 2 u – ( 0.6)

and magnification for the first half is

The ray diagram is as follows: d B1 f = 0.4 m

f = 0.4 m

1

B2

A

(A1, A2) 2

B

0.6 m

0.6 m

Fig. 31.86

0.6 m

Chapter 31

INTRODUCTORY EXERCISE

Refraction of Light — 119

31.6

1. When an object is placed 60 cm in front of a diverging lens, a virtual image is formed 20 cm from the lens. The lens is made of a material of refractive index m = 165 . and its two spherical surfaces have the same radius of curvature. What is the value of this radius?

2. A converging lens has a focal length of 30 cm. Rays from a 2.0 cm high filament that pass through the lens form a virtual image at a distance of 50 cm from the lens. Where is the filament located? What is the height of the image?

3. Show that the focal length of a thin lens is not changed when the lens is rotated so that the left and the right surfaces are interchanged.

4. As an object is moved from the surface of a thin converging lens to a focal point, over what range does the image distance vary?

5. A diverging lens is made of material with refractive index 1.3 and has identical concave surfaces of radius 20 cm. The lens is immersed in a transparent medium with refractive index 1.8. (a) What is now the focal length of the lens? (b) What is the minimum distance that an immersed object must be from the lens so that a real image is formed?

6. An object is located 20 cm to the left of a converging lens with f = 10 cm. A second identical lens is placed to the right of the first lens and then moved until the image it produces is identical in size and orientation to the object. What is the separation between the lenses?

7. Suppose an object has thickness du so that it extends from object distance u to u + du. Prove æ v2ö that the thickness dv of its image is given by çç – 2 ÷÷ du, so the longitudinal magnification è u ø dv 2 = – m , where m is the lateral magnification. du

8. Two thin similar convex glass pieces are joined together front to front, with its rear portion silvered such that a sharp image is formed 0.2 m for an object at infinity. When the air between 4ö æ the glass pieces is replaced by water çm = ÷ , find the position of image. 3ø è

9. When a pin is moved along the principal axis of a small concave mirror, the image position coincides with the object at a point 0.5 m from the mirror. If the mirror is placed at a depth of 0.2 m in a transparent liquid, the same phenomenon occurs when the pin is placed 0.4 m from the mirror. Find the refractive index of the liquid shown in Fig. 31.87.

10. When a lens is inserted between an object and a screen which are a fixed distance apart the size of the image is either 6 cm or

2 cm. Find size of the 3

object.

0.2m 0.2m

Fig. 31.87

11. A lens of focal length 12 cm forms an upright image three times the size of a real object. Find the distance in cm between the object and image.

12. The distance between an object and its upright image is 20 cm. If the magnification is 0.5, what is the focal length of the lens that is being used to form the image?

13. A thin lens of focal length + 10.0 cm lies on a horizontal plane mirror. How far above the lens should an object be held if its image is to coincide with the object?

120 — Optics and Modern Physics

31.7 Total Internal Reflection (TIR) (i) When a ray of light strikes the boundary separating two different media, then part of it is refracted and part is reflected. (ii) If a ray of light is travelling from a denser to a rarer medium with angle of incidence greater than a critical angle ( i > q C ), then no refraction takes place but ray of light is 100% reflected in the same medium. This phenomenon is called TIR. m (iii) sin q C = R mD Here, R stands for rarer medium and D for denser medium. If rarer medium is air, then 1 sin q C = m æm q C = sin -1 çç R è mD

\

ö æ1ö ÷÷ or sin -1 çç ÷÷ èm ø ø

(iv) If value of m increases, then critical angle q C decreases. Therefore, chances of TIR increase in travelling from denser to rarer medium. (v) Rarer Denser

>i

90°

i i

i i

i i

i < qC (a)

i = qC (b)

TIR i > qC (c)

Fig. 31.88

Applying Snell’s law of refraction in Fig. (b), we have m R sin 90° = m D sin q C m or sin q C = R mD æm ö q C = sin -1 çç R ÷÷ è mD ø (vi) In critical case (Fig. b), angle in denser medium is q C and angle in rarer medium is 90°. TIR has following applications (i) Totally reflecting prisms Refractive index of crown glass is 3/2. Hence, or

æ1ö æ2ö q C = sin –1 çç ÷÷ = sin –1 ç ÷ » 42° m è3ø è ø A ray OA incident normally on face PQ of a crown glass prism suffers TIR at face PR since, the angle of incidence in the optically denser medium is 45°. A bright ray AB emerges at right angles

Chapter 31

Refraction of Light — 121

from face QR. The prism thus, reflects the ray through 90°. Light can be reflected through 180° and an erect image can be obtained of an inverted one if the prism is arranged as shown in figure (b). P

45°

45° A

O

45° 45° 45° Q

R 45° B (a)

(b)

Fig. 31.89 Prism reflectors

(ii) Optical fibres Light can be confined within a bent glass rod by TIR and so ‘piped’ along a twisted path as in figure. The beam is reflected from side to side practically without loss (except for that due to absorption in the glass) and emerges only at the end of the rod where it strikes the surface almost normally, i.e. at an angle less than the critical angle. A single, very thin, solid glass fibre behaves in the same way and if several thousands are taped together a flexible light pipe is obtained that can be used, for example in medicine and engineering to illuminate an inaccessible spot. Optical fibres are now a days used to carry telephone, television and computer signals from one place to the other.

Glass rod Light

Fig. 31.90 Principle of an optical fibre

Note As we have seen

æm ö qC = sin –1 çç R ÷÷ èm D ø

Suppose we have two sets of media 1 and 2 and æmR ö æm ö çç ÷÷ < çç R ÷÷ èm D ø1 èm D ø2 then ( qC ) 1 < ( qC ) 2 So, a ray of light has more chances to have TIR in case 1.

Examples of TIR V

Example 31.30 An isotropic point source is placed at a depth h below the water surface. A floating opaque disc is placed on the surface of water so that the source is not visible from the surface. What is the minimum radius of the disc? Take refractive index of water = m .

122 — Optics and Modern Physics Solution R A

B qC

h

i > qC

qC

S Fig. 31.91

As shown in figure light from the source will not emerge out of water if i > qC . m Therefore, minimum radius R corresponds to i = qC 1 sin qC = m In DSAB, qC R = tan qC Öm2 – 1 h \

Fig. 31.92

R = h tan qC h R= 2 m –1

or

1

Ans.

Note Only that portion of light refracts in air which falls on the circle (on the surface of water) with A as centre and AB as radius. V

Example 31.31 A point source of light is placed at a distance h below the surface of a large and deep lake. Show that the fraction f of light that escapes directly from water surface is independent of h and is given by f =

[ 1 – 1 – 1/m 2 ]

2 Solution Due to TIR, light will be reflected back into the water if i > qC . So, only that portion of incident light will escape which passes through the cone of angle q = 2 qC . So, the fraction of light escaping

C A

S

Now, as f depends on qC and which depends only on m, it is independent of h. Proved. cos qC =

m2 – 1 m

= 1 – 1/ m 2

Þ

f=

B

qC qC q C

h

2pR 2 (1 – cos qC ) 1 – cos qC area ACB f= = = Total area of sphere 2 4pR 2

Further

D

Fig. 31.93

1 – 1 – 1/m 2 2

Ans.

Note Area of ACB = 2 pR 2 ( 1 - cos qC ) can be obtained by integration. In the above example, we have seen that light falling on the circle with centre at D and radius DB will only refract in air and in the absence of water surface only that light would fall on surface ACB of sphere.

Chapter 31 V

Refraction of Light — 123

Example 31.32 In the figure shown, m 1 > m 2 . Find minimum value of i so that TIR never takes place at P. P Q

a

b

m2 m1

i

Fig. 31.94

Solution

Here,

Let us take b = qC . Then, a = 90° - b or a = 90° - qC m m sin qC = R = 2 m D m1 æm qC = sin -1 çç 2 è m1

\

ö ÷ ÷ ø

Applying Snell’s law at point Q, m 2 sin i = m 1 sin a = m 1 sin ( 90° - qC ) æm ö = m 1 cos qC = m 1 cos sin -1 çç 2 ÷÷ è m1 ø \

ém æm i = sin -1 ê 1 cos sin -1 çç 2 è m1 ëm 2

öù ÷ú ÷ øû

Ans.

Now, we can see that starting from this value of i, we get b = qC . If i is increased from this value, then a will also increase (becomes > 90° - qC ) and b decreases from qC ( becomes < qC ) and no TIR takes place at P. So far no TIR condition i has to be increased from the above value. Or this is the minimum value of i. V

Example 31.33 Monochromatic light is Medium I ( m 1) incident on a plane interface AB between two D Medium III E media of refractive indices m 1 and m 2 (m 2 > m 1 ) (m3) F G at an angle of incidence q as shown in the B A figure. Medium II q (m2) The angle q is infinitesimally greater than the critical angle for the two media so that total internal reflection takes place. Now if a transparent slab DEFG of uniform thickness Fig. 31.95 and of refractive index m 3 is introduced on the interface (as shown in the figure), show that for any value of m 3 all light will ultimately be reflected back again into medium II. Consider separately the cases. (1986, 6M) (a) m 3 < m 1

(b) m 3 > m 1

124 — Optics and Modern Physics æm Given, q is slightly greater than sin -1 çç 1 èm2 (a) When m 3 < m 1 Solution

m 3 |R2|. Two point objects O1 and O2 are kept at same distance from the lens. Image distance of O1 from the lens will be more compared to the image distance of O2. O2

O1

R1

Reason : directly.

R2

If medium on two sides of the lens is different, we cannot apply lens formulae

11. Assertion : White light is incident on face AB of an isosceles right angled prism as shown. Colours, for which refractive index of material of prism is more than 1.414, will be able to emerge from the face AC. A

C

B

Reason : Total internal reflection cannot take place for the light travelling from a rarer medium to a denser medium.

12. Assertion : Image formed by concave lens is not always virtual. Reason : Image formed by a lens is real if the image is formed in the direction of ray of light with respect to the lens.

13. Assertion : Although the surfaces of goggle lens are curved, it does not have any power. Reason : In case of goggles, both the curved surfaces have equal radii of curvature and have centre of curvature on the same side.

Objective Questions 1. An endoscope is employed by a physician to view the internal parts of body organ. It is based on the principle of (b) reflection (d) dispersion

(a) refraction (c) total internal reflection

2. Refractive index m is given as m = A +

B l2

, where A and B are constants and l is wavelength, then

dimensions of B are same as that of (a) wavelength (c) pressure

(b) volume (d) area

3. A plane glass slab is placed over various coloured letters. The letter which appears to be raised the least is (a) violet

(b) yellow

(c) red

(d) green

164 — Optics and Modern Physics 4. Critical angle of light passing from glass to air is least for (a) red (c) yellow

(b) green (d) violet

5. The power in dioptre of an equi-convex lens with radii of curvature of 10 cm and refractive index 1.6 is (a) +12 (c) +1.2

(b) +18 (d) +1.8

6. The refractive index of water is 4/3. The speed of light in water is (a) 1.50 ´108 m/s (c) 2.25 ´108 m/s

(b) 1.78 ´108 m/s (d) 2.67 ´108 m/s

7. White light is incident from under water on the water-air interface. If the angle of incidence is slowly increased from zero, the emergent beam coming out into the air will turn from (a) white to violet (c) white to black

(b) white to red (d) None of these

8. When light enters from air to water, then its (a) (b) (c) (d)

frequency increases and speed decreases frequency is same, but the wavelength is smaller in water than in air frequency is same but the wavelength in water is greater than in air frequency decreases and wavelength is smaller in water than in air

9. In the figure shown

sin i is equal to sin r m1

m2

m3 r

i

(a)

m 22 m3 m1

(b)

m3 m1

(c)

m3 m1 m 22

(d)

m1 m3

10. In figure, the reflected ray B makes an angle 90° with the ray C. If i , r1 and r2 are the angles of incidence, reflection and refraction, respectively. Then, the critical angle of the medium is B

A i r1 r2

Denser medium Rarer medium C

(a) sin (c) r1

-1

(tan i )

(b) sin -1 (cot i ) (d) r2

11. A prism of apex angle A = 60° has the refractive index m = 2. The angle of incidence for minimum deviation is (a) 30° (c) 60°

(b) 45° (d) None of these

Chapter 31

Refraction of Light — 165

12. A thin equi-convex lens is made of glass of refractive index 1.5 and its focal length is 0.2 m. If it acts as a concave lens of 0.5 m focal length when dipped in a liquid, the refractive index of the liquid is (a)

17 8

(b)

15 8

(c)

13 8

(d)

9 8

13. A ray of light, travelling in a medium of refractive index m , is incident at an angle i on a composite transparent plate consisting of three plates of refractive indices m 1 , m 2 and m 3 . The ray emerges from the composite plate into a medium of refractive index m 4, at angle x. Then, (b) sin x =

(a) sin x = sin i (c) sin x =

m4 sin i m

(d) sin x =

m sin i m4 m1 m3 m

m 2 m 2m 4

sin i

14. The given equi-convex lens is broken into four parts and rearranged as shown. If the initial focal length is f, then after rearrangement the equivalent focal length is

(a) f

1

2

3

4

(b) f /2

(c) f /4

(d) 4f

15. A thin convergent glass lens ( m g =1.5) has a power of + 5.0 D. When this lens is immersed in a liquid of refractive index m l ,it acts as a divergent lens of focal length 100 cm. The value of m l is (a) 4/3 (c) 5/4

(b) 5/3 (d) 6/5

16. Two convex lenses of focal length 10 cm and 20 cm respectively placed coaxially and are separated by some distance d. The whole system behaves like a concave lens . One of the possible value of d is (a) 15 cm (c) 25 cm

(b) 20 cm (d) 40 cm

17. A prism can have a maximum refracting angle of (qC = critical angle for the material of prism) (a) 60° (c) 2 qC

(b) qC (d) slightly less than 180°

18. A ray of light is incident at small angle I on the surface of prism of small angle A and emerges normally from the opposite surface. If the refractive index of the material of the prism is m , the angle of incidence is nearly equal to A m (c) m A (a)

A 2m (d) m A/2 (b)

19. The refractive angle of a prism is A, and the refractive index of the material of the prism is cot ( A / 2). The angle of minimum deviation is (a) 180° – 3 A (c) 90° – A

(b) 180° + 2 A (d) 180° – 2 A

166 — Optics and Modern Physics 20. A prism of refractive index 2 has refractive angle 60°. In order that a ray suffers minimum deviation it should be incident at an angle of (a) 45° (c) 30°

(b) 90° (d) None

21. The focal length of a combination of two lenses is doubled if the separation between them is doubled. If the separation is increased to 4 times, the magnitude of focal length is (b) quadrupled (d) same

(a) doubled (c) halved

22. A convexo-concave convergent lens is made of glass of refractive index 1.5 and focal length 24 cm. Radius of curvature for one surface is double than that of the other. Then, radii of curvature for the two surfaces are (in cm) (a) 6, 12 (c) 3, 6

(b) 12, 24 (d) 18, 36

23. An optical system consists of a thin convex lens of focal length 30 cm and a plane mirror placed 15 cm behind the lens. An object is placed 15 cm in front of the lens. The distance of the final image from the object is (a) 60 cm (c) 75 cm

(b) 30 cm (d) 45 cm

24. In the figure shown, the angle made by the light ray with the normal in the medium of refractive index 2 is 45°

m1 = 1 m2 = Ö 3 m3 = Ö 2 m4 = 2 m5 = 1.6

(a) 30° (c) 90°

(b) 60° (d) None of these

25. For refraction through a small angled prism, the angle of minimum deviation (a) (b) (c) (d)

increases with increase in refractive index of a prism will be 2d for a ray of refractive index 2.4 if it is d for a ray of refractive index 1.2 is directly proportional to the angle of the prism will decrease with increase in refractive index of the prism

26. A ray of light passes from vacuum into a medium of refractive index n. If the angle of incidence is twice the angle of refraction, then the angle of incidence is (a) cos -1 (n / 2) (c) 2 cos -1 (n / 2)

(b) sin -1 (n / 2) (d) 2 sin -1 (n / 2)

27. A thin convex lens of focal length 30 cm is placed in front of a plane mirror. An object is placed at a distance x from the lens (not in between lens and mirror) so that its final image coincides with itself . Then, the value of x is (a) 15 cm (c) 60 cm

(b) 30 cm (d) Insufficient data

Chapter 31

Refraction of Light — 167

28. One side of a glass slab is silvered as shown in the figure. A ray of light is incident on the other side at angle of incidence 45°. Refractive index of glass is given as 2.The deflection suffered by the ray when it comes out of the slab is 45°

(a) 90°

(b) 180°

29. A prism has refractive index

(c) 120°

(d) 45°

3 and refractive angle 90°. Find the minimum deviation 2

produced by prism (a) 60°

(b) 45°

(c) 30°

(d) 15°

30. In figure, an air lens of radius of curvature of each surface equal to 10 cm is cut into a cylinder of glass of refractive index 1.5. The focal length and the nature of lens are (a) 15 cm diverging (c) 10 cm diverging

Air

(b) 15 cm converging (d) 10 cm converging

31. A point object is placed at a distance of 12 cm from a convex lens of focal length 10 cm. On the other side of the lens, a convex mirror is placed at a distance of 10 cm from the lens such that the image formed by the combination coincides with the object itself. The focal length of the convex mirror is (a) 20 cm

(b) 25 cm

(c) 15 cm

(d) 30 cm

32. An object, a convex lens of focal length 20 cm and a plane mirror are arranged as shown in the figure. How far behind the mirror is the second image formed?

12 cm

(a) 30 cm

(b) 20 cm

10 cm

(c) 40 cm

(d) 50 cm

33. Two parallel light rays pass through an isosceles prism of refractive index 3 / 2 as shown in figure. The angle between the two emergent rays is 45°

45°

(a) 15° (c) 45°

(b) 30° (d) 60°

168 — Optics and Modern Physics 34. A prism having refractive index 2 and refractive angle 30° has one of the refractive surfaces polished. A beam of light incident on the other surface will trace its path if the angle of incidence is (a) 0°

(b) 30°

(c) 45°

(d) 60°

35. In Fig. (i), a lens of focal length 10 cm is shown. It is cut into two parts and placed as shown in Fig. (ii). An object AB of height 1 cm is placed at a distance of 7.5 cm. The height of the image will be B

A

(ii)

(i)

(a) 2 cm

(b) 1 cm

(c) 1.5 cm

(d) 3 cm

36. The image for the converging beam after refraction through the curved surface is formed at m=1

m = 3/2

O

P

x

30 cm R = 20 cm

(a) x = 40 cm (c) x = – 40/3 cm

(b) x = 40/3 cm (d) x = 20 cm

37. A concavo-convex lens is made of glass of refractive index 1.5. The radii of curvature of its two surfaces are 30 cm and 50 cm. Its focal length when placed in a liquid of refractive index 1.4 is (a) 200 cm

(b) 500 cm

(c) 800 cm

(d) 1050 cm

38. From the figure shown, establish a relation between m 1 , m 2 and m 3 m1

m3

m2

(a) m 1 < m 2 < m 3

(b) m 3 < m 2 ; m 3 = m 1

(c) m 3 > m 2 ; m 3 = m 1

(d) None of these

39. When light of wavelength l is incident on an equilateral prism, kept on its minimum deviation position, it is found that the angle of deviation equals the angle of the prism itself. The refractive index of the material of the prism for the wavelength l is (a) (c) 2

3

(b)

3 /2

(d)

2

Chapter 31

Refraction of Light — 169

Subjective Questions 1. The laws of reflection or refraction are the same for sound as for light. The index of refraction of a medium (for sound) is defined as the ratio of the speed of sound in air 343 m/ s to the speed of sound in the medium. (a) What is the index of refraction (for sound) of water (v = 1498 m/s )? (b) What is the critical angle q, for total reflection of sound from water?

2. Light from a sodium lamp ( l0 = 589 nm ) passes through a tank of glycerin (refractive index

= 1.47) 20 m long in a time t1. If it takes a time t2 to transverse the same tank when filled with carbon disulfide (index = 1.63), determine the difference t2 - t1.

3. A light beam of wavelength 600 nm in air passes through film 1 ( n1 = 1.2) of thickness

1.0 mm , then through film 2 (air) of thickness 1.5 mm , and finally through film 3 ( n3 = 1.8) of thickness 1.0 mm (a) Which film does the light cross in the least time, and what is that least time? (b) What are the total number of wavelengths (at any instant) across all three films together?

4. A plate with plane parallel faces having refractive index 1.8 rests on a plane mirror. A light ray is incident on the upper face of the plate at 60°. How far from the entry point will the ray emerge after reflection by the mirror. The plate is 6 cm thick?

5. An object is at a distance of d = 2.5 cm from the surface of a glass

sphere with a radius R = 10 cm. Find the position of the final image produced by the sphere. The refractive index of glass is m = 1.5.

60° M

N

Mirror

6. An air bubble is seen inside a solid sphere of glass ( n = 1.5) of 4.0 cm diameter at a distance of 1.0 cm from the surface of the sphere (on seeing along the diameter). Determine the real position of the bubble inside the sphere.

7. Find the position of final image of an object O as shown in figure. O Air 3 cm

10 cm

RI = 3/2 Mirror

8. One face of a rectangular glass plate 6 cm thick is silvered. An object held 8 cm in front of the unsilvered face forms an image 10 cm behind the silvered face. Find the refractive index of glass. Consider all the three steps.

9. A shallow glass dish is 4.00 cm wide at the bottom as shown in figure. When an observer’s eye is positioned as shown, the observer sees the edge of the bottom of the empty dish. When this dish is filled with water, the observer sees the centre of the bottom of the dish. Find the height of the dish m w = 4/ 3.

h 4.00 cm

170 — Optics and Modern Physics 10. A glass prism in the shape of a quarter cylinder lies on a horizontal table. A uniform, horizontal light beam falls on its vertical plane surface as shown in the figure. If the radius of the cylinder is R = 5 cm and the refractive index of the glass is n = 1.5, where on the table beyond the cylinder, will a path of light be found? Light

R n

11. A glass sphere with 10 cm radius has a 5 cm radius spherical hole at its centre. A narrow beam of parallel light is directed into the sphere. Where, if anywhere, will the sphere produce an image? The index of refraction of the glass is 1.50.

12. A glass sphere has a radius of 5.0 cm and a refractive index of 1.6. A paperweight is constructed by slicing through the sphere on a plate that is 2.0 cm from the centre of the sphere and perpendicular to a radius of the sphere that passes through the centre of the circle formed by the intersection of the plane and the sphere. The paperweight is placed on a table and viewed from directly above an observer who is 8.0 cm from the table top, as shown in figure. When viewed through the paperweight, how far away does the table top appear to the observer?

Observer

8.0 cm 3.0 cm 5.0 cm

13. A fish is rising up vertically inside a pond with velocity 4 cm/ s, and notices a bird, which is diving downward and its velocity appears to be 16 cm/ s (to the fish). What is the real velocity of the diving bird, if refractive index of water is 4/ 3?

14. A lens with a focal length of 16 cm produces a sharp image of an object in two positions, which are 60 cm apart. Find the distance from the object to the screen.

15. Two glasses with refractive indices of 1.5 and 1.7 are used to make two identical double convex lenses. (a) Find the ratio between their focal lengths. (b) How will each of these lenses act on a ray parallel to its optical axis if the lenses are submerged into a transparent liquid with a refractive index of 1.6?

16. A converging beam of rays is incident on a diverging lens. Having passed through the lens the rays intersect at a point 15 cm from the lens. If the lens is removed, the point where the rays meet, move 5 cm closer to the mounting that holds the lens. Find the focal length of the lens.

17. A parallel beam of rays is incident on a convergent lens with a focal length of 40 cm. Where a divergent lens with a focal length of 15 cm be placed for the beam of rays to remain parallel after passing through the two lenses.

18. An optical system consists of two convergent lenses with focal lengths f1 = 20 cm andf2 = 10 cm.

The distance between the lenses is d = 30 cm. An object is placed at a distance of 30 cm from the first lens. At what distance from the second lens will the image be obtained?

19. Determine the position of the image produced by an optical system consisting of a concave mirror with a focal length of 10 cm and a convergent lens with a focal length of 20 cm. The distance from the mirror to the lens is 30 cm and from the lens to the object is 40 cm. Consider only two steps. Plot the image.

Chapter 31

Refraction of Light — 171

20. A parallel beam of light is incident on a system consisting of three thin lenses with a common optical axis. The focal lengths of the lenses are equal to f1 = + 10 cm and f2 = - 20 cm , and f3 = + 9 cm respectively. The distance between the first and the second lens is 15 cm and between the second and the third is 5 cm. Find the position of the point at which the beam converges when it leaves the system of lenses.

21. A point source of light S is placed at the bottom of a vessel containing a liquid of refractive index 5/ 3. A person is viewing the source from above the surface. There is an opaque disc of radius 1 cm floating on the surface. The centre of the disc lies vertically above the source S. The liquid from the vessel is gradually drained out through a tap. What is the maximum height of the liquid for which the source cannot at all be seen from above?

22. A ray of light travelling in glass (m g = 3/ 2) is incident on a horizontal glass-air surface at the

critical angle qC . If a thin layer of water (m w = 4/ 3) is now poured on the glass-air surface. At what angle will the ray of light emerges into water at glass-water surface?

23. A ray of light is incident on the left vertical face of glass cube of refractive index n 2 ,as shown in figure. The plane of incidence is the plane of the page, and the cube is surrounded by liquid (refractive index = n1). What is the largest angle of incidence q1 for which total internal reflection occurs at the top surface? n1

n2

q1

æ

3ö 2ø

æ

4ö 3ø

24. Light is incident from glass ç m g = ÷ to water ç m w = ÷. Find the range of the angle of deviation è

è

for refracted light.

25. The angle of minimum deviation for a glass prism with m = 3 equals the refracting angle of the prism. What is the angle of the prism?

26. A ray incident on the face of a prism is refracted and escapes through an adjacent face. What is the maximum permissible angle of the prism, if it is made of glass with a refractive index of m = 1.5?

27. In an equilateral prism of m = 1.5, the condition for minimum deviation

A

is fulfilled. If face AC is polished (a) Find the net deviation. (b) If the system is placed in water what will be the net deviation for same 4 angle of incidence? Refractive index of water = . 3

m = 1.5 B

C

28. In a certain spectrum produced by a glass prism of dispersive power 0.0305, it is found that the refractive index for the red ray is 1.645 and that for the violet ray is 1.665. What is the refractive index for the yellow ray?

29. An achromatic lens-doublet is formed by placing in contact a convex lens of focal length 20 cm and a concave lens of focal length 30 cm. The dispersive power of the material of the convex lens is 0.18. (a) Determine the dispersive power of the material of the concave lens. (b) Calculate the focal length of the lens-doublet.

172 — Optics and Modern Physics 30. An achromatic convergent lens of focal length 150 cm is made by combining flint and crown glass lenses. Calculate the focal lengths of both the lenses and point out which one is divergent if the ratio of the dispersive power of flint and crown glasses is 3 : 2.

31. The index of refraction of heavy flint glass is 1.68 at 434 nm and 1.65 at 671 nm. Calculate the difference in the angle of deviation of blue (434 nm) and red (671 nm) light incident at 65° on one side of a heavy flint glass prism with apex angle 60°.

LEVEL 2 Single Correct Option 1. A bird is flying over a swimming pool at a height of 2 m from the water surface. If the bottom is perfectly plane reflecting surface and depth of swimming pool is 1 m, then the distance of final image of bird from the bird itself is m w = 4/ 3 (a)

11 m 3

(b)

23 m 3

(c)

11 m 4

(d)

11 m 2

2. A parallel narrow beam of light is incident on the surface of a transparent hemisphere of radius R and refractive index m =1.5 as shown. The position of the image formed by refraction at the spherical surface only is R 2 R (c) 3 (a)

45°

(b) 3R (d) 2 R Normal

3. Consider the situation as shown in figure. The point O is the centre . The light ray forms an angle of 60° with the normal. The normal makes an angle 60° with the horizontal and each mirror makes an angle 60° with the normal. The value of refractive index of that spherical portion so that light ray retraces its path is (a)

2

(b)

2 3

(c)

3 2

(d)

Light ray 60° O

3

4. The figure shows an equi-convex lens. What should be the condition of the refractive indices so that the lens becomes diverging? (a) 2 m 3 > m 1 - m 2 (c) 2 m 2 > 2m 1 - m 3

m1

m2

m3

(b) 2 m 2 < m 1 + m 3 (d) None of these

5. An object is kept at a distance of 16 cm from a thin lens and the image formed is real. If the object is kept at a distance of 6 cm from the same lens, the image formed is virtual. If the sizes of the image formed are equal, the focal length of the lens will be (a) 19 cm (c) 21 cm

(b) 17 cm (d) 11 cm

6. The apparent depth of water in cylindrical water tank of diameter 2R cm is reducing at the rate of x cm/min when water is being drained out at a constant rate. The amount of water drained in cc/min is (n1 = refractive index of air, n 2 = refractive index of water) (a)

xpR2n1 n2

(b)

xpR2n2 n1

(c)

2pRn1 n2

(d) pR2x

Chapter 31

Refraction of Light — 173 A

7. A ray of light makes normal incidence on the diagonal face of a right angled prism as shown in figure. If q = 37°, then the angle of deviation after second step (from AB) is (sin 37° = 3/ 5) (a) (b) (c) (d)

53° 74° 106° 90°

Air Air B

90° m = 5/3 90° q C

8. A bird in air looks at a fish directly below it inside in a transparent liquid in a tank. If the distance of the fish as estimated by the bird is h1 and that of the bird as estimated by the fish is h2, then the refractive index of the liquid is

h2 h1 h1 + h2 (c) h1 - h2

h1 h2 h - h2 (d) 1 h1 + h2

(a)

(b)

9. Diameter of the flat surface of a circular plano-convex lens is 6 cm and thickness at the centre is 3 mm. The radius of curvature of the curved part is (b) 20 cm (d) 10 cm

(a) 15 cm (c) 30 cm

10. When the object is at distances u1 and u 2 from the optical centre of a convex lens, a real and a virtual image of the same magnification are obtained . The focal length of the lens is

(a)

u1 - u2 2

(b) u1 + u2

(c)

(d)

u1 u2

11. Two convex lenses placed in contact form the image of a distant object at

u1 + u2 2

A

B

P. If the lens B is moved to the right, the image will (a) (b) (c) (d)

move to the left move to the right remain at P move either to the left or right, depending upon focal lengths of the lenses

P

12. Two light rays 1 and 2 are incident on two faces AB and AC on an isosceles prism as shown in the figure. The rays emerge from the side BC. Then, B 75°

1 A 2

(a) (b) (c) (d)

30°

75° C

minimum deviation of ray 1 > minimum deviation of ray 2 minimum deviation of ray 1 < minimum deviation of ray 2 minimum deviation of ray 1 = minimum deviation of ray 2 Cannot be determined

7 and the angle of prism is 60°. The limiting angle of incidence 3 of a ray that will be transmitted through the prism is approximately

13. Refractive index of a prism is (a) 30°

(b) 45°

(c) 15°

(d) None of these

174 — Optics and Modern Physics 14. A plano-convex thin lens of focal length 10 cm is silvered at its plane surface. The distance d at which an object must be kept in order to get its image on itself is O (a) 5 cm (c) 10 cm

(b) 20 cm (d) 2.5 cm

d

15. There is a small black dot at the centre C of a solid glass sphere of refractive index m. When seen from outside, the dot will appear to be located (a) (b) (c) (d)

Away from the C for all values of m At C for all values of m. At C for m = 1.5, but away from C for m not equal to 1.5 At C for 2 < m < 1.5

16. In the figure ABC is the cross-section of a right angled prism and BCDE is the cross-section of a glass slab. The value of q so that light incident normally on the face AB does not cross the face BC is (Given sin-1 3/ 5 = 37° ) E m = 6/5

m = 3/2

B

q A

(a) q < 37°

C

(b) q < 53°

D

(c) q ³ 37°

(d) q ³ 53°

17. An object O is kept in air in front of a thin plano-convex lens of radius of curvature 10 cm. Its refractive index is 3/2 and the medium towards right of the plane surface is water of refractive index 4/3. What should be distance x of the object so that the rays become parallel finally? mw = 4/3

x O mg = 3/2

(a) 5 cm

(b) 10 cm

(c) 20 cm

(d) 15 cm

18. If a symmetrical bi-concave thin lens is cut into two identical halves, and they are placed in different ways as shown, then Object

(i)

(a) (b) (c) (d)

three images will be formed in case (i) two images will be formed in case (ii) the ratio of focal lengths in (ii) and (iii) is 1 the ratio of focal lengths in (ii) and (iii) is 2

(ii)

(iii)

Chapter 31

Refraction of Light — 175

19. If an object is placed at A (OA > f ); where f is the focal length of the lens, the image is formed at B. A perpendicular is erected at O and C is chosen on it such that the angle ÐBCA is a right angle. Then, the value of f will be C

B

AB OC 2 OC 2 (c) AB (a)

O

A

( AC ) (BC ) OC (OC ) ( AB) (d) AC + BC

(b)

20. An object is seen through a glass slab of thickness 36 cm and refractive index 3/2. The observer, and the slab are dipped in water (m = 4/3). The shift produced in the position of the object is (a) 12 cm (c) 6 cm

(b) 4 cm (d) 8 cm

21. How much water should be filled in a container of height 21 cm, so that it appears half filled to the observer when viewed from the top of the container (m = 4/3). (a) 8 cm (c) 12 cm

(b) 10.5 cm (d) 14 cm

22. Optic axis of a thin equi-convex lens is the x-axis. The co-ordinates of a point object and its image are (– 40 cm, 1 cm) and (50 cm, – 2 cm), respectively. Lens is located at (a) x = 20 cm (c) x = - 10 cm

(b) x = - 30 cm (d) origin

23. A thin plano-convex lens acts like a concave mirror of radius of curvature 20 cm when its plane surface is silvered. The radius of curvature of the curved surface if index of refraction of its material is 1.5 will be (a) 40 cm (c) 10 cm

(b) 30 cm (d) 20 cm

24. A thin lens, made of glass of refractive index 3/2, produces a real and magnified image of an object in air. If the whole system, maintaining the same distance between the object and the lens, is immersed in water (RI = 4/3), then the image formed will be (a) real, magnified (c) virtual, magnified

(b) real, diminished (d) virtual, diminished

25. The maximum value of refractive index of a prism which permits the transmission of light through it when the refracting angle of the prism is 90°, is given by (a) 1.500 (c) 2.000

(b) 1.414 (d) 1.732

26. A glass slab of thickness 4 cm contains the same number of waves as 5 cm of water, when both are traversed by the same monochromatic light. If the refractive index of water is 4/3, then refractive index of glass is (a) 5/3 (c) 16/15

(b) 5/4 (d) 1.5

176 — Optics and Modern Physics 27. If the optic axis of convex and concave lenses are separated by a distance 5 mm as shown in the figure. Find the coordinate of the final image formed by the combination if parallel beam of light is incident on lens. Origin is at the optical centre of convex lens f = 20 cm f = – 10 cm Principal axis of concave lens Principal axis of convex lens

5 mm

30 cm

(a) (25 cm, 0.5 cm) (c) (25 cm, – 0.5 cm)

(b) (25 cm, 0.25 cm) (d) (25 cm, – 2.5 cm)

28. A light source S is placed at the centre of a glass sphere of radius R and refractive index m. The maximum angle q with the x-axis (as shown in the figure) an incident light ray can make without suffering total internal reflection is

S

æ1ö (a) cos -1 ç ÷ èm ø -1 æ 1 ö (c) tan ç ÷ èm ø

æ

q

x

æ1ö (b) sin -1 ç ÷ èm ø (d) there will never be total internal reflection

4ö

29. A sphere ç m = ÷ of radius 1 m has a small cavity of diameter 1 cm at its centre. An observer 3ø è who is looking at it from right, sees the magnification of diameter of the cavity as

4 3 (c) 1 (a)

3 4 (d) 0.5 (b)

30. An equi-convex lens of m = 1.5 and R = 20 cm is cut into two equal parts along its axis. Two parts are then separated by a distance of 120 cm (as shown in figure). An object of height 3 mm is placed at a distance of 30 cm to the left of first half lens. The final image will form at

120 cm

(a) (b) (c) (d)

120 cm to the right of first half lens, 3 mm in size and inverted 150 cm to the right of first half lens, 3 mm in size and erect 120 cm to the right of first half lens, 4 mm in size and inverted 150 cm to the right of first half lens, 4 mm in size and erect

Refraction of Light — 177

Chapter 31

31. As shown in the figure, region BCDEF and ABFG are of refractive index 2.0 and 1.5 respectively. A particle O is kept at the mid of DH. Image of the object as seen by the eye is at a distance B

C

m=1

R = 10 cm O Eye

D

A m = 1.5

H

m=2

G

E 20 cm

50 cm

F

(b) 22.5 cm from point D (d) 20 cm from point D

(a) 10 cm from point D (c) 30 cm from point D

32. A point object O is placed at a distance of 20 cm from a convex lens of focal length 10 cm as shown in the figure. At what distance x from the lens should a convex mirror of focal length 60 cm, be placed so that final image coincide with the object? (a) (b) (c) (d)

O

10 cm 20 cm 40 cm 20 cm Final image can never coincide with the object in the given conditions

x

33. A point object is placed at a distance of 20 cm from a thin plano-convex lens of focal length 15 cm (m = 1.5). The curved surface is silvered. The image will form at A

L

O

20 cm

(a) 60 cm left of AB 20 cm left of AB 7 Note Neglect thickness of lens. (c)

B

(b) 30 cm left of AB (d) 60 cm right of AB

34. A flat glass slab of thickness 6 cm and index 1.5 is placed in front of a plane mirror. An observer is standing behind the glass slab and looking at the mirror. The actual distance of the observer from the mirror is 50 cm. The distance of his image from himself, as seen by the observer is (a) 94 cm (c) 98 cm

(b) 96 cm (d) 100 cm

35. Distance of an object from the first focus of an equi-convex lens is 10 cm and the distance of its real image from second focus is 40 cm. The focal length of the lens is (a) 25 cm (c) 20 cm

(b) 10 cm (d) 40 cm

178 — Optics and Modern Physics 36. A cubical block of glass of refractive index n1 is in contact with the surface of water of refractive index n 2. A beam of light is incident on vertical face of the block. After refraction a total internal reflection at the base and refraction at the opposite face take place. The ray emerges at angle q as shown. The value of q is given by n1

q

n2

(a) sin q < n12 - n22 1 (c) sin q < 2 n1 - n22

(b) cos q < n12 - n22 1 (d) cos q < 2 n1 - n22

37. A concave mirror of focal length 2 cm is placed on a glass slab as shown in the figure. The image of point object O formed due to reflection at mirror and then refraction by the slab (a) (b) (c) (d)

is virtual and at 2 cm from pole of the concave mirror is virtual and on the pole of mirror is real and on the object itself None of the above

1 cm n=1 n = 3/2 9 cm n=1

38. Two refracting media are separated by a spherical interface as shown in the figure. AB is the principal axis, m 1 and m 2 are the refractive indices of medium of incidence and medium of A refraction respectively. Then, (a) (b) (c) (d)

O

m2

m1 B

if m 2 > m 1, then there cannot be a real image of real object if m 2 > m 1, then there cannot be a real image of virtual object if m 1 > m 2, then there cannot be a real image of virtual object if m 1 > m 2, then there cannot be a virtual image of virtual object

39. A concavo-convex lens has refractive index 1.5 and the radii of curvature of its surfaces are 10 cm and 20 cm. The concave surface is upwards and is filled with oil of refractive index 1.6. The focal length of the combination will be (a) 18.18 cm (c) 22 cm

(b) 15 cm (d) 28.57 cm

40. A convex spherical refracting surface separates two media glass and air (m g = 1.5). If the image is to be real, at what minimum distance u should the object be placed in air if R is the radius of curvature (a) u > 3R (c) u < 4R

(b) u > 2R (d) u < R

41. An object is moving towards a converging lens on its axis. The image is also found to be moving towards the lens. Then, the object distance u must satisfy (a) 2 f < u < 4 f (c) u > 4 f

(b) f < u < 2 f (d) u < f

Chapter 31

Refraction of Light — 179

42. Two diverging lenses are kept as shown in figure. The final image formed will be

O d1

(a) (b) (c) (d)

d2

virtual for any value of d1 and d2 real for any value of d1 and d2 virtual or real depends on d1 and d2 only virtual or real depends on d1 and d2 and also on the focal lengths of the lens

43. In the figure shown, a point object O is placed in air on the principal axis. The radius of curvature of the spherical surface is 60 cm. I is the final image formed after all reflections and refractions. ng = 3/2 O x y

(a) (b) (c) (d)

If x = 120 cm, then I is formed on O for any value of y If x = 240 cm, then I is formed on O only if y = 360 cm If x = 240 cm, then I is formed on O for any value of y None of the above

44. In the figure, a point object O is placed in air. A spherical boundary separates two media of radius of curvature 1.0 m. AB is principal axis. The separation between the images formed due to refraction at A spherical surface is (a) 12 m (c) 14 m

1.6

O

B 2m

2.0

(b) 20 m (d) 10 m

More than One Correct Option 1. n number of identical equilateral prisms are kept in contact as shown in figure. If deviation through a single prism is d. Then, ( n , m are integers) 2 1

(a) (b) (c) (d)

4 3

5

if n = 2m, deviation through n prisms is zero if n = 2m + 1, deviation through system of n prisms is d if n = 2m, deviation through system of n prisms is d if n = 2m + 1, deviation through system of n prisms is zero

n

180 — Optics and Modern Physics 2. A ray of monochromatic light is incident on the plane surface of separation between two media x and y with angle of incidence i in the medium x and angle of refraction r in the medium y. The graph shows the relation between sin i and sin r. sin r

30° sin i

(a) The speed of light in the medium y is 3 times than in medium x 1 (b) The speed of light in the medium y is times than in medium x 3 (c) The total internal reflection can take place when the incidence is in x (d) The total internal reflection can take place when the incidence is in y

3. Which of the following statement(s) is/are true? (a) In vacuum the speed of red colour is more than that of violet colour (b) An object in front of a mirror is moved towards the pole of a spherical mirror from infinity, it is found that image also moves towards the pole. The mirror must be convex (c) There exist two angles of incidence in a prism for which angles of deviation are same except minimum deviation (d) A ray travels from a rarer medium to denser medium. There exist three angles of incidence for which the deviation is same

4. A lens of focal length f is placed in between an object and screen at a distance D. The lens forms two real images of object on the screen for two of its different positions, a distance x apart. The two real images have magnifications m1 and m2 respectively ( m1 > m2 ). Choose the correct statement(s). (a) m1m2 = - 1 D 2 - x2 (c) f = 4D

(b) m1m2 = 1 (d) D ³ 4 f

5. A small angled prism of apex angle A = 4° and refractive index m = 1.5 is placed in front of a vertical plane mirror as shown in figure. If the mirror is rotated through an angle q, then the light ray becomes horizontal either after the mirror or after second time passing from the prism in opposite direction. The value of q is

4°

(a) 1° (c) 4°

(b) 2° (d) Not possible

Chapter 31

Refraction of Light — 181

Comprehension Based Questions Passage : (Q. No. 1 to 3) A plano-convex lens P and a concavo-convex lens Q are in contact as shown in figure. The refractive index of the material of the lens P and Q is 1.8 and 1.2 respectively. The radius of curvature of the concave surface of the lens Q is double the radius of curvature of the convex surface. The convex surface of Q is silvered.

P Q

1. An object is placed on the principal axis at a distance 10 cm from the plane surface. The image is formed at a distance 40 cm from the plane surface on the same side. The focal length of the system is (a) – 8 cm

(c) -

(b) 8 cm

40 cm 3

(d)

40 cm 3

2. The radius of curvature of common surface is (b) 24 cm (d) 8 cm

(a) 48 cm (c) 12 cm

3. If the plane surface of P is silvered as shown in figure, the system acts as (a) (b) (c) (d)

convex mirror of focal length 24 cm concave mirror of focal length 8 cm concave mirror of focal length 24 cm convex mirror of focal length 8 cm

P Q

Match the Columns 1. Match the following two columns for a convex lens corresponding to object position shown in Column I. Column I (a) (b) (c) (d)

Between O and F1 Between F1 and 2F1 Between O and F2 Between F2 and 2F2

Column II (p) (q) (r) (s)

Real Virtual Erect Inverted

Note O ® optical centre, F1 ® first focus and F2 ® second focus.

2. Match the following two columns for a concave lens corresponding to object position shown in Column II. Column I (a) (b) (c) (d)

Between O and F1 Between F1 and 2F1 Between O and F2 Between F2 and 2F2

Column II (p) (q) (r) (s)

Real Virtual Erect Inverted

182 — Optics and Modern Physics 3. Match the following two columns corresponding to single refraction from plane surface. In all cases shown in Column I, m 1 > m 2. Column I (a)

Column II (p) Image distance greater than x from plane surface

O x 1 2

(q) Image distance less than x from plane surface

(b) 1 2

x O

(c)

1 2

(r)

Real image

(s)

Virtual image

x O

(d)

O x 1 2

4. Match the following two columns. Column I

(a)

Air

(b)

Air

Column II

O

(p) Real image

O

(c)

(r) May be real or virtual image

Air O

(d)

(q) Virtual image

(s) Image is at infinity Air

O

Chapter 31

Refraction of Light — 183

5. A convex lens L1 and a concave lens L2 have refractive index 1.5. Match the following two columns. Column I

Column II

(a) L1 is immersed in a liquid of refractive (p) Lens will behave as convex lens index 1.4 (b) L1 is immersed in a liquid of refractive (q) Lens will behave as concave lens index 1.6 (c) L 2 is immersed in a liquid of refractive (r) Magnitude of power of lens will index 1.4 increase (d) L 2 is immersed in a liquid of refractive (s) Magnitude of power of lens will index 1.6 decrease

6. Consider a linear extended object that could be real or virtual with its length at right angles to the optic axis of a lens. With regard to image formation by lenses. Column I

Column II

(a) Image of the same size as the object (b) Virtual image of a size greater than the object (c) Real image of a size smaller than the object (d) Real and magnified image

(p) Concave lens in case of real object (q) Convex lens in case of real object (r) Concave lens in case of virtual object (s) Convex lens in case of virtual object

Subjective Questions 1. Figure shows the optical axis of a lens, the point source of light A and its virtual image A¢. Trace the rays to find the position of the lens and of its principal focus. What type of lens is it?

A¢ A

2. Solve the problem similar to the previous one if A and A¢ are interchanged. 3. In figure, a fish watcher watches a fish through a 3.0 cm thick glass wall of a fish tank. The watcher is in level with the fish; the index of refraction of the glass is 8/ 5 and that of the water is 4/ 3. 6.8 cm

8.0 cm Observer Wall

Water

(a) To the fish, how far away does the watcher appear to be? (b) To the watcher, how far away does the fish appear to be?

4. A concave spherical mirror with a radius of curvature of 0.2 m is filled with water. What is the focal length of this system? Refractive index of water is 4/ 3.

184 — Optics and Modern Physics 5. A convexo-convex lens has a focal length of f1 = 10 cm. One of the lens surfaces having a radius

of curvature of R = 10 cm is coated with silver. Construct the image of the object produced by the given optical system and determine the position of the image if the object is at a distance of a = 15 cm from the lens. Refractive index of lens = 1.5.

6. A lens with a focal length of f = 30 cm produces on a screen a sharp image of an object that is at a distance of a = 40 cm from the lens. A plane parallel plate with thickness of d = 9 cm is placed between the lens and the object perpendicular to the optical axis of the lens. Through what distance should the screen be shifted for the image of the object to remain distinct? The refractive index of the glass of the plate is m = 1.8.

7. One side of radius of curvature R2 = 120 cm of a convexo-convex lens of material of refractive index m = 1.5 and focal length f1 = 40 cm is slivered. It is placed on a horizontal surface with silvered surface in contact with it. Another convex lens of focal length f2 = 20 cm is fixed coaxially d = 10 cm above the first lens. A luminous point object O on the axis gives rise to an image coincident with it. Find its height above the upper lens.

8. A small object is placed on the principal axis of concave spherical mirror of radius 20 cm at a distance of 30 cm. By how much will the position of the image alter only after mirror, when a parallel-sided slab of glass of thickness 6 cm and refractive index 1.5 is introduced between the centre of curvature and the object? The parallel sides are perpendicular to the principal axis.

9. A thin glass lens of refractive index m 2 = 1.5 behaves as an interface between two media of

refractive indices m 1 = 1.4 and m 3 = 1.6 respectively. Determine the focal length of the lens for the shown arrangement of radius of curvature of both the surfaces 20 cm.

m1

m3 m2

10. A glass hemisphere of radius 10 cm and m = 1.5 is silvered over its curved surface. There is an air bubble in the glass 5 cms from the plane surface along the axis. Find the position of the images of this bubble seen by observer looking along the axis into the flat surface of the atmosphere.

11. A equilateral prism of flint glass (m g = 3/ 2) is placed inside water (m w = 4/ 3). (a) At what angle should a ray of light fall on the face of the prism so that inside the prism the ray is perpendicular to the bisector of the angle of the prism. (b) Through what angle will the ray turn after passing through both faces of the prism?

12. Rays of light fall on the plane surface of a half cylinder at an angle 45° in the plane perpendicular to the axis (see figure). Refractive index of glass is 2. Discuss the condition that the rays do not suffer total internal reflection.

Chapter 31

Refraction of Light — 185

13. The figure shows an arrangement of an equi-convex lens and a concave mirror. A point object O is placed on the principal axis at a distance 40 cm from the lens such that the final image is also formed at the position of the object. If the radius of curvature of the concave mirror is 80 cm, find the distance d. Also draw the ray diagram. The focal length of the lens in air is 20 cm. O

m2 = 1.5 m3 = 2.0

m1 = 1.2

d

40 cm

14. A convex lens is held 45 cm above the bottom of an empty tank. The image of a point at the bottom of a tank is formed 36 cm above the lens. Now, a liquid is poured into the tank to a depth of 40 cm. It is found that the distance of the image of the same point on the bottom of the tank is 48 cm above the lens. Find the refractive index of the liquid.

15. A parallel beam of light falls normally on the first face of a prism of a small angle. At the second face it is partly transmitted and partly reflected. The reflected beam striking at the first face again emerges from it in a direction making an angle of 6°30¢ with the reversed direction of the incident beam. The refracted beam is found to have undergone a deviation of 1°15 from the original direction. Calculate the refractive index of the glass and the angle of the prism.

16. Two converging lenses of the same focal length f are separated by a distance 2f. The axis of the second lens is inclined at angle q = 60° with respect to the axis of the first lens. A parallel paraxial beam of light is incident from left side of the lens. Find the coordinates of the final image with respect to the origin of the first lens. y 60°

x 60°

2f

17. A cubical vessel with non-transparent walls is so located that the eye of an observer does not see its bottom but sees all of the wall CD. To what height should water be poured into the vessel for the observer to see an object F arranged at a distance of b = 10 cm from corner D? The face of the vessel is a = 40 cm. 4 Refractive index of water is . 3

C

B

F A

b

D

186 — Optics and Modern Physics 18. A spherical ball of transparent material has index of refraction m. A narrow beam of light AB is aimed as shown. What must the index of refraction be in order that the light is focused at the point C on the opposite end of the diameter from where the light entered? Given that x 1 For 0 £ x £ 1, de-Broglie wavelength is l 1 and for x > 1 the de-Broglie wavelength l (JEE 2005) is l 2 . Total energy of the particle is 2E 0 . Find 1 . l2 Solution

For 0 £ x £ 1, PE = E 0

\

Kinetic energy K 1 = Total energy – PE = 2E 0 - E 0 = E 0

\

l1 =

h

…(i)

2mE 0

For x > 1, PE = 0 \

Kinetic energy K 2 = Total energy = 2E 0

\

l2 =

h 4mE 0

From Eqs. (i) and (ii), we have l1 = 2 l2

…(ii)

260 — Optics and Modern Physics INTRODUCTORY EXERCISE

33.1

1. Find the energy and momentum of a photon of ultraviolet radiation of 280 nm wavelength. 2. A small plate of a metal is placed at a distance of 2 m from a monochromatic light source of

wavelength 4.8 ´ 10-7 m and power 1.0 Watt. The light falls normally on the plate. Find the number of photons striking the metal plate per square metre per second.

3. A proton and a deuteron are accelerated by same potential difference. Find the ratio of their de-Broglie wavelengths.

4. A deuteron and an a - particle have same kinetic energy. Find the ratio of their de-Broglie wavelengths.

5. Two protons are having same kinetic energy. One proton enters a uniform magnetic field at right angles to it. Second proton enters a uniform electric field in the direction of field. After some time their de-Broglie wavelengths are l1 and l 2, then (a) l1 = l 2 (b) l1 < l 2 (c) l1 > l 2 (d) some more information is required

6. Find the de-Broglie wavelengths of (a) a 46 g golf ball with a velocity of 30 m/s

(b) an electron with a velocity of 107 m /s.

33.5 Early Atomic Structures Every atom consists of a small nucleus of protons and neutrons with a number of electrons some distance away. In the present article and in the next, our chief concern will be the structure of the atom, since it is this structure that is responsible for nearly all the properties of matter.In nineteenth century many models were present by different Electron scientists, but ultimately the first theory of the atom to meet with any success was put forward in 1913 by Neils Bohr. But before studying Bohr's model of atom let us have a look on other two models of the period one presented by J.J. Thomson Positively charged matter in 1898 and the other by Ernest Rutherford in 1911. J.J. Thomson suggested that atoms are just positively charged Fig. 33.2 The Thomson model of the The Rutherford scattering lumps of matter with electrons embedded in them like raisins in atom. experiment showed it to be incorrect. a fruit cake. Thomson's model called the ‘plum pudding’ model is illustrated in Fig. 33.2. Thomson had played an important role in discovering the electron, his idea was taken seriously. But, the real atom turned out to be quite different.

Rutherford's Nuclear Atom The nuclear atom is the basis of the modern theory of atomic structure and was proposed by Rutherford in 1911. He, with his two assistants Geiger and Marsden did an experiment in which they directed a narrow beam of a-particles onto gold foil about1mm thick and found that while most of the

Chapter 33

Modern Physics - I — 261

particles passed straight through, some were scattered appreciably and a very few about 1 in 8000 suffered deflection of more than 90°. To account for this very surprising result Rutherford suggested that : “All the positive charge and nearly all the mass were concentrated in a very small volume or nucleus at the centre of the atom. The electrons were supposed to move in circular orbits round the nucleus (like planets round the sun). The electrostatic attraction between the two opposite charges being the required centripetal force for such motion. The large angle scattering of a-particles would then be explained by the strong electrostatic repulsion from the nucleus. + Rutherford's model of the atom, although strongly supported by evidence for the nucleus, is inconsistent with classical physics. An electron moving in a circular orbit round a nucleus is accelerating and according to electromagnetic theory it should therefore, emit radiation continuously and thereby lose energy. If this happened the radius of the orbit would decrease and the electron would spiral into the nucleus in a fraction of second. But atoms do not collapse. In 1913, an effort was made by Neils Bohr to overcome this paradox.

Nucleus

Electron e–

Fig. 33.3 An atomic electron should, classically, spiral rapidly into the nucleus as it radiates energy due to its acceleration

33.6 The Bohr Hydrogen Atom After Neils Bohr obtained his doctorate in 1911, he worked under Rutherford for a while. In 1913, he presented a model of the hydrogen atom, which has one electron and one proton. He postulated that an electron moves only in certain circular orbits, called stationary orbits. In stationary orbits, electron does not emit radiation, contrary to the predictions of classical electromagnetic theory. According to Bohr, there is a definite energy associated with each stable orbit and an atom radiates energy only when it makes a transition from one of these orbits to another. The energy is radiated in the form of a photon with energy and frequency given by DE = hf = E i - E f

F

M + e

e– – m vn

rn

Fig. 33.4

…(i)

Bohr found that the magnitude of the electron's angular momentum is quantised, and this magnitude h for the electron must be integral multiple of . The magnitude of the angular momentum is L = mvr 2p for a particle with mass m moving with speed v in a circle of radius r. So, according to Bohr's postulate, nh (n =1, 2, 3 . . . .) mvr = 2p Each value of n corresponds to a permitted value of the orbit radius, which we will denote by rn and the corresponding speed v n . The value of n for each orbit is called principal quantum number for the orbit. Thus, nh …(ii) mv n rn = 2p

262 — Optics and Modern Physics mv 2 is rn needed to the electron which is being provided by the electrical attraction between the positive proton and the negative electron.

According to Newton's second law, a radially inward centripetal force of magnitude F =

2

mv n 1 e2 = 4pe 0 rn 2 rn

Thus,

…(iii)

Solving Eqs. (ii) and (iii), we get rn = vn =

and

e 0n2h2 pme

2

æ nth orbit radius ö ÷÷ çç è in Bohr model ø

…(iv)

ö ÷÷ ø

…(v)

æ nth orbit speed çç è in Bohr model

e2 2e 0 nh

The smallest orbit radius corresponds to n =1. We'll denote this minimum radius, called the Bohr radius as a 0 . Thus, a0 =

e 0h2 pme 2

Substituting values of e 0 , h, p, m and e, we get a 0 = 0.529 ´ 10 -10 m = 0.529 Å

…(vi)

Eq. (iv), in terms of a 0 can be written as rn = n 2 a 0

or

rn µ n 2

Similarly, substituting values of e, e 0 and h with n =1 in Eq. (v), we get c . ´ 10 6 m/s » v1 = 219 137

…(vii)

…(viii)

This is the greatest possible speed of the electron in the hydrogen atom. Which is approximately equal to c/137, where c is the speed of light in vacuum. Eq. (v), in terms of v1 can be written as v 1 …(ix) or vn µ vn = 1 n n Energy levels

Kinetic and potential energies K n and U n in nth orbit are Kn =

and

1 me 4 2 mv n = 2 2 8 e 0 n2h2

Un = -

1 ( e)( e) me 4 =2 4pe 0 rn 4 e 0 n2h2

Modern Physics - I — 263

Chapter 33 The total energy E n is the sum of the kinetic and potential energies. me 4 En = Kn + U n = 2 8 e 0 n2h2

Substituting values of m, e, e 0 and h with n =1, we get the least energy of the atom in first orbit, which is -136 . eV. Hence, …(x) E1 = - 136 . eV E 136 . …(xi) and En = 1 = eV 2 2 n n Substituting n = 2, 3, 4 . . ., etc., we get energies of atom in different orbits. E 2 = - 3.40 eV, E 3 = - 1.51eV, . . . E ¥ = 0 Ionization energy of the hydrogen atom is the energy required to remove the electron completely. In ground state (n =1), energy of atom is –13.6 eV and energy corresponding to n = ¥ is zero. Hence, energy required to remove the electron from ground state is 13.6 eV.

Emission spectrum of hydrogen atom Under normal conditions the single electron in hydrogen atom stays in ground state (n =1). It is excited to some higher energy state when it acquires some energy from external source. But, it hardly stays there for more than 10 -8 second. A photon corresponding to a particular spectrum line is emitted when an atom makes a transition from a state in an excited level to a state in a lower excited level or the ground level. Let ni be the initial and n f the final energy state, then depending on the final energy state following series are observed in the emission spectrum of hydrogen atom. Balmer series (visible light)

n=7 n=6 n=5 n=4 n=3

Paschen series (infrared) Brackett series (infrared) Pfund series (infrared)

Lyman series (ultraviolet)

Lyman series

Paschen Pfund series series –0.28 eV –0.38 eV –0.54 eV –0.85 eV –1.51 eV Brackett series –3.40 eV

n=2 Balmer series

n=1 n=2 n=3 n=4 n=5 n=1

n=6

–13.6 eV

Fig. 33.5

For the Lyman series n f =1, for Balmer series n f = 2 and so on. The relation of the various spectral series to the energy levels and to electron orbits is shown in figure.

264 — Optics and Modern Physics Wavelength of Photon Emitted in De-excitation According to Bohr when an atom makes a transition from higher energy level to a lower energy level it emits a photon with energy equal to the energy difference between the initial and final levels. If E i is the initial energy of the atom before such a transition, E f is its final energy after the transition, and hc the photon's energy is hf = , then conservation of energy gives l hc (energy of emitted photon) …(xii) hf = = Ei - E f l By 1913, the spectrum of hydrogen had been studied intensively. The visible line with longest wavelength, or lowest frequency is in the red and is called H a , the next line, in the blue-green is called Hb and so on. In 1885, Johann Balmer, a swiss teacher found a formula that gives the wave lengths of these lines. This is now called the Balmer series. The Balmer's formula is 1 1 ö æ 1 =Rç ÷ 2 l è2 n2 ø

…(xiii)

Here, n = 3, 4, 5 . . ., etc. R = Rydberg constant = 1.097 ´ 10 7 m -1 and l is the wavelength of light/photon emitted during transition. For n = 3, we obtain the wavelength of H a line. Similarly, for n = 4, we obtain the wavelength of Hb line. For n = ¥, the smallest wavelength (= 3646 Å) of this series is obtained. Using the relation hc E = , we can find the photon energies corresponding to the wavelength of the Balmer series. l Multiplying Eq. (xiii) by hc, we find hc 1 ö Rhc Rhc æ 1 E= = hcR ç ÷ = 2 - 2 = En - E2 2 l è2 n2 ø 2 n This formula suggests that En = -

Rhc n2

, n = 1, 2, 3 . . .

…(xiv)

Comparing this with Eq. (xi) of the same article, we have Rhc =1360 . eV

…(xv)

The wavelengths corresponding to other spectral series (Lyman, Paschen , etc.) can be represented by formula similar to Balmer formula. 1 1 ö æ 1 Lyman series =Rç ÷, n = 2, 3, 4 . . . 2 l è1 n2 ø Paschen series

1 1 ö æ 1 =Rç ÷ , n = 4, 5, 6 . . . 2 l è3 n2 ø

Brackett series

1 1 ö æ 1 =Rç ÷ , n = 5, 6, 7 . . . 2 l è4 n2 ø

Modern Physics - I — 265

Chapter 33 1 1 ö æ 1 =Rç ÷ , n = 6, 7, 8 . . . 2 l è5 n2 ø

Pfund series

The Lyman series is in the ultraviolet, and the Paschen, Brackett and Pfund series are in the infrared region.

33.7 Hydrogen Like Atoms The Bohr model of hydrogen can be extended to hydrogen like atoms, i.e. one electron atoms such as singly ionized helium ( He + ), doubly ionized lithium ( Li +2 ) and so on. In such atoms, the nuclear charge is +Ze, where Z is the atomic number, equal to the number of protons in the nucleus. The effect in the previous analysis is to replace e 2 everywhere by Ze 2 . Thus, the equations for, rn , v n and E n are altered as under rn =

e 0n2h2 pmZe 2

=

n2 a0 Z

rn µ

or

n2 Z

…(i)

a 0 = 0.529 Å

where, vn =

(radius of first orbit of H)

2

Ze Z = v1 2e 0 nh n

vn µ

or

Z n

…(ii)

v1 = 2.19 ´ 10 6 m /s

where,

En = –

mZ 2 e 4 8 e 20 n 2 h 2

=

(speed of electron in first orbit of H)

Z2 n2

E1

or

En µ

E1 = – 13.60 eV

where,

Z2 n2

…(iii)

(energy of atom in first orbit of H)

Fig. 33.6 compares the energy levels of H and He which has Z = 2. H and He + have many spectrum lines that have almost the same wavelengths +

E n=3

E3 = –1.5 eV

n=2

E2 = –3.4 eV

E1 = –13.6 eV

n=1

n=6 n=5 n=4

E6 = –1.5 eV E5 = –2.2 eV E4 = –3.4 eV

n=3

E3 = –6.0 eV

n=2

E2 = –13.6 eV

H E1 = –54.4 eV

n=1 He +

+

Fig. 33.6 Energy levels of H and He . Because of the additional factor Z 2 in the energy expression, the energy of the He+ ion with a given n is almost exactly four times that of the H-atom with the same n. There are small differences (of the order of 0.05%) because of the different masses.

266 — Optics and Modern Physics Extra Points to Remember ˜

Bohr’s theory is applicable for hydrogen and hydrogen like atoms/ions. For such types of atoms/ions number of electron is one. Although, atomic numbers may be different. e.g, For 1H1, atomic number Z = 1, For He + , atomic number Z = 2 and For Li +2 , atomic number Z = 3 But for all three number of electron is one.

˜

In nth orbit

(e ) (Ze ) 1 mv 2 and = × 4p e0 r r2

Ln = mvr =

nh 2p

After solving these two equations, we will get following results. n2 1 Z and r µ (b) v µ and v µ m0 (a) r µ Z n m Z2 (c) E µ 2 and E µ m (d) r1H = 0.529 Å n c (f) E1H = - 13.6 eV (e) v1H = 2.19 ´ 106 m/s = 137 (g) K =|E| and U = 2 E

Note With the help of above results, we can find any value in any orbit of hydrogen like atoms. In the above expressions, m is the mass of electron ˜

˜

Total number of emission lines from some higher energy state n1 to lower energy state n2 (< n1 ) is given by (n1 – n2 ) (n1 – n2 + 1) . 2 n (n – 1) . For example, total number of lines from n1 = n to n2 = 1 are 2 As the principal quantum number n is increased in hydrogen and hydrogen like atoms, some quantities are decreased and some are increased. The table given below shows which quantities are increased and which are decreased.

Table 33.1 Increased

Decreased

Radius

Speed

Potential energy

Kinetic energy

Total energy

Angular speed

Time period

Frequency

Angular momentum ˜

1ö æ Whenever the force obeys inverse square law ç F µ 2 ÷ and potential energy is inversely proportional to r ø è 1ö æ r ç U µ ÷ , kinetic energy (K ), potential energy (U ) and total energy (E ) have the following relationships. rø è |U| U and E = – K = K= 2 2 1 1 If force is not proportional to 2 or potential energy is not proportional to , the above relations do not r r hold good.

Chapter 33 ˜

˜

˜

Modern Physics - I — 267

Total energy of a closed system is always negative and the modulus of this is the binding energy of the system. For instance, suppose a system has a total energy of –100 J. It means that this system will separate if 100 J of energy is supplied to this. Hence, binding energy of this system is 100 J. Thus, total energy of an open system is either zero or greater than zero. Kinetic energy of a particle can’t be negative, while the potential energy can be zero, positive or negative. It basically depends on the reference point where we have taken it zero. It is customary to take zero potential energy when the electron is at infinite distance from the nucleus. In some problem, suppose we take zero potential energy in first orbit (U1 = 0), then the modulus of actual potential energy in first orbit (when reference point was at infinity) is added in U and E in all energy states, while K remains unchanged. See sample example number 33.11. In the transition from n2 to n1( < n1 ). The wavelength of emitted photon can be given by the following shortcut formula, l (in Å ) =

12375 En 2 - En1

where, En1 and En 2 are in eV. In general, En in eV is given by En = - (13.6) ˜

V

Z2 n2

In hydrogen emission spectrum, Balmer series was first discovered as it lies in visible light.

Example 33.7 Using the known values for hydrogen atom, calculate (a) radius of third orbit for Li +2 (b) speed of electron in fourth orbit for He + (c) angular momentum of electron in 3rd orbit of He + Solution (a) Z = 3 for Li +2 . Further we know that rn = Substituting,

n = 3, Z = 3 and

We have r3 for Li +2 =

n2 a0 Z

a 0 = 0.529 Å

2

( 3) ( 0.529) Å = 1.587 Å ( 3)

Ans.

(b) Z = 2 for He + . Also we know that Z v1 n Z = 2 and v1 = 2.19 ´ 106 m / s vn =

n = 4,

Substituting,

æ 2ö We get, v 4 for He + = ç ÷ ( 2.19 ´ 106 ) m/s è4ø = 1.095 ´ 106 m/s

Ans.

æ h ö (c) Ln = n ç ÷ è 2p ø æ h ö For n = 3, L3 = 3ç ÷ è 2p ø Note This result is independent of value of Z.

Ans.

268 — Optics and Modern Physics V

Example 33.8 A doubly ionized lithium atom is hydrogen like with atomic number 3. Find the wavelength of the radiation required to excite the electron in Li ++ from the first to the third Bohr orbit. The ionization energy of the hydrogen atom is 13.6 eV . Solution Q E n = –

Z2 n2

(13.6eV) En = –

By putting Z = 3, we have

E1 = –

122.4 n2 122.4

eV = – 122.4 eV

(1) 2 122.4

and

E3 = –

\

DE = E 3 – E1 = 108.8 eV

( 3) 2

= – 13.6 eV

The corresponding wavelength is 12375 12375 Å= Å DE ( in eV) 108.8

l=

= 113.74 Å V

Example 33.9 Find variation of angular speed and time period of single electron of hydrogen like atoms with n and Z. Solution

Angular speed,

Now, \

wµ

w=

v r

vµ

Z n

(Z / n ) (n 2 / Z ) T=

Time period,

and or

rµ wµ

n2 Z Z2 n3

Ans.

2p 1 or T µ w w Tµ

\ V

Ans.

n3 Z2

Ans.

Example 33.10 Find kinetic energy, electrostatic potential energy and total energy of single electron in 2nd excited state of Li +2 atom. Solution Q E IH = -13.6 eV

Further, E µ

Z2 n2

, For

Li +2 , Z = 3

and for 2nd excited state n = 3

Chapter 33

Modern Physics - I — 269

2

æ 3ö E = - 13.6ç ÷ = -13.6 eV è 3ø

\

Ans.

K = | E | = 13.6 eV

Ans.

U = 2E = - 27.2 eV

Ans.

Note In the above expressions E is the total energy, K is the kinetic energy and U is the potential energy. V

Example 33.11 Find the kinetic energy, potential energy and total energy in first and second orbit of hydrogen atom if potential energy in first orbit is taken to be zero. Solution

E2 =

E1 ( 2) 2

E1 = – 13.60 eV, = – 3.40 eV,

K 1 = – E1 = 13.60eV, U1 = 2 E1 = – 27.20eV

K 2 = 3.40eV

U 2 = – 6.80eV

and

Now, U1 = 0, i.e. potential energy has been increased by 27.20 eV. So, we will increase U and E in all energy states by 27.20 eV, while kinetic energy will remain unchanged. Changed values in tabular form are as under. Table 33.2

V

Orbit

K (eV)

U (eV)

E (eV)

First

13.60

0

13.60

Second

3.40

20.40

23.80

Example 33.12 A small particle of mass m moves in such a way that the potential energy U = ar 2 , where a is constant and r is the distance of the particle from the origin. Assuming Bohr model of quantization of angular momentum and circular orbits, find the radius of nth allowed orbit. Solution

The force at a distance r is

dU = – 2ar dr Suppose r be the radius of nth orbit. Then, the necessary centripetal force is provided by the above force. Thus, F =–

mv 2 = 2ar r

…(i)

Further, the quantization of angular momentum gives nh mvr = 2p

…(ii)

Solving Eqs. (i) and (ii) for r, we get æ n2h2 r =ç ç 8 am p 2 è

1/ 4

ö ÷ ÷ ø

Ans.

270 — Optics and Modern Physics V

Example 33.13 Calculate (a) the wavelength and (b) the frequency of the H b line of the Balmer series for hydrogen. Solution

(a) Hb line of Balmer series corresponds to the transition from n = 4 to n = 2 level.

Using Eq. (xiii), the corresponding wavelength for Hb line is 1 1 ö æ 1 . = (1097 ´ 107 ) ç 2 - 2 ÷ = 0.2056 ´ 107 l è2 4 ø l = 4.9 ´ 10-7 m

\

Ans.

8

f=

(b) V

c 3.0 ´ 10 = = 612 . ´ 1014 Hz l 4.9 ´ 10-7

Ans.

Example 33.14 Find the largest and shortest wavelengths in the Lyman series for hydrogen. In what region of the electromagnetic spectrum does each series lie? Solution

The transition equation for Lyman series is given by 1 1 ö æ 1 = R ç 2 - 2 ÷ , n = 2, 3, ... l è1 n ø

The largest wavelength is corresponding to n = 2 1 æ1 1ö \ = 1.097 ´ 107 ç - ÷ = 0.823 ´ 107 l max è1 4ø l max = 1.2154 ´ 10-7 m = 1215 Å

\

Ans.

The shortest wavelength corresponds to n = ¥ 1 æ1 1 ö = 1.097 ´ 107 ç - ÷ \ l min è1 ¥ ø l min = 0.911´ 10–7 m = 911 Å

or

Ans.

Both of these wavelengths lie in ultraviolet (UV) region of electromagnetic spectrum. V

Example 33.15 In a hypothetical atom, mass of electron is doubled, value of atomic number is Z = 4. Find wavelength of photon when this electron jumps from 3rd excited state to 2 nd orbit. Solution

Eµ

Z2 n2

µm

Mass is doubled, Z = 4 and 3 rd excited state means n = 4, second orbit means n = 2. For these values, we have 2 2 æ4ö æ4ö E 4 = - 13.6 ´ 2ç ÷ = - 27.2 eV and E 2 = - 13.6 ´ 2ç ÷ = - 108.8 eV è4ø è 2ø 12375 12375 12375 l ( in Å ) = = = DE ( in eV) E 4 - E 2 -27.2 + 108.8 \

l = 151.65 Å

Ans.

Chapter 33

INTRODUCTORY EXERCISE

Modern Physics - I — 271

33.2

1. Find the ionisation energy of a doubly ionized lithium atom. 2. A hydrogen atom is in a state with energy –1.51 eV. In the Bohr model, what is the angular momentum of the electron in the atom with respect to an axis at the nucleus?

3. As an electron makes a transition from an excited state to the ground state of a hydrogen like atom/ion (JEE 2015) (a) kinetic energy, potential energy and total energy decrease (b) kinetic energy decreases, potential energy increases but total energy remains same (c) kinetic energy and total energy decrease but potential energy increases (d) its kinetic energy increases but potential energy and total energy decrease

4. The wavelength of the first spectral line in the Balmer series of hydrogen atom is 6561 Å. The wavelength of the second spectral line in the Balmer series of singly ionized helium atom is (a) 1215 Å (b) 1640 Å (c) 2430 Å (d) 4687 Å (JEE 2011)

5. The largest wavelength in the ultraviolet region of the hydrogen spectrum is 122 nm. The smallest wavelength in the infrared region of the hydrogen spectrum (to the nearest integer) is (a) 802 nm (b) 823 nm (c) 1882 nm (d) 1648 nm (JEE 2007)

6. A photon collides with a stationary hydrogen atom in ground state inelastically. Energy of the colliding photon is 10.2 eV. After a time interval of the order of micro second another photon collides with same hydrogen atom inelastically with an energy of 15 eV. What will be observed by the detector? (JEE 2005) (a) 2 photons of energy 10.2 eV (b) 2 photons of energy 1.4 eV (c) One photon of energy 10.2 eV and an electron of energy 1.4 eV (d) One photon of energy 10.2 eV and another photon of energy 1.4 eV

7. A hydrogen atom and a Li 2+ ion are both in the second excited state. If lH and lLi are their respective electronic angular momenta, and EH and ELi their respective energies, then (a) lH > lLi and | EH | > | ELi | (b) lH = lLi and | EH | < | ELi | (JEE 2002) (c) lH = lLi and | EH | > | ELi | (d) lH < lLi and | EH | < | ELi |

8. The transition from the state n = 4 to n = 3 in a hydrogen like atom results in ultraviolet radiation. Infrared radiation will be obtained in the transition (a) 2 ® 1 (b) 3 ® 2 (c) 4 ® 2

(JEE 2001)

(d) 5 ® 4

9. As per Bohr model, the minimum energy (in eV) required to remove an electron from the ground state of doubly ionized Li atom (Z = 3) is (a) 1.51 (b) 13.6

(JEE 1997)

(c) 40.8

(d) 122.4

10. Consider the spectral line resulting from the transition n = 2 ® n = 1in the atoms and ions given below. The shortest wavelength is produced by (a) hydrogen atom (b) deuterium atom (c) singly ionized helium (d) doubly ionized lithium

11. The energy levels of a certain atom are shown in figure. If a photon of frequency f is emitted when there is an electron transition from 5E to E, what frequencies of photons could be produced by other energy level transitions?

12. Find the longest wavelength present in the Balmer series of hydrogen.

(JEE 1983)

5E 4E E

272 — Optics and Modern Physics

33.8 X-Rays Electromagnetic radiation with wavelengths from 0.1 Å to 100 Å falls into the category of X-rays. The boundaries of this category are not sharp. The shorter wavelength end overlaps gamma rays and the longer wavelength end overlaps ultraviolet rays. Photoelectric effect (will be discussed later) provides convincing evidence that photons of light can transfer energy to electrons. Is the inverse process also possible? That is, can part or all of the kinetic energy of a moving electron be converted into a photon? Yes, it is possible. In 1895 Wilhelm Roentgen found that a highly penetrating radiation of unknown nature is produced when fast moving electrons strike a target of high atomic number and high melting point. These radiations were given a name X-rays as their nature was unknown (in mathematics an unknown quantity is normally designated by X). Later, it was discovered that these are high energy photons (or electromagnetic waves). Production of X-Rays Figure shows a diagram of a X-ray tube, called the coolidge tube. A cathode (a plate connected to negative terminal of a battery), heated by a filament through which an electric current is passed, supplies electrons by thermionic emission. The high potential difference V maintained between the cathode and a metallic target accelerate the electrons toward the later. The face of the target is at an angle relative to the electron beam, and the X-rays that leave the target pass through the side of the tube. The tube is evacuated to permit the electrons to get to the target unimpedded. Evacuated tube

X-rays

+

Target

–

Cathode V

Fig. 33.7 An X-ray tube. The higher the accelerating voltage V, the faster the electrons and the shorter the wavelengths of the X-rays

Continuous and characteristic X-rays X-rays so produced by the coolidge tube are of two types, continuous and characteristic. While the former depends only on the accelerating voltage V, the later depends on the target used. Continuous X-rays Electromagnetic theory predicts that an accelerated electric charge will radiate electromagnetic waves, and a rapidly moving electrons when suddenly brought to rest is certainly accelerated (of course negative). X-rays produced under these circumstances is given the German name bremsstrahlung (braking radiation). Energy loss due to bremsstrahlung is more important for electrons than for heavier particles because electrons are more violently accelerated when passing near nuclei in their paths. The continuous X-rays (or bremsstrahlung X-rays) produced at a given accelerating potential V vary in wavelength, but none has a wavelength shorter than a certain value l min . This minimum wavelength corresponds to the maximum energy of the X-rays which in turn is equal to the maximum kinetic energy qV or eV of the striking electrons. Thus, hc hc = eV or l min = l min eV

Chapter 33

Modern Physics - I — 273

After substituting values of h, c and e we obtain the following simple formula for l min . 12375 l min ( in Å) = V ( in volts )

…(i)

Increasing V decreases l min . This wavelength is also known as the cut off wavelength or the threshold wavelength. Characteristic X-rays The X-ray spectrum typically consists of a broad continuous band containing a series of sharp lines as shown in Fig. 33.8. Ka Kb L g Lb

La

Fig. 33.8 X-ray spectrum

As discussed above the continuous spectrum is the result of collisions between incoming electrons and atoms in the target. The kinetic energy lost by the electrons during the collisions emerges as the energy of the X-ray photons radiated from the target. The sharp lines superimposed on the continuous spectrum are known as characteristic X-rays because they are characteristic of the target material. They were discovered in 1908, but their origin remained unexplained until the details of atomic structure, particularly the shell structure of the atom, were discovered. Characteristic X-ray emission occurs when a bombarding electron that collides with a target atom has sufficient energy to remove an inner shell electron from the atom. The vacancy created in the shell is filled when an electron from a higher level drops down into it. This transition is accompanied by the emission of a photon whose energy equals the difference in energy between the two levels. Let us assume that the incoming electron has dislodged an atomic N Lb Kg electron from the innermost shell-the K shell. If the vacancy is filled M La Kb M-series by an electron dropping from the next higher shell the L shell, the L L-series photon emitted has an energy corresponding to the K a characteristic X-ray line. If the vacancy is filled by an electron K dropping from the M shell, the Kb line is produced. An La line is a produced as an electron drops from the M shell to the L-shell, and an Lb line is produced by a transition from the N-shell to the L-shell. K-series

Moseley’s Law for Characteristic Spectrum

Fig. 33.9

Although multi-electron atoms cannot be analyzed with the Bohr model, Henery G.J. Moseley in 1914 made an effort towards this. Moseley measured the frequencies of characteristic X-rays from a large number of elements and plotted the square root of the frequency f against the atomic number

274 — Optics and Modern Physics Z of the element. He discovered that the plot is very close to a straight line. He plotted the square root of the frequency of the K a line versus the atomic number Z. As figure shows, Moseley’s plot did not pass through the origin. Let us see why. It can be understood from Gauss’s law. Consider an atom of atomic number Z in which one of the two electrons in the K-shell has been ejected. Imagine that we draw a Gaussian sphere just inside the most probable radius of the L-electrons. The effective charge inside the Gaussian surface is the positive nuclear charge and one negative charge due to the single K-electron. If we ignore the interactions between L-electrons, a single L electron behaves as if it experiences an electric field due to a charge ( Z – 1) enclosed by the Gaussian surface. 2.5

Ö f (× 109Hz1/2)

2.0 1.5 1.0 0.5 5

10

25 30 15 20 Atomic number, Z

35

40

45

50

Fig. 33.10 A plot of the square root of the frequency of the K a lines versus atomic number using Moseley's data

Thus, Moseley’s law of the frequency of K a line is f K a = a ( Z – 1)

…(ii)

where, a is a constant that can be related to Bohr theory. The above law in general can be stated as under f = a ( Z – b) or

f µ ( Z - b)

1 ö æ 1 DE = hf = Rhc ( Z – 1) 2 ç – ÷ 2 è1 22 ø

For K a line,

3Rc ( Z – 1) 4 3Rc and b =1 a= 4 f =

or or

After substituting values of R and c, we get

a = 4.98 ´ 10 7 ( Hz )1/ 2

Eq. (iii) can also be written as

f = a 2 ( Z – b) 2

For K a line, Hence,

…(iii)

a2 =

3Rc = (2.48 ´ 1015 Hz ) and 4 f K a = (2.48 ´ 1015 Hz ) ( Z – 1) 2

b =1

…(iv)

Chapter 33

Modern Physics - I — 275

Extra Points to Remember ˜

˜

˜

In continuous X-ray spectrum, all wavelengths greater than lmin are obtained. Characteristic X-ray spectrum is discrete. Certain fixed wavelengths like lK a , lK b etc. are obtained and these wavelengths are different for different target elements. The mixed spectrum of continuous and characteristic X-rays is as shown in figure 33.8. In general, if we compare between different series, then EK > EL > EM or fK > fL > fM or

lK < lL < lM

And if we compare between a, b and g, then Ea < Eb < Eg or fa < fb < fg

la > lb > lg

or

In Moseley’s law, f µ (Z - b) b = 1 for K a line b = 7.4 for La line

and Thus, f = 0 at Z = b = 1for K a line and

f = 0 at Z = b = 7.4 for K b line Öf

Ka La

1

Z0

7.4

Z

Fig. 33.11

For lower atomic numbers lines are shown dotted. This is because X-rays are obtained only at high atomic numbers. Further, we can see that for a given atomic number (say Z 0 ). ˜

fK > fL Screening effect The energy levels, in general, depend on the principal quantum number (n) and orbital quantum number (l). Let us take sodium (Z = 11) as an example. According to Gauss’s law, for any spherically symmetric charge distribution the electric field magnitude at a distance r from the centre is 1 qencl , where qencl is the total charge enclosed within a sphere with radius r. Mentally, remove the 4pe0 r 2 outer (valence) electron from a sodium atom. What you have left is a spherically symmetric collection of 10 electrons (filling the K and L shells) and 11 protons. So, qencl = – 10e + 11 e = + e If the eleventh is completely outside this collection of charges, it is attracted by an effective charge of +e, not + 11e. This effect is called screening, the 10 electrons screen 10 of the 11 protons leaving an effective net charge of +e. In general, an electron that spends all its time completely outside a positive charge Z eff e has energy levels given by the hydrogen expression with e 2 replaced by Z eff e 2 . i.e. En = –

2 Zeff

n2

(13.6 eV ) (energy levels with screening)

If the eleventh electron in the sodium atom is completely outside the remaining charge distribution, then Z eff = 1. We can estimate the frequency of K a X-ray photons using the concept of screening. A K a X-ray photon is emitted when an electron in the L-shell (n = 2 ) drops down to fill a hole in the K-shell (n = 1.) As the electron

276 — Optics and Modern Physics drops down, it is attracted by the Z protons in the nucleus screened by one remaining electron in the K-shell. Thus, Zeff = (Z – 1), ni = 2 and

nf = 1

The energy before transition is Ei = –

(Z – 1)2 22

(13.6 eV ) = – (Z – 1)2 (3.4 eV )

and energy after transition is Ef = –

(Z – 1)2 12

(13..6 eV ) = – (Z – 1)2 (13.6 eV )

The energy of the K a X-ray photon is EK a = Ei – Ef = (Z – 1)2 (10.2 eV ) The frequency of K a X-ray photon is therefore, EK (Z – 1)2 (10.2 eV ) fK a = a = h (4.136 ´ 10–15 eV - s) = (2.47 ´ 1015 Hz )(Z - 1)2 This relation agrees almost exactly with Moseley’s experimental law. ˜

˜

˜

V

The target (or anode) used in the Coolidge tube should be of high melting point. This is because less than 0.5% of the kinetic energy of the electrons is converted into X-rays. The rest of the kinetic energy converts into internal energy of the target which simultaneously has to be kept cool by circulating oil or water. Atomic number of the target material should be high. This is because X-rays are high energy photons and as we have seen above energy of the X-rays increases as Z increases. X-rays are basically electromagnetic waves. So, they possess all the properties of electromagnetic waves.

Example 33.16 Find the cut off wavelength for the continuous X-rays coming from an X-ray tube operating at 40 kV . Solution

Cut off wavelength l min is given by 12375 12375 l min ( in Å ) = = V ( in volts ) 40 ´ 103 = 0.31 Å

V

Ans.

Example 33.17 Use Moseley’s law with b = 1 to find the frequency of the K a X-rays of La ( Z = 57) if the frequency of the K a X-rays of Cu ( Z = 29) is known to be 1.88 ´ 1018 Hz. Solution

Using the equation,

f = a (Z – b )

f La æ Z La – 1 ö ÷ =ç fCu çè ZCu – 1 ÷ø

2

or

(b = 1) æ Z – 1ö ÷÷ f La = fCu çç La è ZCu – 1 ø

æ 57 – 1 ö ÷÷ = 1.88 ´ 1018 çç è 29 – 1 ø = 7.52 ´ 1018 Hz

2

2

Ans.

Chapter 33

Modern Physics - I — 277

V

Example 33.18 Electrons with de-Broglie wavelength l fall on the target in an X-ray tube. The cut off wavelength of the emitted X-rays is (JEE 2007) 2 2mcl 2h (b) l 0 = (a) l 0 = h mc 2m 2 c 2 l3 (d) l 0 = l (c) l 0 = h2 Solution Momentum of bombarding electrons, h p= l \ Kinetic energy of bombarding electrons, p2 h2 K= = 2 m 2ml2 This is also maximum energy of X-ray photons. hc h2 Therefore, = l 0 2ml2 2ml2 c or l0 = h \ Correct option is (a).

V

Example 33.19 Electrons with energy 80 keV are incident on the tungsten target of an X-ray tube. K-shell electrons of tungsten have 72.5 keV energy. X-rays emitted by the tube contain only (JEE 2000) (a) a continuous X-ray spectrum (Bremsstrahlung) with a minimum wavelength of » 0.155 Å (b) a continuous X-ray spectrum (Bremsstrahlung) with all wavelengths (c) the characteristic X-ray spectrum of tungsten (d) a continuous X-ray spectrum (Bremsstrahlung) with a minimum wavelength of Å and the characteristic X-ray spectrum of tungsten » 0155 . Solution Minimum wavelength of continuous X-ray spectrum is given by l min (in Å) 12375 = E ( in eV) Here, E = energy of incident electrons (in eV) = energy corresponding to minimum wavelength l min of X-ray E = 80 keV = 80 ´ 103 eV \

l min (in Å) =

12375 80 ´ 103

» 0.155

Also the energy of the incident electrons (80 keV) is more than the ionization energy of the K-shell electrons (i.e. 72.5 keV). Therefore, characteristic X-ray spectrum will also be obtained because energy of incident electron is high enough to knock out the electron from K or L-shells. \ The correct option is (d).

278 — Optics and Modern Physics INTRODUCTORY EXERCISE

33.3

1. If l Cu is the wavelength of Ka , X-ray line of copper (atomic number 29) and lMo is the wavelength of the Ka X-ray line of molybdenum (atomic number 42), then the ratio l Cu / lMo is close to (JEE 2014) (a) 1.99 (b) 2.14 (c) 0.50 (d) 0.48

2. Which one of the following statements is wrong in the context of X-rays generated from an X-ray tube? (JEE 2008) (a) Wavelength of characteristic X-rays decreases when the atomic number of the target increases (b) Cut off wavelength of the continuous X-rays depends on the atomic number of the target (c) Intensity of the characteristic X-rays depends on the electrical power given to the X-ray tube (d) Cut off wavelength of the continuous X-rays depends on the energy of the electrons in the X-ray tube

3. Ka wavelength emitted by an atom of atomic number Z = 11 is l. Find the atomic number for an atom that emits Ka radiation with wavelength 4l (a) Z = 6 (b) Z = 4 (c) Z = 11 (d) Z = 44

(JEE 2005)

4. The intensity of X-rays from a coolidge tube is plotted against wavelength l as shown in the figure. The minimum wavelength found is l c and the wavelength of the Ka line is l k . As the accelerating voltage is increased (JEE 2001) I

lc

lk

l

Fig. 33.12

(a) l k - l c increases (c) l k increases

(b) l k - l c decreases (d) l k decreases

5. X-rays are produced in an X-ray tube operating at a given accelerating voltage. The wavelength of the continuous X-rays has values from (a) 0 to ¥ (b) l min to ¥ , where l min > 0 (c) 0 to l max, where l max < ¥ (d) l min to l max, where 0 < l min < l max < ¥

(JEE 1998)

6. Characteristic X-rays of frequency 4.2 ´ 1018 Hz are produced when transitions from L-shell to K-shell take place in a certain target material. Use Mosley’s law to determine the atomic number of the target material. Given Rydberg constant R = 1.1 ´ 107 m -1. (JEE 2003)

Chapter 33

Modern Physics - I — 279

33.9 Emission of Electrons At room temperature the free electrons move randomly within the conductor, but they don’t leave the surface of the conductor due to attraction of positive charges. Some external energy is required to emit electrons from a metal surface. Minimum energy is required to emit the electrons which are just on the surface of the conductor. This minimum energy is called the work-function (denoted by W) of the conductor. Work-function is the property of the metallic surface. The energy required to liberate an electron from metal surface may arise from various sources such as heat, light, electric field etc. Depending on the nature of source of energy, the following methods are possible : (i) Thermionic emission The energy to the free electrons can be given by heating the metal. The electrons so emitted are known as thermions. (ii) Field emission When a conductor is put under strong electric field, the free electrons on it experience an electric force in the opposite direction of field. Beyond a certain limit, electrons start coming out of the metal surface. Emission of electrons from a metal surface by this method is called the field emission. (iii) Secondary emission Emission of electrons from a metal surface by the bombardment of high speed electrons or other particles is known as secondary emission. (iv) Photoelectric emission Emission of free electrons from a metal surface by falling light (or any other electromagnetic wave which has an energy greater than the work-function of the metal) is called photoelectric emission. The electrons so emitted are called photoelectrons.

33.10 Photoelectric Effect When light of an appropriate frequency (or correspondingly of an appropriate wavelength) is incident on a metallic surface, electrons are liberated from the surface. This observation is known as photoelectric effect. Photoelectric effect was first observed in 1887 by Hertz. For photoemission to take place, energy of incident light photons should be greater than or equal to the work-function of the metal. or

E ³W

\

hf ³ W W f ³ h

or

…(i)

W is the minimum frequency required for the emission of electrons. This is known as threshold h frequency f 0 . W Thus, (threshold frequency) …(ii) f0 = h Here,

Further, Eq. (i) can be written as hc ³W l

or l £

hc W

280 — Optics and Modern Physics hc is the largest wavelength beyond which photoemission does not take place. This is called W the threshold wavelength l 0 . hc …(iii) (threshold wavelength) Thus, l0 = W Here,

Hence, for the photoemission to take place either of the following conditions must be satisfied. E ³W

or

f ³ f0

or l £ l 0

…(iv)

Stopping Potential and Maximum Kinetic Energy of Photoelectrons When the frequency f of the incident light is greater than the threshold frequency, some electrons are emitted from the metal with substantial initial speeds. Suppose E is the energy of light incident on a metal surface and W (< E ) the work-function of metal. As minimum energy is required to extract electrons from the surface, they will have the maximum kinetic energy which is E – W. K max = E – W

Thus,

…(v)

This value K max can experimentally be found by keeping the metal plate P (from which electrons are emitting) at higher potential relative to an another plate Q placed in front of P. Some electrons after emitting from plate P, reach the plate Q despite the fact that Q is at lower potential and it is repelling the electrons from reaching in itself. This is because the electrons emitted from plate P possess some kinetic energy and due to this energy they reach the plate Q and current i flows in the circuit in the direction shown in figure. Light

i P

Q

G

V

Fig. 33.13

As the potential V is increased, the force of repulsion to the electrons gets increased and less number of electrons reach the plate Q and current in the circuit gets decreased. At a certain valueV0 electrons having maximum kinetic energy (K max ) also get stopped and current in the circuit becomes zero. This is called the stopping potential. As an electron moves from P to Q, the potential decreases byV0 and negative work – eV0 is done on the (negatively charged) electron, the most energetic electron leaves plate P with kinetic energy K max =

1 2

2 and has zero kinetic energy at Q. Using the work energy theorem, we have mv max

Wext = – eV0 = DK = 0 – K max or

K max =

1 2 mv max = eV0 2

…(vi)

Modern Physics - I — 281

Chapter 33

Photoelectric Current Figure shows an apparatus used to study the variation of photocurrent i with the intensity and frequency of light falling on metal plate P. Photoelectrons are emitted from plate P which are being attracted by the positive plate Q and a photoelectric current i flows in the circuit, which can be measured by the galvanometer G.

P

i

Q G

Fig. 33.14

Figure 33.15 (a) shows graphs of photocurrent as a function of potential difference VQP for light of constant frequency and two different intensities. When VQP is sufficiently large and positive the current becomes constant, showing that all the emitted electrons are being collected by the anode plate Q. The stopping potential difference –V0 needed to reduce the current to zero is shown. If the intensity of light is increased, (or we can say the number of photons incident per unit area per unit time is increased) while its frequency is kept the same, the current becomes constant at a higher value, showing that more electrons are being emitted per unit time. But the stopping potential is found to be the same. f is constant

2I I

f2 > f1 f2

–V0

VQP

0

I is constant f1

–V02 – V01 0

VQP

(b) Photocurrent i as a function of the potential VQP Photocurrent i as a function of the potential VQP of the anode with respect to a cathode for two of the anode with respect to the cathode for a different light frequencies f1 and f2 with the same constant light frequency f, the stopping potential intensity. The stopping potential V0 (and therefore V0 is independent of the light intensity I. the maximum kinetic energy of the photoelectrons) increases linearly with frequency. Z Fig. 33.15 (a)

Figure 33.15 (b) shows current as a function of potential difference for two different frequencies with the same intensity in each case. We see that when the frequency of the incident monochromatic light is increased, the stopping potentialV0 gets increased. Of course,V0 turn out to be a linear function of the frequency f.

Graph between K max and f Let us plot a graph between maximum kinetic energy K max of photoelectrons and frequency f of incident light. The equation between K max and f is K max = hf – W

282 — Optics and Modern Physics Comparing it with y = mx + c, the graph between K max and f is a straight line with positive slope and negative intercept. 1

Kmax

2

f

(f0)1 (f0)2 W1 W2

Fig. 33.16

From the graph, we can note the following points (i) K max = 0 at f = f 0 (ii) Slope of the straight line is h, a universal constant, i.e. if graph is plotted for two different metals 1 and 2, slope of both the lines is same. (iii) The negative intercept of the line is W, the work-function which is characteristic of a metal, i.e. intercepts for two different metals will be different. Further, W2 > W1 \

[as W = hf 0 ]

( f 0 )2 > ( f 0 ) 1 f 0 = threshold frequency

Here,

Graph between V 0 and f V0

Let us now plot a graph between the stopping potential V0 and the incident frequency f. The equation between them is

or

Again comparing with y = mx + c, the graph between h V0 and f is a straight line with positive slope e W (a universal constant) and negative intercept (which e depends on the metal). The corresponding graph is shown in figure.

2 (Slope)1 = (Slope)2 =

eV0 = hf – W æhö æW ö V0 = ç ÷ f – ç ÷ èeø è e ø

1

(f0)1

W2 e

W1 e

(f0)2

Fig. 33.17

f

h e

Chapter 33

Modern Physics - I — 283

Extra Points to Remember ˜

˜

The major features of the photoelectric effect could not be explained by the wave theory of light which were later explained by Einstein’s photon theory. (i) Wave theory suggests that the kinetic energy of the photoelectrons should increase with the increase in intensity of light. However equation, Kmax = eV0 suggests that it is independent of the intensity of light. (ii) According to wave theory, the photoelectric effect should occur for any frequency of the light, provided that the light is intense enough. However, the above equation suggests that photoemission is possible only when frequency of incident light is either greater than or equal to the threshold frequency f0 . (iii) If the energy to the photoelectrons is obtained by soaking up from the incident wave, it is not likely that the effective target area for an electron in the metal is much more than a few atomic diameters. (see example 33.20) between the impinging of the light on the surface and the ejection of the photoelectrons. During this interval the electron should be “soaking up” energy from the beam until it had accumulated enough energy to escape. However, no detectable time lag has ever been measured. Einstein’s photon theory Einstein succeeded in explaining the photoelectric effect by making a remarkable assumption, that the energy in a light beam travels through space in concentrated bundles, called photons. The energy E of a single photon is given by E = hf Applying the photon concept to the photoelectric effect, Einstein wrote hf = W + Kmax

(already discussed)

Consider how Einstein’s photon hypothesis meets the three objections raised against the wave theory interpretation of the photoelectric effect. As for objection 1 (the lack of dependence of Kmax on the intensity of illumination), doubling the light intensity merely doubles the number of photons and thus doubles the photoelectric current, it does not change the energy of the individual photons Objection 2 (the existence of a cutoff frequency) follows from equation hf = W + Kmax . If Kmax equals zero, We have hf0 = W which asserts that the photon has just enough energy to eject the photoelectrons and none extra to appear as kinetic energy. The quantity W is called the work-function of the substance. If f is reduced below f0 , the individual photons, no matter how many of them there are (that is, no matter how intense the illumination), will not have enough energy to eject photo electrons. Objection 3 (the absence of a time lag) follows from the photon theory because the required energy is supplied in a concentrated bundle. It is not spread uniformly over a large area, as in the wave theory. Although, the photon hypothesis certainly fits the facts of photoelectricity, it seems to be in direct conflict with the wave theory of light. Out modern view of the nature of light is that it has a dual character, behaving like a wave under some circumstances and like a particle or photon under others. V

Example 33.20 A metal plate is placed 5 m from a monochromatic light source whose power output is 10 –3 W . Consider that a given ejected photoelectron may collect its energy from a circular area of the plate as large as ten atomic diameters ( 10 –9 m) in radius. The energy required to remove an electron through the metal surface is about 5.0 eV . Assuming light to be a wave, how long would it take for such a ‘target’ to soak up this much energy from such a light source. The target area is S 1 = p (10–9 ) 2 = p ´ 10–18 m 2 . The area of a 5 m sphere centered on the light source is, S 2 = 4p ( 5) 2 = 100 p m 2 . Thus, if the light source radiates uniformly in all directions the rate P at which energy falls on the target is given by Solution

284 — Optics and Modern Physics æ p ´ 10–18 ö æS ö ÷ = 10–23 J/s P = (10–3 watt ) çç 1 ÷÷ = (10–3 ) ç ç 100 ´ p ÷ S è 2ø ø è Assuming that all power is absorbed, the required time is æ 5 eV ö æ 1.6 ´ 10–19 J ö ÷ » 20 h ÷ç t = ç –23 ÷ ç 10 J/s ÷ ç 1 eV ø øè è V

Ans.

Example 33.21 The photoelectric work-function of potassium is 2.3 eV . If light having a wavelength of 2800 Å falls on potassium, find (a) the kinetic energy in electron volts of the most energetic electrons ejected. (b) the stopping potential in volts Solution

\

Given, W = 2.3eV, l = 2800 Å 12375 12375 E ( in eV) = = = 4.4 eV l ( in Å) 2800

(a) K max = E – W = ( 4.4 – 2.3) eV= 2.1 eV

Ans.

(b) K max = eV0 \ V

or V0 = 2.1 volt

2.1eV = eV0

Ans.

Example 33.22 When a beam of 10.6 eV photons of intensity 2.0 W / m2 falls on a platinum surface of area 1.0 ´ 10 –4 m2 and work-function 5.6 eV , 0.53 % of the incident photons eject photoelectrons. Find the number of photoelectrons emitted per second and their minimum and maximum energies ( in eV ). Take 1 eV = 1.6 ´ 10 –19 J . Solution

Number of photoelectrons emitted per second (Intensity ) (Area) 0.53 = ´ (Energy of each photon ) 100 =

( 2.0) (1.0 ´ 10–4 ) (10.6 ´ 1.6 ´ 10

–19

´ )

0.53 100

11

= 6.25 ´ 10

Ans.

Minimum kinetic energy of photoelectrons, K min = 0 and maximum kinetic energy is,

K max = E – W = (10.6 – 5.6) eV = 5.0 eV

V

Ans.

Example 33.23 Maximum kinetic energy of photoelectrons from a metal surface is K 0 when wavelength of incident light is l. If wavelength is decreased to l /2, the maximum kinetic energy of photoelectrons becomes (a) = 2K 0

(b) > 2K 0

(c) < 2K 0

Chapter 33 Solution

Modern Physics - I — 285

Using the equation, K max = E - W , we have hc K0 = -W l

K (i)

l with wavelength , suppose the maximum kinetic energy is K 0¢ , then 2 hc hc K 0¢ = -W = 2 -W l l/ 2 æ hc ö = 2ç - W ÷ + W è l ø but

hc - W is K 0 . Therefore, l K 0¢ = 2K 0 + W K 0¢ > 2K 0

or \ The correct option is (b). V

Ans.

Example 33.24 Intensity and frequency of incident light both are doubled. Then, what is the effect on stopping potential and saturation current. By increasing the frequency of incident light energy of incident light will increase. So, maximum kinetic energy of photoelectrons will also increase. Hence, stopping potential will increase. Further, by doubling the frequency of incident light energy of each photon will be doubled. So, intensity itself becomes two times without increasing number of photons incident per unit area per unit time. Therefore, saturation current will remain unchanged. Solution

V

Example 33.25 When a monochromatic point source of light is at a distance of 0.2 m from a photoelectric cell, the cut off voltage and the saturation current are respectively 0.6 V and 18.0 mA. If the same source is placed 0.6 m away from the photoelectric cell, then (JEE 1992) (a) the stopping potential will be 0.2 V (b) the stopping potential will be 0.6 V (c) the saturation current will be 6.0 mA (d) the saturation current will be 2.0 mA (b) Stopping potential depends on two factors – one the energy of incident light and the other the work-function of the metal. By increasing the distance of source from the cell, neither of the two change. Therefore, stopping potential remains the same. (d) Saturation current is directly proportional to the intensity of light incident on cell and for a point source, intensity I µ 1/ r 2 Solution

When distance is increased from 0.2 m to 0.6 m (three times), the intensity and hence the saturation current will decrease 9 times, i.e. the saturation current will be reduced to 2.0 mA. \ The correct options are (b) and (d).

286 — Optics and Modern Physics V

Example 33.26 The threshold wavelength for photoelectric emission from a material is 5200 Å. Photoelectrons will be emitted when this material is illuminated with monochromatic radiation from a (1982, 3M) (b) 1 W infrared lamp (a) 50 W infrared lamp (c) 50 W ultraviolet lamp (d) 1 W ultraviolet lamp Solution For photoemission to take place, wavelength of incident light should be less than the threshold wavelength. Wavelength of ultraviolet light < 5200 Å while that of infrared radiation > 5200 Å. \ The correct options are (c) and (d).

INTRODUCTORY EXERCISE

33.4

1. Light of wavelength 2000 Å is incident on a metal surface of work-function 3.0 eV. Find the minimum and maximum kinetic energy of the photoelectrons.

2. Is it correct to say that Kmax is proportional to f ? If not, what would a correct statement of the relationship between Kmax and f ?

3. When a metal is illuminated with light of frequency f, the maximum kinetic energy of the photoelectrons is 1.2 eV. When the frequency is increased by 50%, the maximum kinetic energy increases to 4.2 eV. What is the threshold frequency for this metal?

4. A metal surface is illuminated by light of two different wavelengths 248 nm and 310 nm. The maximum speeds of the photoelectrons corresponding to these wavelengths are u1 and u 2, respectively. If the ratio u1 : u 2 = 2 : 1 and hc = 1240 eV nm, the work-function of the metal is nearly (JEE 2014) (a) 3.7 eV (b) 3.2 eV (c) 2.8 eV (d) 2.5 eV

5. The figure shows the variation of photocurrent with anode potential for a photosensitive surface for three different radiations. Let I a, I b and I c be the intensities and fa, fb and fc be the frequencies for the curves a, b and c, respectively (JEE 2004) (a) fa = fb and I a ¹ I b (b) fa = fc and I a = I c (c) fa = fb and I a = I b (d) fb = fc and I b = I c

I

c

b a

6. The work-function of a substance is 4.0 eV. The longest wavelength of light that can cause photoelectron emission from this substance is approximately (a) 540 nm (b) 400 nm (c) 310 nm

V

(JEE 1998)

(d) 220 nm

7. The maximum kinetic energy of photoelectrons emitted from a surface when photons of energy 6 eV fall on it is 4 eV. The stopping potential in volt is (a) 2 (b) 4 (c) 6

8. Photoelectric effect supports quantum nature of light because

(JEE 1997)

(d) 10 (JEE 1987)

(a) there is a minimum frequency of light below which no photoelectrons are emitted (b) the maximum kinetic energy of photoelectrons depends only on the frequency of light and not on its intensity (c) even when the metal surface is faintly illuminated, the photoelectrons leave the surface immediately (d) electric charge of the photoelectrons is quantised

Chapter 33

Modern Physics - I — 287

Final Touch Points 1. In hydrogen and hydrogen like atoms : Circumference of n th orbit = n (wavelength of single electron in that orbit) Proof According to Bohr’s assumption,

or \ or

æh ö Ln = n ç ÷ è 2p ø æh ö mvr = n ç ÷ è 2p ø æ h ö ( 2pr ) = n ç ÷ è mv ø circumference = n (de-Broglie wavelength of electron)

2. Rhc = 13.6 eV Therefore, Rhc has the dimensions of energy or [Rhc ] = [ML2T -2 ]

3. 1 Rhc is also called 1 Rydberg. It is not Rydberg constant R.

Solved Examples TYPED PROBLEMS Type 1. Based on de-Broglie wavelength V

Example 1 A proton is fired from very far away towards a nucleus with charge Q = 120 e, where e is the electronic charge. It makes a closest approach of 10 fm to the nucleus. The de-Broglie wavelength (in units of fm) of the proton at its start is [Take the proton mass, m p = ( 5 /3) ´ 10 -27 kg, h /e = 4.2 ´ 10 -15 J-s/C, 1 = 9 ´ 10 9 m/F, 1 fm = 10 -15 m] (JEE 2013) 4pe 0 Solution r = closest distance = 10 fm

Ze

From energy conservation, we have

+

Ki + Ui = K f + U f 1 qq K + 0 =0 + . 1 2 4pe0 r

or

K =

or

K

e Å

r

1 (120 e) (e) . 4pe0 r

…(i)

de-Broglie wavelength, l=

h 2Km

…(ii)

Substituting the given values in above two equations, we get l = 7 ´ 10-15 m = 7 fm V

Example 2 Find de-Broglie wavelength of single electron in 2 nd orbit of hydrogen atom by two methods. Solution Method 1 Kinetic energy of single electron in 2nd orbit is 3.4 eV using the equation, l= Method 2 \ or Now,

150 150 = = 6.64 Å KE (in eV ) 3.4

Ans.

Circumference of nth orbit = nl 2 pr = 2 l l = pr r µ n2 r1 = 0.529 Å

\

l = p (0.529)(2)2Å = 6.64 Å

Ans.

Chapter 33

Modern Physics - I — 289

Type 2. Based on Bohr’s atomic models V

Example 3 The electric potential between a proton and an electron is given by r , where r0 is a constant. Assuming Bohr model to be applicable, V = V 0 ln r0 write variation of rn with n, being the principal quantum number. (JEE 2003) (a) rn µ n

(b) rn µ

1 n

(c) rn µ n 2

(d) rn µ

1 n2

ærö

Solution Q U = eV = eV 0 ln çç ÷÷ è r0 ø |F |= -

dU eV 0 = dr r

This force will provide the necessary centripetal force. Hence,

mv2 eV 0 = r r

eV 0 m nh mvr = 2p v=

or Moreover,

…(i) …(ii)

Dividing Eq. (ii) by Eq. (i), we have æ nh ö mr = ç ÷ è 2p ø V

m eV 0

or

rn µ n

Example 4 Imagine an atom made up of proton and a hypothetical particle of double the mass of the electron but having the same charge as the electron. Apply the Bohr atom model and consider all possible transitions of this hypothetical particle to the first excited level. The longest wavelength photon that will be emitted has wavelength l (given in terms of the Rydberg constant R for the hydrogen atom) equal to (JEE 2000) (a) 9/5R

(b) 36/5R

(c) 18/5R

(d) 4/R

Rhc Solution In hydrogen atom, E n = - 2 n Also, En µ m where, m is the mass of the electron. Here, the electron has been replaced by a particle whose mass is double of an electron. Therefore, for this hypothetical atom energy in nth orbit will be given by 2Rhc En = n2 The longest wavelength l max (or minimum energy) photon will correspond to the transition of particle from n = 3 to n = 2 . 1ö hc æ1 = E3 - E 2 = 2Rhc ç 2 - 2 ÷ \ l max 3 ø è2 This gives, l max = 18 / 5R \ The correct option is (c).

290 — Optics and Modern Physics V

Example 5 The recoil speed of a hydrogen atom after it emits a photon is going from n = 5 state to n = 1 state is ........ m/s. (JEE 1997) Solution From conservation of linear momentum, |Momentum of recoil hydrogen atom| = |Momentum of emitted photon| DE or mv = c é 1 1ù Here, DE = E5 - E1 = – 13.6 ê 2 - 2 ú eV ë5 1 û

and \

= (13.6) (24 / 25) eV = 13.056 eV = 13.056 ´ 1.6 ´ 10-19 J = 2.09 ´ 10-18 J m = mass of hydrogen atom = 1.67 ´ 10-27 kg DE 2.09 ´ 10-18 v= = mc (1.67 ´ 10-27 ) (3 ´ 108 ) v » 4.17 m/s

V

Example 6 A hydrogen like atom (described by the Bohr model) is observed to emit six wavelengths, originating from all possible transitions between a group of levels. These levels have energies between – 0.85 eV and –0.544 eV (including both these values). (JEE 2002) (a) Find the atomic number of the atom. (b) Calculate the smallest wavelength emitted in these transitions. (Take, hc = 1240 eV-nm, ground state energy of hydrogen atom = -13.6 eV) Solution (a) Total 6 lines are emitted. Therefore, n (n - 1) =6 2

or

n =4

So, transition is taking place between mth energy state and (m + 3) th energy state. æ Z 2ö Em = - 0.85 eV or -13.6 çç 2 ÷÷ = - 0.85 èm ø Z or = 0.25 m Similarly, or or

…(i)

Em + 3 = - 0.544 eV -13.6

Z2 = - 0.544 (m + 3)2 Z = 0.2 (m + 3)

…(ii)

Solving Eqs. (i) and (ii) for Z and m, we get Ans. m = 12 and Z = 3 (b) Smallest wavelength corresponds to maximum difference of energies which is obviously Em + 3 - Em \ D E max = - 0.544 - (- 0.85) = 0.306 eV hc 1240 Ans. l min = = = 4052.3 nm \ DE max 0.306

Chapter 33 V

Modern Physics - I — 291

Example 7 A hydrogen like atom of atomic number Z is in an excited state of quantum number 2n. It can emit a maximum energy photon of 204 eV. If it makes a transition to quantum state n, a photon of energy 40.8 eV is emitted. Find n, Z and the ground state energy (in eV) of this atom. Also calculate the minimum energy (in eV) that can be emitted by this atom during de-excitation. Ground state energy of hydrogen atom is –13.6 eV. (JEE 2000) Solution Let ground state energy (in eV) be E1. Then, from the given condition E 2n - E1 = 204 eV E1 - E1 = 204 eV 4n 2 æ 1 ö E1 ç 2 - 1÷ = 204 eV è 4n ø

or or and

E 2n - E n = 40.8 eV

or

E1 E - 1 = 40.8 eV 4n 2 n 2

or

æ -3 ö E1 ç 2 ÷ = 40.8 eV è 4n ø

From Eqs. (i) and (ii), we get 1 4n 2 = 5 or 1 = 1 + 15 3 4n 2 4n 2 2 4n 4 = 1 or n = 2 n2

1-

or From Eq. (ii),

4 2 n (40.8) eV 3 4 = - (2)2 (40.8) eV 3

E1 = -

or

E1 = - 217.6 eV E1 = - (13.6) Z 2

\ \

Z2 =

E1 -217.6 = = 16 -13.6 -13.6

Z =4 E min = E 2n - E 2n - 1 =

E1 E1 2 4n (2n - 1)2

E1 E1 7 =E1 16 9 144 æ 7 ö =-ç ÷ (- 217.6) eV è 144 ø =

\

E min = 10.58 eV

…(i)

…(ii)

292 — Optics and Modern Physics V

Example 8 A hydrogen like atom (atomic number Z) is in a higher excited state of quantum number n. The excited atom can make a transition to the first excited state by successively emitting two photons of energy 10.2 eV and 17.0 eV, respectively. Alternatively, the atom from the same excited state can make a transition to the second excited state by successively emitting two photons of energies 4.25 eV and 5.95 eV, respectively. (JEE 1994) Determine the values of n and Z. (Ionization energy of H-atom = 13.6 eV) Solution From the given conditions,

and

E n - E 2 = (10.2 + 17) eV = 27.2 eV

…(i)

E n - E3 = (4.25 + 5.95) eV = 10.2 eV

…(ii)

Eq. (i) - Eq. (ii) gives E3 - E 2 = 17.0 eV

æ1 1ö Z 2 (13.6) ç - ÷ = 17.0 è4 9ø

Z 2 (13.6) (5 / 36) = 17.0 Z 2 = 9 or Z = 3

Þ Þ From Eq. (i),

or

or

æ1 1 ö æ1 1 ö Z 2 (13.6) ç - 2 ÷ = 27.2 or (3)2 (13.6) ç - 2 ÷ = 27.2 4 n è ø è4 n ø 1 1 = 0.222 or 1 / n 2 = 0.0278 4 n2 n 2 = 36 Þ

or

n =6

Type 3. Based on X-rays V

Example 9 Determine the energy of the characteristic X-ray ( K b ) emitted from a tungsten ( Z = 74) target when an electron drops from the M-shell ( n = 3) to a vacancy in the K-shell ( n = 1). Solution Energy associated with the electron in the K-shell is approximately EK = – (74 – 1)2 (13.6 eV ) = – 72474 eV An electron in the M-shell is subjected to an effective nuclear charge that depends on the number of electrons in the n = 1 and n = 2 states because these electrons shield the M electrons from the nucleus. Because there are eight electrons in the n = 2 state and one remaining in the n = 1 state, roughly nine electrons shield M electrons from the nucleus. So, Z eff = z – 9 Hence, the energy associated with an electron in the M-shell is 2 –13.6 Z eff –13.6 (Z – 9)2 eV = eV 2 3 32 (13.6) (74 – 9)2 =– eV = – 6384 eV 9

EM =

Therefore, emitted X-ray has an energy equal to EM – EK = {–6384 – (–72474)} eV = 66090 eV

Ans.

Chapter 33 V

Example 10 The potential difference applied to an X-ray tube is 5 kV and the current through it is 3.2 mA. Then, the number of electrons striking the target per second is (JEE 2002) (a) 2 ´ 1016

(b) 5 ´ 106

(c) 1 ´ 1017

(d) 4 ´ 1015

Solution Q i =

q ne = t t it e

\

n=

Substituting

i = 3.2 ´ 10-3 A, e = 1.6 ´ 10-19 C and t = 1 s n = 2 ´ 1016

We get, \ The correct answer is (a). V

Modern Physics - I — 293

Example 11 X-rays are incident on a target metal atom having 30 neutrons. The ratio of atomic radius of the target atom and 42 He is ( 14)1/3 . (JEE 2005) (a) Find the mass number of target atom. (b) Find the frequency of K a line emitted by this metal. Hint : Radius of a nucleus (r) has the following relation with mass number (A). r µ A1/ 3 (R = 1.1 ´ 107 m -1 , c = 3 ´ 108 m/s) Solution (a) From the relation r µ A1/3 , r2 æ A2 ö =ç ÷ r1 çè A1 ÷ø

We have, æ A2 ö ç ÷ è 4 ø

or \

1/3

1/3

= (14)1/3 A2 = 56

Z 2 = A2 - number of neutrons

(b) \

= 56 - 30 = 26 1 ö 3Rc æ1 f = Rc (Z - 1)2 ç 2 - 2 ÷ = (Z - 1)2 4 1 2 è ø

Substituting the given values of R, c and Z, we get f = 1.55 ´ 1018 Hz V

Example 12 Stopping potential of 24, 100, 110 and 115 kV are measured for photoelectrons emitted from a certain element when it is radiated with monochromatic X-ray. If this element is used as a target in an X-ray tube, what will be the wavelength of K a -line? Solution Stopping potentials are 24, 100, 110 and 115 kV, i.e. if the electrons are emitted

from conduction band, maximum kinetic energy of photoelectrons would be 115 ´ 103 eV. If they are emitted from next inner shell, maximum kinetic energy of photoelectrons would be 110 ´ 103 eV and so on.

294 — Optics and Modern Physics For photoelectrons of L- shell it would be 100 ´ 103 eV and for K-shell it is 24 ´ 103 eV. Therefore, difference between energy of L-shell and K-shell is DE = EL – EK = (100 – 24) ´ 103 eV = 76 ´ 103 eV \ Wavelength of K a -line (transition of electron from L-shell to K-shell) is, 12375 12375 lK a (in Å ) = = DE (in eV ) 76 ´ 103 = 0.163 Å V

Ans.

Example 13 In Moseley’s equation f = a ( Z – b), a and b are constants. Find their values with the help of the following data.

Solution

Element

Z

Wavelength of Ka X-rays

Mo

42

0.71 Å

Co

27

1.785 Å

f = a (Z – b)

or

c = a (Z 1 – b) l1

…(i)

and

c = a (Z 2 – b) l2

…(ii)

é 1 1 ù cê – ú = a (Z 1 – Z 2) l2 û ë l1

…(iii)

From Eqs. (i) and (ii), we have

c = 3.0 ´ 108 m/s, l1 = 0.71 ´ 10–10 m l 2 = 1.785 ´ 10 m, Z 1 = 42 and Z 2 = 27, we get and a = 5 ´ 107 (Hz )1/ 2 b = 1.37

Solving above three equations with

–10

Ans.

Type 4. Based on photoelectric effect V

Example 14 A monochromatic light source of frequency f illuminates a metallic surface and ejects photoelectrons. The photoelectrons having maximum energy are just able to ionize the hydrogen atoms in ground state. When the whole 5 experiment is repeated with an incident radiation of frequency f, the 6 photoelectrons so emitted are able to excite the hydrogen atom beam which then emits a radiation of wavelength 1215 Å. (a) What is the frequency of radiation? (b) Find the work-function of the metal. Solution (a) Using Einstein’s equation of photoelectric effect, K max = hf – W

Chapter 33 K max = 13.6 eV hf – W = 13.6 eV 12375 æ5 ö h ç f÷ – W = = 10.2 eV 1215 è6 ø

Here, \ Further,

Solving Eqs. (i) and (ii), we have hf = 3.4 eV 6

or

f =

…(i) …(ii)

(6) (3.4) (1.6 ´ 10–19 ) (6.63 ´ 10–34 )

= 4.92 ´ 1015 Hz W = hf – 13.6 = 6 (3.4) – 13.6 = 6.8 eV

(b)

V

Modern Physics - I — 295

Ans. [from Eq. (i)] Ans.

Example 15 The graph between 1/l and stopping potential (V ) of three metals having work-functions f 1 , f 2 and f 3 in an experiment of photoelectric effect is plotted as shown in the figure. Which of the following statement(s) is/are correct? (Here, l is the wavelength of the incident ray). (JEE 2006) V Metal 1

q 0.001 0.002

Metal 2

Metal 3

0.004 1/l nm–1

(a) Ratio of work-functions f1 : f2 : f3 = 1 : 2 : 4 (b) Ratio of work-functions f1 : f2 : f3 = 4 : 2 : 1 (c) tan q is directly proportional to hc / e, where h is Planck constant and c is the speed of light (d) The violet colour light can eject photoelectrons from metals 2 and 3 Solution From the relation, hc æ hc ö æ 1 ö f - f or V = ç ÷ ç ÷ l è e ø è lø e hc This is equation of straight line. Slope is tan q = . e hc Further V = 0 at f = l hc hc hc 1 1 1 =1:2:4 \ f1 : f2 : f3 = : : = : : l 01 l 02 l 03 l 01 l 02 l 03 1 = 0.001 nm-1 or l 01 = 10000 Å l 01 1 = 0.002 nm-1 or l 02 = 5000 Å l 02 eV =

1 = 0.004 nm-1 l 03

or

l 03 = 2500 Å

Violet colour has wavelength 4000 Å. So, violet colour can eject photoelectrons from metal 1 and metal 2. \ The correct options are (a) and (c).

296 — Optics and Modern Physics V

Example 16 A beam of light has three wavelengths 4144 Å, 4972 Å and 6216 Å with a total intensity of 3.6 ´ 10 -3 Wm -2 equally distributed amongst the three wavelengths. The beam falls normally on an area 1.0 cm2 of a clean metallic surface of work-function 2.3 eV. Assume that there is no loss of light by reflection and that each energetically capable photon ejects one electron. Calculate the number of photoelectrons liberated in two seconds. (JEE 1989) Solution Energy of photon having wavelength 4144 Å, E1 =

Similarly,

12375 eV 4144

= 2.99 eV 12375 E2 = eV 4972 = 2.49 eV and 12375 E3 = eV 6216 = 1.99 eV

Since, only E1 and E 2 are greater than the work-function W = 2.3 eV , only first two wavelengths are capable for ejecting photoelectrons. Given intensity is equally distributed in all wavelengths. Therefore, intensity corresponding to each wavelength is 3.6 ´ 10-3 = 1.2 ´ 10-3 W/m2 3 Or energy incident per second in the given area ( A = 1.0 cm2 = 10-4 m2) is P = 1.2 ´ 10-3 ´ 10-4 = 1.2 ´ 10-7 J/s Let n1 be the number of photons incident per unit time in the given area corresponding to first wavelength. Then, P n1 = E1 =

Similarly,

1.2 ´ 10-7 2.99 ´ 1.6 ´ 10-19

= 2.5 ´ 1011 P n2 = E2 =

1.2 ´ 10-7 2.49 ´ 1.6 ´ 10-19

= 3.0 ´ 1011 Since, each energetically capable photon ejects one electron, total number of photoelectrons liberated in 2 s. = 2(n1 + n2) = 2 (2.5 + 3.0) ´ 1011 = 1.1 ´ 1012

Ans.

Miscellaneous Examples V

Example 17 Two metallic plates A and B each of area 5 ´ 10 -4 m2 , are placed parallel to each other at a separation of 1 cm. Plate B carries a positive charge of 33.7 ´ 10 -12 C. A monochromatic beam of light, with photons of energy 5 eV each, starts falling on plate A at t = 0 so that 1016 photons fall on it per square metre per second. Assume that one photoelectron is emitted for every 10 6 incident photons. Also assume that all the emitted photoelectrons are collected by plate B and the work-function of plate A remains constant at the value 2 eV. Determine (JEE 2002) (a) the number of photoelectrons emitted upto t = 10 s, (b) the magnitude of the electric field between the plates A and B at t = 10 s and (c) the kinetic energy of the most energetic photoelectrons emitted at t = 10s when it reaches plate B. Neglect the time taken by the photoelectron to reach plate B. (Take, e 0 = 8.85 ´ 10-12C 2 / N -m 2). Solution Area of plates, = 5 ´ 10-4 m2 Distance between the plates, d = 1 cm = 10-2 m (a) Number of photoelectrons emitted upto t = 10 s are n=

(number of photons falling on unit area in unit time) ´ (area ´ time) 106 1 = 6 [(10)16 ´ (5 ´ 10-4 ) ´ (10)] 10 = 5.0 ´ 107

(b) At time t = 10 s, Charge on plate A,

qA = + ne = (5.0 ´ 107 ) (1.6 ´ 10-19 ) = 8.0 ´ 10-12 C

and charge on plate B, qB = (33.7 ´ 10-12 - 8.0 ´ 10-12) = 25.7 ´ 10-12 C (q - qA ) \ Electric field between the plates, E = B 2 Ae0 or

E=

(25.7 - 8.0) ´ 10-12 2 ´ (5 ´ 10-4 ) (8.85 ´ 10-12)

= 2 ´ 103 N/C (c) Energy of most energetic photoelectrons at plate A, = E - W = (5 - 2) eV = 3 eV

Ans.

298 — Optics and Modern Physics Increase in energy of photoelectrons = (eEd ) joule = (Ed ) eV = (2 ´ 103 ) (10-2) eV = 20 eV Energy of photoelectrons at plate B = (20 + 3) eV = 23 eV V

Ans.

Example 18 Photoelectrons are emitted when 400 nm radiation is incident on a surface of work-function 1.9 eV. These photoelectrons pass through a region containing a-particles. A maximum energy electron combines with an a-particle to form a He + ion, emitting a single photon in this process. He + ions thus formed are in their fourth excited state. Find the energies in eV of the photons lying in the 2 to 4 eV range, that are likely to be emitted during and after the combination. [Take, h = 4.14 ´ 10 -15 eV-s] (JEE 1999) Solution Given work-function W = 1.9 eV Wavelength of incident light, l = 400 nm hc = 3.1 eV \ Energy of incident light, E = l (Substituting the values of h, c and l) Therefore, maximum kinetic energy of photoelectrons K max = E - W = (3.1 - 1.9) = 1.2 eV Now, the situation is as shown in figure. e– K max = 1.2eV

n=5 E 5 = – 2.2 eV

a-particles He+ in fourth excited state or n=5 (Z = 2)

He+

Energy of electron in 4th excited state of He+ (n = 5) will be Z2 eV n2 (2)2 E5 = - (13.6) 2 = - 2.2 eV (5) E5 = - 13.6

Þ

Therefore, energy released during the combination = 1.2 - (-2.2) = 3.4 eV Similarly, energies in other energy states of He+ will be (2)2 = - 3.4 eV (4)2 (2)2 E3 = - 13.6 = - 6.04 eV (3)2 E 4 = - 13.6

E 2 = - 13.6

(2)2 = - 13.6 eV (2)2

Chapter 33

Modern Physics - I — 299

The possible transitions are DE5 ® 4 = E5 DE5 ® 3 = E5 DE5 ® 2 = E5 DE 4 ® 3 = E 4 DE 4 ® 2 = E 4

- E 4 = 1.2 eV < 2 eV - E3 = 3.84 eV - E 2 = 11.4 eV > 4 eV - E3 = 2.64 eV - E 2 = 10.2 eV > 4 eV

Hence, the energy of emitted photons in the range of 2 eV and 4 eV are 3.4 eV during combination and 3.84 eV and 2.64 after combination. V

Example 19 Light from a discharge tube containing hydrogen atoms falls on the surface of a piece of sodium. The kinetic energy of the fastest photoelectrons emitted from sodium is 0.73 eV. The work-function for sodium is 1.82 eV. (JEE 1992) Find (a) the energy of the photons causing the photoelectrons emission. (b) the quantum numbers of the two levels involved in the emission of these photons. (c) the change in the angular momentum of the electron in the hydrogen atom, in the above transition, and (d) the recoil speed of the emitting atom assuming it to be at rest before the transition. (Ionization potential of hydrogen is 13.6 eV.) Solution (a) From Einstein’s equation of photoelectric effect, Energy of photons causing the photoelectric emission = Maximum kinetic energy of emitted photons + work-function or

E = K max + W = (0.73 + 1.82) eV E = 2.55 eV

or

Ans.

(b) In case of a hydrogen atom, E1 = - 13.6 eV , E 2 = - 3.4 eV , E3 = - 1.5 eV E 4 = - 0.85 eV E 4 - E 2 = 2.55 eV

Since,

Therefore, quantum numbers of the two levels involved in the emission of these photons are 4 and 2(4 ® 2). (c) Change in angular momentum in transition from 4 to 2 will be æhö æhö DL = L 2 - L 4 = 2 ç ÷ - 4 ç ÷ è 2p ø è 2p ø

or

DL = -

h p

(d) From conservation of linear momentum

or or

|Momentum of hydrogen atom|=|Momentum of emitted photon| E (m = mass of hydrogen atom) mv = c v=

E (2.55 ´ 1.6 ´ 10-19 J) = mc (1.67 ´ 10-27 kg ) (3.0 ´ 108 m/s )

v = 0.814 m/s

Ans.

300 — Optics and Modern Physics V

Example 20 If an X-ray tube operates at the voltage of 10 kV , find the ratio of the de-Broglie wavelength of the incident electrons to the shortest wavelength of X-rays produced. The specific charge of electron is 1.8 ´ 1011 C / kg . Solution de-Broglie wavelength when a charge q is accelerated by a potential difference of V volts is lb =

h 2qVm

For cut off wavelength of X-rays, we have

qV =

hc lm

or

lm =

From Eqs. (i) and (ii), we get

lb = lm

For electron

…(ii)

qV 2m c

q = 1.8 ´ 1011 C/kg (given). Substituting the values the desired ratio is m lb = lm

V

hc qV

…(i)

1.8 ´ 1011 ´ 10 ´ 103 2 = 0.1 3 ´ 108

Ans.

Example 21 The wavelength of the first line of Lyman series for hydrogen is identical to that of the second line of Balmer series for some hydrogen like ion x. Calculate energies of the first four levels of x. Solution Wavelength of the first line of Lyman series for hydrogen atom will be given by the equation 1 1 ö 3R æ1 = R ç 2 – 2÷ = l1 4 2 ø è1 The wavelength of second Balmer line for hydrogen like ion x is 1 1 ö 3RZ 2 æ1 = RZ 2ç 2 – 2 ÷ = 16 l2 4 ø è2 1 1 Given that = l1 = l 2 or l1 l 2 i.e.

…(i)

…(ii)

3R 3RZ 2 = 4 16

\ Z =2 i.e. x ion is He+ . The energies of first four levels of x are

and

E1 = – (13.6) Z 2 = – 54.4 eV E E 2 = 12 = – 13.6 eV (2) E E3 = 12 = – 6.04 eV (3) E E 4 = 12 = – 3.4 eV (4)

Ans.

Chapter 33 V

Modern Physics - I — 301

Example 22 A moving hydrogen atom makes a head on collision with a stationary hydrogen atom. Before collision both atoms are in ground state and after collision they move together. What is the minimum value of the kinetic energy of the moving hydrogen atom, such that one of the atoms reaches one of the excited state? Solution Let K be the kinetic energy of the moving hydrogen atom and K ¢, the kinetic energy of combined mass after collision. n=2 K¢

K m

m

DE = 10.2 eV

2m

n=1

From conservation of linear momentum, p = p¢ or

2Km = 2K ¢ (2m) K = 2K ¢ K = K ¢ + DE K Solving Eqs. (i) and (ii), we get DE = 2 Now, minimum value of DE for hydrogen atom is 10.2 eV. or DE ³ 10.2 eV K ³ 10.2 \ 2

V

or From conservation of energy,

…(i) …(ii)

\ K ³ 20.4 eV Therefore, the minimum kinetic energy of moving hydrogen is 20.4 eV.

Ans.

Example 23 An imaginary particle has a charge equal to that of an electron and mass 100 times the mass of the electron. It moves in a circular orbit around a nucleus of charge + 4e. Take the mass of the nucleus to be infinite. Assuming that the Bohr model is applicable to this system. (a) Derive an expression for the radius of nth Bohr orbit. (b) Find the wavelength of the radiation emitted when the particle jumps from fourth orbit to the second orbit. Solution (a) We have

m pv2 rn

=

1 Ze2 4pe0 rn2

…(i)

The quantization of angular momentum gives m pvrn =

nh 2p

Solving Eqs. (i) and (ii), we get r= Substituting

n 2h 2e0 Zpm pe2

m p = 100 m

…(ii)

302 — Optics and Modern Physics where, m = mass of electron and Z = 4 rn =

We get,

n 2h 2e0 400 p me2

Ans.

E1H = – 13.60 eV

(b) As we know,

æ Z 2ö E n µ çç 2 ÷÷ m èn ø

and

E4 =

For the given particle,

(–13.60) (4)2 ´ 100 (4)2

= – 1360 eV E2 =

and

(–13.60) (4)2 ´ 100 (2)2

= – 5440 eV DE = E 4 – E 2 \

= 4080 eV 12375 l (in Å ) = DE (in eV ) =

12375 4080

= 3.0 Å V

Ans.

Example 24 The energy levels of a hypothetical one electron atom are given by En = -

18.0 n2

eV

where n = 1, 2, 3, ¼ (a) Compute the four lowest energy levels and construct the energy level diagram. (b) What is the first excitation potential (c) What wavelengths (Å) can be emitted when these atoms in the ground state are bombarded by electrons that have been accelerated through a potential difference of 16.2 V? (d) If these atoms are in the ground state, can they absorb radiation having a wavelength of 2000 Å? (e) What is the photoelectric threshold wavelength of this atom? Solution (a) E1 =

and

–18.0 = – 18.0 eV (1)2 E2 =

–18.0 = – 4.5 eV (2)2

E3 =

–18.0 = – 2.0 eV (3)2

E4 =

–18.0 = – 1.125 eV (4)2

Chapter 33

Modern Physics - I — 303

The energy level diagram is shown in figure. E4 = –1.125 eV E3 = –2.0 eV E2 = – 4.5 eV

E1 = –18.0 eV

(b) E 2 - E1 = 13.5 eV \ First excitation potential is 13.5 V. (c) Energy of the electron accelerated by a potential difference of 16.2 V is 16.2 eV. With this energy the electron can excite the atom from n = 1 to n = 3 as E 4 – E1 = – 1.125 – (–18.0) = 16.875 eV > 16.2 eV and Now,

E3 – E1 = – 2.0 – (–18.0) = 16.0 eV < 16.2 eV 12375 12375 l32 = = E3 – E 2 – 2.0 – (– 4.5) l31

and

= 4950 Å 12375 12375 = = = 773 Å E3 – E1 16

l 21 =

Ans. Ans.

12375 12375 = E 2 – E1 – 4.5 – (–18.0)

= 917 Å

Ans.

(d) No, the energy corresponding to l = 2000 Å is 12375 E= = 6.1875 eV 2000

Ans.

The minimum excitation energy is 13.5 eV (n = 1 to n = 2). (e) Threshold wavelength for photoemission to take place from such an atom is 12375 l min = 18 = 687.5 Å V

Ans.

Example 25 In a photocell the plates P and Q have a separation of 5 cm, which are connected through a galvanometer without any cell. Bichromatic light of wavelengths 4000 Å and 6000 Å are incident on plate Q whose work-function is 2.39 eV . If a uniform magnetic field B exists parallel to the plates, find the minimum value of B for which the galvanometer shows zero deflection. Solution Energy of photons corresponding to light of wavelength l1 = 4000 Å is E1 =

12375 = 3.1 eV 4000

and that corresponding to l 2 = 6000 Å is E2 =

12375 = 2.06 eV 6000

304 — Optics and Modern Physics As, E 2 < W and E1 > W Photoelectric emission is possible with l1 only.

(W = work-function)

P B G

r

u

d

Q

Photoelectrons experience magnetic force and move along a circular path. The galvanometer will indicate zero deflection if the photoelectrons just complete semicircular path before reaching the plate P. Thus, d = r = 5 cm \

r = 5 cm = 0.05 m mv 2Km r= = Bq Bq

Further, \

Bmin =

2Km rq

K = E1 – W = (3.1 – 2.39) = 0.71 eV

Here, Substituting the values, we have Bmin =

2 ´ 0.71 ´ 1.6 ´ 10–19 ´ 9.109 ´ 10–31 (0.05) (1.6 ´ 10–19 )

= 5.68 ´ 10–5 T

Ans.

Exercises LEVEL 1 Assertion and Reason Directions : Choose the correct option. (a) If both Assertion and Reason are true and the Reason is correct explanation of the Assertion. (b) If both Assertion and Reason are true but Reason is not the correct explanation of Assertion. (c) If Assertion is true, but the Reason is false. (d) If Assertion is false but the Reason is true. 1. Assertion : X-rays cannot be deflected by electric or magnetic fields. Reason : These are electromagnetic waves. 2. Assertion : If wavelength of light is doubled, energy and momentum of photons are reduced to half. Reason : By increasing the wavelength, speed of photons will decrease. 3. Assertion : We can increase the saturation current in photoelectric experiment without increasing the intensity of light. Reason : Intensity can be increased by increasing the frequency of incident photons.

4. Assertion : Photoelectric effect proves the particle nature of light. Reason : Photoemission starts as soon as light is incident on the metal surface, provided frequency of incident light is greater than or equal to the threshold frequency.

5. Assertion : During de-excitation from n = 6 to n = 3, total six emission lines may be obtained. Reason :

From n = n to n = 1, total

n ( n - 1) emission lines are obtained. 2

6. Assertion : If frequency of incident light is doubled, the stopping potential will also become two times. Reason : Stopping potential is given by h (n - n 0 ) e 7. Assertion : X-rays cannot be obtained in the emission spectrum of hydrogen atom. Reason : Maximum energy of photons emitted from hydrogen spectrum is 13.6 eV. V0 =

8. Assertion : If applied potential difference in coolidge tube is increased, then difference between K a wavelength and cut off wavelength will increase. Reason : Cut off wavelength is inversely proportional to the applied potential difference in coolidge tube. 9. Assertion : In n = 2, energy of electron in hydrogen like atoms is more compared to n = 1. Reason : Electrostatic potential energy in n = 2 is more. 10. Assertion : In continuous X-ray spectrum, all wavelengths can be obtained. Reason : Accelerated (or retarded) charged particles radiate energy. This is the cause of production of continuous X-rays.

306 — Optics and Modern Physics Objective Questions 1. According to Einstein’s photoelectric equation, the plot of the maximum kinetic energy of the emitted photoelectrons from a metal versus frequency of the incident radiation gives a straight line whose slope (a) (b) (c) (d)

depends on the nature of metal used depends on the intensity of radiation depends on both intensity of radiation and the nature of metal used is the same for all metals and independent of the intensity of radiation

2. The velocity of the electron in the first Bohr orbit as compared to that of light is about (a) 1/300 (c) 1/137

3.

86 A

222

(b) 1/500 (d) 1/187

® 84B210. In this reaction, how many a and b particles are emitted?

(a) 6 a , 3 b (c) 4 a , 3 b

(b) 3 a , 4 b (d) 3 a , 6 b

4. An X-ray tube is operated at 20 kV. The cut off wavelength is (a) 0.89 Å (c) 0.62 Å

(b) 0.75 Å (d) None of these

5. An X-ray tube is operated at 18 kV. The maximum velocity of electron striking the target is (a) 8 ´107 m/s (c) 5 ´107 m/s

(b) 6 ´107 m/s (d) None of these

6. What is the ratio of de-Broglie wavelength of electron in the second and third Bohr orbits in the hydrogen atoms? (a) 2/3 (c) 4/3

(b) 3/2 (d) 3/4

7. The energy of a hydrogen like atom (or ion) in its ground state is – 122.4 eV. It may be (a) hydrogen atom (c) Li 2+

(b) He + (d) Be3 +

8. The operating potential in an X-ray tube is increased by 2%. The percentage change in the cut off wavelength is (a) 1% increase (c) 2% decrease

(b) 2% increase (d) 1% decrease

9. The energy of an atom or ion in the first excited state is –13.6 eV. It may be (a) He + (c) hydrogen

(b) Li + + (d) deuterium

10. In order that the short wavelength limit of the continuous X-ray spectrum be 1 Å, the potential difference through which an electron must be accelerated is (a) 124 kV (c) 12.4 kV

(b) 1.24 kV (d) 1240 kV

11. The momentum of an X-ray photon with l = 0.5 Å is (a) (b) (c) (d)

13.26 ´ 10-26 kg-m/s 1.326 ´ 10-26 kg-m/s 13.26 ´ 10-24 kg-m/s 13.26 ´ 10-22 kg-m/s

Chapter 33

Modern Physics - I — 307

12. The work-function of a substance is 1.6 eV. The longest wavelength of light that can produce photoemission from the substance is (a) 7750 Å (c) 5800 Å

(b) 3875 Å (d) 2900 Å

13. Find the binding energy of an electron in the ground state of a hydrogen like atom in whose spectrum the third Balmer line is equal to 108.5 nm. (a) 54.4 eV (c) 112.4 eV

(b) 13.6 eV (d) None of these

14. Let the potential energy of hydrogen atom in the ground state be zero. Then, its energy in the first excited state will be (a) 10.2 eV (c) 23.8 eV

(b) 13.6 eV (d) 27.2 eV

15. Light of wavelength 330 nm falling on a piece of metal ejects electrons with sufficient energy with required voltage V 0 to prevent them from reaching a collector. In the same set up, light of wavelength 220 nm ejects electrons which require twice the voltage V 0 to stop them in reaching a collector. The numerical value of voltage V 0 is 16 V 15 15 (c) V 8 (a)

15 V 16 8 (d) V 15

(b)

16. Maximum kinetic energy of a photoelectron is E when the wavelength of incident light is l. If energy becomes four times when wavelength is reduced to one-third, then work-function of the metal is 3 hc l hc (c) l

(a)

hc 3l hc (d) 2l

(b)

17. If the frequency of K a X-ray emitted from the element with atomic number 31 is f, then the frequency of K a X-ray emitted from the element with atomic number 51 would be 5f 3 9f (c) 25

(a)

51 f 31 25 f (d) 9 (b)

18. According to Moseley’s law, the ratio of the slope of graph between f and Z for K b and K a is (a) (c)

32 27 5 36

(b) (d)

27 32 36 5

19. If the electron in hydrogen orbit jumps from third orbit to second orbit, the wavelength of the emitted radiation is given by R 6 36 (c) l = 5R (a) l =

5 R 5R (d) l = 36 (b) l =

308 — Optics and Modern Physics 20. A potential of 10000 V is applied across an X-ray tube. Find the ratio of de-Broglie wavelength associated with incident electrons to the minimum wavelength associated with X-rays. (Given, e/m =1.8 ´ 1011 C/kg for electrons) (b) 20 (d) 1/20

(a) 10 (c) 1/10

21. When a metallic surface is illuminated with monochromatic light of wavelength l, the stopping potential is 5 V 0. When the same surface is illuminated with the light of wavelength 3l, the stopping potential is V 0. Then, the work-function of the metallic surface is (a) hc/6l (c) hc/4l

(b) hc/5l (d) 2hc/4l

22. The threshold frequency for a certain photosensitive metal is n 0. When it is illuminated by light

of frequency n = 2 n 0 , the stopping potential for photoelectric current is V 0. What will be the stopping potential when the same metal is illuminated by light of frequency n = 3n 0? (b) 2 V 0 (d) 3 V 0

(a) 1.5 V 0 (c) 2.5 V 0

23. The frequency of the first line in Lyman series in the hydrogen spectrum is n. What is the frequency of the corresponding line in the spectrum of doubly ionized Lithium? (a) n (c) 9 n

(b) 3 n (d) 2 n

24. Which energy state of doubly ionized lithium (Li + + ) has the same energy as that of the ground state of hydrogen? (a) n = 1 (c) n = 3

(b) n = 2 (d) n = 4

25. Two identical photo-cathodes receive light of frequencies n1 and n 2. If the velocities of the photoelectrons (of mass m) coming out are v1 and v2 respectively, then é 2h ù (a) v1 - v2 = ê (n1 - n 2)ú m ë û

1/ 2

é 2h ù (c) v1 + v2 = ê (n1 - n 2)ú m ë û

(b) v12 - v22 =

2h (n1 - n 2) m

(d) v12 + v22 =

2h (n1 - n 2) m

1/ 2

26. The longest wavelength of the Lyman series for hydrogen atom is the same as the wavelength

of a certain line in the spectrum of He+ when the electron makes a transition from n ® 2. The value of n is

(a) 3

(b) 4

(c) 5

(d) 6

27. The wavelength of the K a - line for the uranium is (Z = 92) (R = 1.0973 ´107 m-1 ) (a) 1.5 Å

(b) 0.5 Å

(c) 0.15 Å

(d) 2.0 Å

28. The frequencies of K a , K b and La X-rays of a material are g1 , g 2 and g3 respectively. Which of the following relation holds good? (a) g 2 = g1 g3 g + g3 (c) g 2 = 1 2

(b) g 2 = g1 + g3 (d) g3 = g1 g 2

29. A proton and an a-particle are accelerated through same potential difference. Then, the ratio of de-Broglie wavelength of proton and a-particle is (a)

2

(b)

1 2

(c) 2 2

(d) None of these

Chapter 33

Modern Physics - I — 309

30. If E1 , E2 and E3 represent respectively the kinetic energies of an electron , an a-particle and a proton each having same de-Broglie wavelength, then (a) E1 > E3 > E 2 (c) E1 > E 2 > E3

(b) E 2 > E3 > E1 (d) E1 = E 2 = E3

31. If the potential energy of a hydrogen atom in the ground state is assumed to be zero, then total energy of n = ¥ is equal to (a) 13.6 eV (c) zero

(b) 27.2 eV (d) None of these

32. A 1000 W transmitter works at a frequency of 880 kHz. The number of photons emitted per second is (a) 1.7 ´1028 (c) 1.7 ´1023

(b) 1.7 ´1030 (d) 1.7 ´1025

33. Electromagnetic radiation of wavelength 3000 Å is incident on an isolated platinum surface of work-function 6.30 eV. Due to the radiation, the (a) (b) (c) (d)

sphere becomes positively charged sphere becomes negatively charged sphere remains neutral maximum kinetic energy of the ejected photoelectrons would be 2.03 eV

34. The energy of a hydrogen atom in its ground state is – 13.6 eV. The energy of the level corresponding to the quantum number n = 5 is (a) – 0.54 eV (c) – 0.85 eV

(b) – 5.40 eV (d) – 2.72 eV

35. Ultraviolet radiation of 6.2 eV falls on an aluminium surface (work-function = 4.2 eV). The kinetic energy in joule of the fastest electrons emitted is (a) 3.2 ´ 10-21 (c) 3.2 ´ 10-17

(b) 3.2 ´ 10-19 (d) 3.2 ´ 10-15

36. What should be the velocity of an electron so that its momentum becomes equal to that of a photon of wavelength 5200 Å? (a) 700 m/s (c) 1400 m/s

(b) 1000 m/s (d) 2800 m/s

37. Photoelectric work-function of a metal is 1 eV. Light of wavelength l = 3000 Å falls on it. The photoelectrons come out with maximum velocity (a) 10 m/s (c) 104 m/s

(b) 103 m/s (d) 106 m/s

Subjective Questions Note You can take approximations in the answers. h = 6.62 ´ 10 -34 J- s, c = 3.0 ´ 10 8 m/ s, me = 9.1 ´ 10 -31 kg and 1 eV = 1.6 ´ 10 -19 J

1. For a given element the wavelength of the K a -line is 0.71 nm and of the K b-line it is 0.63 nm. Use this information to find wavelength of the La -line.

2. The energy of the n = 2 state in a given element is E2 = – 2870 eV. Given that the wavelengths of the K a and K b-lines are 0.71 nm and 0.63 nm respectively, determine the energies E1 and E3 .

3. 1.5 mW of 400 nm light is directed at a photoelectric cell. If 0.1% of the incident photons produce photoelectrons, find the current in the cell.

310 — Optics and Modern Physics 4. A photon has momentum of magnitude 8.24 ´ 10-28 kg-m/ s. (a) What is the energy of this photon? Give your answer in joules and in electron volts. (b) What is the wavelength of this photon? In what region of the electromagnetic spectrum does it lie?

5. A 75 W light source emits light of wavelength 600 nm. (a) Calculate the frequency of the emitted light. (b) How many photons per second does the source emit?

6. An excited nucleus emits a gamma-ray photon with energy of 2.45 MeV. (a) What is the photon frequency? (b) What is the photon wavelength?

7. (a) A proton is moving at a speed much less than the speed of light. It has kinetic energy K 1

and momentum p1. If the momentum of the proton is doubled, so p2 = 2 p1 , how is its new kinetic energy K 2 related to K 1?

(b) A photon with energy E1 has momentum p1. If another photon has momentum p2 that is twice p1 , how is the energy E 2 of the second photon related to E1?

8. A parallel beam of monochromatic light of wavelength 500 nm is incident normally on a perfectly absorbing surface. The power through any cross-section of the beam is 10 W. Find (a) the number of photons absorbed per second by the surface and (b) the force exerted by the light beam on the surface.

9. A beam of white light is incident normally on a plane surface absorbing 70% of the light and reflecting the rest. If the incident beam carries 10 W of power, find the force exerted by it on the surface.

10. A parallel beam of monochromatic light of wavelength 663 nm is incident on a totally reflecting plane mirror. The angle of incidence is 60° and the number of photons striking the mirror per second is 1.0 ´ 1019. Calculate the force exerted by the light beam on the mirror.

11. Wavelength of Bullet. Calculate the de-Broglie wavelength of a 5.00 g bullet that is moving at 340 m/ s. Will it exhibit wave like properties?

12. (a) An electron moves with a speed of 4.70 ´ 106 m/ s. What is its de-Broglie wavelength? (b) A proton moves with the same speed. Determine its de-Broglie wavelength.

13. An electron has a de-Broglie wavelength of 2.80 ´ 10-10 m. Determine (a) the magnitude of its momentum, (b) its kinetic energy (in joule and in electron volt).

14. Find de-Broglie wavelength corresponding to the root-mean square velocity of hydrogen molecules at room temperature (20°C).

15. An electron, in a hydrogen like atom, is in excited state. It has a total energy of –3.4 eV, find the de-Broglie wavelength of the electron.

16. In the Bohr model of the hydrogen atom, what is the de-Broglie wavelength for the electron when it is in (a) the n = 1 level? (b) the n = 4 level? In each case, compare the de-Broglie wavelength to the circumference 2prn of the orbit.

17. The binding energy of an electron in the ground state of He atom is equal to E0 = 24.6 eV. Find the energy required to remove both electrons from the atom.

Chapter 33

Modern Physics - I — 311

18. Hydrogen atom in its ground state is excited by means of monochromatic radiation of wavelength 1023 Å. How many different lines are possible in the resulting spectrum? Calculate the longest wavelength among them. You may assume the ionization energy of hydrogen atom as 13.6 eV.

19. A doubly ionized lithium atom is hydrogen like with atomic number 3. Find the wavelength of

the radiation required to excite the electron in Li+ + from the first to the third Bohr orbit (ionization energy of the hydrogen atom equals 13.6 eV).

20. Find the quantum number n corresponding to nth excited state of He+ ion if on transition to the ground state the ion emits two photons in succession with wavelengths 108.5 nm and 30.4 nm. The ionization energy of the hydrogen atom is 13.6 eV.

21. A hydrogen like atom (described by the Bohr model) is observed to emit ten wavelengths, originating from all possible transitions between a group of levels. These levels have energies between –0.85 eV and –0.544 eV (including both these values). (a) Find the atomic number of the atom. (b) Calculate the smallest wavelength emitted in these transitions. (Take ground state energy of hydrogen atom = - 13.6 eV)

22. The energy levels of a hypothetical one electron atom are shown in the figure.

µ n=5 n=4

(a) Find the ionization potential of this atom. (b) Find the short wavelength limit of the series terminating at n = 3 n = 2. n=2 (c) Find the excitation potential for the state n = 3. n=1 (d) Find wave number of the photon emitted for the transition n = 3 to n = 1.

0 eV –0.80 eV –1.45 eV –3.08 eV –5.30 eV –15.6 eV

23. (a) An atom initially in an energy level with E = – 6.52 eV absorbs a photon that has wavelength 860 nm. What is the internal energy of the atom after it absorbs the photon? (b) An atom initially in an energy level with E = - 2.68 eV emits a photon that has wavelength 420 nm. What is the internal energy of the atom after it emits the photon?

24. A silver ball is suspended by a string in a vacuum chamber and ultraviolet light of wavelength 2000 Å is directed at it. What electrical potential will the ball acquire as a result? Work function of silver is 4.3 eV. 1 25. A small particle of mass m moves in such a way that the potential energy U = m 2w2r 2, 2 where wis a constant and r is the distance of the particle from the origin. Assuming Bohr model of quantization of angular momentum and circular orbits, show that radius of the nth allowed orbit is proportional to n.

26. Wavelength of K a -line of an element is l0. Find wavelength of K b-line for the same element. 27. X-rays are produced in an X-ray tube by electrons accelerated through an electric potential difference of 50.0 kV. An electron makes three collisions in the target coming to rest and loses half its remaining kinetic energy in each of the first two collisions. Determine the wavelength of the resulting photons. (Neglecting the recoil of the heavy target atoms).

28. From what material is the anode of an X-ray tube made, if the K a - line wavelength of the characteristic spectrum is 0.76 Å?

29. A voltage applied to an X-ray tube being increased h = 1.5 times, the short wave limit of an X-ray continuous spectrum shifts by Dl = 26 pm. Find the initial voltage applied to the tube.

312 — Optics and Modern Physics 30. The K a X-rays of aluminium ( Z = 13) and zinc ( Z = 30) have wavelengths 887 pm and 146 pm, respectively. Use Moseley’s equation iron ( Z = 26).

n = a( Z - b) to find the wavelength of the K a X-ray of

31. Characteristic X-rays of frequency 4.2 ´ 1018 Hz are produced when transitions from L shell take place in a certain target material. Use Moseley’s law and determine the atomic number of the target material. Given, Rydberg constant is R = 1.1 ´ 107 m -1.

32. The electric current in an X-ray tube operating at 40 kV is 10 mA. Assume that on an average 1% of the total kinetic energy of the electrons hitting the target are converted into X-rays. (a) What is the total power emitted as X-rays and (b) How much heat is produced in the target every second?

33. The stopping potential for the photoelectrons emitted from a metal surface of work-function 1.7 eV is 10.4 V. Find the wavelength of the radiation used. Also, identify the energy levels in hydrogen atom, which will emit this wavelength.

34. What will be the maximum kinetic energy of the photoelectrons ejected from magnesium (for which the work-function W = 3.7 eV) when irradiated by ultraviolet light of frequency 1.5 ´ 1015 s-1.

35. A metallic surface is irradiated with monochromatic light of variable wavelength. Above a wavelength of 5000 Å, no photoelectrons are emitted from the surface. With an unknown wavelength, stopping potential of 3 V is necessary to eliminate the photocurrent. Find the unknown wavelength.

36. A graph regarding photoelectric effect is shown between the maximum kinetic energy of electrons and the frequency of the incident light . On the basis of data as shown in the graph, calculate Kmax (eV) 8 6 4 2 0 –2 –4 C

(a) threshold frequency,

A 10

D 20 30 f (´ 1014 Hz)

(b) work-function,

(c) planck constant

37. A metallic surface is illuminated alternatively with light of wavelengths 3000 Å and 6000 Å. It is observed that the maximum speeds of the photoelectrons under these illuminations are in the ratio 3 : 1. Calculate the work-function of the metal and the maximum speed of the photoelectrons in two cases.

38. Light of wavelength 180 nm ejects photoelectrons from a plate of metal whose work-function is 2 eV. If a uniform magnetic field of 5 ´ 10–5 T be applied parallel to the plate, what would be the radius of the path followed by electrons ejected normally from the plate with maximum energy.

39. Light described at a place by the equation E = (100 V/ m ) [sin( 5 ´ 1015 s –1 )t + sin( 8 ´ 1015 s –1 )t ] falls on a metal surface having work-function 2.0 eV. Calculate the maximum kinetic energy of the photoelectrons.

40. The electric field associated with a light wave is given by E = E0 sin [(1.57 ´ 107 m -1 ) ( x - ct )]. Find the stopping potential when this light is used in an experiment on photoelectric effect with a metal having work-function 1.9 eV.

LEVEL 2 Single Correct Option 1. If we assume only gravitational attraction between proton and electron in hydrogen atom and the Bohr quantization rule to be followed, then the expression for the ground state energy of the atom will be (the mass of proton is M and that of electron is m.) G 2M 2m2 h2 2p 2GM 2m3 (c) h2

(a)

(b) -

2p 2G 2M 2m3 h2

(d) None of these

2. An electron in a hydrogen atom makes a transition from first excited state to ground state. The magnetic moment due to circulating electron (a) increases two times (c) increases four times

(b) decreases two times (d) remains same

3. The excitation energy of a hydrogen like ion to its first excited state is 40.8 eV. The energy needed to remove the electron from the ion in the ground state is (a) 54.4 eV (c) 72.6 eV

(b) 62.6 eV (d) 58.6 eV

4. An electron in a hydrogen atom makes a transition from first excited state to ground state. The equivalent current due to circulating electron (a) increases 4 times (c) increases 8 times

(b) decreases 4 times (d) decreases 8 times

5. In a sample of hydrogen like atoms all of which are in ground state, a photon beam containing photons of various energies is passed. In absorption spectrum, five dark lines are observed. The number of bright lines in the emission spectrum will be (assume that all transitions take place) (a) 21 (c) 15

(b) 10 (d) None of these

6. Let An be the area enclosed by the nth orbit in a hydrogen atom. The graph of ln ( An / A1 ) against ln ( n ) (a) will not pass through origin (c) will be a rectangular hyperbola

(b) will be a straight line with slope 4 (d) will be a parabola

7. In the hydrogen atom, an electron makes a transition from n = 2 to n = 1. The magnetic field produced by the circulating electron at the nucleus (a) decreases 16 times (c) decreases 4 times

(b) increases 4 times (d) increases 32 times

8. A stationary hydrogen atom emits photon corresponding to the first line of Lyman series. If R is the Rydberg constant and M is the mass of the atom, then the velocity acquired by the atom is 3Rh 4M Rh (c) 4M

(a)

4M 3Rh 4M (d) Rh (b)

314 — Optics and Modern Physics 9. Light wave described by the equation 200 V/m sin (1.5 ´ 1015 s-1 ) t cos ( 0.5 ´ 1015 s-1 ) t falls on metal surface having work-function 2.0 eV. Then, the maximum kinetic energy of photoelectrons is (b) 2.2 eV (d) None of these

(a) 3.27 eV (c) 2.85 eV

10. A hydrogen like atom is excited using a radiation. Consequently, six spectral lines are observed in the spectrum. The wavelength of emission radiation is found to be equal or smaller than the radiation used for excitation. This concludes that the gas was initially at (a) ground state (c) second excited state

(b) first excited state (d) third excited state

11. The time period of the electron in the ground state of hydrogen atom is two times the time period of the electron in the first excited state of a certain hydrogen like atom (Atomic number Z). The value of Z is (a) 2 (c) 4

(b) 3 (d) None of these

12. The wavelengths of K a X-rays from lead isotopes Pb 204 , Pb 206 and Pb 208 are l1 , l2 and l3 respectively. Choose the correct alternative. (a) l1 < l 2 < l3 (c) l1 = l 2 = l3

(b) l1 > l 2 > l3 (d) None of these

13. In case of hydrogen atom, whenever a photon is emitted in the Balmer series, (a) there is a probability of emitting another photon in the Lyman series (b) there is a probability of emitting another photon of wavelength 1213 Å (c) the wavelength of radiation emitted in Lyman series is always shorter than the wavelength emitted in the Balmer series (d) All of the above

14. An electron of kinetic energy K collides elastically with a stationary hydrogen atom in the ground state. Then, (b) K > 10.2 eV (d) data insufficient

(a) K > 13.6 eV (c) K < 10.2 eV

15. In a stationary hydrogen atom, an electron jumps from n = 3 to n = 1. The recoil speed of the hydrogen atom is about (a) 4 m/s (c) 4 mm/s

(b) 4 cm/s (d) 4 ´ 10-4 m/s

16. An X-ray tube is operating at 150 kV and 10 mA. If only 1% of the electric power supplied is converted into X-rays, the rate at which the target is heated in calorie per second is (a) 3.55 (c) 355

(b) 35.5 (d) 3550

17. An electron revolves round a nucleus of atomic number Z. If 32.4 eV of energy is required to excite an electron from the n = 3 state to n = 4 state, then the value of Z is (a) 5 (c) 4

(b) 6 (d) 7

18. If the de-Broglie wavelength of a proton is 10-13 m, the electric potential through which it must have been accelerated is (a) 4.07 ´ 104 V

(b) 8.15 ´ 104 V

(c) 8.15 ´ 103 V

(d) 4.07 ´ 105 V

Chapter 33

Modern Physics - I — 315

19. If En and Ln denote the total energy and the angular momentum of an electron in the nth orbit of Bohr atom, then (a) E n µ L n (c) E n µ L2n

1 Ln 1 (d) E n µ 2 Ln

(b) E n µ

20. An orbital electron in the ground state of hydrogen has the magnetic moment m 1. This orbital electron is excited to 3rd excited state by some energy transfer to the hydrogen atom. The new magnetic moment of the electron is m 2, then (a) m 1 = 4m 2 (c) 16 m 1 = m 2

(b) 2 m 1 = m 2 (d) 4 m 1 = m 2

21. A moving hydrogen atom makes a head-on collision with a stationary hydrogen atom. Before collision, both atoms are in ground state and after collision they move together. The minimum value of the kinetic energy of the moving hydrogen atom, such that one of the atoms reaches one of the excitation state is (a) 20.4 eV (c) 54.4 eV

(b) 10.2 eV (d) 13.6 eV

22. In an excited state of hydrogen like atom an electron has total energy of – 3.4 eV. If the kinetic energy of the electron is E and its de-Broglie wavelength is l, then (a) l = 6.6 Å (c) Both are correct

(b) E = 3.4 eV (d) Both are wrong

More than One Correct Options 1. If the potential difference of coolidge tube producing X-ray is increased, then choose the correct option (s). (a) (b) (c) (d)

the interval between lKa and lKb increases the interval between lKa and l 0 increases the interval between lKb and l 0 increases l 0 does not change

Here, l0 is cut off wavelength and lK a and lK b are wavelengths of K a and K b characteristic X-rays.

2. In Bohr model of the hydrogen atom, let R , v and E represent the radius of the orbit, speed of the electron and the total energy of the electron respectively. Which of the following quantities are directly proportional to the quantum number n? (a) vR v (c) E

(b) RE R (d) E

3. The magnitude of angular momentum, orbital radius and time period of revolution of an electron in a hydrogen atom corresponding to the quantum number n are L, r and T respectively. Which of the following statement(s) is/are correct? rL is independent of n T T (c) µn r

(a)

L 1 µ T n2 1 (d) Lr µ 3 n

(b)

316 — Optics and Modern Physics 4. In which of the following cases the heavier of the two particles has a smaller de-Broglie wavelength? The two particles (a) (b) (c) (d)

move with the same speed move with the same linear momentum move with the same kinetic energy have the same change of potential energy in a conservative field

5. Hydrogen atom absorbs radiations of wavelength l0 and consequently emit radiations of

6 different wavelengths, of which two wavelengths are longer than l0. Choose the correct alternative(s).

(a) (b) (c) (d)

The final excited state of the atoms is n = 4 The initial state of the atoms is n = 2 The initial state of the atoms is n = 3 There are three transitions belonging to Lyman series

6. In coolidge tube, if f and l represent the frequency and wavelength of K a -line for a metal of atomic number Z, then identify the statement which represents a straight line (a)

f versus Z

(c) f versus Z

1 versus Z l (d) l versus Z (b)

Comprehension Based Questions Passage I (Q. No. 1 to 3) When a surface is irradiated with light of wavelength 4950 Å, a photocurrent appears which vanishes if a retarding potential greater than 0.6 volt is applied across the phototube. When a second source of light is used, it is found that the critical retarding potential is changed to 1.1 volt. 1. The work-function of the emitting surface is (a) 2.2 eV (c) 1.9 eV

(b) 1.5 eV (d) 1.1 eV

2. The wavelength of the second source is (a) 6150 Å (c) 4125 Å

(b) 5150 Å (d) 4500 Å

3. If the photoelectrons (after emission from the source) are subjected to a magnetic field of 10 tesla, the two retarding potentials would (a) uniformly increase (c) remain the same

(b) uniformly decrease (d) None of these

Passage II (Q. No. 4 to 6) In an experimental set up to study the photoelectric effect a point source of light of power 3.2 ´ 10-3 W was taken. The source can emit monoenergetic photons of energy 5 eV and is located at a distance of 0.8 m from the centre of a stationary metallic sphere of work-function 3.0 eV . The radius of the sphere is r = 8 ´ 10-3 m. The efficiency of photoelectric emission is one for every 106 incident photons. Based on the information given above answer the questions given below. (Assume that the sphere is isolated and photoelectrons are instantly swept away after the emission).

Chapter 33

Modern Physics - I — 317

4. de-Broglie wavelength of the fastest moving photoelectron is (a) 6.63 Å (c) 2 Å

(b) 8.69 Å (d) 5.26 Å

5. It was observed that after some time emission of photoelectrons from the sphere stopped. Charge on the sphere when the photon emission stops is (a) 16pe0r coulomb (c) 15pe0r coulomb

(b) 8pe0r coulomb (d) 20pe0r coulomb

6. Time after which photoelectric emission stops is (a) 100 s (c) 111 s

(b) 121 s (d) 141 s

Match the Columns 1. Match the following two columns for hydrogen spectrum. Column I

Column II

(a) Lyman series

(p) infrared region

(b) Balmer series

(q) visible region

(c) Paschen series

(r) ultraviolet region

(d) Brackett series

(s) X-rays

2. Ionization energy from first excited state of hydrogen atom is E. Match the following two columns for He+ atom.

Column I

Column II

(a) Ionization energy from ground state

(p) 4 E

(b) Electrostatic potential energy in first excited state.

(q) – 16 E

(c)

Kinetic energy of electron in (r) – 8 E ground state.

(d) Ionization energy from first excited state.

3.

(s) 16 E

Kmax

v0 1

2

f

–Y1

f

–Y2

Maximum kinetic energy versus frequency of incident light and stopping potential versus frequency of incident light graphs are shown in figure. Match the following two columns.

318 — Optics and Modern Physics Column I (a) (b) (c) (d)

Slope of line-1 Slope of line-2 Y1 Y2

Column II h/e h W W/e

(p) (q) (r) (s)

Here, h = Planck constant, e = 1.6 ´ 10-19 C and W = work-function.

4. For hydrogen and hydrogen type atoms, match the following two columns. Column I (a) (b) (c) (d)

Time period Angular momentum Speed Radius

Column II Proportional to n/Z Proportional to n 2/Z Proportional to n3 /Z 2 None of these

(p) (q) (r) (s)

5. In hydrogen atom wavelength of second line of Balmer series is l. Match the following two columns corresponding to the wavelength. Column I

Column II

(a) First line of Balmer series

(p) (27/20) l

(b) Third line of Balmer series

(q) (l/4)

(c) First line of Lyman series

(r) (25/12) l

(d) Second line of Lyman series (s) None of these

6. Match the following (Give most appropriate one matching) Column I (a) Characteristic X-ray (b) X-ray production (c) Cut off wavelength (d) Continuous X-ray

Column II (p) Inverse process of photoelectric effect (q) Potential difference (r) Moseley’s law (s) None of these

7. In a photoelectric effect experiment. If f is the frequency of radiations incident on the metal surface and I is the intensity of the incident radiations, then match the following columns. Column I (a) If f is increased keeping I and work-function constant (b) If distance between cathode and anode is increased (c) If I is increased keeping f and work-function constant (d) Work-function is decreased keeping f and I constant

Column II (p)

Stopping potential increases

(q) Saturation current increases (r) Maximum kinetic energy of photoelectron increases (s) Stopping potential remains same

Chapter 33

Modern Physics - I — 319

Subjective Questions 1. The wavelength for n = 3 to n = 2 transition of the hydrogen atom is 656.3 nm. What are the wavelengths for this same transition in (a) positronium, which consists of an electron and a positron (b) singly ionized helium (Note: A positron is a positively charged electron).

2. (a) Find the frequencies of revolution of electrons in n = 1 and n = 2 Bohr orbits. (b) What is the frequency of the photon emitted when an electron in an n = 2 orbit drops to an n = 1 hydrogen orbit? (c) An electron typically spends about 10–8s in an excited state before it drops to a lower state by emitting a photon. How many revolutions does an electron in an n = 2 Bohr hydrogen orbit make in 1.00 ´ 10-8 s?

3. A muon is an unstable elementary particle whose mass is 207 me and whose charge is either + e or – e. A negative muon (m – ) can be captured by a nucleus to form a muonic atom. (a) A proton captures a m – . Find the radius of the first Bohr orbit of this atom. (b) Find the ionization energy of the atom.

4. (a) A gas of hydrogen atoms in their ground state is bombarded by electrons with kinetic energy 12.5 eV. What emitted wavelengths would you expect to see? (b) What if the electrons were replaced by photons of same energy?

5. A source emits monochromatic light of frequency 5.5 ´ 1014 Hz at a rate of 0.1 W. Of the photons given out, 0.15% fall on the cathode of a photocell which gives a current of 6 mA in an external circuit. (a) Find the energy of a photon. (b) Find the number of photons leaving the source per second. (c) Find the percentage of the photons falling on the cathode which produce photoelectrons.

6. The hydrogen atom in its ground state is excited by means of monochromatic radiation. Its resulting spectrum has six different lines. These radiations are incident on a metal plate. It is observed that only two of them are responsible for photoelectric effect. If the ratio of maximum kinetic energy of photoelectrons in the two cases is 5 then find the work-function of the metal.

7. Electrons in hydrogen like atoms ( Z = 3) make transitions from the fifth to the fourth orbit and from the fourth to the third orbit. The resulting radiations are incident normally on a metal plate and eject photoelectrons. The stopping potential for the photoelectrons ejected by the shorter wavelength is 3.95 volts. Calculate the work-function of the metal and the stopping potential for the photoelectrons ejected by the longer wavelength.

8. Find an expression for the magnetic dipole moment and magnetic field induction at the centre of Bohr’s hypothetical hydrogen atom in the n th orbit of the electron in terms of universal constant.

9. An electron and a proton are separated by a large distance and the electron approaches the proton with a kinetic energy of 2 eV. If the electron is captured by the proton to form a hydrogen atom in the ground state, what wavelength photon would be given off?

10. Hydrogen gas in the atomic state is excited to an energy level such that the electrostatic potential energy of H-atom becomes –1.7 eV. Now, a photoelectric plate having work-function W = 2.3 eV is exposed to the emission spectra of this gas. Assuming all the transitions to be possible, find the minimum de-Broglie wavelength of the ejected photoelectrons.

320 — Optics and Modern Physics 11. A gas of hydrogen like atoms can absorb radiation of 68 eV. Consequently, the atom emits radiations of only three different wavelengths. All the wavelengths are equal or smaller than that of the absorbed photon. (a) Determine the initial state of the gas atoms. (b) Identify the gas atoms. (c) Find the minimum wavelength of the emitted radiations. (d) Find the ionization energy and the respective wavelength for the gas atoms.

12. A photon with energy of 4.9 eV ejects photoelectrons from tungsten. When the ejected electron enters a constant magnetic field of strength B = 2.5 mT at an angle of 60° with the field direction, the maximum pitch of the helix described by the electron is found to be 2.7 mm. Find the work-function of the metal in electron-volt. Given that specific charge of electron is 1.76 ´ 1011 C/ kg.

13. For a certain hypothetical one-electron atom, the wavelength (in Å) for the spectral lines for transitions originating at n = p and terminating at n = 1 are given by l=

1500 p2 p2 - 1

, where

p = 2, 3, 4

(a) Find the wavelength of the least energetic and the most energetic photons in this series. (b) Construct an energy level diagram for this element showing the energies of the lowest three levels. (c) What is the ionization potential of this element?

14. A photocell is operating in saturation mode with a photocurrent 4.8 mA when a monochromatic radiation of wavelength 3000 Å and power of 1 mW is incident. When another monochromatic radiation of wavelength 1650 Å and power 5 mW is incident, it is observed that maximum velocity of photoelectron increases to two times. Assuming efficiency of photoelectron generation per incident photon to be same for both the cases, calculate (a) the threshold wavelength for the cell (b) the saturation current in second case (c) the efficiency of photoelectron generation per incident photon

15. Wavelengths belonging to Balmer series for hydrogen atom lying in the range of 450 nm to 750 nm were used to eject photoelectrons from a metal surface whose work-function is 2.0 eV. Find (in eV) the maximum kinetic energy of the emitted photoelectrons.

16. Assume that the de-Broglie wave associated with an electron can form a standing wave between the atoms arranged in a one-dimensional array with nodes at each of the atomic sites. It is found that one such standing wave is formed if the distance d between the atoms of the array is 2 Å. A similar standing wave is again formed if d is increased to 2.5 Å but not for any intermediate value of d. Find the energy of the electron in eV and the least value of d for which the standing wave of the type described above can form.

17. The negative muon has a charge equal to that of an electron but a mass that is 207 times as great. Consider hydrogen like atom consisting of a proton and a muon. (a) What is the reduced mass of the atom? (b) What is the ground-level energy (in eV)? (c) What is the wavelength of the radiation emitted in the transition from the n = 2 level to the n = 1 level?

Chapter 33

Modern Physics - I — 321

18. Assume a hypothetical hydrogen atom in which the potential energy between electron and proton at separation r is given by U = [k ln r – ( k/ 2)], where k is a constant. For such a hypothetical hydrogen atom, calculate the radius of nth Bohr orbit and energy levels.

19. An electron is orbiting in a circular orbit of radius r under the influence of a constant magnetic field of strength B. Assuming that Bohr postulate regarding the quantisation of angular momentum holds good for this electron, find (a) the allowed values of the radius r of the orbit. (b) the kinetic energy of the electron in orbit (c) the potential energy of interaction between the magnetic moment of the orbital current due to the electron moving in its orbit and the magnetic field B. (d) the total energy of the allowed energy levels. (e) the total magnetic flux due to the magnetic field B passing through the nth orbit. (Assume that the charge on the electron is – e and the mass of the electron is m).

20. A mixture of hydrogen atoms (in their ground state) and hydrogen like ions (in their first excited state) are being excited by electrons which have been accelerated by same potential difference V volts. After excitation when they come directly into ground state, the wavelengths of emitted light are found in the ratio 5 : 1. Then, find (a) the minimum value of V for which both the atoms get excited after collision with electrons. (b) atomic number of other ion. (c) the energy of emitted light.

21. When a surface is irradiated with light of l = 4950 Å a photocurrent appears which vanishes if a retarding potential 0.6 V is applied. When a different source of light is used, it is found that critical retarding potential is changed to 1.1 volt. Find the work-function of emitting surface and wavelength of second source. If photoelectrons after emission from surface are subjected to a magnetic field of 10 tesla, what changes will be observed in the above two retarding potentials?

22. In an experiment on photoelectric effect light of wavelength 400 nm is incident on a metal plate at the rate of 5 W. The potential of the collector plate is made sufficiently positive with respect to emitter so that the current reaches the saturation value. Assuming that on the average one out of every 106 photons is able to eject a photoelectron, find the photocurrent in the circuit.

23. A light beam of wavelength 400 nm is incident on a metal of work-function 2.2 eV. A particular electron absorbs a photon and makes 2 collisions before coming out of the metal (a) Assuming that 10% of existing energy is lost to the metal in each collision find the final kinetic energy of this electron as it comes out of the metal. (b) Under the same assumptions find the maximum number of collisions, the electron should suffer before it becomes unable to come out of the metal.

Answers Introductory Exercise 33.1 kg-m

1. 4.6 eV, 2.45 ´ 10 -27 4.

2. 4.82 ´ 1016 per m2 -s

s

2

6. (a) 4.81 ´ 10 -34 m (b) 7.12 ´ 10 -11m

5. (c)

2

3.

Introductory Exercise 33.2 1. 122.4 eV

2. 3.16 ´ 10 –34 kg- m2 /s

3. (d)

4. (a)

6. (c)

7. (b)

8. (d)

9. (d)

10. (d)

11.

3. (a)

4. (a)

5. (b)

6. 42

1. Zero, 3.19 eV

2. Kmax µ (f - f0 )

3. 1.16 ´ 1015 Hz

5. (a)

6. (c)

7. (b)

5. (b) 3f f , 4 4

12. 651 nm

Introductory Exercise 33.3 1. (b)

2. (b)

Introductory Exercise 33.4 4. (a) 8. (a)

Exercises LEVEL 1 Assertion and Reason 1. (a)

2. (c)

3. (a or b)

4. (a)

5. (a or b)

4. 14. 24. 34.

5. 15. 25. 35.

6. (d)

7. (a)

8. (b)

9. (b)

8. (c) 18. (a) 28. (b)

9. (a) 19. (c) 29. (c)

10. (d)

Objective Questions 1. 11. 21. 31.

(d) (c) (a) (b)

2. 12. 22. 32.

(c) (a) (b) (b)

3. 13. 23. 33.

(b) (a) (c) (c)

(c) (c) (c) (a)

(a) (c) (b) (b)

6. 16. 26. 36.

(a) (b) (b) (c)

7. 17. 27. 37.

(c) (d) (c) (d)

10. (c) 20. (c) 30. (a)

Subjective Questions 1. 5.59 nm

2. E1 = - 4613 eV, E3 = - 2650 eV

3. 0.48 mA

4. (a) 2.47 ´ 10 –19 J = 1.54 eV (b) 804 nm, infrared 5. (a) 5.0 ´ 1014 Hz (b) 2.3 ´ 10 20 photons /s 7. (a) K 2 = 4K1 (b) E 2 = 2E1

8. (a) 2.52 ´ 1019

6. (a) 5.92 ´ 10 20 Hz (b) 5.06 ´ 10 –13 m (b) 3.33 ´ 10 –8 N

9. 4.3 ´ 10 –8 N

10. 10 –8 N

12. (a) 1.55 ´ 10 –10 m (b) 8.44 ´ 10 –14 m 11. 3.90 ´ 10 –34 m, No kg-m 13. (a) 2.37 ´ 10 –24 (b) 3.07 ´ 10 –18 J = 19.2 eV 14. 1.04 Å 15. 6.663 Å s 16. (a) 3.32 ´ 10 –10 m (b) 1.33 ´ 10 –9 m 17. 79 eV 18. 3, 6513 Å 19. 113.74 Å 20. n = 5

21. (a) Z = 4 (b) lmin = 40441 Å

22. (a) 15.6 volt (b) 2335 Å

(c) 12.52 V (d) 1.01 ´ 107 m–1

23. (a) –5.08 eV (b) –5.63 eV

Chapter 33 27 l0 32 31. Z = 42

24. 1.9 V

26.

30. 198 pm 34. 2.51 eV

27. 49.5 pm, 99.0 pm

28. Z » 41

32. (a) 4 W

33. 1022 Å, n = 3 to n = 1

15

35. 2260 Å

Modern Physics - I — 323

36. (a) 10

(b) 396 J/s

29. 15865 V

Hz (b) 4 eV (c) 6.4 ´ 10 –34 J- s

37. 1.81 eV, 9.0 ´ 10 5 m /s, 3.0 ´ 10 5 m /s 38. 0.148 m

39. 3.27 eV

40. 1.2 V

LEVEL 2 Single Correct Option 1. (b)

2. (b)

3. (a)

4. (c)

5. (c)

6. (b)

7. (d)

8. (a)

9. (d)

10. (c)

11. (c)

12. (c)

13. (d)

14. (c)

15. (a)

16. (c)

17. (d)

18. (b)

19. (d)

20. (d)

21. (a)

22. (c)

More than One Correct Options 1.(b,c)

2.(a,c)

3.(a,b,c)

4.(a,c)

5.(a,b,d)

6.(a,b)

Comprehension Based Questions 1. (c)

2. (c)

3. (c)

4. (b)

5. (b)

6. (c)

Match the Columns 1. (a) ® r

(b) ® q

(c) ® p

(d) ® p

2. (a) ® s

(b) ® r

(c) ® s

(d) ® p

3. (a) ® q

(b) ® p

(c) ® r

(d) ® s

4. (a) ® r

(b) ® s

(c) ® s

(d) ® q

5. (a) ® p

(b) ® s

(c) ® q

(d) ® s

6. (a) ® r

(b) ® p

(c) ® q

(d) ® q

7. (a) ® p,r

(b) ® s

(c) ® q,s

(d) ® p,r

Subjective Questions 1. (a) 1.31 mm (b) 164 nm 2. (a) 6.58 ´ 1015 Hz, 0.823 ´ 1015 Hz (b) 2.46 ´ 1015 Hz (c) 8.23 ´ 10 6 revolutions 3. (a) 2.55 ´ 10 -13 m (b) 2.81 keV 5. (a) 2.27 eV

(b) 2.75 ´ 1017

4. (a) 102 nm, 122 nm, 651 nm (b) No lines 6. W = 11.925 eV

(c) 9%

neh m 0 pm2e7 8. , 4pm 8e0 h5 n5

9. 793.3 Å

11. (a) ni = 2 (b) Z = 6 (c) 28.43 Å 13. (a) 2000 Å, 1500 Å 14. (a) 4125 Å

(b) K =

15.

16.

0.55 eV

nhBe 4pm

(c) U =

nheB 4pm

(d) E =

23. (a) 0.31 eV

(b) 4

nheB 2pm

(e)

150 eV, 0.5 Å

18. rn =

(b) –2.53 keV (c) 0.653 nm

20. (a) 10.2 volt (b) Z = 2 (c) 10.2 eV and 51 eV 22. 1.6 mA

12. 4.5 eV

(d) 489.6 eV, 25.3 Å

(b) 34 mA (c) 5.1%

nh 2pBe

10. 3.8 Å

(b) E1 = - 8.25 eV, E 2 = - 2.05 eV and E3 = - 0.95 eV (c) 8.25 V

17. (a) 1.69 ´ 10 –28 kg 19. (a) rn =

7. 2 eV, 0.754 V

nh ì nh ü , E n = k ln í ý 2p mk î 2p mk þ

nh 2e

21. 1.9 eV, 4125 Å,

No change is observed

34.1

Nuclear Stability and Radioactivity

34.2

The radioactive decay law

34.3

Successive disintegration

34.4

Equivalence of mass and energy

34.5

Binding energy and nuclear stability

34.6

Nuclear fission (Divide and conquer)

34.7

Nuclear fusion

326 — Optics and Modern Physics

34.1 Nuclear Stability and Radioactivity Among about 1500 known nuclides, less than 260 are stable. The others are unstable that decay to form other nuclides by emitting a and b-particles and g-electromagnetic waves. This process is called radioactivity. It was discovered in 1896 by Henry Becquerel.

For heavier nuclei, instability caused by electrostatic repulsion between the protons is minimized when there are more neutrons than protons. Figure shows a plot of N versus Z for the stable nuclei. For mass numbers upto about A = 40, we see that N » Z. 40 Ca is the heaviest stable nucleus for which N = Z. For larger values of Z, the (short range) nuclear force is unable to hold the nucleus together against the (long range) electrical repulsion of the protons unless the number of neutrons exceeds the number of protons. At Bi ( Z = 83, A = 209), the neutron excess is N – Z = 43. There are no stable nuclides with Z > 83. The nuclide

209 Bi is 83

the heaviest stable nucleus.

N

Number of neutrons

Whilst the chemical properties of an atom are governed entirely by the number of protons in the nucleus (i.e. the proton number Z), the stability of an atom appears to depend on both the number of protons and the number of neutrons. For light nuclei, the greatest stability is achieved when the numbers of protons and neutrons are approximately equal ( N » Z ).

130 120 110 100 90 80 70 60 50 40 30 20 10

Line of stability

N=Z

Z

0 10 20 30 40 50 60 70 80 90 Number of protons

Fig. 34.1 The stable nuclides plotted on a graph of neutron number, N, versus proton number, Z. Note that for heavier nuclides, N is larger relative to Z. The stable nuclides group along a curve called the line of stability.

Atoms are radioactive if their nuclei are unstable and spontaneously (and randomly) emit various particles, the a , b and or g radiations. When naturally occurring nuclei are unstable, we call the phenomena natural radioactivity. Other nuclei can be transformed into radioactive nuclei by various means, typically involving irradiation by neutrons, this is called artificial radioactivity. A radioactive nucleus is called a parent nucleus, the nucleus resulting from its decay by particle emission is called daughter nucleus. Daughter nuclei also might be granddaughter nuclei, and so on. There are no son or grandson nuclei. For unstable nuclides and radioactivity, the following points can be made. (i) Disintegrations tend to produce new nuclides near the stability line and continue until a stable nuclide is formed. (ii) Radioactivity is a nuclear property, i.e. a , b and g emission take place from the nucleus. (iii) Nuclear processes involve huge amount of energy so the particle emission rate is independent of temperature and pressure. The rate depends solely on the concentration of the number of atoms of the radioactive substance. (iv) A radioactive substance is either an a-emitter or a b-emitter, g-rays emit with both.

Chapter 34

Modern Physics - II — 327

Alpha Decay An alpha particle is a helium nucleus. Thus, a nucleus emitting an alpha particle loses two protons and two neutrons. Therefore, the atomic number Z decreases by 2, the mass number A decreases by 4 and the neutron number N decreases by 2. The decay can be written as A Z

X =

A-4 Y Z -2

+ 42 He

where, X is the parent nucleus and Y the daughter nucleus. As examples U 238 and Ra 226 are both alpha emitters and decay according to 238 234 4 92 U ¾® 90Th + 2 He 226 88 Ra

¾®

222 86 Rn

+ 42 He

N=A–Z

238 U As a general rule, in any decay sum of mass numbers A and atomic numbers Z must be the same on both sides. 234 Th 234 Pa Note that a nuclide below the stability line in Fig. 34.2 234 U disintegrates in such a way that its proton number 140 230 Th decreases and its neutron to proton ratio increases. In 226 heavy nuclides, this can occur by alpha emission. Ra 222 If the original nucleus has a mass number A that is 4 Rn 218 times an integer, the daughter nucleus and all those in Po 214 Pb the chain will also have mass numbers equal to 4 times 214 Bi an integer. (Because in a-decay A decreases by 4 and in 214 130 210 Po Ti b-decay it remains the same). Similarly, if the mass 210 210 Pb Bi number of the original nucleus is 4n + 1, where n is an 210 integer, all the nuclei in the decay chain will have mass Po 206 numbers given by 4n + 1 with n decreasing by 1 in each Pb a-decay a-decay. We can see therefore, that there are four b-decay possible a-decay chains, depending on whether A 80 84 88 92 equals 4n, 4n + 1, 4n + 2 or 4n + 3, where n is an integer. Z Series 4n + 1 is now not found. Because its longest lived Fig. 34.2 The uranium decay series member (other than the stable end product Bi 209 ) is (A = 4n + 2). The decay of 214 83Bi may proceed Np 237 which has a half-life of only 2 ´ 10 6 years. either by alpha emission and then beta emission or in the reverse order. Because this is much less than the age of the earth, this series has disappeared. Figure shows the uranium ( 4n + 2) series. The series branches at Bi 214, which decays either by a-decay to Ti 210 or b-decay to Po 214 . The branches meet at the lead isotope Pb 210 . Table 34.1 lists the four radioactive series.

Table 34.1 Four Radioactive Series Mass numbers

Series

Parent

Half-life, Years

Stable product

4n

Thorium

232 90 Th

1.39 ´ 1010

208 82 Pb

4n + 1

Neptunium

237 93 Np

2.25 ´ 106

209 83 Bi

4n + 2

Uranium

238 92 U

4.47 ´ 109

206 82 Pb

4n + 3

Actinium

235 92 U

7.07 ´ 108

207 82 Pb

328 — Optics and Modern Physics Beta Decay Beta decay can involve the emission of either electrons or positrons. A positron is a form of antimatter which has a charge equal to + e and mass equal to that of an electron. The electrons or positrons emitted in b-decay do not exist inside the nucleus. They are only created at the time of emission, just as photons are created when an atom makes a transition from a higher to a lower energy state. In b – decay a neutron in the nucleus is transformed into a proton, an electron and an antineutrino. n ¾® p + e – + n To conserve energy and momentum in the process, the emission of an antineutrino ( n ) (alongwith proton and electron) was first suggested by W. Pauli in 1930, but it was first observed experimentally in 1957. Thus, a parent nucleus with atomic number Z and mass number A decays by b – emission into a daughter with atomic number Z +1 and the same mass number A. A Z

b–

A Y Z +1

X ¾¾®

b – decay occurs in nuclei that have too many neutrons. An example of b – decay is the decay of carbon-14 into nitrogen, 14 14 – 6 C ¾® 7 N + e + n In b + decay, a proton changes into a neutron with the emission of a positron (and a neutrino) p ¾® n + e + + n Positron ( e + ) emission from a nucleus decreases the atomic number Z by 1 while keeping the same mass number A. A ZX

b+

¾¾®

A Y Z -1

B + decay occurs in nuclei that have too few neutrons. A typical b + decay is 13 7 N

¾®

13 6 C+

e+ + n

Electron capture Electron capture is competitive with positron emission since both processes lead to the same nuclear transformation. This occurs when a parent nucleus captures one of its own orbital electrons and emits a neutrino. A Z

X + e –1 ¾®

A Y Z -1

+n

In most cases, it is a K-shell electron that is captured, and for this reason the process is referred to as K-capture. One example is the capture of an electron by 4 Be 7 7 4 Be

+ e – ¾®

7 3 Li

+n

Gamma Decay Very often a nucleus that undergoes radioactive decay (a or b-decay) is left in an excited energy state (analogous to the excited states of the orbiting electrons, except that the energy levels associated with

Chapter 34

Modern Physics - II — 329

the nucleus have much larger energy differences than those involved with the atomic electrons). The typical half-life of an excited nuclear state is 10 -10 s. The excited nucleus ( X * ) then undergoes to a lower energy state by emitting a high energy photon, called the g-ray photon. The following sequence of events represents a typical situation in which g-decay occurs. 12 12 * 5 B ¾® 6 C 12 * 6C

+ e– + n

12 6C +

¾®

g

12 5B

e e

– 13.4 MeV

– 12 6

C*

g 4.4 MeV 12

6C Gamma decay

Fig. 34.3 12

Figure shows decay of B nucleus, which undergoes b-decay to either of two levels of C12 . It can either decay directly to the ground state of C12 by emitting a 13.4 MeV electron or undergo b-decay to an excited state of 126 C* followed by g-decay to the ground state. The later process results in the emission of a 9.0 MeV electron and a 4.4 MeV photon. The various pathways by which a radioactive nucleus can undergo decay are summarized in Table 34.2. Note In both a and b-decay, the Z value of a nucleus changes and the nucleus of one element becomes the nucleus of a different element. In g-decay, the element does not change, the nucleus merely goes from an excited state to a less excited state.

Table 34.2 Various Decay Pathways Alpha decay

A ZX

¾®

A -4 Z - 2Y

+ 42 He

Beta decay ( b – )

A ZX

¾®

A Z + 1Y

+ e– + n

Beta decay ( b + )

A ZX

¾®

A Z -1 Y

+ e+ + n

Electron capture Gamma decay

A ZX

+ e – ¾® A * ZX

¾®

A Z - 1Y A ZX

+ n

+ g

Extra Points to Remember ˜

˜

After emission of one alpha particle and two beta particles isotopes are produced. This is because after the emission of one alpha particle, atomic number decreases by 2. Further, after the emission of two beta particles atomic number increases by 2. So, finally atomic number remains unchanged. From beta emission mass number does not change. Therefore, isobars will be produced.

330 — Optics and Modern Physics V

Example 34.1 Mass number of a nucleus X is A and atomic number is Z. Find mass number and atomic number of the new nucleus (say Y) after the emission of m- alpha particles and n - beta particles. By the emission of one alpha particle, atomic number decreases by 2 and by the emission of one beta particle atomic number increases by 1. Therefore, the atomic number of nucleus Y is Z = Z - 2m + n Further, by the emission of one alpha particle mass number decreases by 4 and by the emission of beta particle, mass number does not change. Therefore, the mass number of Y is Solution

A¢ = A - 4m

34.2 Radioactive Decay Law Radioactive decay is a random process. Each decay is an independent event and one cannot tell when a particular nucleus will decay. When a particular nucleus decays, it is transformed into another nuclide, which may or may not be radioactive. When there is a very large number of nuclei in a sample, the rate of decay is proportional to the number of nuclei, N , that are present æ dN ö ç– ÷µ N è dt ø æ dN ö or ç– ÷ = lN è dt ø dN where, l is called the decay constant. This equation may be expressed in the form = – ldt and N integrated, N dN t òN 0 N = – l ò0 dt æ N ö ÷÷ = – lt or ln çç è N0 ø where, N 0 is the initial number of parent nuclei at t = 0.The number that survives at time t is therefore, N = N 0 e – lt

…(i)

This function is plotted in Fig. 34.4. N N0

0.5 N0 0.37 N0 t1/2

tav

Fig. 34.4

t

Modern Physics - II — 331

Chapter 34

Half-life The time required for the number of parent nuclei to fall to 50% is called half-life t1/ 2 and may be related to l as follows. 0.5 N 0 = N 0 e – lt 1/ 2 lt1/ 2 = ln (2) = 0.693

We have

t1/ 2 =

\ Mean life

ln (2) 0.693 = l l

…(ii)

The average or mean life t av is the reciprocal of the decay constant.

1 …(iii) l The mean life is analogous to the time constant in the exponential decrease in the charge on a capacitor in an RC circuit. After a time equal to the mean life time, the number of radioactive nuclei 1 decreases to times or approximately 37% of their original values. e t av =

Activity of a Radioactive Substance The decay rate R of a radioactive substance is the number of decays per second. And as we have seen above dN dN or – µN – = lN dt dt dN or Thus, R =– RµN dt or or R = lN R = l N 0 e – lt R = R 0 e – lt

or

…(iv)

where, R 0 = lN 0 is the activity of the radioactive substance at time t = 0. The activity versus time graph is shown in Fig. 34.5. R

R0

0.5 R0 0.37 R0 t1/2

tav

t

Fig. 34.5

Thus, the number of nuclei and hence the activity of the radioactive substance also decreases exponentially with time. Units of activity The SI unit for the decay rate is the Becquerel (Bq), but the curie (Ci) and rutherford (rd) are often used in practice. 1 Bq = 1 decays/s, 1 Ci = 3.7 ´ 1010 Bq and 1 rd = 10 6 Bq

332 — Optics and Modern Physics Extra Points to Remember ˜

After n half-lives, æ 1ö (a) number of nuclei left = N0 ç ÷ è2 ø

n

n

æ 1ö (b) fraction of nuclei left = ç ÷ and è2 ø 2

˜

æ 1ö (c) percentage of nuclei left = 100 ç ÷ è2 ø Number of nuclei decayed after time t,

Number of nuclei decayed

= N0 – N

N0

= N0 – N0e – lt = N0 (1 – e – lt ) ˜

The corresponding graph is as shown in Fig. 34.6. Probability of a nucleus for survival upto time t, P (survival ) =

N e – lt N = 0 = e – lt N0 N0

Time

Fig. 34.6

The corresponding graph is shown in Fig. 34.7. P(Survival) 1

Time

Fig. 34.7 ˜

Probability of a nucleus to disintegrate in time t is, P (disintegration) = 1 – P (survival ) = 1 – e

˜

The corresponding graph is as shown in Fig. 34.8. Half-life and mean life are related to each other by the relation, t 1/ 2 = 0.693 t av

˜

P (disintegration) – lt

1

or t av = 1.44 t 1/ 2

As we discussed above number of nuclei decayed in time t are N0 (1 – e – lt ). This expression involves power of e.

Time

Fig. 34.8

So, to avoid it we can use DN = lNDt where, DN are the number of nuclei decayed in time Dt at the instant when total number of nuclei are N. But, this can be applied only when Dt l x , therefore, y will decay at a faster rate than x. \ The correct option is (c).

Modern Physics - II — 335

Chapter 34 V

Example 34.8 A radioactive sample consists of two distinct species having equal number of atoms initially. The mean life of one species is t and that of the other is 5 t . The decay products in both cases are stable. A plot is made of the total number of radioactive nuclei as a function of time. Which of the following figure best represents the form of this plot? (JEE 2001) N

N

(a)

(b)

t

t

t

t

N

N

(c)

(d)

t

t

t

t

Fig. 34.9

The total number of atoms can neither remain constant (as in option a) nor can ever increase (as in options b and c). They will continuously decrease with time. Therefore, (d) is the appropriate option. Solution

INTRODUCTORY EXERCISE

34.1

1. The decay constant of a radioactive sample is l. The half-life and mean life of the sample are respectively given by (a) 1/ l and (ln 2) / l (c) l (ln 2) and 1/ l

(JEE 1989)

(b) (ln 2) / l and 1/ l (d) l / (ln 2) and 1/ l

2. Consider a-particles, b-particles and g-rays each having an energy of 0.5 MeV. In increasing order of penetrating powers, the radiations are (a) a , b, g (b) a, g, b (c) b, g, a (d) g, b, a

3. Which of the following is a correct statement?

(JEE 1994)

(JEE 1999)

(a) Beta rays are same as cathode rays (b) Gamma rays are high energy neutrons (c) Alpha particles are singly ionized helium atoms (d) Protons and neutrons have exactly the same mass

4. The electron emitted in beta radiation originates from (a) inner orbits of atom (b) free electrons existing in nuclei (c) decay of a neutron in a nucleus (d) photon escaping from the nucleus

(JEE 2001)

336 — Optics and Modern Physics 5. During a negative beta decay,

(JEE 1987)

(a) an atomic electron is ejected (b) an electron which is already present within the nucleus is ejected (c) a neutron in the nucleus decays emitting an electron (d) a part of the binding energy of the nucleus is converted into an electron

6. A freshly prepared radioactive source of half-life 2 h emits radiation of intensity which is 64 times the permissible safe level. The minimum time after which it would be possible to work safely with this source is (JEE 1988) (a) 6 h (b) 12 h (c) 24 h (d) 128 h

7. A radioactive sample S1 having an activity of 5 mCi has twice the number of nuclei as another sample S2 which has an activity of 10 mCi. The half-lives of S1 and S2 can be (a) 20 yr and 5 yr, respectively (b) 20 yr and 10 yr, respectively (c) 10 yr each (d) 5 yr each

(JEE 2008)

8. Half-life of a radioactive substance A is 4 days. The probability that a nucleus will decay in two half-lives is 1 (a) 4

(JEE 2006)

3 (b) 4

1 (c) 2

(d) 1

9. After 280 days, the activity of a radioactive sample is 6000 dps. The activity reduces to 3000 dps after another 140 days. The initial activity of the sample in dps is (a) 6000 (b) 9000 (c) 3000

10. The half-life of

215

(JEE 2004)

(d) 24000

At is 100 ms. The time taken for the activity of a sample of

1 th of its initial value is 16 (a) 400 ms (c) 40 ms

215

At to decay to (JEE 2002)

(b) 63 ms (d) 300 ms

11. The half-life of the radioactive radon is 3.8 days. The time, at the end of which 1/20 th of the radon sample will remain undecayed, is (given log10 e = 0.4343) (a) 3.8 days (b) 16.5 days (c) 33 days (d) 76 days

(JEE 1981)

12. Activity of a radioactive substance decreases from 8000 Bq to 1000 Bq in 9 days. What is the half-life and average life of the radioactive substance?

13. A radioactive substance has a half-life of 64.8 h. A sample containing this isotope has an initial activity (t = 0) of 40 mCi. Calculate the number of nuclei that decay in the time interval between t1 = 10.0 h and t 2 = 12.0 h.

14. A freshly prepared sample of a certain radioactive isotope has an activity of 10 mCi. After 4.0 h its activity is 8.00 mCi. (a) Find the decay constant and half-life (b) How many atoms of the isotope were contained in the freshly prepared sample? (c) What is the sample's activity 30.0 h after it is prepared?

15. A radioactive substance contains 1015 atoms and has an activity of 6.0 ´ 1011 Bq. What is its half-life?

16. Two radioactive elements X and Y have half-life periods of 50 minutes and 100 minutes, respectively. Initially, both of them contain equal number of atoms. Find the ratio of atoms left NX /NY after 200 minutes.

Chapter 34

Modern Physics - II — 337

34.3 Successive Disintegration Suppose a parent radioactive nucleus A (decay constant = l a ) has number of atoms N 0 at time t = 0. After disintegration it converts into a nucleus B (decay constant = l b ) which is further radioactive. Initially (t = 0), number of atoms of B are zero. We are interested in finding N b , the number of atoms of B at time t. la N a

lb N b

A

B

Fig. 34.10

B 0 Nb = ?

A N0 N a = N 0 e– la t

At t = 0 At t = t

At time t, net rate of formation of B = rate of disintegration of A – rate of disintegration of B dN b \ =la Na – lb Nb dt dN b (as N a = N 0 e – l a t ) or = l a N 0 e– lat – l b N b dt or

dN b + l b N b dt = l a N 0 e – l a t

Multiplying this equation by e l b t , we have e l b t dN b + e l b t l b N b dt = l a N 0 e ( l b - l a ) t d {N b e l b t } = l a N 0 e ( l b - l a ) t dt

\ Integrating both sides, we get

æ la N b e l b t = çç èlb –la

ö ÷÷ N 0 e ( l b – l a ) t + C ø

…(i)

where, C is the constant of integration, which can be found as under. At time, t = 0, \

Nb =0 æ la C = – çç èlb –la

ö ÷÷ N 0 ø

Substituting this value in Eq. (i), we have N 0l a (e – l a t – e – l b t ) Nb = lb –la

…(ii)

Now, the following conclusions may be drawn from the above discussion. 1. From Eq. (ii) we can see that N b = 0 at time t = 0 (it was given) and at t = ¥ (because B is also radioactive)

338 — Optics and Modern Physics 2. N a will continuously decrease with time while N b will first increase (until l a N a > l b N b ),

reaches to a maximum value (when l a N a = l b N b ) and then decreases (when l b N b > l a N a ). The two graphs for N a and N b with time are shown below. Nb

Na

laNa = lbNb

t

t laNa>lbNb

lbNb>laNa

Fig. 34.11 V

Example 34.9 A radio nuclide X is produced at constant rate a. At time t = 0, number of nuclei of X are zero. Find (a) the maximum number of nuclei of X. (b) the number of nuclei at time t. Decay constant of X is l. Solution

(a) Let N be the number of nuclei of X at time t. Rate = a

Rate = lN X

Fig. 34.12

Rate of formation of X = a (given) Rate of disintegration = lN Number of nuclei of X will increase until both the rates will become equal. Therefore, a = lN max a N max = l

\ (b) Net rate of formation of X at time t is N a/l

t

Fig. 34.13

dN = a – lN dt

Ans.

Chapter 34

Modern Physics - II — 339

dN = dt a – lN

\

Integrating with proper limits, we have N

ò0 or

t dN = ò dt 0 a – lN a N = (1 – e – lt ) l

Ans.

This expression shows that number of nuclei of X are increasing exponentially from 0 to V

a . l

Example 34.10 In the above problem if each decay produces E 0 energy, then find (a) power produced at time t (b) total energy produced upto time t a Solution (a) Q N = (1 - e - lt ) l At time t, number of decays per second = lN = a (1- e - lt ). Each decay produces E 0 energy. Therefore, energy produced per second or power. = (number of decays per second) (energy produced in each decay) = ( lN ) E 0 = aE 0 (1 - e - lt ) or

P = aE 0 (1 - e - lt )

Ans.

(b) Power is a function of time. Therefore, total energy produced upto time t can be obtained by integrating this power or t

E Total = ò Pdt 0

Alternate Method Energy is produced only in decay. Upto time t total at nuclei are produced and N nuclei are left. So, total number of nuclei decayed. a N d = at - N = at - (1 - e - lt ) l é 1 ù = a êt - (1 - e - lt )ú ë l û Each decay produces E 0 energy. Therefore, total energy produced upto time t, E Total = N d E 0 é 1 = aE 0 êt - (1 - e - lt ë l

ù )ú û

Ans.

340 — Optics and Modern Physics

34.4 Equivalence of Mass and Energy In 1905, while developing his special theory of relativity, Einstein made the suggestion that energy and mass are equivalent. He predicted that if the energy of a body changes by an amount E, its mass changes by an amount m given by the equation, E = mc 2 where, c is the speed of light. Everyday examples of energy gain are much too small to produce detectable changes of mass. But in nuclear physics this plays an important role. Mass appears as energy and the two can be regarded as equivalent. In nuclear physics, mass is measured in unified atomic mass units (u), 1 u being one-twelfth of the mass of carbon-12 atom and equals 1.66 ´10 –27 kg. It can readily be shown using E = mc 2 that, 1 u mass has energy 931.5 MeV. 1 u º 931.5 MeV/c 2 or c 2 = 931.5 MeV/ u

Thus,

A unit of energy may therefore be considered to be a unit of mass. For example, the electron has a rest mass of about 0.5 MeV. If the principle of conservation of energy is to hold for nuclear reactions it is clear that mass and energy must be regarded as equivalent. The implication of E = mc 2 is that any reaction producing an appreciable mass decrease is a possible source of energy. V

Example 34.11 Find the increase in mass of water when 1.0 kg of water absorbs 4.2 ´ 103 J of energy to produce a temperature rise of 1 K. Solution

m=

E c2

=

4.2 ´ 103 ( 3.0 ´ 108 ) 2

kg

= 4.7 ´ 10-14 kg

Ans.

34.5 Binding Energy and Nuclear Stability The existence of a stable nucleus means that the nucleons (protons and neutrons) are in a bound state. Since, the protons in a nucleus experience strong electrical repulsion, there must exist a stronger attractive force that holds the nucleus together. The nuclear force is a short range interaction that extends only to about 2 fm. (In contrast, the electromagnetic interaction is a long-range interaction). An important feature of the nuclear force is that it is essentially the same for all nucleons, independent of charge. The binding energy ( E b ) of a nucleus is the energy required to completely separate the nucleons. The origin of the binding energy may be understood with the help of mass-energy relation, DE = Dmc 2 , where Dm is the difference between the total mass of the separated nucleons and the mass of the stable nucleus. The mass of the stable nucleus is less than the sum of the mass of its nucleons. The binding energy of a nuclide Z X A is thus, E b = [ZmP + ( A – Z ) m N – m X ] c 2 where, mP = mass of proton, m N = mass of neutron and m X = mass of nucleus

…(i)

Chapter 34 Note

Modern Physics - II — 341

(i) Dm = [ ZmP + ( A – Z ) mN – mX ] is called the mass defect. This much mass is lost during the formation of a nucleus. Energy DE = ( Dm) c 2 is liberated during the making of the nucleus. This is the energy due to which nucleons are bound together. So, to break the nucleus in its constituent nucleons this much energy has to be given to the nucleus. (ii) Stability : Although nuclides with Z values upto Z = 92 (uranium) occur naturally, not all of these are stable. The nuclide 209 83 Bi is the heaviest stable nucleus. Even though uranium is not stable, however, its long lived isotope 238 U, has a half-life of some 4 billion year.

Binding energy per nucleon (MeV)

Binding energy per nucleon If the binding energy of a nucleus is divided by its mass number, the binding energy per nucleon is obtained. A plot of binding energy per nucleon E b / A as a function of mass number A for various stable nuclei is shown in figure. Eb /A 12

C

56

Fe 208

8 4

6

7

Pb

He

Li

4 2 2

0

H 40

80

120 160 200

A

Mass number

Fig. 34.14 The binding energy per nucleon, E b /A, as a function of the mass number A

Note That it is the binding energy per nucleon which is more important for stability of a nucleus rather than the total binding energy.

Following conclusions can be drawn from the above graph. 1. The greater the binding energy per nucleon the more stable is the nucleus. The curve reaches a 238 maximum of about 8.75 MeV in the vicinity of 56 26 Fe and then gradually falls to 7.6 MeV for 92 U. 2. In a nuclear reaction energy is released if total binding energy is increasing. Let us take an

example. Suppose a nucleus X, which has total binding energy of 100 MeV converts into some another nucleus Y which has total binding energy 120 MeV. Then, in this process 20 MeV energy will be released. This is because 100 MeV energy has already been released during the formation of X while in case of Y it is 120 MeV. So, the remaining 20 MeV will be released now. Energy is released if SE b is increasing. 3. SE b in a nuclear process is increased if binding energy per nucleon of the daughter products gets

increased. Let us take an example. Consider a nucleus X (A X =100) breaks into lighter nuclei Y ( AY = 60) and Z ( AZ = 40). X ®Y + Z Binding energy per nucleon of these three are say, 7 MeV, 7.5 MeV and 8.0 MeV. Then, total binding energy of X is 100 ´ 7 = 700 MeV and that of Y + Z is (60 ´ 7.5) + ( 40 ´ 8.0) = 770 MeV. So, in this process 70 MeV energy will be released.

342 — Optics and Modern Physics 4. Binding energy per nucleon is increased if two or more lighter nuclei combine to form a heavier

nucleus. This process is called nuclear fusion. +

+ + E

+ E Fusion

Fission

Fig. 34.15

In nuclear fission a heavy nucleus splits into two or more lighter nuclei of almost equal mass.

+

Eb/A

Fission

Fusion +

A

Fig. 34.16

In both the processes E b / A is increasing. Thus, energy will be released.

34.6 Nuclear Fission (Divide and Conquer) As we saw in the above article nuclear fission occurs when a heavy nucleus such as 235 U , splits into two lighter nuclei. In nuclear fission, the combined mass of the daughter nuclei is less than the mass of the parent nucleus. The difference is called the mass defect. Fission is initiated when a heavy nucleus captures a thermal neutron (slow neutrons). Multiplying the mass defect by c 2 gives the numerical value of the released energy. Energy is released because the binding energy per nucleon of the daughter nuclei is about 1 MeV greater than that of the parent nucleus. The fission of 235 U by thermal neutrons can be represented by the equation, 1 0n

+

235 92 U

¾®

236 92 U *

¾® X + Y + neutrons

where, 236 U * is an intermediate excited state that lasts only for10 –12 s before breaking into nuclei X and Y, which are called fission fragments. In any fission equation there are many combinations of X and Y that satisfy the requirements of conservation of energy and charge with uranium, for example, there are about 90 daughter nuclei that can be formed. Fission also results in the production of several neutrons, typically two or three. On the average, about 2.5 neutrons are released per event. A typical fission reaction for uranium is 1 0n

+

235 92 U

¾®

141 56 Ba

+

92 36 Kr

+ 310 n

About 200 MeV is released in the fission of a heavy nucleus. The fission energy appears mostly as kinetic energy of the fission fragments (e.g. barium and krypton nuclei) which fly apart at great speed. The kinetic energy of the fission neutrons also makes a slight contribution. In addition one or both of the large fragments are highly radioactive and small amount of energy takes the form of beta and gamma radiation.

Chapter 34

Modern Physics - II — 343

Chain Reaction Shortly after nuclear fission was discovered, it was realized that, the fission neutrons can cause further fission of 235 U and a chain reaction can be maintained. Neutron U235

Fission fragment

Fission fragment U235

U235

U235

U235

U235

U235

Fig. 34.17 A chain reaction

In practice only a proportion of the fission neutrons is available for new fissions since, some are lost by escaping from the surface of the uranium before colliding with another nucleus. The ratio of neutrons escaping to those causing fission decreases as the size of the piece of uranium-235 increases and there is a critical size (about the size of a cricket ball) which must be attained before a chain reaction can start. In the ‘atomic bomb’ an increasing uncontrolled chain reaction occurs in a very short time when two pieces of uranium-235 are rapidly brought together to form a mass greater than the critical size. Nuclear Reactors In a nuclear reactor the chain reaction is steady and controlled so that on average only one neutron from each fission produces another fission. The reaction rate is adjusted by inserting neutron absorbing rods of boron steel into the uranium 235. Graphite core

Steel

Uranium rods Concrete shield

Boron steel control rods

Fig. 34.18 Nuclear reactor

Graphite core is used as a moderator to slow down the neutrons. Natural uranium contains over 99% of 238 U and less than 1% of 235 U. The former captures the medium speed fission neutrons without fissioning. It fissions with very fast neutrons. On the other hand 235 U (and plutonium-239) fissions with slow neutrons and the job of moderator is to slow down the fission neutrons very quickly so that most escape capture by 238 U and then cause the fission of 235 U. A bombarding particle gives up most energy when it has an elastic collision with a particle of similar mass. For neutrons, hydrogen atoms would be most effective but they absorb the neutrons. But deuterium (in heavy water) and carbon (as graphite) are both suitable as moderator. To control the power level control rods are used. These rods are made of materials such as cadmium, that are very efficient in absorbing neutrons. The first nuclear reactor was built by Enrico Fermi and his team at the University of Chicago in 1942.

344 — Optics and Modern Physics

34.7 Nuclear Fusion Binding energy for light nuclei (A < 20) is much smaller than the binding energy for heavier nuclei. This suggests a process that is the reverse of fission. When two light nuclei combine to form a heavier nucleus, the process is called nuclear fusion. The union of light nuclei into heavier nuclei also lead to a transfer of mass and a consequent liberation of energy. Such a reaction has been achieved in ‘hydrogen bomb’ and it is believed to be the principal source of the sun’s energy. A reaction with heavy hydrogen or deuterium which yields 3.3 MeV per fusion is 2 2 3 1 H + 1H ® 2

He + 10 n

By comparison with the 200 MeV per fission of 235 U this seems small, but per unit mass of material it is not. Fusion of two deuterium nuclei, i.e. deuterons, will only occur if they overcome their mutual electrostatic repulsion. This may happen, if they collide at very high speed when, for example, they are raised to a very high temperature (10 8 - 10 9 K). So much high temperature is obtained by using an atomic (fission) bomb to trigger off fusion. If a controlled fusion reaction can be achieved, an almost unlimited supply of energy will become available from deuterium in the water of the oceans.

Extra Points to Remember ˜

Q-value of a nuclear reaction (optional) Consider a nuclear reaction in which a target nucleus X is

bombarded by a particle ‘a’ resulting in a daughter nucleus Y and a particle b. a+ X ®Y + b

Sometimes this reaction is written as X (a, b )Y The reaction energy Q associated with a nuclear reaction is defined as the total energy released as a result of the reaction. Thus, Q = (M a + M X – M Y – M b ) c 2 A reaction for which Q is positive is called exothermic. A reaction for which Q is negative is called endothermic. In an exothermic reaction, the total mass of incoming particles is greater than that of the outgoing particles and the Q-value is positive. If the total mass of the incoming particles is less than that of the outgoing particles, energy is required for reaction to take place and the reaction is said to be endothermic. Thus, an endothermic reaction does not occur unless the bombarding particle has a kinetic energy greater than|Q|. The minimum energy necessary for such a reaction to occur is called threshold energy Kth . The threshold energy is somewhat greater than|Q | because the outgoing particles must have some kinetic energy to conserve momentum. Thus,

K th >|Q | (in endothermic reaction) X

Y E Fig. 34.19

Consider a bombarding particle X of mass m1 and a target Y of mass m2 (at rest). The threshold energy of X for endothermic reaction (negative value of Q) to take place is æm ö K th = |Q |çç 1 + 1÷÷ m è 2 ø

Chapter 34 V

Modern Physics - II — 345

Example 34.12 In the fusion reaction 21 H + 21 H ¾® 32 He + 10 n, the masses of deuteron, helium and neutron expressed in amu are 2.015, 3.017 and 1.009 respectively. If 1 kg of deuterium undergoes complete fusion, find the amount of total energy released. 1 amu º 931.5 MeV /c2 . Solution Dm = 2 ( 2.015) – ( 3.017 + 1.009) = 0.004 amu \ Energy released = ( 0.004 ´ 931.5) MeV = 3.726 MeV 3.726 Energy released per deuteron = = 1.863 MeV 2

Number of deuterons in 1 kg =

6.02 ´ 1026 = 3.01´ 1026 2

\ Energy released per kg of deuterium fusion = ( 3.01´ 1026 ´ 1.863) = 5.6 ´ 1026 MeV » 9.0 ´ 1013 J

Example 34.13 A nucleus with mass number 220 initially at rest emits an a-particle. If the Q-value of the reaction is 5.5 MeV, calculate the kinetic energy of the a-particle. (JEE 2003) (a) 4.4 MeV (b) 5.4 MeV (c) 5.6 MeV (d) 6.5 MeV Solution Given that K 1 + K 2 = 5.5 MeV …(i) From conservation of linear momentum, p1 = p 2 or 2K 1 ( 216 m ) = 2K 2 ( 4 m ) as p = 2Km \ Solving Eqs. (i) and (ii), we get \ The correct option is (b).

V

K 2 = 54 K 1 K 2 = KE of a-particle = 5.4 MeV

Example 34.14 Binding energy per nucleon versus mass number curve for nuclei is shown in figure. W, X, Y and Z are four nuclei indicated on the curve. The process that would (JEE 1999) release energy is

Binding energy/nucleon in MeV

V

Ans.

8.5 8.0 7.5 5.0

…(ii)

Y X W Z

(a) Y ® 2Z (b) W ® X + Z 30 60 90 120 0 (c) W ® 2Y Mass number of nuclei Fig. 34.20 (d) X ® Y + Z Solution Energy is released in a process when total binding energy of the nucleus (= binding energy per nucleon ´ number of nucleons) is increased or we can say, when total binding energy of products is more than the reactants. By calculation we can see that only in option (c), this happens. Given, W ® 2Y Binding energy of reactants = 120 ´ 7.5 = 900 MeV and binding energy of products = 2 ( 60 ´ 8.5) = 1020 MeV > 900 MeV \ The correct option is (b).

346 — Optics and Modern Physics V

Example 34.15 A star initially has 1040 deuterons. It produces energy via the processes 1 H 2 + 1 H 2 ® 1 H 3 + p and 1 H 2 + 1 H 3 ® 2 He4 + n. If the average power radiated by the star is 1016 W, the deuteron supply of the star is (JEE 1993) exhausted in a time of the order of (a) 10 6 s

(b) 10 8 s

(c) 1012 s

(d) 1016 s

The masses of the nuclei are as follows M ( H 2 ) = 2.014 amu; M ( n ) = 1.008 amu, M( p) = 1.007 amu; M ( He4 ) = 4.001 amu Solution

The given reactions are 2 1H 2 1H

+ 1 H3 ¾® 2 He 4 + n 3 1 H2 ¾® 2 He 4 + n + p

Þ Mass defect,

+ 1 H2 ¾® 1 H3 + p

Dm = ( 3 ´ 2.014 - 4.001 - 1.007 - 1.008) amu = 0.026 amu Energy released = 0.026 ´ 931 MeV = 0.026 ´ 931´ 1.6 ´ 10-13 J = 3.87 ´ 10-12 J

This is the energy produced by the consumption of three deuteron atoms. \ Total energy released by 1040 deuterons =

1040 ´ 3.87 ´ 10-12 J = 1.29 ´ 1028 J 3

The average power radiated is P = 1016 W or 1016 J/s. Therefore, total time to exhaust all deuterons of the star will be 1.29 ´ 1028 t= = 1.29 ´ 1012 s » 1012 s 1016 \ The correct option is (c). V

Example 34.16 Assume that the nuclear binding energy per nucleon ( B / A) versus mass number (A) is as shown in the figure. Use this plot to choose the correct choice(s) given below. (JEE 2008)

B/A 8 6 4

(a) Fusion of two nuclei with mass numbers lying in the 2 range of 1 < A < 50 will release energy. 0 (b) Fusion of two nuclei with mass numbers lying in the 100 200 range of 51 < A < 100 will release energy. Fig. 34.21 (c) Fission of a nucleus lying in the mass range of 100 < A < 200 will release energy when broken into two equal fragments. (d) Fission of a nucleus lying in the mass range of 200 < A < 260 will release energy when broken into two equal fragments.

A

Chapter 34

Modern Physics - II — 347

In fusion, two or more lighter nuclei combine to make a comparatively heavier nucleus. In fission, a heavy nucleus breaks into two or more comparatively lighter nuclei. Further, energy will be released in a nuclear process if total binding energy increases. \ The correct options are (b) and (d). Solution

INTRODUCTORY EXERCISE

34.2

1. If a star can convert all the He nuclei completely into oxygen nuclei. The energy released per oxygen nuclei is (Mass of the helium nucleus is 4.0026 amu and mass of oxygen nucleus is 15.9994 amu) (JEE 2005) (a) 7.6 MeV (b) 56.12 MeV (c) 10.24 MeV (d) 23.4 MeV

2. Fast neutrons can easily be slowed down by

(JEE 1994)

(a) the use of lead shielding (b) passing them through heavy water (c) elastic collisions with heavy nuclei (d) applying a strong electric field

3. During a nuclear fusion reaction, (a) (b) (c) (d)

(JEE 1987)

a heavy nucleus breaks into two fragments by itself a light nucleus bombarded by thermal neutrons breaks up a heavy nucleus bombarded by thermal neutrons breaks up two light nuclei combine to give a heavier nucleus and possibly other products

4. The equation

4 11 H ¾®

4 2+ 2 He

+ 2e– + 26 MeV represents

(JEE 1983)

(b) g-decay (d) fission

(a) b-decay (c) fusion

5. (a) How much mass is lost per day by a nuclear reactor operated at a 109 watt power level? (b) If each fission releases 200 MeV, how many fissions occur per second to yield this power level?

6. Find energy released in the alpha decay, 238 92 U

4 ¾® 234 90 Th + 2 He

M ( 238 92 U ) = 238.050784 u

Given,

M ( 234 90 Th ) = 234.043593 u M (42He) = 4.002602 u

7. Complete the nuclear reactions. (a) (c)

6 7 1 3 Li + ? ¾® 4 Be + 0 n 9 4 4 4 Be + 2 He ¾® 3 ( 2 He ) + ?

8. Consider the reaction

2 2 H+ H 1 1

(b) (d) =

4 He + 2

35 32 4 17 Cl + ? ¾® 16 S + 2 He 79 2 1 35 Br + 1 H ¾® ? + 2 ( 0 n )

Q . Mass of the deuterium atom = 2.0141u. Mass of

helium atom = 4.0024 u. This is a nuclear ........ reaction in which the energy Q released is ........ MeV. (JEE 1996)

9. The binding energies per nucleon for deuteron (1H2 ) and helium ( 2He4 ) are 1.1 MeV and 7.0 MeV respectively. The energy released when two deuterons fuse to form a helium nucleus ( 2He4 ) is ......... . (JEE 1988)

348 — Optics and Modern Physics

Final Touch Points 1. Classification of nuclei The nuclei have been divided in isotopes, isobars and isotones on the basis of number of protons (atomic number) or the total number of nucleons (mass number). Isotopes The elements having the same number of protons but different number of neutrons are called isotopes. In other words, isotopes have same value of atomic number ( Z ) but different values of mass number ( A ). Almost every element has isotopes. Because of the same atomic number isotopes of an element have the same place in the periodic table. The isotopes of some elements are given below. Element

Hydrogen

Its isotopes

Number of protons

Number of neutrons

1 1H

1

0

1H

2

1

1

1H

3

1

2

16 8O

8

8

17

8

9

18

8

10

8O

Oxygen

8O

Chlorine

Uranium

17Cl

35

17

18

17Cl

37

17

20

92 U

235

92

143

92 U

238

92

146

In nature, the isotopes of chlorine (17 Cl35 and 17 Cl37) are found in the ratio 75.4% and 24.6%. When chlorine is prepared in laboratory, its atomic mass is found to be M = ( 35 ´ 0.754) + ( 37 ´ 0.246) = 35.5

Note Since, the isotopes have the same atomic number, they have the same chemical properties. Their physical properties are different as they have different mass numbers. Two isotopes, thus cannot be separated by chemical method, but they can be separated from the physical methods. Isobars The elements having the same mass number ( A ) but different atomic number ( Z ) are called isobars. They have different places in periodic table. Their chemical (as well as physical) properties are different. 1H

Isotones

3

and 2He3, 8O17 and 9F17are examples of isobars.

Elements having the equal number of neutrons ( A – Z ) are called isotones. 3 Li

7

and 4Be8, 1H3 and 2He4 are examples of isotones.

2. Nuclear forces In nucleus the positively charged protons and the uncharged neutrons are held

together in an extremely small space ( » 10-15 m ) in spite of the strong electrostatic repulsion between the protons. Obviously, there are some strong attractive forces operating within the nucleus between the nucleons. The nuclear forces are non-electric and non gravitational forces. These forces are extremely short-range forces. They become operative only when the distance between two nucleons is a small multiple of 10-15 m. They do not exist when the distance is appreciably larger than10-15 m and become repulsive when the distance is appreciably smaller than 10-15 m. Nuclear forces

Chapter 34

Modern Physics - II — 349

between protons and protons between neutrons and neutrons and between protons and neutrons are all essentially the same in magnitude. Thus, we can say that nuclear forces are charge independent. Yukawa’s meson theory of nuclear forces A Japanese scientist Yukawa in 1935 suggested that the nuclear forces are ‘exchange forces. Which are produced by the exchange of new particles called p-mesons between nucleons. These particles were later on actually discovered in cosmic radiation. There are three types of p-mesons, p + , p - and p 0. There is a continuous exchange of p-mesons between protons and neutrons due to which they continue to be converted into one another. When a p + - meson jumps from a proton to a neutron, the proton is converted into a neutron and the neutron is converted into a proton. p - p + ¾® n n + p + ¾® p

and -

Conversely, when a p - meson jumps from a neutron to a proton, then neutron is converted into a proton and the proton is converted into a neutron. Thus, n - p - ¾® p p + p - ¾® n

and +

-

The exchange of p and p - mesons between protons and neutrons is responsible for the origin of nuclear forces between them. Similarly, nuclear forces between two protons and between two neutrons are generated by a continuous exchange of p 0-mesons between them. Thus, the basis of nuclear forces is the exchange of mesons and hence these are called ‘exchange forces’.

3. Size and shape of the nucleus The Rutherford scattering experiment established that mass of an atom is concentrated within a small positively charged region at the centre which is called the nucleus of the atom. The nuclear radius is given by R = R 0 A1 / 3 Here, A is the mass number of the particular nucleus and R 0 = 1.3 fm (fermi) = 1.3 ´ 10-15 m. This means that the nucleus radius is of the order of10-15 m. Here, R 0 = 1.3 fm is the distance of closest approach to the nucleus and is also known as nuclear unit radius.

4. Nuclear density Let us consider the nucleus of an atom having the mass number A. Mass of nucleus » A ´ 1.67 ´ 10-27 kg 4 Volume of the nucleus = pR 3 3 4 4 = p (R 0A1 / 3 )3 = pR 03A 3 3 mass \ Density of the nucleus, r = volume or

r=

A ´ 1.67 ´ 10-27 4 ´ p ´ (1.3 ´ 10-15 )3 ´ A 3

= 1.8 ´ 1017 kg/m 3 Thus, density of a nucleus is independent of the mass number A and of the order of 1017 kg/ m 3.

5. Magic numbers We know that the electrons in an atom are grouped in ‘shells’ and ‘sub-shells’. Atoms with 2, 10, 18, 36, 54 and 86 electrons have all of their shells completely filled. Such atoms are unusually stable and chemically inert. A similar situation exists with nuclei also. Nuclei having 2, 8, 20, 28, 50, 82 and 126 nucleons of the same kind (either protons or neutrons) are more stable than nuclei of neighbouring mass numbers. These numbers are called as ‘magic numbers’.

350 — Optics and Modern Physics 6. Fundamental particles The particles which are not constituted by any other particles are called fundamental particles. A brief discussion of important fundamental particles is as follows. (i) Electron It was discovered in 1897 by Thomson. Its charge is - e and mass is 9.1 ´ 10-31 kg. Its symbol is e - (or -1b 0). (ii) Proton It was discovered in 1919 by Rutherford in artificial nuclear disintegration. It has a positive charge + e and its mass is 1836 times (1.673 ´ 10-27 kg) the mass of electron. In free state, the proton is a stable particle. Its symbol is p + . It is also written as 1H1. (iii) Neutron It was discovered in 1932 by Chadwick. Electrically, it is a neutral particle. Its mass is 1839 times (1.675 ´ 10-27 kg) the mass of electron. In free state the neutron is unstable (mean life » 17 minutes) but it constitutes a stable nucleus with the proton. Its symbol is n or 0n1. (iv) Positron It was discovered by Anderson in 1932. It is the antiparticle of electron, i.e. its charge is +e and its mass is equal to that of the mass of electron. Its symbol is e + (or +1b 0). (v) Antiproton It is the antiparticle of proton. It was discovered in 1955. Its charge is -e and its mass is equal to that of the mass of proton. Its symbol is p - . (vi) Antineutron It was discovered in 1956. It has no charge and its mass is equal to the mass of neutron. The only difference between neutron and antineutron is that if they spin in the same direction, their magnetic momenta will be in opposite directions. The symbol for antineutron is n . (vii) Neutrino and antineutrino The existence of these particles was predicted in 1930 by Pauli while explaining the emission of b-particles from radioactive nuclei, but these particles were actually observed experimentally in 1956. Their rest mass and charge are both zero but they have energy and momentum. These are mutually antiparticles of each other. They have the symbol n and n. (viii) Pi-mesons The existence of pi-mesons was predicted by Yukawa in 1935, but they were actually discovered in 1947 in cosmic rays. Nuclear forces are explained by the exchange of pi-mesons between the nucleons. pi-mesons are of three types, positive p-mesons ( p + ), negative pi-mesons ( p - ) and neutral p-mesons ( p 0 ). Charge on p ± is ± e. Whereas mass of p ± is 274 times the mass of electron. p 0 has mass nearly 264 times the electronic mass. (ix) Mu-Mesons These were discovered in 1936 by Anderson and Neddermeyer. These are found in abundance in the cosmic rays at the ground level. There are two types of mu-mesons. Positive mu-meson (m + ) and negative mu-meson (m - ). There is no neutral mu-meson. Both the mu-mesons have the same rest mass 207 times the rest mass of the electron. (x) Photon These are bundles of electromagnetic energy and travel with the speed of light. Energy hn and momentum of a photon of frequency n are hn and , respectively. c Antiparticles For every fundamental particle there exists an identical fundamental particle just opposite in some property. For example electron and positron are identical in all respects, except that charges on them are opposite. The following table shows various particles and their antiparticles. Some particles are their own antiparticles. For example p 0 and g. Name of particle

Symbol

Antiparticle

Mass in comparison to mass of electron

Average life (in seconds) for the unstable particles

Electron

e-

e +1

1

stable

Proton

p+

p-

1836

stable

Neutron

n

n

1839

1010

Neutrino

n

n

0

stable

Chapter 34 Name of particle

Modern Physics - II — 351

Symbol

Antiparticle

Mass in comparison to mass of electron

Average life (in seconds) for the unstable particles

p+

p-

274

2.6 ´ 10 -8

p0

p0

264

0.9 ´ 10 -16

Mu-Mesons

m-

m+

207

2.2 ´ 10 -6

Photon

g

g

0

stable

Pi-Mesons

7. If an unstable nucleus decays by two different processes and decay constants in two processes are l1 and l 2, then effective value of l is l = l1 + l 2 Now, the above equation can also be written as ln 2 ln 2 ln 2 = + T T1 T2 or

1 1 1 = + T T1 T2

Þ

T =

Proof

or \ or

(T = half-life)

TT 1 2 T1 + T2

Suppose at some instant, the unstable nucleus has N number of nuclei, then net rate of decay = decay in process 1 + decay in process 2 dN æ dN ö æ dN ö = ç÷ ÷ + çdt dt è ø 1 è dt ø 2 lN = l1 N + l 2 N l = l1 + l 2

Hence Proved.

Solved Examples TYPED PROBLEMS Type 1. Based on radioactivity V

Example 1 At a given instant there are 25% undecayed radioactive nuclei in a sample. After 10 s the number of undecayed nuclei reduces to 12.5%. Calculate (a) mean life of the nuclei, (JEE 1996) (b) the time in which the number of undecayed nuclei will further reduce to 6.25% of the reduced number. Solution (a) In 10 s, number of nuclei has been reduced to half (25% to 12.5%). Therefore, its half-life is t1/ 2 = 10 s Relation between half-life and mean life is t 10 tmean = 1/ 2 = s ln 2 0.693 tmean = 14.43 s (b) From initial 100% to reduction till 6.25%, it takes four half-lives.

Ans.

t1/ 2 t1/ 2 t1/ 2 t1/ 2 100% ¾® 50% ¾® 25% ¾® 12.5% ¾® 6.25% \

t = 4 t1/ 2 = 4 (10)s = 40 s t = 40 s

V

Ans.

Example 2 A radioactive element decays by b-emission. A detector records n beta particles in 2 s and in next 2 s it records 0.75 n beta particles. Find mean life correct to nearest whole number. Given ln |2| = 06931 . , ln |3| = 1.0986. (JEE 2003) Solution Let n0 be the number of radioactive nuclei at time t = 0. Number of nuclei decayed in

time t are given by n0 (1 - e-lt ), which is also equal to the number of beta particles emitted during the same interval of time. For the given condition, n = n0 (1 - e-2l ) -4l

(n + 0.75n ) = n0 (1 - e

)

…(i) …(ii)

Dividing Eq. (ii) by Eq. (i), we get 1.75 = or \ Let us take e-2l = x

1 - e-4l 1 - e-2l

1.75 - 1.75 e-2l = 1 - e-4l 3 1.75 e-2l - e -4l = 4

…(iii)

Chapter 34

Modern Physics - II — 353

Then, the above equation is x2 - 1.75 x + 0.75 = 0 1.75 ± (1.75)2 - (4) (0.75) 2 3 x = 1 and 4 x=

or or \ From Eq. (iii) either

e-2l = 1 3 e-2l = 4

or

but e-2l = 1 is not acceptable because which means l = 0. 3 Hence, e-2l = 4 -2l ln (e) = ln (3) - ln (4) = ln (3) - 2 ln (2) 1 l = ln (2) - ln (3) 2

or \ Substituting the given values,

l = 0.6931 \ Mean life, tmean =

1 ´ (1.0986) = 0.14395 s-1 2

1 = 6.947 s l

\ The correct answer is 7. V

Ans.

Example 3 A small quantity of solution containing Na 24 radio nuclide ( half - life = 15 h ) of activity 1.0 microcurie is injected into the blood of a person. A sample of the blood of volume 1 cm3 taken after 5h shows an activity of 296 disintegrations per minute. Determine the total volume of the blood in the body of the person. Assume that the radioactive solution mixes uniformly in the blood of the person. (JEE 1994) (1 curie = 3.7 ´ 1010 disintegrations per second) Solution Q l = Disintegration constant 0.693 0.693 -1 = h = 0.0462 h -1 t1/ 2 15 Let R0 = initial activity = 1 microcurie = 3.7 ´ 104 disintegrations per second r = Activity in 1 cm3 of blood at t = 5 h 296 disintegration per second = 60 = 4.93 disintegration per second, and R = Activity of whole blood at time t = 5 h Total volume of blood should be V =

R R e-lt = 0 r r

354 — Optics and Modern Physics Substituting the values, we have æ 3.7 ´ 104 ö - ( 0. 0462) (5 ) ÷e cm3 V = çç ÷ è 4.93 ø V = 5.95 ´ 103 cm3 or V = 5.95 L V

Ans.

Example 4 A radioactive nucleus X decays to a nucleus Y with a decay constant l X = 0.1 s -1 , Y further decays to a stable nucleus Z with a decay constant l Y = 1 / 30 s -1 . Initially, there are only X nuclei and their number is N 0 = 1020 . Set up the rate equations for the populations of X, Y and Z . The population of Y nucleus as a function of time is given by N Y ( t) = { N 0 l X /( l X - l Y )} [exp ( -l Y t) - exp ( -l X t) ]. Find the time at which N Y is maximum and determine the populations X and Z at that instant. (JEE 2001) Solution (a) Let at time t = t, number of nuclei of Y and Z are N Y and N Z . Then, Rate equations of the populations of X, Y and Z are æ dN X ö ç ÷ = - lX N X è dt ø æ dN Y ö ç ÷ = l X N X - lY N Y è dt ø æ dN Z ö and ç ÷ = lY N Y è dt ø N 0 lX (b) Given, N Y (t) = [e- lY t - e- l X t ] l X - lY

…(i) …(ii) …(iii)

For N Y to be maximum dN Y (t ) =0 dt i.e or or

…(iv) l X N X = lY N Y N l 0 X l X (N 0 e-l X t ) = lY [e- lY t - e- l X t ] l X - lY l X - lY e - lY t = -l t - 1 lY e X lX = e( l X lY

or or

[from Eq. (ii)]

- lY )t

æl ö (l X - lY ) t ln (e) = ln çç X ÷÷ è lY ø t=

æl ö 1 ln çç X ÷÷ l X - lY è lY ø

Substituting the values of l X and lY , we have 1 æ 0.1 ö ln ç t= ÷ = 15 ln (3) (0.1 - 1 / 30) è 1 / 30 ø or

t = 16.48 s

Ans.

Chapter 34

Modern Physics - II — 355

(c) The population of X at this moment, N X = N 0 e- l X t = (1020 ) e- ( 0.1) (16. 48) N X = 1.92 ´ 1019 N l NY = X X lY = (1.92 ´ 1019 )

NZ

[From Eq. (iv) ] (0.1) (1 / 30)

= 5.76 ´ 1019 = N 0 - N X - NY = 1020 - 1.92 ´ 1019 - 5.76 ´ 1019

or

N Z = 2.32 ´ 1019

Type 2. Based on nuclear physics V

Example 5 In a nuclear reactor 235 U undergoes fission liberating 200 MeV of energy. The reactor has a 10% efficiency and produces 1000 MW power. If the reactor is to function for 10 yr, find the total mass of uranium required. (JEE 2001) Solution The reactor produces 1000 MW power or 109J/s. The reactor is to function for 10 yr. Therefore, total energy which the reactor will supply in 10 yr is E = (power) (time) = (109 J/s ) (10 ´ 365 ´ 24 ´ 3600 s ) = 3.1536 ´ 1017 J But since the efficiency of the reactor is only 10%, therefore actual energy needed is 10 times of it or 3.1536 ´ 1018J. One uranium atom liberates 200 MeV of energy or 200 ´ 1.6 ´ 10-13 J or 3.2 ´ 10-11 J of energy. So, number of uranium atoms needed are 3.1536 ´ 1018 = 0.9855 ´ 1029 3.2 ´ 10-11 or number of kg-moles of uranium needed are 0.9855 ´ 1029 n= = 163.7 6.02 ´ 1026 Hence, total mass of uranium required is or or

V

m = (n )M = (163.7) (235) kg m » 38470 kg m = 3.847 ´ 104 kg

13 Example 6 The element curium 248 96 Cm has a mean life of 10 s. Its primary decay modes are spontaneous fission and a-decay, the former with a probability of 8% and the later with a probability of 92%, each fission releases 200 MeV of energy. The masses involved in decay are as follows (JEE 1997) 4 = 248.072220 u , 244 94 Pu = 244.064100 u and 2 He = 4.002603 u. Calculate the 20 power output from a sample of 10 Cm atoms. ( 1 u = 931 MeV / c 2 ) 248 96 Cm

356 — Optics and Modern Physics Solution The reaction involved in a-decay is Mass defect,

Dm = mass of

248 244 4 96 Cm ® 94 Pu + 2 He 248 244 96 Cm - mass of 94 Pu

- mass of 42 He

= (248.072220 - 244.064100 - 4.002603) u = 0.005517 u Therefore, energy released in a-decay will be E a = (0.005517 ´ 931) MeV = 5.136 MeV Similarly, E fission = 200 MeV Mean life is given as tmean = 1013 s = 1 / l \ Disintegration constant l = 10-13 s-1 Rate of decay at the moment when number of nuclei are 1020

(given)

= lN = (10-13 ) (1020 ) = 107 disintegration per second Of these, 8% are in fission and 92% are in a-decay. Therefore, energy released per second = (0.08 ´ 107 ´ 200 + 0.92 ´ 107 ´ 5.136) MeV = 2.074 ´ 108 MeV = energy released per second (J/s)

\ Power output (in watt)

= (2.074 ´ 108 ) (1.6 ´ 10-13 ) = 3.32 ´ 10-5 Js -1 \ Power output = 3.32 ´ 10-5 W V

Example 7 A nucleus X, initially at rest, undergoes alpha-decay according to (JEE 1991) the equation. A 92 X

®

228 ZY

+a

(a) Find the values of A and Z in the above process. (b) The alpha particle produced in the above process is found to move in a circular track of radius 0.11m in a uniform magnetic field of 3 T. Find the energy (in MeV) released during the process and the binding energy of the parent nucleus X. Given that m (Y) = 228.03 u , m (10 n ) = 1.009 u

m ( 42 He) = 4.003 u , m( 11 H ) = 1.008 u Solution (a) A - 4 = 228 \

A = 232 92 - 2 = Z Z = 90

or (b) From the relation, r= =

2Km Bq

Þ

Ka =

r 2B2q2 2m

(0.11)2(3)2 (2 ´ 1.6 ´ 10-19 )2 MeV 2 ´ 4.003 ´ 1.67 ´ 10-27 ´ 1.6 ´ 10-13

= 5.21 MeV

Chapter 34

Modern Physics - II — 357

From the conservation of momentum, pY = pa or 2K YmY = 2K ama æm ö 4.003 ´ 5.21 K Y = çç a ÷÷ K a = 228.03 m Y ø è

\

= 0.09 MeV \ Total energy released = K a + K Y = 5.3 MeV Total binding energy of daughter products = [92 ´ (mass of proton ) + (232 - 92) (mass of neutron ) - (mY ) - (ma )] ´ 931.48 MeV = [(92 ´ 1.008) + (140) (1.009) - 228.03 - 4.003] 931.48 MeV = 1828.5 MeV \ Binding energy of parent nucleus = binding energy of daughter products – energy released = (1828.5 - 5.3) MeV = 1823.2 MeV V

Example 8 It is proposed to use the nuclear fusion reaction, 2 1H

+ 12 H ® 42 He

in a nuclear reactor 200 MW rating. If the energy from the above reaction is used with a 25 per cent efficiency in the reactor, how many grams of deuterium fuel will be needed per day? (The masses of 12 H and 42 He are 2.0141 atomic mass units and 4.0026 atomic mass units respectively.) (JEE 1990) Solution Mass defect in the given nuclear reaction, Dm = 2 ( mass of deuterium) - ( mass of helium) = 2 (2.0141) - (4.0026) = 0.0256 Therefore, energy released DE = (Dm) (931.48) MeV = 23.85 MeV = 23.85 ´ 1.6 ´ 10-13 J = 3.82 ´ 10-12 J Efficiency is only 25%, therefore, æ 25 ö -12 25% of DE = ç ÷ (3.82 ´ 10 ) J è 100 ø = 9.55 ´ 10-13 J i.e. by the fusion of two deuterium nuclei, 9.55 ´ 10-13 J energy is available to the nuclear reactor. Total energy required in one day to run the reactor with a given power of 200 MW, E Total = 200 ´ 106 ´ 24 ´ 3600 = 1.728 ´ 1013 J \ Total number of deuterium nuclei required for this purpose, E 2 ´ 1.728 ´ 1013 n = Total = DE /2 9.55 ´ 10-13 = 0.362 ´ 1026 \ Mass of deuterium required = (Number of g-moles of deuterium required) × 2 g æ 0.362 ´ 1026 ö ÷ ´ 2 = 120.26 g = çç 23 ÷ è 6.02 ´ 10 ø

Miscellaneous Examples V

Example 9 Find the minimum kinetic energy of an a-particle to cause the reaction 14 N ( a , p)17 O. The masses of 14 N , 4 He, 1 H and 17 O are respectively 14.00307 u, 4.00260 u, 1.00783 u and 16.99913 u. Solution Since, the masses are given in atomic mass units, it is easiest to proceed by finding the mass difference between reactants and products in the same units and then multiplying by 931.5 MeV/u. Thus, we have MeV ö æ Q = (14.00307 u + 4.00260 u – 1.00783 u – 16.99913 u ) ç931.5 ÷ u ø è = – 1.20 MeV Q-value is negative. It means reaction is endothermic. So, the minimum kinetic energy of a-particle to initiate this reaction would be ö æm æ 4.00260 ö K min = |Q | çç a + 1÷÷ = (1.20) ç + 1÷ m 14.00307 è ø ø è N = 1.54 MeV

V

Ans.

Example 10 Neon-23 decays in the following way, 23 10 Ne

23 ¾® 11 Na +

0 e –1

+n

Find the minimum and maximum kinetic energy that the beta particle ( –10 e) can have. The atomic masses of respectively.

23

Ne and

23

Na are 22.9945 u and 22.9898 u,

Solution Here, atomic masses are given (not the nuclear masses), but still we can use them for calculating the mass defect because mass of electrons get cancelled both sides. Thus, Mass defect Dm = (22.9945 – 22.9898) = 0.0047 u \

Q = (0.0047 u ) (931.5 MeV /u )

= 4.4 MeV Hence, the energy of beta particles can range from 0 to 4.4 MeV. V

Example 11 The mean lives of an unstable nucleus in two different decay processes are 1620 yr and 405 yr, respectively. Find out the time during which three-fourth of a sample will decay. Solution Let at some instant of time t, number of nuclei are N. Then, æ – dN ö æ – dN ö æ – dN ö =ç ç ÷ ÷ +ç ÷ è dt ø net è dt ø 1 è dt ø 2 If the effective decay constant is l, then lN = l1N + l 2 N or

l = l1 + l 2 =

1 1 1 + = year -1 1620 405 324

Ans.

Chapter 34 N0 = N 0 e– lt 4 æ1ö – lt = ln ç ÷ = – 1.386 è4ø

Now, \

æ 1 ö ç ÷ t = 1.386 è 324 ø

or \ V

Modern Physics - II — 359

t = 449 yr

Ans.

Example 12 In the chemical analysis of a rock the mass ratio of two radioactive isotopes is found to be 100 : 1. The mean lives of the two isotopes are 4 ´ 10 9 years and 2 ´ 10 9 years, respectively. If it is assumed that at the time of formation the atoms of both the isotopes were in equal proportional, calculate the age of the rock. Ratio of the atomic weights of the two isotopes is 1.02 : 1 . Solution At the time of observation (t = t ), m1 100 = m2 1

(given)

A1 1.02 = A2 1

Further it is given that

m A N 1 m1 A2 100 = ´ = N 2 m2 A1 1.02 N =

Number of atoms, \

…(i)

Let N 0 be the number of atoms of both the isotopes at the time of formation, then N 1 N 0 e– l1 t = = e( l 2 – l1 ) t N 2 N 0 e– l 2 t

…(ii)

Eq. (i) and Eq. (ii), we have e( l 2 – l1 ) t = or

100 1.02

(l 2 - l1 )t = ln 100 - ln 1.02

\

t=

ln 100 – ln 1.02 æ 1 1 ö÷ ç ç 2 ´ 109 – 4 ´ 109 ÷ è ø

Substituting the values, we have t = 1.834 ´ 1010 yr V

Ans.

Example 13 A proton is bombarded on a stationary lithium nucleus. As a result of the collision, two a-particles are produced. If the direction of motion of the a-particles with the initial direction of motion makes an angle cos –1 ( 1/4), find the kinetic energy of the striking proton. Given, binding energies per nucleon of Li 7 and He4 are 5.60 and 7.06 MeV , respectively. (Assume mass of proton » mass of neutron).

360 — Optics and Modern Physics Solution Q-value of the reaction is Q = (2 ´ 4 ´ 7.06 – 7 ´ 5.6) MeV = 17.28 MeV Applying conservation of energy for collision, K p + Q = 2 Ka (Here, K p and K a are the kinetic energies of proton and a-particle respectively) Li7

…(i)

a

P

Þ

q q

a

From conservation of linear momentum, 2 m pK p = 2 2 ma K a cos q æ1ö K p = 16K a cos 2 q = (16 K a ) ç ÷ è4ø

\

…(ii) 2

\ Ka = K p Solving Eqs. (i) and (iii) with Q = 17.28 MeV We get K p = 17.28 MeV V

(as ma = 4 m p) …(iii) Ans.

Example 14 A 7 Li target is bombarded with a proton beam current of 10 –4 A for 1 hour to produce 7 Be of activity 1.8 ´ 10 8 disintegrations per second. Assuming that one 7 Be radioactive nucleus is produced by bombarding 1000 protons, determine its half-life. Solution At time t, let say there are N atoms of 7 Be (radioactive). Then, net rate of formation of 7 Be nuclei at this instant is dN 10–4 = – lN dt 1.6 ´ 10–19 ´ 1000 dN = 6.25 ´ 1011 – lN dt

or or

N0

ò0

3600 dN =ò dt 11 0 6.25 ´ 10 – lN

where, N 0 are the number of nuclei at t = 1 h or 3600 s. æ 6.25 ´ 1011 – l N 0 ö 1 ÷ = 3600 \ – ln çç ÷ l 6.25 ´ 1011 è ø l N 0 = activity of 7 Be at t = 1 h = 1.8 ´ 108 disintegrations/s 1 æ 6.25 ´ 1011 – 1.8 ´ 108 ö÷ \ – ln çç ÷ = 3600 l è 6.25 ´ 1011 ø \ Therefore, half-life

l = 8.0 ´ 10–8 sec-1 0.693 = 8.66 ´ 106 s t1/ 2 = 8.0 ´ 10–8 = 100.26 days

Ans.

Chapter 34 V

Example 15 A

118

Modern Physics - II — 361

Cd radio nuclide goes through the transformation chain. 118 118 Cd ¾¾® ln ¾¾® Sn ( stable ) 30 min 45 min

118

The half-lives are written below the respective arrows. At time t = 0 only Cd was present. Find the fraction of nuclei transformed into stable over 60 minutes. Solution At time t = t, N 1 = N 0 e– l1 t \

and N 2 =

N 0l1 (e– l1 t – e– l 2t ) l 2 – l1

(see Article 34.3)

N3 = N 0 – N1 – N 2 é ù l1 = N 0 ê1 – e– l1 t – (e– l1 t – e– l 2t )ú l 2 – l1 ë û l N3 1 (e– l1 t – e– l 2t ) = 1 – e– l1 t – l 2 – l1 N0

\

0.693 = 0.0231 min -1 30 0.693 l2 = = 0.0154 min -1 45 l1 =

t = 60 min N3 0.0231 = 1 – e– 0.0231 ´ 60 – (e– 0.0231 ´ 60 – e–0.0154 ´60 ) N0 0.0154 – 0.0231

and \

= 1 – 0.25 + 3 (0.25 – 0.4) = 0.31 V

Example 16 Natural uranium is a mixture of three isotopes 238 92 U

Ans. 235 234 92 U , 92 U

and

with mass percentage 0.01%, 0.71% and 99.28 % respectively. The half-life of

three isotopes are 2.5 ´ 105 yr, 7.1 ´ 10 8 yr and 4.5 ´ 10 9 yr respectively. Determine the share of radioactivity of each isotope into the total activity of the natural uranium. Solution Let R1 , R2 and R3 be the activities of U234 , U235 and U238 respectively. Total activity, Share of U

234

,

R = R1 + R2 + R3 R1 l1N 1 = R l1N 1 + l 2N 2 + l3 N 3

Let m be the total mass of natural uranium. 0.01 0.71 99.28 Then, m1 = m , m2 = m and m3 = m 100 100 100 m m m Now, and N1 = 1 , N2 = 2 N3 = 3 M1 M2 M3 where M1, M 2 and M3 are atomic weights. \

æ m1 ö 1 ÷ ç çM ÷ T R1 è 1ø 1 = m1 1 m 1 m 1 R + 2. + 3 . M1 T1 M 2 T2 M3 T3

362 — Optics and Modern Physics

=

æ 0.01 / 100 ö ç ÷ è 234 ø

æ ö 1 ç ÷ ç 2.5 ´ 105 ÷ è ø

= 0.648 » 64.8 % Similarly, share of and of V

(0.01 /100) 1 ´ 234 2.5 ´ 105 years ö æ 99.28/100 ö 1 æ 0.71 /100 ö æç ÷+ç +ç ÷ç ÷ 8÷ 235 è ø è 7.1 ´ 10 ø è 238 ø U235 = 0.016 % U238 = 35.184 %

æ ö 1 ç ÷ ç 4.5 ´ 109 ÷ è ø

Ans.

Example 17 Uranium ores on the earth at the present time typically have a composition consisting of 99.3% of the isotope 92 U 238 and 0.7% of the isotope 235 . The half-lives of these isotopes are 4.47 ´ 10 9 yr and 7.04 ´ 10 8 yr, 92 U respectively. If these isotopes were equally abundant when the earth was formed, estimate the age of the earth. Solution Let N 0 be number of atoms of each isotope at the time of formation of the earth (t = 0) and N 1 and N 2 the number of atoms at present (t = t ). Then, N 1 = N 0e– l1 t and N 2 = N 0e– l 2t N1 \ = e( l 2 – l1 ) t N2

…(i) …(ii) …(iii)

Further it is given that N 1 99.3 = N 2 0.7

…(iv)

Equating Eqs. (iii) and (iv) and taking log on both sides, we have æ 99.3 ö (l 2 – l1 ) t = ln ç ÷ è 0.7 ø æ 1 ö 99.3 ö ÷ ln æç \ t = çç ÷ ÷ l – l 0.7 ø è 1ø è 2 Substituting the values, we have t=

or

1 æ 99.3 ö ln ç ÷ 0.693 0.693 è 0.7 ø – 7.04 ´ 108 4.47 ´ 109

t = 5.97 ´ 109 yr

Ans.

Exercises LEVEL 1 Assertion and Reason Directions : Choose the correct option. (a) If both Assertion and Reason are true and the Reason is correct explanation of the Assertion. (b) If both Assertion and Reason are true but Reason is not the correct explanation of Assertion. (c) If Assertion is true, but the Reason is false. (d) If Assertion is false but the Reason is true. 1. Assertion : Rate of radioactivity cannot be increased or decreased by increasing or decreasing pressure or temperature. Reason : Rate depends on the number of nuclei present in the radioactive sample.

2. Assertion : Only those nuclei which are heavier than lead are radioactive. Reason :

Nuclei of elements heavier than lead are unstable.

3. Assertion : After emission of one a-particle and two b-particles, atomic number remains unchanged. Reason : Mass number changes by four.

4. Assertion : g-rays are produced by the transition of a nucleus from some higher energy state to some lower energy state. Reason : Electromagnetic waves are always produced by the transition process.

5. Assertion : During b-decay a proton converts into a neutron and an electron. No other particle is emitted. Reason : During b-decay linear momentum of system should remain constant.

6. Assertion : If we compare the stability of two nuclei, then that nucleus is more stable whose total binding energy is more. Reason : More the mass defect during formation of a nucleus more will be the binding energy.

7. Assertion : In a nuclear process energy is released if total binding energy of daughter nuclei is more than the total binding energy of parent nuclei. Reason : If energy is released then total mass of daughter nuclei is less than the total mass of parent nuclei.

8. Assertion : Binding energy per nucleon is of the order of MeV. Reason :

1 MeV = 1.6 ´ 10-13 J.

9. Assertion : 1 amu is equal to 931.48 MeV. 1 th the mass of C12 atom. 12 10. Assertion : Between a , b and g radiations, penetrating power of g-rays is maximum. Reason : Ionising power of g-rays is least. Reason :

1 amu is equal to

364 — Optics and Modern Physics 11. Assertion : The nuclear energy can be obtained by the nuclear fission of heavier nuclei as well as by fusion of lighter nuclei. Reason : As the mass number increases, the binding energy per nucleon, first increases and then decreases.

Objective Questions 1. For uranium nucleus how does its mass vary with volume? (a) m µ V (c) m µ V

(JEE 2003)

(b) m µ 1 / V (d) m µ V 2

2. Order of magnitude of density of uranium nucleus is (m p = 1.67 ´ 10-27 kg) (a) 1020 kg/m3 (c) 1014 kg/m3

(JEE 1999)

(b) 1017 kg/m3 (d) 1011 kg/m3

3. During a beta decay, (a) (b) (c) (d)

an atomic electron is ejected an electron present inside the nucleus is ejected a neutron in the nucleus decays emitting an electron a part of the binding energy is converted into electron

4. In the nucleus of helium if F1 is the net force between two protons, F2 is the net force between two neutrons and F3 is the net force between a proton and a neutron. Then, (a) F1 = F2 = F3 (c) F2 > F3 > F1

(b) F1 > F2 > F3 (d) F2 = F3 > F1

5. What are the respective number of a and b-particles emitted in the following radioactive decay? 200 90 X

(a) 6 and 8 (c) 8 and 8

6. If an atom of and an atom

® 168 80 Y (b) 6 and 6 (d) 8 and 6

235 92 U, after 94 of 38 Sr, the

absorbing a slow neutron, undergoes fission to form an atom of 138 54 Xe other particles produced are

(a) one proton and two neutrons (c) two neutrons

(b) three neutrons (d) one proton and one neutron

7. Nucleus A is converted into C through the following reactions, A®B+ a B ® C + 2b then, (a) A and B are isotopes (c) A and B are isobars

(b) A and C are isobars (d) A and C are isotopes

8. The binding energy of a-particle is (if m p = 1.00785 u, mn = 1.00866 u and ma = 4.00274 u) (a) 56.42 MeV

9.

(b) 2.821 MeV

(c) 28.21 MeV

(d) 32.4 MeV

7 th of the active nuclei present in a radioactive sample has decayed in 8 s. The half-life of the 8 sample is (a) 2 s

(b) 1 s

(c) 7 s

(d)

8 s 3

Chapter 34

Modern Physics - II — 365

10. A radioactive element disintegrates for a time interval equal to its mean life. The fraction that has disintegrated is 1 e 0.693 (c) e

(b) 1 -

(a)

1 e

1ö æ (d) 0.693 ç1 - ÷ eø è

11. Starting with a sample of pure

66

Cu,

3 of it decays into Zn in 15 minutes. The corresponding 4

half-life is (a) 5 minutes (c) 10 minutes

(b) 7.5 minutes (d) 3.5 minutes

12. A sample of radioactive substance loses half of its activity in 4 days. The time in which its activity is reduced to 5% is (b) 8.3 days (d) None of these

(a) 12 days (c) 17.3 days

13. On bombardment of U 235 by slow neutrons, 200 MeV energy is released. If the power output of atomic reactor is 1.6 MW, then the rate of fission will be (a) 5 ´ 1016 per second (c) 15 ´ 1016 per second

(b) 10 ´ 1016 per second (d) 20 ´ 1016 per second

14. Atomic masses of two heavy atoms are A1 and A2. Ratio of their respective nuclear densities will be approximately (a)

æA ö (b) çç 1 ÷÷ è A2 ø

A1 A2

1/3

æA ö (c) çç 2 ÷÷ è A1 ø

1/3

(d) 1

15. A radioactive element is disintegrating having half-life 6.93 s. The fractional change in number of nuclei of the radioactive element during 10 s is (a) 0.37 (c) 0.25

(b) 0.63 (d) 0.50

16. The activity of a radioactive sample goes down to about 6% in a time of 2 hour. The half-life of the sample in minute is about (a) 30

(b) 15

(c) 60

(d) 120

17. What is the probability of a radioactive nucleus to survive one mean life? (a)

1 e

(b)

1 e+1

(c) 1 -

1 e

(d)

1 -1 e

Subjective Questions Note You can take approximations in the answers.

1. The disintegration rate of a certain radioactive sample at any instant is 4750 disintegrations per minute. Five minutes later the rate becomes 2700 per minute. Calculate (a) decay constant and

(b) half-life of the sample 15

2. A radioactive sample contains 1.00 ´ 10

atoms and has an activity of 6.00 ´ 1011 Bq. What is

its half-life?

3. Obtain the amount of half-life of

60

60

Co necessary to provide a radioactive source of 8.0 Ci strength. The Co is 5.3 years?

366 — Optics and Modern Physics 9 4. The half-life of 238 92 U against alpha decay is 4.5 ´ 10 year. How much disintegration per second

occurs in 1 g of

238 92 U

?

5. What is the probability that a radioactive atom having a mean life of 10 days decays during the fifth day?

6. In an ore containing uranium, the ratio of 238U to 206Pb nuclei is 3. Calculate the age of the ore, assuming that all the lead present in the ore is the final stable product of half-life of

238

238

U. Take the

9

U to be 4.5 ´ 10 years..

7. The half-lives of radioisotopes P 32 and P 33 are 14 days and 25 days respectively. These radioisotopes are mixed in the ratio of 4 : 1 of their atoms. If the initial activity of the mixed sample is 3.0 mCi, find the activity of the mixed isotopes after 60 years.

8. Complete the following reactions. (a)

226 88 Ra

®a+

(b)

19 19 8 O® 9 F+

(c)

25 25 13 AI ® 12 Mg

(b)

238 206 92 U ® 82 Pb

+

9. Consider two decay reactions. 236 (a) 92 U ® 206 82 Pb + 10 protons + 20 neutrons Are both the reactions possible?

+ 8 42He + 6 electrons

10. Obtain the binding energy of a nitrogen nucleus from the following data : mH = 1.00783 u, mN = 1.00867 u, m (14 7 N ) = 14.00307 u Give your answer in units of MeV. [ Remember 1 u = 931.5 MeV/ c2 ]

11. 8 protons and 8 neutrons are separately at rest. How much energy will be released if we form 16 8 O

nucleus?

Given : Mass of 16 8 O atom = 15.994915 u Mass of neutron = 1.008665 u Mass of hydrogen atom = 1.007825 u

12. Assuming the splitting of U 235 nucleus liberates 200 MeV energy, find (a) the energy liberated in the fission of 1 kg of U235 and (b) the mass of the coal with calorific value of 30 kJ/g which is equivalent to 1 kg of U235 .

13.

212 83 Bi decays

as per following equation. 212 83 Bi

®

208 82 Ti

+ 42 He

The kinetic energy of a-particle emitted is 6.802 MeV. Calculate the kinetic energy of Ti recoil atoms. nucleus, usable energy of 185 MeV is released. If 92U 235 reactor is continuously operating it at a power level of 100 MW power, how long will it take for 1 kg of uranium to be consumed in this reactor?

14. In a neutron induced fission of

92U

235

15. Calculate the Q-values of the following fusion reactions : (a)

2 1H

+ 12H ® 31 H + 11H (b)

Atomic masses are

2 1H

+ 12H ® 32He + n (c)

2 1H

+ 31 H ® 42He + n

m (12H ) = 2.014102 u, m (31 H ) = 3.016049 u, m (32He ) = 3.016029 u, m( 42He) = 4.002603 u, m (11H ) = 1.007825 u

Chapter 34

Modern Physics - II — 367

16. Calculate the Q-value of the fusion reaction, 4

He + 4He ® 8 Be

Is such a fusion energetically favourable? Atomic mass of 8 Be is 8.0053 u and that of 4He is 4.0026 u.

17. When fission occurs, several neutrons are released and the fission fragments are beta radioactive, why?

LEVEL 2 Single Correct Option 1. The count rate observed from a radioactive source at t second was N 0 and at 4t second it was N0 æ 11 ö . The count rate observed at ç ÷ t second will be 16 è 2ø

N0 128 N0 (c) 32

(b)

(a)

N0 64

(d) None of these

2. The half-lives of a radioactive sample are 30 years and 60 years for two decay processes. If the sample decays by both the processes simultaneously. The time after which, only one-fourth of the sample will remain is (a) 10 years (c) 40 years

(b) 20 years (d) 60 years

3. Consider the nuclear fission reaction W ® X + Y . What is the Q-value (energy released) of the

Binding energy per nucleon

reaction? Z E3

X Y

E2

W

E1

N3 N2 N1

Mass number

(a) E1N 1 - (E 2N 2 + E3 N 3 )

(b) (E 2N 2 + E3 N 3 - E1N 1 )

(c) E 2N 2 + E1N 1 - E3 N 3

(d) E1N 1 + E3 N 3 - E 2N 2

4. Consider the following nuclear reaction, X 200 ® A110 + B90 + Energy If the binding energy per nucleon for X, A and B are 7.4 MeV, 8.2 MeV and 8.2 MeV respectively, the energy released will be (a) 90 MeV (c) 200 MeV

(b) 110 MeV (d) 160 MeV

368 — Optics and Modern Physics 5. The energy released by the fission of a single uranium nucleus is 200 MeV. The number of fissions of uranium nucleus per second required to produce 16 MW of power is (Assume efficiency of the reactor is 50%) (a) 2 ´ 106 (c) 5 ´ 106

(b) 2.5 ´ 106 (d) None of these

6. A radioactive isotope is being produced at a constant rate A. The isotope has a half-life T. Initially, there are no nuclei, after a time t > > T , the number of nuclei becomes constant. The value of this constant is A ln 2 T AT (d) ln 2

(a) AT

(b)

(c) AT ln 2

7. A bone containing 200 g carbon-14 has a b-decay rate of 375 decay/min. Calculate the time that has elapsed since the death of the living one. Given the rate of decay for the living organism is equal to 15 decay per min per gram of carbon and half-life of carbon-14 is 5730 years. (a) 27190 years (c) 17190 years

(b) 1190 years (d) None of these

8. Two identical samples (same material and same amount) P and Q of a radioactive substance having mean life T are observed to have activities AP and AQ respectively at the time of observation. If P is older than Q, then the difference in their age is æA ö (a) T ln ç P ÷ ç AQ ÷ è ø æ AP ö ÷ (c) T ç ç AQ ÷ è ø

æ AQ ö ÷÷ (b) T ln çç è AP ø æ AQ ö ÷÷ (d) T çç è AP ø

9. A star initially has 1040 deuterons. It produces energy via the processes 12H + 12H ® 31H + p and 2 1H

+ 31H ® 42He + n. Where the masses of the nuclei are

m ( 2H ) = 2.014 amu, m( p) = 1.007 amu, m( n ) = 1.008 amu and m( 4He ) = 4.001 amu. If the average power radiated by the star is 1016 W, the deuteron supply of the star is exhausted in a time of the order of (a) 106 s

(b) 108 s

(c) 1012 s

(d) 1016 s

10. Two radioactive samples of different elements (half-lives t1 and t2 respectively) have same number of nuclei at t = 0. The time after which their activities are same is

(a)

t1t2 t ln 2 0.693 (t2 - t1 ) t1

(b)

t1t2 t ln 2 0.693 t1

(c)

t1t2 t ln 2 0.693 (t1 + t2) t1

(d) None of these

11. A nucleus X initially at rest, undergoes alpha decay according to the equation 232 Z X

®

A 90Y

+a

What fraction of the total energy released in the decay will be the kinetic energy of the alpha particle? (a)

90 92

(b)

228 232

(c)

228 232

(d)

1 2

Chapter 34

Modern Physics - II — 369

12. A stationary nucleus of mass 24 amu emits a gamma photon. The energy of the emitted photon is 7 MeV. The recoil energy of the nucleus is (a) 2.2 keV (c) 3.1 keV

(b) 1.1 keV (d) 22 keV

13. A radioactive material of half-life T was kept in a nuclear reactor at two different instants. The quantity kept second time was twice of that kept first time. If now their present activities are A1 and A2 respectively, then their age difference equals

T 2 A1 ln ln 2 A2 T A2 (c) ln ln 2 2 A1

(a)

A1 A2 A (d) T ln 2 2 A1 (b) T ln

More than One Correct Options 1. At t = 0, number of radioactive nuclei of a radioactive substance are x and its radioactivity is y. Half-life of radioactive substance is T. Then, x is constant throughout y x (b) >T y (a)

(c) value of xy remains half after one half-life (d) value of xy remains one fourth after one half-life

2. Choose the correct options. (a) (b) (c) (d)

Isotopes have same number of atomic number Isobars have same atomic weight Isotones have same number of neutrons In neutral isotope atoms number of electrons are same

3. Choose the correct options. (a) (b) (c) (d)

By gamma radiations atomic number is not changed By gamma radiations mass number is not changed By the emission of one a and two b particles isotopes are produced By the emission of one a and four b particles isobars are produced

4. Two radioactive substances have half-lives T and 2T . Initially, they have equal number of nuclei. After time t = 4T , the ratio of their number of nuclei is x and the ratio of their activity is y. Then, (a) x = 1/8 (c) y = 1/2

(b) x = 1/4 (d) y = 1/4

5. Regarding the nuclear forces, choose the correct options. (a) They are short range forces (c) They are not electromagnetic forces

(b) They are charge independent forces (d) They are exchange forces

6. Regarding a nucleus choose the correct options. (a) (b) (c) (d)

Density of a nucleus is directly proportional to mass number A Density of all the nuclei is almost constant of the order of 1017 kg/m3 Nucleus radius is of the order of 10-15 m Nucleus radius µ A

370 — Optics and Modern Physics Comprehension Based Questions Passage : (Q. No. 1 to 3) The atomic masses of the hydrogen isotopes are Hydrogen m1H1 = 1.007825 amu Deuterium m1H 2 = 2.014102 amu Tritium m1H3 = 3.016049 amu

1. The energy released in the reaction, 1H

2

+ 1H 2 ® 1H3 + 1H1 is nearly

(a) 1 MeV (c) 4 MeV

(b) 2 MeV (d) 8 MeV

2. The number of fusion reactions required to generate 1 kWh is nearly (a) 108 (c) 1028

(b) 1018 (d) 1038

3. The mass of deuterium, 1H 2 that would be needed to generate 1 kWh (a) 3.7 kg (c) 3.7 ´ 10-5 kg

(b) 3.7 g (d) 3.7 ´ 10-8 kg

Match the Columns 1. At t = 0, x nuclei of a radioactive substance emit y nuclei per second. Match the following two columns. Column I

Column II

(a) Decay constant l

(p) (ln 2) (x/y)

(b) Half-life

(q) x/y

1 (c) Activity after time t = l

(r) y/e (s) None of these

(d) Number of nuclei after 1 time t = l

Binding energy per nucleon

2. Corresponding to the graph shown in figure, match the following two columns.

Q

R

P

50

100

150

Mass number

Mass number

Chapter 34 Column I

Modern Physics - II — 371

Column II

(a) P + P = Q

(p) energy is released

(b) P + P + P = R

(q) energy is absorbed

(c) P + R = 2Q

(r) No energy transfer will take place

(d) P + Q = R

(s) data insufficient

3. In the following chain, A® B®C A and B are radioactive, while C is stable. Initially, we have only A and B nuclei. There is no nucleus of C. As the time passes, match the two columns. Column I

Column II

(a) Nuclei of ( A + B)

(p) will increase continuously

(b) Nuclei of B

(q) will decrease continuously

(c) Nuclei of (C + B)

(r) will first increase then decrease

(d) Nuclei of ( A + C )

(s) data insufficient

4. Match the following two columns. Column I

Column II

(a) After emission of one a and one b particles

(p) atomic number will decrease by 3.

(b) After emission of two a and one b particle

(q) atomic number will decrease by 2

(c) After emission of one a and two b particles

(r) mass number will decrease by 8

(d) After emission of two a and two b-particles.

(s) mass number will decrease by 4

5. Match the following two columns. Column I

Column II

(a) The energy of air molecules at room temperature

(p) 0.02 eV

(b) Binding energy of heavy nuclei per nucleon

(q) 2 eV

(c) X-ray photon energy

(r) 10 keV

(d) Photon energy of visible light

(s) 7 MeV

372 — Optics and Modern Physics Subjective Questions 1. A F32 radio nuclide with half-life T = 14.3 days is produced in a reactor at a constant rate

q = 2 ´ 109 nuclei per second. How soon after the beginning of production of that radio nuclide will its activity be equal to R = 109 disintegration per second? 2. Consider a radioactive disintegration according to the equation A ® B ® C. Decay constant of A and B is same and equal to l. Number of nuclei of A, B and C are N 0 , 0, 0 respectively at t = 0. Find (a) number of nuclei of B as function of time t. (b) time t at which the activity of B is maximum and the value of maximum activity of B.

3. Nuclei of a radioactive element A are being produced at a constant rate a. The element has a decay constant l. At time t = 0, there are N 0 nuclei of the element. (a) Calculate the number N of nuclei of A at time t. (b) If a = 2N 0l, calculate the number of nuclei of A after one half-life of A, and also the limiting value of N as t ® ¥.

4. A solution contains a mixture of two isotopes A (half-life = 10 days) and B (half-life = 5 days).

Total activity of the mixture is 1010 disintegration per second at time t = 0. The activity reduces to 20% in 20 days. Find (a) the initial activities of A and B, (b) the ratio of initial number of their nuclei. 5. A radio nuclide with disintegration constant l is produced in a reactor at a constant rate a nuclei per second. During each decay energy E0 is released. 20% of this energy is utilized in increasing the temperature of water. Find the increase in temperature of m mass of water in time t. Specific heat of water is s. Assume that there is no loss of energy through water surface. 6. A stable nuclei C is formed from two radioactive nuclei A and B with decay constant of l1 and l2 respectively. Initially, the number of nuclei of A is N 0 and that of B is zero. Nuclei B are produced at a constant rate of P. Find the number of the nuclei of C after time t. 4 206 7. Polonium ( 210 84 Po) emits 2He particles and is converted into lead ( 82 Pb). This reaction is used for 210 producing electric power in a space mission. Po has half-life of 138.6 days. Assuming an efficiency of 10% for the thermoelectric machine, how much 210Po is required to produce 1.2 ´ 107 J of electric energy per day at the end of 693 days. Also find the initial activity of the material. Given : Masses of nuclei 210 Po = 209.98264 amu, 206Pb = 205.97440 amu, 42He = 4.00260 amu, 1 amu = 931 MeV/ c2 and Avogadro’s number = 6 ´ 1023 / mol

8. A radio nuclide consists of two isotopes. One of the isotopes decays by a-emission and other by b-emission with half-lives T1 = 405 s and T2 = 1620 s, respectively. At t = 0, probabilities of getting a and b-particles from the radio nuclide are equal. Calculate their respective probabilities at t = 1620 s. If at t = 0, total number of nuclei in the radio nuclide are N 0. Calculate the time t when total number of nuclei remained undecayed becomes equal to N 0 / 2. log10 2 = 0.3010, log10 5.94 = 0.7742 and x 4 + 4x – 2.5 = 0, x = 0.594

9. Find the amount of heat generated by 1 mg of Po210 preparation during the mean life period of these nuclei if the emitted alpha particles are known to possess kinetic energy 5.3 MeV and practically all daughter nuclei are formed directly in the ground state. 10. In an agricultural experiment, a solution containing 1 mole of a radioactive material (T1/ 2 = 14.3 days) was injected into the roots of a plant. The plant was allowed 70 h to settle down and then activity was measured in its fruit. If the activity measured was 1 mCi, what percentage of activity is transmitted from the root to the fruit in steady state?

Answers Introductory Exercise 34.1 1. (b) 6.(b) 11. (b)

2. (a) 3. (a) 7. (a) 8. (b) 12. 3 days, 4.32 days

14. (a) 1.55 ´ 10 –5 / s, 12.4 h 1 16. 4

4. (c) 5. (c) 9. (d) 10. (a) 13. 9.47 ´ 10 9 nuclei

(b) 2.39 ´ 1013 atoms (c) 1.87 mCi

15. 1.16 ´ 103 s

Introductory Exercise 34.2 1. (c)

2. (b)

5. (a) 9.6 ´ 10 7. (a)

2 1H

–4

3. (d) 19

kg (b) 3.125 ´ 10

(b) 11 H (c)

1 0

n (d)

79 36

4. (c)

6. 4.27 MeV 8. Fusion, 24

Kr

9. 23.6 MeV

Exercises LEVEL 1 Assertion and Reason 1. (b)

2. (d)

3. (b)

4. (c)

5. (d)

6. (d)

7. (a or b)

8. (b)

9. (d)

10. (b)

11. (a or b)

Objective Questions 1. (a)

2. (b)

3. (c)

4. (a)

5. (d)

11. (b)

12. (c)

13. (a)

14. (d)

15. (b)

6. (b) 16.

(a)

7. (d) 17.

8. (c)

9. (d)

10. (b)

(a)

Subjective Questions 1. (a) 0.113 min–1 (b) 6.132 min

2. 19.25 min

3. 7.11 ´ 10 –3 g

4. 1.23 ´ 10 dps

5. 0.39

6. 1.88 ´ 10 9 yr

7. 0.205 mCi

8. (a)

4

9. Reaction (a) is possible (b) is not possible

222 86Rn

10. 104.72 MeV

12. (a) 8.19 ´ 1013 J (b) 2.7 ´ 10 6 kg

13.

(b) e + n (c) e + + n 11. 127.6 MeV

0.1308 MeV 14.

8.78 day

15. (a) 4.05 MeV (b) 3.25 MeV (c) 17.57 MeV 16. – 93.1 keV, No 17. N/P ratio required for stability decreases with decreasing A, hence there is an excess of neutrons when fission occurs. Some of the excess neutrons are released directly, and others change to protons by beta decay in the fission products.

LEVEL 2 Single Correct Option 1.(b)

2.(c)

3.(b)

11.(b)

12.(b)

13.(c)

4.(d)

5.(d)

6.(d)

7.(c)

8.(b)

9.(c)

10.(a)

374 — Optics and Modern Physics More than One Correct Options 1.(a,b,d)

2.(a,b,c,d)

3.(a,b,c)

4.(b,c)

5.(a,b,c,d)

6.(b,c)

Comprehension Based Questions 1.(c)

2.(b)

3.(d)

Match the Columns 1. (a) ® s

(b) ® p

(c) ® r

(d) ® s

2. (a) ® p

(b) ® p

(c) ® p

(d) ® s

3. (a) ® q

(b) ® s

(c) ® p

(d) ® s

4. (a) ® s

(b) ® p,r

(c) ® s

(d) ® q, r

5. (a) ® p

(b) ® s

(c) ® r

(d) ® q

Subjective Questions 2. (a) NB = lN0 (te – lt ) (b) t =

1. 14.3 h 3. (a)

lN0 1 , Rmax = l e

1 3 [ a - (a - lN0 ) e – lt ] (b) N0 , 2N0 l 2 10

4. (a) 0.73 ´ 10

10

dps, 0.27 ´ 10

dps (b) 5.4

5.

a 0.2 E 0 é at - (1 - e – lt )ù êë úû l ms

– l2t

æ e - 1ö ÷ 6. Nc = N0 (1 - e - l1t ) + P ç t + ç ÷ l 2 è ø 1 8 8. , , 1215 s 9 9 10. 1.26 ´ 10 –11%

7. 10 g, 4.57 ´ 10 21 disintegrations/day 9. 1.55 ´ 10 6 J

35.1

Introduction

35.2

Energy Bands in Solids

35.3

Intrinsic and Extrinsic Semiconductors

35.4

p-n Junction Diode

35.5

Junction Diode as a Rectifier

35.6

Applications of p-n Junction Diodes

35.7

Junction Transistors

35.8

Transistor as an Amplifier

35.9

Digital Electronics and Logic Gates

376 — Optics and Modern Physics

35.1 Introduction Solids can be classified in three types as per their electrical conductivity. (i) conductors, (ii) insulators and (iii) semiconductors. In a conductor, large number of free electrons are present. They are always in zig-zag motion inside the conductor. In an insulator, all the electrons are tightly bound to the nucleus. If an electric field is applied inside a conductor, the free electrons experience force due to the field and acquire a drift speed. This results in an electric current. The conductivity of a conductor such as copper decreases as the temperature is increased. This is because as the temperature is increased, the random collisions of the free electrons with the particles in the conductor become more frequent. This results in a decrease in the drift speed and hence the conductivity decreases. In insulators, almost zero current is obtained unless a very high electric field is applied. Semiconductors conduct electricity when an electric field is applied, but the conductivity is very small as compared to the usual metallic conductors. Silicon, germanium, carbon etc., are few examples of semiconductors. Conductivity of silicon is about 1011 times smaller than that of copper and is about 1013 times larger than that of fused quartz. Conductivity of a semiconductor increases as the temperature is increased.

Extra Points to Remember ˜

˜

Before the discovery of transistors (in 1948) mostly vacuum tubes (also called valves) were used in all electrical circuits. 1ö æ The order of electrical conductivity (s) and resistivity ç r = ÷ of metals, semiconductors and insulators are sø è given below in tabular form.

Table 35.1 r (W -

)

s (W -1 -

-1

)

S.No

Types of solid

1.

Metals

10-2 - 10-8

102 - 108

2.

Semiconductors

10-5 - 106

105 - 10-6

3.

Insulators

1011 - 1019

10-11 - 10-19

35.2 Energy Bands In Solids To understand the energy bands in solids, let us consider the electronic configuration of sodium atom which has 11 electrons. The configuration is (1s) 2 , (2s) 2 , (2 p) 6 and (3s)1 . The levels 1s, 2s and 2p are completely filled. The level 3s is half filled and the levels above 3s are empty. Consider a group of N sodium atoms all in ground state separated from each other by large distances such as in sodium vapour. There are total 11N electrons. Each atom has two energy states in 1s energy level. So, there are 2N identical energy states lebelled 1s and all them are filled from 2N electrons. Similarly, energy level 2p has 6N identical energy states which are also completely filled. In 3s energy levels N of the 2N states are filled by the electrons and the remaining N states are empty.

Chapter 35

Semiconductors — 377

These ideas are shown in the table given below. Table 35.2 Energy level

Total available energy states

Total occupied states

1s

2N

2N

2s

2N

2N

2p

6N

6N

3s

2N

N

3p

6N

0

Total states = 2N In the above discussion, we have assumed that N sodium Occupied = N 3s atoms are widely spread and hence the electrons of one Empty = N atom do not interact with others. As a result energy states of different states (e.g. 1s) are identical. When atoms are Total states = 6N Occupied = 6N drawn closer to one another, electron of one atom starts 2p interacting with the electrons of the neighbouring atoms of the same energy states. For example 1s electrons of one 2s Total states = 2N Occupied = 2N atom interact with 1s electrons of the other. Due to interaction of electrons, the energy states are not identical, Total states = 2N but a sort of energy band is formed. These bands are shown 1s Occupied = 2N in figure. Fig. 35.1 The difference between the highest energy in a band and the lowest energy in the next higher band is called the band gap between the two energy bands. Thus, we can conclude that energy levels of an electron in a solid consists of bands of allowed states. There are regions of energy, called gaps, where no states are possible. In each allowed band, the energy levels are very closely spaced. Electrons occupy states which minimize the total energy. Depending on the number of electrons and on the arrangement of the bands, a band may be fully occupied or partially occupied. Now, electrical conductivity of conductors, insulators and semiconductors can be explained by these energy bands.

} } (a)

Conduction band

}

Conduction band

}

Eg ~ 6 eV Eg ~ 1 eV

}

Valence band (b)

Valence band

Conduction band

}

Valence band

(c)

Fig 35.2 Energy band diagram for a (a) metal, (b) insulator and (c) semiconductor. Note that one can have a metal either when the conduction band is partially filled or when the conduction and valence bands overlap in energy.

378 — Optics and Modern Physics Conductors The energy band structure of a conductor is shown in figure (a). The last occupied band of energy level (called conduction band) is only partially filled. In conductors, this band overlaps with completely filled valence band. Insulators The energy band structure of an insulator is shown in figure (b). The conduction band is separated from the valence band by a wide energy gap (e.g. 6 eV for diamond). But at any non-zero temperature, some electrons can be excited to the conduction band. Semiconductors The energy band structure of a semiconductor is shown in figure (c). It is similar to that of an insulator but with a comparatively small energy gap. At absolute zero temperature, the conduction band of semiconductors is totally empty, and all the energy states in the valence band are filled. The absence of electrons in the conduction band at absolute zero does not allow current to flow under the influence of an electric field. Therefore, they are insulators at low temperatures. However at room temperatures some valence electrons acquire thermal energy greater than the energy gap E g and move to the conduction band where they are free to move under the influence of even a small electric field. Thus, a semiconductor originally an insulator at low temperatures becomes slightly conducting at room temperature. Unlike conductors the resistance of semiconductors decreases with increasing temperature. We are generally concerned with only the highest valence band and the lowest conduction band. So, when we say valence band, it means the highest valence band. Similarly, when we say conduction band, it means the lowest conduction band. V

Example 35.1 What is the energy band gap of : germanium? Solution

V

(i) silicon and (ii)

The energy band gap of silicon is 1.1 eV and of germanium is about 0.7 eV.

Example 35.2 In a good conductor, what is the energy gap between the conduction band and the valence band. In a good conductor, conduction band overlaps with the valence band. Therefore, the energy gap between them is zero. Solution

35.3 Intrinsic and Extrinsic Semiconductors As discussed above, in semiconductors the conduction band and the valence band are separated by a relatively small energy gap. For silicon, this gap is 1.1 eV and for germanium it is 0.7 eV. Silicon has an atomic number 14 and electronic configuration 1s 2 , 1s 2 , 2 p 6 , 3s 2 , 3 p 2. The chemistry of silicon tells us that it has a valency 4. Each silicon atom makes covalent bonds with the four neighbouring silicon atoms. On the basis of bonds the atoms make with their neighbouring atoms, semiconductors are divided in two groups. Intrinsic Semiconductors A pure (free from impurity) semiconductor which has a valency 4 is called an intrinsic semiconductor. Pure germanium, silicon or carbon in their natural state are intrinsic semiconductors. As discussed above, each atom makes four covalent bonds with their neighbouring atoms. At temperature close to zero, all valence electrons are tightly bound and

Si

Si

Fig. 35.3 Free electron

Si

Si

Hole Fig. 35.4

Chapter 35

Semiconductors — 379

so no free electrons are available to conduct electricity through the crystal. At room temperature, however a few of the covalent bonds are broken due to thermal agitation and thus some of the valence electrons become free. Thus, we can say that a valence electrons is shifted to conduction band leaving a hole (vacancy of electron) in valence band. In intrinsic semiconductors, Number of holes = Number of free electrons or

nh = ne

Extrinsic Semiconductors The conductivity of an intrinsic semiconductor is very poor (unless the temperature is very high). At ordinary temperature, only one covalent bond breaks in 10 9 atoms of Ge. Conductivity of an intrinsic (pure) semiconductor is significantly increased, if some pentavalent or trivalent impurity is mixed with it. Such impure semiconductors are called extrinsic or doped semiconductors. Extrinsic semiconductors are again of two types (i) p-type and (ii) n-type. (i) p-type semiconductors When a trivalent (e.g. boron, aluminium, gallium or indium) is added to a germanium or silicon crystal it replaces one of the germanium or silicon atom. Its three valence Si Al electrons form covalent bonds with neighbouring three Ge (or Si) atoms while the fourth valence electron of Ge (or Si) is not able to form the bond. Thus, there remains a hole (an empty space) on one Fig. 35.5 side of the impurity atom. The trivalent impurity atoms are called acceptor atoms because they create holes which accept electrons. Following points are worthnoting regarding p-type semiconductors. (a) Holes are the majority charge carriers and electrons are minority charge carriers in case of p-type semiconductors or number of holes are much greater than the number of electrons. nh >> ne (b) p-type semiconductor is electrically neutral. (c) p-type semiconductor can be shown as

– –

– –

–

Hole or

Fig. 35.6

(ii) n-type semiconductors When a pentavalent impurity atom (antimony, phosphorus or arsenic is added to a Ge (or Si) crystal it replaces a Ge (or Si) atom. Four of the five valence electrons of the Si P impurity atom form covalent bonds with four neighbouring Ge (or Si) atoms and the fifth valence electron becomes free to move inside the crystal lattice. Thus, by doping pentavalent impurity number of free Fig. 35.7 electrons increases. The impurity (pentavalent) atoms are called donor atoms because they donate conduction electrons inside the crystal. Following points are worthnoting regarding n-type semiconductors, (a) Electrons are the majority charge carriers and holes are minority or number of electrons are much greater than the number of holes ne >> nh

380 — Optics and Modern Physics (b) n-type semiconductor is also electrically neutral. (c) n-type semiconductor can be shown as

+ +

+ +

+

or

Fig. 35.8

Electrical Conduction through Semiconductors When a battery is connected across a semiconductor (whether intrinsic or extrinsic) a potential difference is developed across its ends. Due to the potential difference an electric field is produced inside the semiconductor. A current (although very small) starts flowing through the semiconductor. This current may be due to the motion of (i) free electrons and (ii) holes. Electrons move in opposite direction of electric field while holes move in the same direction. The motion of holes towards right (in the figure) take place because electrons from right hand side come to fill this hole, creating a new hole in their own position. Thus, we can say that holes are moving from left to right. Thus, current in a semiconductor can be written as, i = ie + ih

E

Si

Si

Al

Fig. 35.9

But it should be noted that mobility of holes is less than the mobility of electrons. V

Example 35.3 C, Si and Ge have same lattice structure. Why is C insulator while Si and Ge intrinsic semiconductors? Solution The energy gap between conduction band and valence band is least for Ge, followed

by Si and highest for C. Hence, number of free electrons are negligible for C. This is why carbon is insulator. V

Example 35.4 In an n-type silicon, which of the following statements is true? (a) Electrons are majority carriers and trivalent atoms are the dopants. (b) Electrons are minority carriers and pentavalent atoms are the dopants. (c) Holes are minority carriers and pentavalent atoms are the dopants. (d) Holes are majority carriers and trivalent atoms are the dopants. Solution (c) Holes are minority charge carriers and pentavalent atoms are the dopants in an

n - type silicon.

Chapter 35 V

Semiconductors — 381

Example 35.5 Which of the statements given in above example is true for p-type semiconductors? (d) Holes are majority carriers and trivalent atoms are the dopants in an p -type semiconductors. Solution

INTRODUCTORY EXERCISE

35.1

1. Carbon, silicon and germanium have four valence electrons each. These are characterised by valence and conduction bands separated by energy band-gap respectively equal to (Eg )C, (Eg )Si and (Eg )Ge. Which of the following statements is true? (a) (Eg )Si < (Eg )Ge < (Eg )C (b) (Eg )C < (Eg )Ge > (Eg ) Si (c) (Eg )C > (Eg )Si > (Eg )Ge (d) (Eg )C = (Eg )Si = (Eg )Ge

35.4 p-n Junction Diode

Acceptor ion

Junction

Donor ion

A p-type or n-type silicon crystal can be made by Electron adding appropriate impurity as discussed above. – + – – – + + + These crystals are cut into thin slices called the Hole – + – – – + + + wafer. Semiconductor devices are usually made of – + these wafers. – – – + + + – + If on a wafer of n-type silicon, an aluminium film is – – – + + + – + placed and heated to a high temperature, aluminium p-type n-type Depletion diffuses into silicon. region In this way, a p-type semiconductor is formed on an (a) Formation of p-n junction n-type semiconductor. Such a formation of p-region V on n-region is called the p-n-junction. Another way to make a p-n V junction is by diffusion of phosphorus into a p-type semiconductor. p n Such p-n junctions are used in a host of semiconductor devices of practical applications. The simplest of the semiconductor devices is a p-n junction diode. Biasing of a diode In a p-n junction diode, holes are majority carriers on p-side and electrons on n-side. Holes, thus diffuse to (b) Forward biased p-n junction n-side and electrons to p-side. V This diffusion causes an excess positive charge in the n region and V an excess negative charge in the p region near the junction. This p n double layer of charge creates an electric field which exerts a force on the electrons and holes, against their diffusion. In the equilibrium position, there is a barrier, for charge motion with the n-side at a higher potential than the p-side. The junction region has a very low density of either p or n-type (c) Reverse biased p-n junction carriers, because of inter diffusion. It is called depletion region. Fig. 35.10 There is a barrierVB associated with it, as described above. This is called potential barrier. B

B

382 — Optics and Modern Physics Now suppose a DC voltage source is connected across the p-n junction. The polarity of this voltage can lead to an electric field across the p-n junction that is opposite to the already present electric field. The potential drop across the junction decreases and the diffusion of electrons and holes is thereby increased, resulting in a current in the circuit. This is called forward biasing. The depletion layer effectively becomes smaller. In the opposite case, called reverse biasing the barrier increases, the depletion region becomes larger, current of electrons and holes is greatly reduced. Thus, the p-n junction allows a much larger current flow in forward biasing than in reverse biasing. This is crudely, the basis of the action of a p-n junction as a rectifier. The symbol of p-n junction diode n) is ( p

Diffusion Current and Drift Current Because of concentration difference, holes try to diffuse from the p-side to the n-side at the p-n junction. This diffusion give rise to a current from p-side to n-side called diffusion current. Because of thermal collisions, electron-hole pair are created at every part of a diode. However, if an electron-hole pair is created in the depletion region, the electron is pushed by the electric field towards the n-side and the hole towards the p-side. This gives rise to a current from n-side to p-side called the drift current. Thus, I df ¾® from p-side to n-side I dr ¾® from n-side to p-side When diode is unbiased I df = I dr or I net = 0. When diode is forward biased I df > I dr or I net is from p-side to n-side. When diode is reverse biased I dr > I df or I net is from n-side to p-side.

Characteristic Curve of a p-n Junction Diode (a) Circuit for obtaining the characteristics of a forward biased diode and (b) Circuit for obtaining the characteristics of a reverse bias diode. V

p

V

n

p

mA

mA

S +

n

S

– V

– + V

Fig. 35.11

When the diode is forward biased i.e. p-side is kept at higher potential, the current in the diode changes with the voltage applied across the diode. The current increases very slowly till the voltage across the diode crosses a certain value.

Chapter 35 After this voltage, the diode current increases rapidly, even for very small increase in the diode voltage. This voltage is called the threshold voltage or cut-off voltage. The value of the cut-off voltage is about 0.2 V for a germanium diode and 0.7 V for a silicon diode. When the diode is reverse biased, a very small current (about a few micro amperes) produces in the circuit which remains nearly constant till a characteristic voltage called the breakdown voltage, is reached. Then the reverse current suddenly increases to a large value. This phenomenon is called avalanche breakdown. The reverse voltage beyond which current suddenly increases is called the breakdown voltage. V

Semiconductors — 383 I (mA)

100 80 60 40 20 100 80 60 40 20 Vbr

0.2 0.4 0.6 0.8 1.6

V (Volt)

10 20 30 I (mA)

Fig. 35.12

Example 35.6 Can we take one slab of p-type semiconductor and physically join it to another n-type semiconductor to get p-n junction? No. Any slab will have some roughness. Hence continuous contact at the atomic level will not be possible. For the charge carriers, the junction will behave as a discontinuity. Solution

V

Example 35.7 Find current passing through 2 W and 4 W resistance in the circuit shown in figure. Solution In the given circuit diode D1 is forward biased and D2 reverse biased. Hence, D1 will conduct but D2 not. Therefore, current through 4 W resistance will be zero while through 2W resistance will be, 10 = 5 A. 2

INTRODUCTORY EXERCISE

D1 D2

Fig. 35.13

35.2

free electrons in the n-region attract them they move across the junction by the potential difference hole concentration in p-region is more as compared to n-region All of the above

2. When a forward bias is applied to a p - n junction. It (a) (b) (c) (d)

raises the potential barrier reduces the majority carrier current to zero lowers the potential barrier All of the above

4W

10V

1. In an unbiased p-n junction, holes diffuse from the p-region to n-region because (a) (b) (c) (d)

2W

384 — Optics and Modern Physics

35.5 Junction Diode as a Rectifier A rectifier is a device which converts an alternating current (or voltage) into a direct (or unidirectional) current (or voltage). A p-n junction diode can work as an excellent rectifier. It offers a low resistance for the current to flow when it is forward biased, but a very high resistance when reverse biased. Thus, it allows current through it only in one direction and acts as a rectifier. The junction diode can be used either as an half-wave rectifier or as a full-wave rectifier. (i) p-n junction diode as half-wave rectifier A simple rectifier circuit called the half-wave rectifier, using only one diode is shown in figure. p

n

Mains

A

RL Primary B

Y

Voltage across RL

Secondary

AC voltage at point A

Input AC waveform X

t

O

Output DC waveform

t

O

Fig. 35.14

Fig. 35.15

When the voltage at A is positive, the diode is forward biased and it conducts and when the voltage at A is negative, the diode is reverse biased and does not conduct. Since, the diode conducts only in the positive half cycles, the voltage between X and Y or across R L will be DC but in pulses. When this is given to a circuit called filter (normally a capacitor), it will smoothen the pulses and will produce a rather steady DC voltage. (ii) p-n junction diode as full-wave rectifier Figure shows a circuit which is used in full-wave rectification. Two diodes are used for this purpose. The secondary coil of the transformer is wound in two parts and the junction is called a Centre-Tap (CT ). During one-half cycle D1 is forward biased and D2 is reverse biased. Therefore, D1 conducts but D2 does not, current flows from X to Y through load resistance R L . During another half cycle D2 is forward biased and D1 reverse biased. Therefore, D2 conducts and D1 does not. In this half cycle also current through R L flows from X to Y. Thus, current through R L in both the half cycles is in one direction, i.e. from X to Y. AC waveform at A

A Secondary D1 CT

t

X

Y RL

Primary

t

B

(a) Ful-wave rectifier

D2

AC waveform at B

(b) AC voltage waveforms at points A and B Fig. 35.16

Output AC waveform

t

(c) Output DC waveforms of a full-wave rectifier.

Chapter 35

Semiconductors — 385

Bridge rectifier Another full-wave rectifier called the bridge rectifier which uses four diodes is shown in figure. For one-half cycle diodes D1 and D3 are forward biased and D2 and D4 are reverse biased. So, D1 and D3 conduct but D2 and D4 don't. Current through R L flows from X to Y. In another half cycle D2 and D4 are forward biased and D1 and D3 are reverse biased. So, in this half cycle D2 and D4 conduct but D1 and D3 do not. Current again flows from X to Y through R L . Thus, we see that current through R L always flows in one direction from X to Y. Secondary D2

Primary

D1

D2 and D4 conducting D1 and D3 conducting

X D4 D3

RL t

O (a)

Y

(b)

Fig. 35.17 (a) Bridge rectifier and (b) output waveforms for a bridge rectifier

Note Even after rectification ripples are present in the output which can be removed upto great extent by a filter circuit. A filter circuit consists of a capacitor. V

Example 35.8 In half-wave rectification, what is the output frequency, if the input frequency is 50 Hz? What is the output frequency of a full-wave rectifier for the same input frequency? A half-wave rectifier conducts once during a cycle. Therefore frequency of AC output is also the frequency of AC input i.e. 50 Hz. A full-wave rectifier rectifies both the half cycles of the AC output i.e. it conducts twice during a cycle. So, Frequency of AC output = 2 ´ frequency of AC input Solution

= 2 ´ 50 = 100 Hz V

Ans.

Example 35.9 In the figure, the input is across the terminals A and C and the output is across B and D. Then the output is (a) zero (b) same as the input (c) full-wave rectified (d) half-wave rectified

B

A

C

D

Fig. 35.18

Solution (c) During the half cycle when V M > V N , D1 and D3 are forward

biased. Hence, the path of current is MABPQDCNM. In the second half cycle, when V N > V M , D2 and D4 are forward biased while D1 and D3 are reverse biased. Hence, the path of current is NCBPQDAMN. Therefore, in both half cycles current flows from P to Q from load resistance R L . Or, it is a full-wave rectifier.

B D1 A

P D2

M ~ N D4

RL C

D3 D Fig. 35.19

Q

386 — Optics and Modern Physics

35.6 Applications of p-n Junction Diodes Zener Diode A diode meant to operate under reverse bias in the breakdown region is called an avalanche diode or a zener diode. Such diode is used as a voltage regulator. The symbol of zener diode is shown in figure. Fig. 35.20

Once the breakdown occurs, the potential difference across the diode does not increase even if, there is large change in the current. Figure shows a zener diode in reverse biasing. I (mA)

Reverse bias Vz

Forward bias V (V)

I (mA) Fig. 35.21

An input voltage Vi is connected to the zener diode through a series resistance R such that the zener diode is reverse biased. If the input voltage increases, the current through R and zener diode also increases. This increases the voltage drop across R without any change in the voltage across the zener diode. Similarly, if the input voltage decreases the current through R and zener diode also decreases. The voltage drop across R decreases without any change in the voltage across the zener diode. Thus any increase/decrease in the input voltage results in increase/decrease of the voltage drop across R without any change in voltage across the zener diode (and hence across load resistance R L ). Thus, the zener diode acts as a voltage regulator. We have to select the zener diode according to the required output voltage and accordingly the series resistance R. R Vi

Zener diode

RL Vo

Fig. 35.22

Optoelectronic Devices Semiconductor diodes in which carriers are generated by photons (photo excitation) are called optoelectronic devices. Examples of optoelectronic devices are, photodiodes, Light Emitting Diodes (LED) and photovoltaic devices, etc.

Chapter 35

Semiconductors — 387

(a) Photodiodes Photodiodes are used as photodetector to detect optical signals. They are operated in reverse biased connections. hf

p-side

n-side

mA

Fig. 35.23

When light of energy greater than the energy gap falls on the depletion region of the diode, electron-hole pairs are generated. Due to the electric field of junction, electrons and holes are separated before they recombine. Electrons reach n-side and holes reach p-side giving rise to an emf. When an external load is connected, current flows. The magnitude of the photocurrent depends on the intensity of incident light. (b) Light Emitting Diode (LED) It is heavily doped p-n junction diode which under forward bias emits spontaneous radiation. LEDs that can emit red, yellow, orange, green and blue light are commercially available. These LEDs find extensive use in remote controls, burglar alarm systems, optical communications, etc. Extensive research is being done for developing white LEDs which can replace incandescent lamps. LED have the following advantages over conventional incandescent power lamps. (i) Long life (ii) Low operational voltage and less power (iii) No warm up time is required. So fast on-off switching capability. (c) Solar Cell It works on the same principle as the photodiode. It is basically a p - n junction which generates emf when solar radiation falls on the p-n junction. The difference between a photodiode and a solar cell is that no external bias is applied and the junction area is kept much larger for solar radiation to be incident because we require more power. hf

Top surface

n p

Metallised finger electrode

n p Back contact

Fig. 35.24 Typical p-n junction Solar cell

388 — Optics and Modern Physics The generation of emf by a solar cell (when light falls on it) is due to the following three processes. (i) Generation Generation of electron-hole pairs due to light ( hf > E g ) falling on it. (ii) Separation Separation of electrons and holes due to electric field of the depletion region. (iii) Collection Electrons are swept to n - side and holes to p-side. Thus, p-side becomes positive and n-side becomes negative giving rise to photovoltage. Solar cells are used to power electronic devices in satellites and space vehicles and also as power supply to some calculators. V

Example 35.10 In a zener regulated power supply a zener diode with V Z = 6.0 V is used for regulation. The load current is to be 4.0 mA and the unregulated input is 10.0 V. What should be the value of series resistor R? Solution Zener current I Z should be sufficiently larger than load current I L . R

4.0 mA

24.0 mA

20.0 mA RL

Fig. 35.25 .

Given, I L = 4.0 mA So, let us take I Z to be five times I L or I Z = 20 mA Total current I = I Z + I L = 24.0 mA Input voltage Vin = 10 V Zener diode voltage VZ = 6 V \ Voltage drop across resistance, VR = Vin - VZ or VR = (10 - 6) V = 4 V V 4 Now, R= R = = 167 W I R 24 ´ 10-3 The nearest value of carbon resistor is 150W. So, a series resistor of 150W is appropriate. V

Example 35.11 The current in the forward bias is known to be more ( in mA) than the current in the reverse bias ( in mA). What is the reason then to operate the photodiodes in reverse bias ? Solution Let us take an example of p-type semiconductor. Without illumination number of holes ( n h ) >> number of electrons ( n e ) ...(i) This is because holes are the majority charge carriers in p-type semiconductor.

Chapter 35

Semiconductors — 389

On illumination, let Dn e and Dn h are the excess electrons and holes generated. Dn e = Dn h

...(ii)

From Eqs. (i) and (ii), we can see that Dn e Dn >> h ne nh From here, we can say that the fractional change due to illumination on the minority carrier dominated reverse bias current is more easily measurable than the fractional change in the forward bias current.

35.7 Junction Transistors A junction transistor is formed by sandwiching a thin wafer of one type of semiconductor between two layers of another type. The n-p-n transistor has a p-type wafer between two n-type layers. Similarly, the p-n-p transistor has a n-type wafer between two p-type layers.

p-n-p Transistor Figure shows a p-n-p transistor, in which a thin layer of n-type semiconductor is sandwiched between two p-type semiconductors. The middle layer (called the base) is very thin (of the order of 1mm) as compared to the widths of the two layers at the sides. Base is very lightly doped. One of the side layer (called emitter) is heavily doped and the other side layer (called collector) is moderately doped. Figure (c) shows the symbol of p-n-p transistor. Emitter

Base

Collector

Emitter Base

Collector

Collector Base

p

n

p

Emitter

(c)

(b)

(a)

Fig. 35.26

n-p-n Transistor In n-p-n transistor, p-type semiconductor is sandwiched between two n-type semiconductors. Symbol of n - p -n transistor is shown in figure (f). Emitter

Base

Collector

Emitter

Base

Collector

Collector Base

n

p

(d)

n

(e)

(f)

Emitter

Fig. 35.27

More points about a transistor A transistor is basically a three-terminal device. Terminals come out from the emitter, base and the collector for external connections. In normal operation of a transistor, the emitter-base junction is always forward biased and collector-base junction is reverse biased.

390 — Optics and Modern Physics The arrow on the emitter-base line shows the direction of current between emitter and base. In an n-p-n transistor for example, there are a large number of conduction electrons in the emitter and a large number of holes in the base. If the junction is forward biased the electrons will diffuse from emitter to the base and holes will diffuse from the base to the emitter. The direction of electric current at this junction is therefore from the base to the emitter. A transistor can be operated in three different modes. (i) Common emitter (or grounded-emitter) (ii) Common collector (or grounded-collector) and (iii) Common base (or grounded-base) In common emitter mode, emitter is kept at zero potential. Similarly in common collector mode collector is at zero potential and so on.

Working of a p-n-p Transistor Let us consider the working of a p-n-p transistor in common base mode. In emitter (p-type) holes are in majority. Since, emitter-base is forward biased, holes move toward base. Few of them combine with electrons in the base and rest go to the collector. Since, base-collector is reverse biased, holes coming from base move toward the terminal of collector. They combine with equal number of electrons entering from collector terminal. p

Higher

n

Lower

ie

p

Higher ib

VEB

Lower ie

VCB

Fig. 35.28

Let us take an example with some numerical values. Suppose 5 holes enter from emitter to base. This deficiency of 5 holes in emitter is compensated when 5 electrons emit from emitter and give rise to ie . One out of five holes which reach the base combine with one electron entering from base (the equivalent current is ib ). Rest four holes enter the collector and move towards its terminal. On the other hand, 5 electrons which leave the emitter (as ie ) come to the base, emitter and collector junctions. One electron of it goes to base and rest four to collector. These four electrons give rise to ic (the collector current) and combine with the four holes coming from the base, and thus circuit is complete. From the figure, we can see that, ie = ic + ib Note that ib is only about 2% of ie , or roughly around 2% of holes coming from emitter to base combine with the electrons. Rest 98% move to collector.

Chapter 35

Semiconductors — 391

Working of n-p-n Transistor A common base circuit of an n-p-n transistor is shown in figure. Majority charge carriers in the emitter (n-type) are electrons. Since, emitter-base circuit is forward biased. The electrons rush from emitter to base. Few of them leave the base terminal (comprising ib ) and rest move to collector. These electrons finally leave the collector terminal (give rise to ic ). Electrons coming from base and from collector meet at junction O and they jointly move to emitter, which gives rise to ie . ie = ib + ic

Thus, here also we can see that n

p

n

ie

Lower

Higher

Lower

Higher

ib

VEB

ie

O

ic

VCB

Fig. 35.29

Note that although the working principle of p-n-p and n-p-n transistors are similar but the current carriers in p-n-p transistor are mainly holes whereas in n-p-n transistors the current carriers are mainly electrons. Mobility of electrons are however more than the mobility of holes, therefore n-p-n transistors are used in high frequency and computer circuits where the carriers are required to respond very quickly to signals.

a and b-parameters: a and b-parameters of a transistor are defined as, a = ic / ie

and

b = ic / ib

As ib is about 1 to 5% of ie , a is about 0.95 to 0.99 and b is about 20 to 100. By simple mathematics we can prove that, a b= 1-a

35.8 Transistor As An Amplifier A transistor can be used for amplifying a weak signal. When a transistor is to be operated as amplifier, three different basic circuit connections are possible. These are (i) common base, (ii) common emitter and (iii) common collector circuits. Whichever circuit configuration, the emitter-base junction is always forward biased, while the collector-base junction is always reverse biased. (a) Common base amplifier using a p-n-p transistor In common base amplifier, the input signal is applied across the emitter and the base, while the amplified output signal is taken across the collector and the base. This circuit provides a very low input resistance, a very high output resistance and a current gain of just less than 1. Still it provides a good voltage and power amplification. There is no phase difference between input and output signals.

392 — Optics and Modern Physics The common base amplifier circuit using a p-n-p transistor is shown in figure. The emitter base input circuit is forward biased by a low voltage battery VEB . The collector base output circuit is reversed biased by means of a high voltage batteryVCC . Since, the input circuit is forward biased, resistance of input circuit is small. Similarly, output circuit is reverse biased, hence resistance of output circuit is high. ie

p-n-p E

C ic

B

RL VCB

ib

Input AC signal

–

VCC +

ie

+ – VEB

ic Output AC signal

ic

Fig. 35.30

The weak input AC voltage signal is superimposed onVEB and the amplified output signal is obtained across collector-base circuit. In the figure we can see that, VCB = VCC - ic R L The input AC voltage signal changes net value ofVEB . Due to fluctuations inVEB , the emitter current ie also fluctuates which in turn fluctuates ic . In accordance with the above equation there are fluctuations in VCB , when the input signal is applied and an amplified output is obtained.

Current gain, Voltage gain and Power gain (i) Current gain Also called AC current gain (a ac ), is defined as the ratio of the change in the collector current to the change in the emitter current at constant collector-base voltage. Di Thus, ( VCB = constant) a ac or simply a= c Die As stated earlier also, a is slightly less than 1. (ii) Voltage gain It is defined as the ratio of change in the output voltage to the change in the input voltage. It is denoted by AV . Thus, Di ´ R out AV = c Die ´ R in Di but c = a, the current gain. Die a R out \ AV = R in Since, R out >> R in , AV is quite high, although a is slightly less than 1. (iii) Power gain It is defined as the change in the output power to the change in the input power. Since P = Vi Therefore, power gain = current gain ´ voltage gain R or Power gain = a 2 × out R in

Semiconductors — 393

Chapter 35 Extra Points to Remember ˜ ˜

The output voltage signal is in phase with the input voltage signal. The common base amplifier is used to amplify high (radio)-frequency signals and to match a very low source impedance (~20 W) to a high load impedance (~100 k W).

(b) Common emitter amplifier using a p-n-p transistor Figure shows a p-n-p transistor as an amplifier in common emitter mode. The emitter is common to both input and output circuits. The input (base-emitter) circuit is forward biased by a low voltage battery VBE . The output (collector-emitter) circuit is reverse biased by means of a high voltage battery VCC . Since, the base-emitter circuit is forward biased, input resistance is low. Similarly, collector-emitter circuit is reverse biased, therefore output resistance is high. The weak input AC signal is superimposed on VBE and the amplified output signal is obtained across the collector-emitter circuit. ic

C ib

B

p-n-p

RL

E

ic

VCE – Input AC signal

ie

VCC

Output AC signal

+ – + VBE

iC

ib

Fig. 35.31

In the figure we can see that, VCE = VCC - ic R L When the input AC voltage signal is applied across the base-emitter circuit, it fluctuatesVBE and hence the emitter current ie . This in turn changes the collector current ic consequently VCE varies in accordance with the above equation. This variation in VCE appears as an amplified output.

Current Gain, Voltage Gain and Power Gain (i) Current gain Also called ac current gain (b ac ), is defined as the ratio of the collector current to the base current at constant collector to emitter voltage. æ Di b ac or simply b = çç c è Dib

ö ÷÷ ø

(VCE = constant)

(ii) Voltage gain It is defined as the ratio of the change in the output voltage to the change in the input voltage. It is denoted by AV . Thus, AV =

Dic ´ R out Dib ´ R in

or

æR AV = b çç out è R in

ö ÷÷ ø

(iii) Power gain It is defined as the ratio of change in output power to the change in the input power. Since, P = Vi Therefore, power gain = current gain ´ voltage gain

or

æR Power gain = b 2 çç out è Rin

ö ÷÷ ø

394 — Optics and Modern Physics Extra Points to Remember ˜

The value of current gain b is from 15 to 50 which is much greater than a.

˜

The voltage gain in common-emitter amplifier is larger compared to that in common base amplifier.

˜

˜

The power gain in common-emitter amplifier is extremely large compared to that in common base amplifier. The output voltage signal is 180° out of phase with the input voltage signal in the common-emitter amplifier.

Transconductance (g m ) There is one more term called transconductance ( g m ) in common-emitter mode. It is defined as the ratio of the change in the collector current to the change in the base to emitter voltage at constant collector to emitter voltage. Thus,

The unit of g m is W -1

æ Dic ö g m = çç ÷÷ è DVBE ø or siemen (S). By simple calculation we can prove that, gm =

(VCE = constant)

b R in

Advantages of a transistor over a triode valve A transistor is similar to a triode valve in the sense that both have three elements. While the elements of a triode are, cathode, plate and grid. The three elements of a transistor are emitter, collector and base. Emitter of a transistor can be compared with the cathode of the triode, the collector with the plate and the base with the grid. Transistor has following advantages over a triode valve (i) A transistor is small and cheap as compared to a triode valve. They can bear mechanical shocks. (ii) A transistor has much longer life as compared to a triode valve. (iii) Loss of power in a transistor is less as it operates at a much lower voltage. (iv) In a transistor no heating current is required. So, unlike a triode valve, a transistor starts functioning immediately as soon as the switch is opened. In case of valves, they come in operation after some time of opening the switch (till cathode gets heated). Drawbacks of a transistor over a triode valve Transistor have following drawbacks as compared to valves. (i) Since, the transistors are made of semiconductors they are temperature sensitive. We cannot work on transistors at high temperatures. (ii) In transistors noise level is high. Keeping all the factors into consideration, transistors have replaced the valve from most of the modern electronic devices. V

Example 35.12 The current gain of a transistor in a common base arrangement in 0.98. Find the change in collector current corresponding to a change of 5.0 mA in emitter current. What would be the change in base current? Solution Given, a = 0.98 and Di e = 5.0 mA Di From the definition of, a= c Di e Change in collector current, Di c = ( a ) ( Di e ) = ( 0.98) ( 5.0) mA = 4.9 mA Further, change in base current, Ans. Di b = Di e - Di c = 0.1 mA

Chapter 35 V

Semiconductors — 395

Example 35.13 A transistor is connected in common emitter configuration. The collector supply is 8 V and the voltage drop across a resistor of 800 W in the collector circuit is 0.5 V. If the current gain factor ( a ) is 0.96, find the base current. Solution

b=

a 0.96 = = 24 1 - a 1 - 0.96

The collector current is, voltage drop across collector resistor 0.5 = A = 0.625 ´ 10-3 A ic = 800 resistance i From the definition of b= c ib ib =

the base current

i c 0.625 ´ 10-3 A = b 24

= 26mA V

Ans.

Example 35.14 In a common emitter amplifier, the load resistance of the output circuit is 500 times the resistance of the input circuit. If a = 0.98, then find the voltage gain and power gain. Solution

Given a = 0.98 and

(i) Voltage gain = (b )

R out a 0.98 = 500 Þ b = = = 49 R in 1 - a 1 - 0.98

R out = ( 49) ( 500) = 24500 R in

(ii) Power gain = (b 2 )

R out = ( 49) 2 ( 500) = 1200500 R in

INTRODUCTORY EXERCISE

35.3

1. For transistor action, which of the following statements are correct? (a) Base, emitter and collector regions should have similar size and doping concentrations (b) The base region must be very thin and lightly doped (c) The emitter junction is forward biased and collector junction is reverse biased (d) Both the emitter junction as well as the collector junction are forward biased

2. For a transistor amplifier, the voltage gain (a) (b) (c) (d)

remains constant for all frequencies is high at high and low frequencies and constant in the middle frequency range is low at high and low frequencies and constant at mid frequencies None of the above

3. For a CE-transistor amplifier, the audio signal voltage across the collector resistance of 2 kW is 2 V. Suppose the current amplification factor of the transistor is 100. Find the input signal voltage and base current, if the base resistance is 1 kW.

396 — Optics and Modern Physics

35.9 Digital Electronics and Logic Gates (i) Binary system There are a number of questions which have only Bulb two answers Yes or No. A statement can be either True or False. A switch can be either ON or OFF. These values may be represented S2 by two symbols 0 and 1. In a number system, in which we have only S1 two digits is called a binary system. (decimal system for example has ten digits). Source In binary system usually we write 1 for positive response (e.g. when Fig. 35.32 a switch is ON) and 0 for negative (when switch is OFF). (ii) Truth table To understand the concept of truth table let us take an example. A bulb is connected to an AC source via two switches S 1 and S 2 . In binary system, we will write 0, if the switch (or bulb) is off and write 1 if it is on. Further let us write A for state of switch S 1 B for state of switch S 2 and C for state of the bulb. Now, let us make a table (called truth table) which is self explanatory. Table 35.3 Switch S 1

Switch S 2

Bulb

A

B

C

Off

On

Off

0

1

0

On

Off

Off

1

0

0

Off

Off

Off

0

0

0

On

On

On

1

1

1

Exercise Make a truth table corresponding to the circuit shown in figure. S1 S2

Source

Bulb

Fig. 35.33

Table 35.4 Switch S 1

Switch S 2

Bulb

A

B

C

On

Off

On

1

0

1

Off

On

On

0

1

1

Off

Off

Off

0

0

0

On

On

On

1

1

1

Chapter 35

Semiconductors — 397

(iii) Logical function A variable (e.g. state of a switch or state of a bulb) which can assume only two values (0 and 1) is called a logical variable. A function of logical variables is called a logical function. AND, OR and NOT represent three basic operations on logical variables. ‘AND’ function Suppose C is a function of A and B, then it will be said an ‘AND’ function when C has value 1 when both A and B have value 1. Truth table corresponding to Table 35.3 is an example of ‘AND’ function. The function is written as, C = A and B AND function is also denoted as C = A×B ‘OR’ function C, a function of A and B will be said an ‘OR’ function when C has value 1 when either of A or B has value 1. Truth table corresponding to Table 35.4 is an example of ‘OR’ function. The function is written as, C = A OR B OR function is also denoted as, C = A+B ‘NOT’ function ‘NOT’ function is a function of a single variable. Switch Source A bulb is short circuited by a switch. If the switch is open, Bulb the current goes through the bulb and it is on. If the switch is closed the current goes through the switch and the bulb is Fig. 35.34 off. The truth table corresponding to the above situation (or NOT function) is as under. Table 35.5 Switch

Bulb

A

B

Open

On

0

1

Closed

Off

1

0

‘NOT’ function is denoted as, B = NOT A or B = A V

Example 35.15 Write the truth table for the logical function D = ( A OR B ) AND B. Solution

A OR B is a logical function, say it is equal to X, i.e.,

Now, The corresponding truth table is as under.

X = A OR B D = X AND B

Table 35.6 A

B

X = A OR B

D = ( A OR B ) AND B

1

0

1

0

0

1

1

1

0

0

0

0

1

1

1

1

398 — Optics and Modern Physics Note that the given function can also be written as,

D = (A + B ) × B (iv) Logic gates Logic gates are important building blocks in digital electronics. These are circuits with one or more inputs and one output. The basic gates are OR, AND, NOT, NAND, NOR and XOR. As we know, in digital electronics only two voltage levels are present. Conventionally, these are 5V and 0V, referred to as 1 and 0 respectively or vice-versa. They are also referred as high and low.Figure given are the symbols of six basic gates.

OR gate

NAND gate

AND gate

NOT gate

NOR gate

XOR gate

Fig. 35.35

OR gate The truth table of ‘OR’ gate is given below. Table 35.7 A

B

X

0 0 1 1

0 1 0 1

0 1 1 1

The output X will be 1 (i.e., 5V) when the A input is 1, OR when the B input is 1, OR when both are 1. This is written as, X = A+B Figure shows construction of an OR gate using two diodes A

A B

X = A+B

Fig. 35.36

X=A+B

D1 B

R

D2

Fig. 35.37

.

When either of point A or point B (or both) has potential +5V, diodes D1 or D2 (or both) are forward biased and the potential at X is the same as the common potential at A and B which is 5V. AND gate The truth table of ‘AND’ gate is given below. Table 35.8 A

B

Y

0 0 1 1

0 1 0 1

0 0 0 1

Chapter 35

Semiconductors — 399

The output X will be 1 (i.e. 5V ) when both the inputs A and B is 1. This is A X = A .B B written as, Fig. 35.38 X = A×B Figure shows construction for an AND gate using two A D1 X=A AND B diodes D1 and D2 . When potentials at A and B both are zero, then both the B R D2 diodes are forward biased and offer no resistance. The 5V potential at X in this position is equal to the potential at A or B i.e. 0. Thus X = 0, when both A and B are zero. Now suppose potential at A is zero but at B is 5V, then D1 is Fig. 35.39 forward biased. In this situation potential at X is also zero. Thus, X = 0 when A = 0. Similarly, we can see that X = 0 when B = 0. Lastly when potentials at both A and B are 5V, so that both the diodes are unbiased and there will be no current through R and the potential at X will be equal to 5V. Thus, X =1 when A and B both are 1. NOT gate This has one input and one output. The output is the inverse of the input. When the input A is 1, the output X will be 0 and vice-versa. The truth table for ‘NOT’ gate is given below. Table 35.9 A

X

0

1

1

0

A

X= A

Fig. 35.40

Note A NOT gate cannot be constructed with diodes. Transistor is used for realisation of a NOT gate, but at this stage students do not require it. A NOT gate is written as X = A .

NAND gate The function, X = NOT (A and B) of two logical variables A and B is called NAND function. It is written as X = A NAND B. It is also written as,

A

X = A. B

B

Fig. 35.41

X = A×B X = AB

or

The truth table of a ‘NAND’ gate is given below. Table 35.10 A

B

A× B

X = A× B

0

0

0

1

0

1

0

1

1

0

0

1

1

1

1

0

NOR gate The function X = NOT (A OR B) is called a NOR function and is written as X = A NOR B. It is also written as, X = A + B . The truth table for a NOR gate is given below.

A B

X = A+ B

Fig. 35.42

400 — Optics and Modern Physics Table 35.11 A

B

A+ B

X = A+ B

0

0

0

1

0

1

1

0

1

0

1

0

1

1

1

0

XOR gate It is also called the exclusive OR function. It is a function of two logical variables A and B which evaluates to 1 if one of two variables is 0 and the other is 1. The function is zero, if both the variables are 0 or 1. A

X= A.B + B .A

B

Fig. 35.43

A XOR B = A × B + A × B The truth table for XOR is given below. Table 35.12 A

V

B

A

B

A× B

A×B

A = A× B + A × B

0

0

1

1

0

0

0

0

1

1

0

0

1

1

1

0

0

1

1

0

1

1

1

0

0

0

0

0

Example 35.16 Construct the truth table for the function X of A and B represented by figure shown here. X

A B

Fig. 35.44

The output X in terms of the input A and B can be written as, X = A × ( A + B ) Let us make the truth table corresponding to this function.

Solution

Table 35.13 A

B

A+ B

X = A × ( A + B)

0

0

0

0

0

1

1

0

1

0

1

1

1

1

1

1

Chapter 35 V

Semiconductors — 401

Example 35.17 Make the output waveform (Y ) of the OR gate for the following inputs A and B. Table 35.14

Solution

Time

A

B

For t < t 1

0

0

From t 1 to t 2

1

0

From t 2 to t 3

1

1

From t 3 to t 4

0

1

From t 4 to t 5

0

0

From t 5 to t 6

1

0

For t > t 6

0

1

Output value Y corresponding to OR gate is given in the following table. Table 35.15

Time

A

B

Y=A+B

For t < t 1

0

0

0

From t 1 to t 2

1

0

1

A (Input)

From t 2 to t 3

1

1

1

B

From t 3 to t 4

0

1

1

From t 4 to t 5

0

0

0

From t 5 to t 6

1

0

1

For t > t 6

0

1

1

t1 t2

t3 t4 t5

t6

Y (Output)

Fig. 35.45

Therefore, the waveform Y will be as shown in the figure. V

Example 35.18 Take A and B inputs similar to that in above example. Sketch the output waveform obtained from AND gate. Solution

Output value, Y corresponding to AND gate is given in the following table. Table 35.16 Time

A

B

Y = A ×B

For t < t 1

0

0

0

From t 1 to t 2

1

0

0

From t 2 to t 3

1

1

1

From t 3 to t 4

0

1

0

From t 4 to t 5

0

0

0

From t 5 to t 6

1

0

0

For t > t 6

0

1

0

402 — Optics and Modern Physics t2

t3

t1

t4

t6

t5

Fig. 35.46

Based on the above table, the output waveform Y for AND gate can be drawn as in figure 35.46.

INTRODUCTORY EXERCISE

35.4

1. Make the output waveform Y of the NAND gate for the following inputs A and B. Table 35.17 Time

A

B

For t < t 1

1

1

From t 1 to t 2

0

0

From t 2 to t 3

0

1

From t 3 to t 4

1

0

From t 4 to t 5

1

1

From t 5 to t 6

0

0

For t > t 6

0

1

2. You are given two circuits. Identify the logic operation carried out by the two circuits. A

A

Y

B

Y B (b)

(a)

Fig. 35.47

Chapter 35

Semiconductors — 403

Final Touch Points 1. Integrated Circuits The short form of integrated circuit is IC. It is revolutionised the electronics technology. The entire electronic circuit (consisting of many passive components like R and C and active devices like diode and transistor) is fabricated on a small single block (called chip) of a semiconductor. Such circuits are more reliable and less shock proof compared to conventional circuits used before. The chip dimensions are as small as 1 mm ´ 1 mm or it could be even smaller than this. Depending on the nature of input signals, ICs are of two types. (i) Linear or analogue IC (ii) Digital IC Linear ICs process analogue signals which change over a range of values between a maximum and a minimum. The digital ICs process signals that have only two values. They contain circuits such as logic gates. IC is the heart of all computer systems. It is used in almost all electronic devices like, cell phones, televisions, cars etc. It was first invented in 1958 by Jack Kilky and he was awarded Nobel prize for this in the year 2000. Growth of semiconductor industry is very fast. From current trends it is expected that by 2020 computers will operate at 40 GHz and would be much smaller, more efficient and less expensive than present day computers.

2. Feedback amplifier and transistor oscillator In an amplifier, a sinusoidal input is given which gets amplified as an output. Hence, an external input is necessary to sustain AC signal in the output. In an oscillator, we get AC output without any external input signal. A portion of the output power is returned back (feedback) to the input (in phase) with the starting power. In other words, the output in an oscillator is self sustained. Input Transistor Amplifier

Output

Feedback network

Principle of a transistor amplifier with positive feedback working as an oscillator

3. In transistors, the base region is narrow and lightly doped, otherwise the electrons or holes coming from the input side (say emitter in CE-configuration) will not be able to reach the collector.

Solved Examples V

Example 1 Sn, C, Si and Ge are all group XIV elements. Yet, Sn is a conductor, C is an insulator while Si and Ge are semiconductors. Why ? Solution It all depends on energy gap between valence band and conduction band. The energy gap for Sn is 0 eV, for C is 5.4 eV, for Si is 1.1 eV and for Ge is 0.7 eV.

V

Example 2 Three photodiodes D1 , D2 and D3 are made of semiconductors having band gaps of 2.5 eV, 2 eV and 3 eV, respectively. Which one will be able to detect o

light of wavelength 6000 A ? Solution Energy of incident light E (in eV) =

12375 o

l (in A ) E=

12375 eV 6000

or E = 2.06 eV For the incident radiation to be detected by the photodiode energy of incident radiation should be greater than the band gap. This is true only for D2. Therefore only D2 will detect this radiation. V

Example 3 What is the range of energy gap ( E g ) in insulators, semiconductors and conductors? Solution For insulators E g > 3 eV, for semiconductors, E g = 0 . 2 eV to 3eV while for conductors (or metals) E g = 0.

V

Example 4 n-type extrinsic semiconductor is negatively charged, while p-type extrinsic semiconductor is positively charged. Is this statement true or false? Solution False. Intrinsic as well as extrinsic semiconductors are electrically neutral.

V

Example 5 What is resistance of an intrinsic semiconductor at 0K ? Solution At 0K number of holes (or number of free electrons) in an intrinsic semiconductor become zero. Therefore, resistance of an intrinsic semiconductor becomes infinite at 0 K.

V

Example 6 Consider an amplifier circuit using a transistor. The output power is several times greater than the input power. Where does the extra power come from? Solution The extra power required for amplified output is obtained from the DC source.

V

Example 7 A piece of copper and the other of germanium are cooled from the room temperature to 80 K. What will happen to their resistance?

Chapter 35

Semiconductors — 405

Solution Copper is conductor and germanium is semiconductor. With decrease in temperature resistance of a conductor decreases and that of semiconductor increases. Therefore resistance of copper will decrease and that of semiconductor will increase. V

Example 8 A transistor has three impurity regions, emitter, base and collector. Arrange them in order of increasing doping levels. Solution The order of increasing doping levels is base > collector > emitter.

V

Example 9 Name two gates which can be used repeatedly to produce all the basic or complicated gates. Solution NAND and NOR gates can be used repeatedly to produce all the basic or complicated gates. This is why these gates are called digital building blocks.

V

Example 10 A change of 8.0 mA in the emitter current brings a change of 7.9 mA in the collector current. How much change in the base current is required to have the same change 7.9 mA in the collector current ? Find the values of a and b. Solution We know that, i e = ib + i c \ Di e = Dib + Di c or Dib = Di e - Di c Substituting the given values of the question, We have Dib = (8.0 - 7.9) mA = 0.1 mA Hence, a change of 0.1 mA in the base current is required to have a change of 7.9 mA in the collector current. i Di 7.9 Further, a = c or c = ie Di e 8.0 = 0.99 Di c 7.9 ic = b = or Dib 0.1 ib = 79

V

Ans.

Ans.

Example 11 A transistor is used in common-emitter mode in an amplifier circuit. When a signal of 20 mV is added to the base-emitter voltage, the base current changes by 20 mA and the collector current changes by 2 mA. The load resistance is 5 kW. Calculate (a) the factor b (b) the input resistance R in (c) the transconductance and (d) the voltage gain. Solution (a) Factor b b=

Di c Dib

Substituting the given values, we have b=

2 ´ 10-3 = 100 20 ´ 10-6

Ans.

406 — Optics and Modern Physics (b) Input Resistance Rin Rin =

DV BE 20 ´ 10-3 = Dib 20 ´ 10-6

= 103 W = 1 kW

Ans.

(c) Transconductance gm gm =

Di c 2 ´ 10-3 = DV BE 20 ´ 10-3

= 0.1 mho

Ans.

(d) Voltage Gain AV æR ö A V = b çç out ÷÷ è Rin ø Substituting the values we have, æ 5 ´ 103 ö ÷ A V = (100) çç 3÷ è 1 ´ 10 ø = 500 V

Ans.

Example 12 An n-p-n transistor is connected in common-emitter configuration in which collector supply is 8V and the voltage drop across the load resistance of 800 W connected in the collector circuit is 0.8 V . If current amplification factor is 25, determine collector-emitter voltage and base current. If the internal resistance of the transistor is 200 W, calculate the voltage gain and the power gain. Solution The corresponding circuit is shown in figure. iC iB RL

0.8 V

iE

8V

Voltage across RL = i c RL = 0.8 V (given) 0.8 0.8 \ ic = = A = 1 mA RL 800 Further it is given that, b = 25 = \

ib =

ic ib

ic = 40 mA 25

Ans.

Collector-Emitter Voltage (VCE ) Applying Kirchhoff’s second law in emitter-collector circuit, we have VCE = (8 - 0.8) V = 7.2 V

Ans.

Chapter 35

Semiconductors — 407

Voltage gain (AV ) æR ö A V = b çç out ÷÷ è Rin ø æ 800 ö A V = 25 ç ÷ = 100 è 200 ø

Voltage gain, or

Ans.

Power gain æR ö æ 800 ö Power gain = b 2 çç out ÷÷ = ( 25)2 ç ÷ = 2500 è 200 ø è Rin ø

Ans.

Note Kirchhoff’s laws can be applied in a transistor circuit in the similar manner as is done in normal circuits. V

Example 13 An n-p-n transistor in a common-emitter mode is used as a simple voltage amplifier with a collector current of 4 mA. The positive terminal of a 8 V battery is connected to the collector through a load resistance R L and to the base through a resistance R B . The collector-emitter voltage VCE = 4 V , the base-emitter voltage V BE = 0.6 V and the current amplification factor b = 100. Calculate the values of R L and R B . Solution Given, ic = 4 mA ib RB

RL 2

C

1

B E ie 8V

Applying Kirchhoff’s second law in loop 1, we have \

Further, \ Now, \

VCE = 8 - i c RL 8 -4 8 - VCE RL = = ic 4 ´ 10-3 = 1000 W = 1 kW i b= c ib ib =

Ans.

i c 4 ´ 10-3 = A = 40 mA b 100

V BE = 8 - ib RB 8 - V BE RB = ib =

8 - 0.6 40 ´ 10-6

= 1.85 ´ 105 W

Ans.

408 — Optics and Modern Physics V

Let X = A × BC. Evaluate X for

Example 14

(a) A = 1, B = 0, C = 1,

(b) A = B = C = 1 and

(c) A = B = C = 0.

Solution (a) When, A = 1, B = 0 and C = 1 BC = 0 \

BC = 1

or

A × BC = 1 A = B =C =1

(b) When,

BC = 1

Then, or

BC = 0

\

A × BC = 0

Ans.

A = B =C =0

(c) When,

BC = 0

Then,

V

Ans.

\

BC = 1

or

A × BC = 0

Ans.

Example 15 Show that given circuit (a) acts as OR gate while the given circuit (b) acts as AND gate. A

A

Y

B

Y B (b)

(a)

Solution (a) The first gate is NOR gate then NOT gate A

Y

X

B

Thus, X = A + B and Y = X The truth table can be made as under.7

Table 35.18 A

B

A+B

X = A+ B

Y= X

1

0

1

0

1

0

1

1

0

1

1

1

1

0

1

0

0

0

1

0

The last column of Y is similar to third column of A + B which is the A truth table corresponding to OR gate. (b) First two gates are NOT gates and the last gate is NOR gate. B Thus, C = A and D = B X =C + D

C X D

Chapter 35

Semiconductors — 409

The truth table corresponding to this can be made as under.

Table 35.19 A

B

A× B

C= A

D= B

C+ D

X =C+ D

1

0

0

0

1

1

0

0

1

0

1

0

1

0

1

1

1

0

0

0

1

0

0

0

1

1

1

0

The last column of X is similar to third column of A × B, which is the truth table corresponding to AND gate. V

Example 16 Write the truth table for the circuit given in figure consisting of NOR gates. Identify the logic operations (OR, AND, NOT) performed by the the circuits.

Solution The truth table corresponding to given circuit of logic gates is A E

B C

X

F

D

Table 35.20 A

B

A+ B

E = A+ B

C

D

C+D

F =C+ D

E+F

X=E+F

1

1

1

0

1

1

1

0

0

1

0

0

0

1

0

0

0

1

1

0

Corresponding to input columns of A, B, C and D we can see that output column of X is of AND gate, X = A+ B+C + D

Exercises Single Correct Option 1. The conductivity of a semiconductor increases with increase in temperature because (a) number density of free current carriers increases (b) relaxation time increases (c) both number density of carriers and relaxation time decreases but effect of decrease in relaxation time is much less than increase in number density. (d) number density of current carriers increases, relaxation time decreases but effect of decrease in relaxation time is much less than increase in number density

2. In figure, assuming the diodes to be ideal, -10V

A

R

D1

D2 B

(a) D1 is forward biased and D2 is reverse biased and hence current flows from A to B. (b) D2 is forward biased and D1 is reverse biased and hence no current flows from B to A and vice-versa. (c) D1 and D2 are both forward biased and hence current flows from A to B. (d) D1 and D2 are both reverse biased and hence no current flows from A to B and vice-versa.

3. Hole is (a) (b) (c) (d)

an anti-particle of electron a vacancy created when an electron leaves a covalent bond absence of free electrons an artificially created particle

4. A 220 V AC supply is connected between points A and B. What will be the potential difference V across the capacitor? A

B

(a) 220 V (c) 0V

(b) 110 V (d) 220 2 V

Chapter 35

Semiconductors — 411

More than One Correct Options 5. When an electric field is applied across a semiconductor, (a) electrons move from lower energy level to higher energy level in the conduction band. (b) electrons move from higher energy level to lower energy level in the conduction band. (c) holes in the valence band move from higher energy level to lower energy level. (d) holes in the valence band move from lower energy level to higher energy level.

6. Consider an n-p-n transistor with its base-emitter junction forward biased and collector base junction reverse biased. Which of the following statements are true? (a) Electrons crossover from emitter to collector. (b) Holes move from base to collector. (c) Electrons move from emitter to base. (d) Electrons from emitter move out of base without going to the collector.

7. In an n-p-n transistor circuit, the collector current is 10 mA. If 95 per cent of the electrons emitted reach the collector, which of the following statements are true? (a) The emitter current will be 8 mA. (c) The base current will be 0.53 mA.

(b) The emitter current will be 10.53 mA. (d) The base current will be 2 mA.

8. In the depletion region of a diode, (a) there are no mobile charges (b) equal number of holes and electrons exist, making the region neutral (c) recombination of holes and electrons has taken place (d) immobile charged ions exist.

9. What happens during regulation action of a Zener diode? (a) The current and voltage across the Zener remains fixed. (b) The current through the series resistance (R) changes. (c) The Zener resistance is constant. (d) The resistance offered by the Zener changes.

10. The breakdown in a reverse biased p-n junction diode is more likely to occur due to (a) large velocity of the minority charge carriers if the doping concentration is small (b) large velocity of the minority charge carriers if the doping concentration is large (c) strong electric field in a depletion region if the doping concentration is small (d) strong electric field in the depletion region if the doping concentration is large.

Subjective Questions 11. Can the potential barrier across a p-n junction be measured by simply connecting a voltmeter across the junction?

12. Two car garages have a common gate which needs to open automatically when a car enters either of the garages or cars enter both. Devise a circuit that resembles this situation using diodes for this situation.

13. Two amplifiers are connected one after the other in series (cascaded). The first amplifier has a voltage gain of 10 and the second has a voltage gain of 20. If the input signal is 0.01 V, calculate the output AC signal.

14. A p -n photodiode is fabricated from a semiconductor with band-gap of 2.8 eV. Can it detect a wavelength of 6000 nm?

412 — Optics and Modern Physics 15. (i) Name the type of a diode whose characteristics are shown in figure. (ii) What does the point P in figure represent? I

(mA) P V (volt)

16. If the resistance R1 is increased, how will the readings of the ammeter and voltmeter change? A V

R2

R1

17. How would you set up a circuit to obtain NOT gate using a transistor? 18. Write the truth table for the circuit shown in figure. Name the gate that the circuit resembles. + 5V D1 A

V0

B D2

19. A Zener of power rating 1 W is to be used as a voltage

R

regulator. If Zener has a breakdown of 5V and it has to regulate voltage which fluctuated between 3V and 7V, Unregulated voltage what should be the value of R for safe operation.

Regulated voltage

X

20. If each diode in figure has a forward bias resistance of 25W and infinite resistance in reverse bias, what will be the values of the current I1 , I 2 , I3 and I 4? A C E I1 G

I4

125 W

I3

125 W

I2

125 W

B D F

25 W H 5V

Chapter 35

Semiconductors — 413

21. In the circuit shown in figure when the input voltage of the base resistance is 10 V. Find the values of I b , I c and b. 10 V Rc

3 kW

Rb 10 V 400 kW

22. For the transistor circuit shown in figure, evaluate V E , RB and RE . Given IC = 1 mA, VCE = 3 V , V BE = 0.5 V, VCC = 12 V and b = 100.

VCC = 12 V RC = 7.8 kW VC

RB

A

RE

20 kW

23. In the circuit shown in figure, find the value of RC . 12 V A

RC

100 kW

C E

20 kW

b = 100 VBE = 0.5 V VCE = 3V

RE =1 kW

Answers Introductory Exercise 35.1 1. (c)

Introductory Exercise 35.2 1. (c)

2. (c)

Introductory Exercise 35.3 1. (b,c)

3. Vi = 0.01 V , ib = 10 mA

2. (c)

Introductory Exercise 35.4

1. Y (Output) t1

t2

t3

t4

t5

t6

2. (a) AND (b) OR

Exercises 1. (d) 7. (b,c) 13. 2 V

2. (b)

3. (b)

8. (a,b,c)

9. (b,d)

4. (d) 10. (a,d)

14. No

15. (i) Zener junction diode and solar cell (ii) Zener breakdown voltage 16. Both readings will decrease 18. AND gate 19. 10 W 20. I1 = 0.05 A, I2 = 0.025 A, I3 = 0, I4 = 0.025 A 21. IB = 25 mA, IC = 3.33 mA, b = 133 22. VE = 1. 2 V , RB = 108 kW , RE = 1. 2 kW 23. 0.56 kW

5. (a,c) 11. No

6. (a,c) 12. OR gate

36.1

Introduction

36.2

Different Terms Used in Communication System

36.3

Bandwidth of Signals

36.4

Bandwidth of Transmission Medium

36.5

Propagation of Electromagnetic Waves or Communication Channels

36.6

Modulation

36.7

Amplitude Modulation

36.8

Production of Amplitude Modulated Wave

36.9

Detection of Amplitude Modulated Wave

416 — Optics and Modern Physics

36.1 Introduction Communication refers to the transfer of information or message from one point to another point. In modern communication systems, the information is first converted into electrical signals and then sent electronically. This has the advantage of speed, reliability and possibility of communicating over long distances. We are using these every day such as telephones, TV and radio transmission, satellite communication etc. Historically, long distance communication started with the advent of telegraphy in early nineteenth century. The milestone in trans-atlantic radio transmission in 1901 is credited to Marconi. However, the concept of radio transmission was first demonstrated by Indian physicist JC Bose. Satellite communication started in 1962 with the launching of Telstar satellite. The first geostationary satellite Early Bird was launched in 1965. Around 1970, optical fibre communication entered in USA, Europe and Japan. The basic units of any communication systems are shown in Fig. 36.1 The transmitter is located at one place. The receiver is located at Information source some other place. Transmission channel connects the transmitter and the receiver. A channel may be in the form of wires or cables Transmitter or it may be wireless. Transmitter converts message signals produced by the source of information into a form suitable for Transmission transmission through the channel. Noise channel In any communication system, a non-electrical signal (like voice signal) is first converted into an electrical signal by a device called Receiver transducer. Most of the speech or information signal cannot be Fig. 36.1 Block diagram of directly transmitted to long distances. For this an intermediate step of communication system modulation is necessary in which the information signal is loaded or superimposed on a high frequency wave which acts as a carrier wave.

Extra Points to Remember ˜

˜

˜

There are basically two communication modes : point to point and broadcast. Point to point In this mode, communication takes place between a single receiver and transmitter. For example : telephonic call between two persons is a point to point communication. Broadcast In this mode, there are a large number of receivers corresponding to a single transmitter. Radio and television are examples of this type of communication.

36.2 Different Terms Used in Communication System Following basic terminology is used in any communication system. Now let us discuss them in detail. Electrical Transducer As discussed earlier also a transducer converts a non-electrical signal (like a voice signal) into an electrical signal. Signal Any information in electrical form suitable for transmission is called a signal. Signals can be either analog or digital. Analog signals are continuous variations of voltage or current. Sine functions of time are fundamental analog signal. Digital signals are those which can take only discrete values. Binary system is extensively used in digital electronics. In binary system 0 corresponds to low level and 1 corresponds to high level of voltage or current.

Chapter 36

Communication System — 417

Noise Unwanted signals which are mixed with the main signals are referred as noise. Transmitter A transmitter makes the incoming message signal suitable for transmission through a channel. Receiver The signal sent by transmitter through channels is received by the receiver. Attenuation When the signal propagates from transmitter to receiver it loses some strength and it becomes weaker. This is known as attenuation. Amplification The signal received by receiver is weaker than the signal sent by transmitter (due to attenuation). The amplitude of this signal is increased by an amplifier. The energy needed for additional signal is obtained from a DC power source. Range This is the largest distance from the transmitter up to which signal can be received with sufficient strength. Bandwidth This is the width of the range of frequencies that an electronic signal uses on a given transmission medium. It is expressed in terms of the difference between the highest frequency signal component and the lowest frequency signal component. Modulation The low frequency message signals cannot be transmitted to long distances by their own. They are superimposed on a high frequency wave (also called a carrier wave). This process is called modulation. Demodulation This is reverse process of modulation. At the receiver end information is retrieved from the carrier wave. This process is known as demodulation. Repeater Repeaters are used to extend the range of a communication system. It is a combination of a receiver and a transmitter. Receiver (or a repeater) first receives the original signals, then amplifies it and retransmits it to other places (sometimes with a different carrier frequency).

36.3 Bandwidth of Signals Message signals (such as voice, picture or computer data) have different range of frequencies. The type of communication system depends on the bandwidth (discussed in the above article). Some frequency range and their corresponding bandwidth are given below. (i) For telephonic communication A bandwidth of 2800 Hz is required. As, the signals range from 300 Hz to 3100 Hz and their difference is 2800 Hz. (ii) For music channels A bandwidth of approximately 20 kHz is required. Because, the audible range of frequencies extends from 20 Hz to 20 kHz and their difference is approximately 20 kHz. (iii) For TV signals A TV signal consists both audio and video. A bandwidth of approximately 6 MHz is required for its transmission.

36.4 Bandwidth of Transmission Medium Like bandwidths of message signals different types of transmission media offer different bandwidths. Commonly used transmission media are optical fibres, free space and wire. The International Telecommunication Union (ITU) administers the present system of frequency allocations. (i) Coaxial cables offers a bandwidth of approximately 750 MHz. (ii) Optical fibres offers a frequency range of 1 THz to 1000 THz.

418 — Optics and Modern Physics (iii) Communication through free space (using radio waves) offers a bandwidth varying from few hundreds of kHz to a few GHz. These frequencies are further subdivided for various services as given in following table. Table 36.1 S.No.

Service

Frequency Bands

1.

AM radio broadcast

540 - 1600 kHz

2.

FM radio broadcast

88 - 108 MHz

3.

Television

54 - 890 MHz

4.

Cellular Phones

840 - 935 MHz

5.

Satellite communication

37 . - 6.425 GHz

36.5 Propagation of Electromagnetic Waves or

Communication Channels Physical medium through which signals propagate between transmitting and receiving station is called the communication channel. There are basically two types of communications. (i) Space communication (ii) Line communication As per syllabus, we are here discussing only space communication.

Space Communication Consider two friends playing with a ball in a closed room. One friend throws the ball (transmitter) and the other receives the ball (receiver). There are three ways in which the ball can be sent to the receiver. (a) By rolling it along the ground (b) Throwing directly and (c) Throwing towards roof and then reflected towards the receiver. Similarly, there are three ways of transmitting an information from one place to the other using physical space around the earth.

Line Communication (a) Along the ground (ground waves). (b) Directly in a straight line through intervening topographic space (space wave, or tropospheric wave or surface wave) and (c) Upwards in sky followed by reflection from the ionosphere (sky wave). These three modes are discussed below. (i) Ground Wave or Surface Wave Propagation Information can be transmitted through this mode when the transmitting and receiving antenna are close to the surface of the earth. The radio waves which progress along the surface of the earth are called ground waves or surface waves. These waves are vertically polarised in order to prevent short-circuiting of the electric component. The electrical field due to the wave induce charges in the earth's surface as shown in figure. As the wave travels, the induced charges in the earth also travel along it. This constitutes a

Chapter 36

Communication System — 419

current in the earth's surface. As the ground wave passes over the surface of the earth, it is weakened as a result of energy absorbed by the earth. Due to these losses, the ground waves are not suited for very long range communication. Further these losses are higher for high frequency. Hence, ground wave propagation can be sustained only at low frequencies (500 kHz to 1500 kHz). + + + + + E + + + + + -

+ + + + + + + +

+ +

+ + +

-

-

+

-

l/2 ++ ++ ++ + ++ + ++ + ++ + ++ + ++ ++ ++ + ++

+ + + + +

+ +

Þ Direction of travel - -EARTH

Fig. 36.2. Vertically polarised wave travelling over the surface of the earth. The solid lines represent the electric field (E) of the electromagnetic wave.

Space Wave Propagation or Tropospheric Wave Propagation Television signal (80 MHz to 200 MHz) waves neither follow the curvature of the earth nor get reflected by ionosphere. Surface wave or sky wave cannot be employed in television communication. Television signals can be reflected from geostationary satellite or tall receiver antennas. Q

d

d P

S 90° R

R

T

R

O Fig. 36.3

Height of Transmitting Antenna The transmitted waves, travelling in a straight line, directly reach the receiver end and are then picked up by the receiving antenna as shown in figure. Due to finite curvature of the earth, such waves cannot be seen beyond the tangent points S and T. Suppose h is the height of antenna PQ. Let R be the radius of earth. Further, let QT = QS = d, PQ = h, OQ = R + h From the right angled triangle OQT, OQ 2 = OT 2 + QT 2

420 — Optics and Modern Physics ( R + h) 2 = R 2 + d 2

\ \

d 2 = h 2 + 2Rh

Since,

R >> h, h 2 + 2Rh » 2Rh

\

d » 2Rh

This distance is of the order of 40 km. Area covered for TV transmission A = pd 2 = 2 pRh If height of receiving antenna is also given in the question, then the maximum line of sight distance d M is given by d M = 2RhT + 2RhR where,

hT = height of transmitting antenna

and

hR = height of receiving antenna population covered = population density ´ area covered.

Further

dT

dM

hT

hR

Fig. 36.4 Line of sight communication by space waves V

Example 36.1 A TV tower has a height of 60 m. What is the maximum distance and area up to which TV transmission can be received? (Take radius of earth as 6.4 ´ 10 6 m.) Solution

(i) Distance d = 2Rh = 2 ´ 6.4 ´ 106 ´ 60 m = 27.7 km 2

Ans. 6

(ii) Area covered = pd = 2pRh = ( 2 ´ 3.14 ´ 6.4 ´ 10 ´ 60) m

2

= 2411 km 2

Sky Wave Propagation or Ionospheric Propagation If one wishes to send signals at far away stations, then either repeater transmitting stations are necessary or height of the antenna is to be increased. However much before the advent of satellites, radio broadcast covered long distances by the reflection of signals from the ionosphere. This mode of transmission is called ionospheric propagation or sky wave propagation. T ® Transmitter, R ® Receiver The ionosphere extends from a height of 80 km to 300 km. The refractive index of ionosphere is less than its free space value. That is, it behaves as a rare medium. As, we go deep into the

Ans. Ionosphere

R1 T

R2 R3

Earth

Fig. 36.5

Chapter 36

Communication System — 421

ionosphere, the refractive index keeps on decreasing. The bending of beam (away from the normal) will continue till it reaches critical angle after which it will be reflected back. The different points on earth receive signals reflected from different depths of the ionosphere. There is a critical frequency f c (5 to 100 MHz ) beyond which the waves cross the ionosphere and do not return back to earth. Communication satellite Ionosphere

Transmitter

Receiver

Fig. 36.6 Principle of satellite communication Satellite-1 Satellite Communication Long distance communication Geostationary orbit beyond 10 to 20 MHz was not possible before 1960 because all the three modes of communication discussed above failed (ground waves due to conduction losses, space wave due to Earth limited line of sight and sky wave due to the penetration of the ionosphere by the high frequencies beyond f c ). Satellite communication made this possible. The basic principle of satellite communication is shown in Satellite-3 Satellite-2 figure. A communication satellite is a spacecraft placed in an orbit around the earth. The frequencies used in satellite Fig. 36.7 communication lie in UHF/microwave regions. These waves can cross the ionosphere and reach the satellite. Polar orbit For steady, reliable transmission and reception it is Highly elliptical (circular) (inclined) orbit preferred that satellite should be geostationary. A geostationary satellite is one that appears to be stationary relative to the earth. It has a circular orbit lying in the equatorial plane of the earth at an approximate height of 36,000 km. Its time period is 24 hours. If we use three geostationary satellites placed at the Geostationary orbit(circular) vertices of an equilateral triangle as shown in figure. The entire earth can be covered by the communication Fig. 36.8 A schematic diagram of various satellite orbits used in satellite communication network. In addition to geostationary equatorial orbits, there are two more orbits which are being used for communication. These are (a) Polar circular orbit This orbit passes over or very close to the poles. It is approximately at a height of 1000 km from earth. (b) Highly elliptical (inclined) orbits

422 — Optics and Modern Physics Remote Sensing Remote sensing is an application of satellite communication. It is the art of obtaining information about an object or area acquired by a sensor that is not in direct contact with the target of investigation. Any photography is a kind of remote sensing. If we want to cover large areas for which information is required, we have to take photographs from larger distances. This is called aerial photography. Town and country planning can also be done by remote sensing. A satellite equipped with appropriate sensors is used for remote sensing. Taking photograph of any object relies on the reflected wave from the object. We use visible light in normal photography. In principle, waves of any wavelength in the electromagnetic spectrum can be used for this purpose by using suitable sensors. Some applications of remote sensing include meteorology (development of weather systems and weather forecasting), climatology (monitoring climate changes), and oceanography etc.

36.6 Modulation In this section, we will discuss in detail about modulation. What is it ? What is the need of modulation or how the modulation is done etc. No signal in general is a single frequency signal but it spreads over a range of frequencies called the signal bandwidth. Suppose we wish to transmit an electronic signal in the Audio Frequency (20 Hz-20 kHz) range over a long distance. Can we do it ? No it cannot because of the following problems. (i) Size of antenna For transmitting a signal we need an antenna. This antenna should have a size comparable to the wavelength of the signal. For an electromagnetic wave of frequency 20 kHz, the wavelength is 15 km. Obviously such a long antenna is not possible and hence direct transmission of such signal is not practical. 1 (ii) Effective power radiated by antenna Power radiated by an antenna µ × (l ) 2 Therefore, power radiated by large wavelength would be small. For good transmission, we require high power and hence need of high frequency transmission is required. Time period T (iii) Mixing up of signal from different transmitters w = 2p T Another problem in transmitting baseband signals Amplitude directly is of intermixing of different signals. Time Suppose many people are talking at the same time or (a) Sinusoidal Pulse many transmitters are transmitting baseband duration information signals simultaneously. All these signals will get mixed and there is no simple way to Pulse Pulse Pulse fall distinguish between them. A possible solution to all rise amplitude above problems is using communication at high frequencies and allotting a band of frequencies to (b) Pulse shaped signals each message signal for its transmission. Fig. 36.9 Thus in the process of modulation the original low frequency information signal is attached with the high frequency carrier wave. The carrier wave may be continuous (sinusoidal) or in the form of pulses as shown in figure.

Chapter 36

Communication System — 423

Modulation Types Different types of modulation depend upon the specific characteristic of the carrier wave which is being varied in accordance with the message signal. We know that a sinusoidal carrier wave can be expressed as E = E 0 sin (wt + f ) 1 0 –1 0

0.5

1

1.5

2

2.5

3

2.5

3

(a) A sinusoidal carrier wave 1 0 –1 0

0.5

1

1.5

2

(b) A modulating signal 1 0 –1 0

0.5

2 1 1.5 (c) Amplitude modulation

2.5

3

0.5

2 1 1.5 (d) Frequency modulation, and

2.5

3

1 0 –1 0 1 0 –1 0

0.5

1

1.5

2

2.5

time

3

(e) Phase modulation

Fig. 36.10 Modulation of a carrier wave

The three distinct characteristics are Amplitude ( E 0 ), angular frequency (w) and phase angle (f ). Either of these three characteristics can be varied in accordance with the signal. The three types of modulation are, amplitude modulation, frequency modulation and phase modulation. Similarly, the characteristics of a pulse are, Pulse amplitude, pulse duration or pulse width and pulse position (time of rise or fall of the pulse amplitude). Hence, different types of pulse modulation are, Pulse Amplitude Modulation (PAM), Pulse Duration Modulation (PDM) or Pulse Width Modulation (PWM) and Pulse Position Modulation (PPM). In this chapter, we shall confine to amplitude modulation (of continuous wave or sinusoidal wave) only.

36.7 Amplitude Modulation In this type of modulation, the amplitude of the carrier signal varies in accordance with the information signal. The high frequency carrier wave (Fig.a) is superimposed on low frequency information signal (Fig. b). As a result, in the amplitude modulated carrier wave, amplitude no longer

424 — Optics and Modern Physics remains constant, but its envelope has similar sinusoidal variation as that of the low frequency or modulating signal. The carrier wave frequency ranges from 0.5 to 2.0 MHz. AM signals are noisy because electrical noise signals significantly affect this. Let C = Ac sin wc t represents a carrier wave and S = A s sin ws t represents the signal wave

(a)

Then after making calculations we see that the modulated signal wave equation can be written as mA (b) m = Ac sin wc t + c cos (wc - ws ) t 2 m Ac ...(i) cos (wc + ws ) t 2 A where, m = s is called the modulation index. In practice, Ac (c) m is kept £ 1 to avoid distortion. Fig. 36.11 In Eq. (i), wc - ws and wc + ws are respectively called the lower side and upper side frequencies. The modulated signal therefore consists of the carrier wave of frequency wc plus two sinusoidal waves each with a frequency slightly different from wc , known as side bands. Ac Amplitude mA c 2

(wc – ws )

wc

(wc + ws )

w in radians

Fig. 36.12 A plot of amplitude versus w for an amplitude modulated signal V

Example 36.2 A message signal of frequency 10 kHz and peak voltage of 10 V is used to modulate a carrier wave of frequency 1 MHz and peak voltage of 20 V . Determine (a) modulation index, (b) the side bands produced. Solution (a) Modulation Index

(b) Side Bands

m=

A s 10 = = 0.5 A c 20

Ans.

1 MHz = 1000 kHz

The side bands are, ( wc - ws ) and ( wc + ws ) or we can write ( f c - f s ) and ( f c + f s ) \

( f c - f s ) = (1000 - 10) = 990 kHz and ( f c + f s ) = (1000 + 10) = 1010 kHz

Ans.

Chapter 36

Communication System — 425

36.8 Production of Amplitude Modulated Wave Amplitude modulation can be produced by a variety of methods. A simple method is shown in the block diagram of Fig.36.13. Carrier Wave Ac sin wct

Modulating Signal As sin wst x

Square law device y AM wave means amplitude modulated wave

Bandpass filter AM wave Power Amplifier

To Transmitting Antenna

Fig. 36.13

C ( Ac2 + As2 ) and sinusoids of 2 frequencies ws , 2ws , 2wc , wc - ws and wc + ws . As shown in block diagram, this signal y is passed through a band pass filter which rejects DC and the sinusoids of frequencies ws , 2ws and 2wc . After bandpass filter, the frequencies remaining are wc , wc - ws and wc + ws . The output (AM wave) of the band pass filter therefore is of the same form as Eq. (i) of previous article. In the y function shown in block diagram there is a DC term

It is further to be noticed that this modulated signal cannot be transmitted as such. This signal is passed through a power amplifier and then the signal is fed to transmitter antenna.

36.9 Detection of Amplitude Modulated Wave Signal received from the receiving antenna is first passed through an amplifier because the signal becomes weak in travelling from transmitting antenna to receiving antenna. For further processing, the signal is passed through intermediate frequency (IF) stage preceding the detection. At this stage, the carrier frequency is usually changed to a lower frequency. The output signal from detector may not be strong enough. So, it is further passed through an amplifier for final use. The block diagram of all steps is shown in Fig. 36.14. Receiving antenna

Received signal

Amplifier

IF Stage

Detector

Fig. 36.14 Block diagram of a receiver

Amplifier

Output for final use

426 — Optics and Modern Physics Inside the Detector Detection is the process of recovering the signal from the carrier wave. In the previous two articles, we have seen that the modulated carrier wave contains the frequencies wc, wc + ws and wc - ws . In order to obtain the original message signal S ( = As sin ws t ) of angular frequency ws a simple method is shown in the form of a block diagram as shown below. From IF stage

Envelope detector

Rectifier

To amplifier

Time

Time AM input wave

S(t)

Rectified wave

Time Output

Fig. 36.15 Block diagram of a detector for AM signal

Note that the quantity on y-axis can be current or voltage.

Extra Points to Remember ˜

˜

The Internet Everyone is well aware with internet. It has billions of users worldwide. It was started in 1960's and opened for public use in 1990's. Its applications include, E-mail, file transfer, website, E-commerce and chatting etc. Facsimile (FAX) It first scans the image of contents of a document. Then those are converted into electronic signals. These electronic signals are sent to another FAX machine using telephone lines. At the destination, signals are reconverted into a replica of the original document.

Solved Examples V

Example 1 Name the device fitted in the satellite which receives signals from Earth station and transmits them in different directions after amplification. Solution Transponder.

V

Example 2 An electromagnetic wave of frequency 28 MHz passes through the lower atmosphere of Earth and gets incident on the ionosphere. Shall the ionosphere reflects these waves? Solution Yes. The ionosphere reflects back electromagnetic waves of frequency less than 30 MHz.

V

Example 3 Which waves constitute amplitude-modulated band? Solution Electromagnetic waves of frequency less than 30 MHz constitute amplitude-modulated band.

V

Example 4 Give the frequency ranges of the following (i) High frequency band (HF) (ii) Very high frequency band (VHF) (iii) Ultra high frequency band (UHF) (iv) Super high frequency band (SHF). Solution

(i) 3 MHz to 30 MHz (ii) 30 MHz to 300 MHz (iii) 300 MHz to 3000 MHz (iv) 3000 MHz to 30,000 MHz. V

Example 5 State the two functions performed by a modem. Solution (i) Modulation (ii) Demodulation.

V

Example 6 Why is the transmission of signals using ground waves restricted up to a frequency of 1500 kHz? Solution This is because at frequencies higher than 1500 kHz, there is an increase in the absorption of signal by the ground.

V

Example 7 How does the effective power radiated by an antenna vary with wavelength? æ1ö è lø

2

Solution Power radiated by an antenna µ ç ÷ . V

Example 8 Why is it necessary to use satellites for long distance TV transmission? Solution Television signals are not properly reflected by the ionosphere. So, reflection is affected by satellites.

V

Example 9 Why long distance radio broadcasts use shortwave bands? Solution This is because ionosphere reflects waves in these bands.

V

Example 10 What is a channel bandwidth? Solution

Channel bandwidth is the range of frequencies that a system can transmit with efficient fidelity.

428 — Optics and Modern Physics V

Example 11 Give any one difference between FAX and e-mail systems of communication. Solution

Electronic reproduction of a document at a distant place is known as FAX. In e-mail system, message can be created, processed and stored. Such facilities are not there in Fax system. V

Example 12 Why ground wave propagation is not suitable for high frequency? Solution At high frequency, the absorption of the signal by the ground is appreciable. So, ground wave propagation is not suitable for high frequency.

V

Example 13 What is the purpose of modulating a signal in transmission? Solution A low frequency signal cannot be transmitted to long distances because of many practical difficulties. On the other hand, effective transmission is possible at high frequencies. So, modulation is always done in communication systems.

V

Example 14

What is a transducer?

Solution A device which converts energy in one form to another is called a transducer. V

Example 15 Why do we need a higher bandwidth for transmission of music compared to that for commercial telephone communication? Solution

As compared to speech signals in telephone communication, the music signals are more complex and correspond to higher frequency range. V

Example 16 From which layer of the atmosphere, radio waves are reflected back? Solution The electromagnetic waves of radio frequencies are reflected by ionosphere.

V

Example 17 Why sky waves are not used in the transmission of television signals? Solution The television signals have frequencies in 100-200 MHz range. As the ionosphere cannot reflect radio waves of frequency greater than 40 MHz back to the earth, the sky waves cannot be used in the transmission of TV signals.

V

Example 18 Why are short waves used in long distance broadcasts? Solution

The short waves (wavelength less than 200 m or frequencies greater than 1,00 kHz) are absorbed by the earth due to their high frequency but are effectively reflected by Flayer in ionosphere. After reflection from the ionosphere, the short waves reach the surface of earth back only at a large distance from the transmitter. For this reason, short waves are used in long distance transmission. V

Example 19 Define the term critical frequency in relation to sky wave propagation of electromagnetic waves. Solution

The highest value of the frequency of radio waves, which on being radiated towards the ionosphere at some angle are reflected back to the earth is called critical frequency.

V

Example 20 What mode of communication is employed for transmission of TV signals? Solution Space wave communication.

Exercises Single Correct Option 1. Three waves A, B and C of frequencies 1600 kHz, 5 MHz and 60 MHz, respectively are to be transmitted from one place to another. Which of the following is the most appropriate mode of communication? (a) (b) (c) (d)

A is transmitted via space wave while B and C transmitted via sky wave. A is transmitted via ground wave, B via sky wave and C via space wave. B and C are transmitted via ground wave while A is transmitted via sky wave. B is transmitted via ground wave while A and C are transmitted via space wave.

2. A 100 m long antenna is mounted on a 500 m tall building. The complex can become a transmission tower for waves with l (a) ~ 400 m (c) ~ 150 m

(b) ~ 25 m (d) ~ 2400 m

3. A speech signal of 3 kHz is used to modulate a carrier signal of frequency 1 MHz, using amplitude modulation. The frequencies of the side bands will be (a) 1.003 MHz and 2.997 MHz (c) 1003 kHz and 1000 kHz

(b) 3001 kHz and 2997 kHz (d) 1 MHz and 0.997 MHz

4. A message signal of frequency wm is superposed on a carrier wave of frequency wc to get an amplitude modulated wave (AM). The frequency of the AM wave will be (a) wm w + wm (c) c 2

(b) wc w - wm (d) c 2

5. A basic communication system consists of (A) transmitter (B) information source (C) user of information (D) channel (E) receiver Choose the correct sequence in which these are arranged in a basic communication system. (a) ABCDE (b) BADEC (c) BDACE (d) BEADC

6. Which of the following frequencies will be suitable for beyond the horizon communication using sky waves? (a) 10 kHz (c) 1 GHz

(b) 10 MHz (d) 1000 GHz

7. Frequencies in the UHF range normally propagate by means of (a) ground waves (c) surface waves

8. Digital signals (i) do not provide a continuous set of values (ii) represent values as discrete steps (iii) can utilize binary system and (iv) can utilize decimal as well as binary systems

(b) sky waves (d) space waves

430 — Optics and Modern Physics Which of the above statements are true? (a) (i) and (ii) only (c) (i), (ii) and (iii) but not (iv)

(b) (ii) and (iii) only (d) All of (i), (ii), (iii) and (iv)

More than One Correct Options 9. A TV transmission tower has a height of 240 m. Signals broadcast from this tower will be received by LOS communication at a distance of (assume the radius of earth to be 6.4 ´ 106m) (a) 100 km (c) 55 km

(b) 24 km (d) 50 km

10. An audio signal of 15 kHz frequency cannot be transmitted over long distances without modulation because (a) (b) (c) (d)

the size of the required antenna would be at least 5 km which is not convenient the audio signal cannot be transmitted through sky waves the size of the required antenna would be at least 20 km, which is not convenient effective power transmitted would be very low, if the size of the antenna is less than 5 km

11. Audio sine waves of 3 kHz frequency are used to amplitude modulate a carrier signal of 1.5 MHz. Which of the following statements are true? (a) (b) (c) (d)

The sideband frequencies are 1506 kHz and 1494 kHz The bandwidth required for amplitude modulation is 6 kHz The bandwidth required for amplitude modulation is 3 MHz The sideband frequencies are 1503 kHz and 1497 kHz

12. In amplitude modulation, the modulation index m, is kept less than or equal to 1 because (a) m > 1 will result in interference between carrier frequency and message frequency, resulting into distortion. (b) m > 1 will result in overlapping of both sidebands resulting into loss of information. (c) m > 1 will result in change in phase between carrier signal and message signal. (d) m > 1 indicates amplitude of message signal greater than amplitude of carrier signal resulting into distortion.

Subjective Questions 13. Compute the LC product of a tuned amplifier circuit required to generate a carrier wave of 1 MHz for amplitude modulation.

14. A carrier wave of peak voltage 12 V is used to transmit a message signal. What should be the peak voltage of the modulating signal in order to have a modulation index of 75%?

15. Which of the following would produce analog signals and which would produce digital signals? (i) A vibrating tuning fork (iii) Light pulse

(ii) Musical sound due to a vibrating sitar string (iv) Output of NAND gate

16. Two waves A and B of frequencies 2 MHz and 3 MHz, respectively are beamed in the same direction for communication via sky wave. Which one of these is likely to travel longer distance in the ionosphere before suffering total internal reflection?

17. The maximum amplitude of an AM wave is found to be 15 V while its minimum amplitude is found to be 3V. What is the modulation index?

18. Why is an AM signal likely to be more noisy than a FM signal upon transmission through a channel?

Chapter 36

Communication System — 431

19. Is it necessary for a transmitting antenna to be at the same height as that of the receiving antenna for line of sight communication? A TV transmitting antenna is 81 m tall. How much service area can it cover, if the receiving antenna is at the ground level?

20. A TV transmission tower antenna is at a height of 20 m. How much service area can it cover if the receiving antenna is (i) at ground level, (ii) at a height of 25 m? Calculate the percentage increase in area covered in case (ii) relative to case (i).

21. If the whole earth is to be connected by LOS communication using space waves (no restriction of antenna size or tower height), what is the minimum number of antennas required? Calculate the tower height of these antennas in terms of earth's radius?

22. The maximum frequency for reflection of sky waves from a certain layer of the ionosphere is found to be fmax = 9 ( N max )1 2, where N max is the maximum electron density at that layer of the ionosphere. On a certain day it is observed that signals of frequencies higher than 5 MHz are not received by reflection from the F1 layer of the ionosphere while signals of frequencies higher than 8 MHz are not received by reflection from the F2 layer of the ionosphere. Estimate the maximum electron densities of the F1 and F2 layers on that day.

23. A 50 MHz sky wave takes 4.04 ms to reach a receiver via re-transmission from a satellite 600 km above earth's surface. Assuming re-transmission time by satellite negligible, find the distance between source and receiver. If communication between the two was to be done by Line of Sight (LOS) method, what should be the size of transmitting antenna?

Answers 1. (b)

2. (a)

8. (c)

9. (b,c,d)

12. (b,d)

3. (a) 10. (a,b,d)

13. 2.53 ´ 10 -14 s 2

4. (b)

5. (b)

11. (b,d) 14. 9 V

15. (i) analog (ii) analog (iii) digital (iv) digital 16. 3 MHz 17. 2/3 19. No, 3258 km2 20. (i) 804 km2 21. Six, h = radius of earth 23. 170 km, 565 m

(ii) 3608 km2 11

22. 3.086 ´ 10

-3

(iii) 349 %

m , 7.9 ´ 1011 m-3

6. (b)

7. (d)

29. Electromagnetic Waves INTRODUCTORY EXERCISE

29.1

df id = e0 E dt d æ Aö = e0 çE ÷ dt è 2 ø eA d = 0 (E ) 2 dt e A d æ sö ç ÷ = 0 2 dt çè e0 ÷ø

1.

2. There is a change in potential difference between the plates of capacitor. Therefore, change in electric field between its plates. Now, change in electric field produces a magnetic field. E 3. 0 = c B0 E 810 \ B0 = 0 = c 3 ´ 108 = 2.7 ´ 10-6 T

e0 A d æ q ö ç ÷ 2 dt çè Ae0 ÷ø 1 dq = 2 dt i = 2

= 2.7 mT 1 uE = uB = e0E 2 2

=

INTRODUCTORY EXERCISE

1. c =

4.

2

æ dq ö = i÷ çQ è dt ø

29.2

1 = speed of light in vacuum e0m 0

\ Unit of

1 æ E0 ö e0 ç ÷ 2 è 2ø 1 = e0E02 4 1 = (8.86 ´ 10-12 ) (50)2 4 =

= 5.54 ´ 10-9 J/m 3

1 should be the unit of c or speed e0 m 0

Total energy = (uE + uB ) (Given volume) = 2 ´ 5.54 ´ 10-9 ´ (10 ´ 10-4 ) (0.50) = 5.55 ´ 10-12 J

or m/s.

Exercises Single Correct Option 1. 11 eV energy radiation lies in UV range. 4. In case of perfectly non-reflecting surface, Dp = where,

E c E = 20 ´ 30 ´ 30 ´ 60 = 1.08 ´ 106 J

\

1.08 ´ 106 Dp = 3 ´ 108 = 36 ´ 10-4 kg-m/s

More than One Correct Options 5. E, B and velocity of electromagnetic waves are mutually perpendicular.

6. Every accelerated charged particle produces electromagnetic waves. c 3 ´ 108 8. l = = = 0.3 m f 109 This wavelength lies in radio waves region.

Subjective Questions c 3 ´ 108 = 10 m = f 30 ´ 106 c 12. l = f

11. l =

l1 =

3 ´ 108 = 40 m 7.5 ´ 106

l2 =

3 ´ 108 = 25 m 12 ´ 106

Chapter 29 13. c =

E0 B0

Now, E0 = cB0 = (3 ´ 108 ) (510 ´ 10-9 )

= 153 V/m e0 A e0 (pR 2 ) 14. (a) C = = d d (8.86 ´ 10-12 ) (3.14) (0.12)2 = 5 ´ 10-2 = 8 ´ 10-12 F = 8 pF q V = C dV 1 æ dq ö i = ç ÷= \ dt C è dt ø C 0.15 = 8 ´ 10-12 = 1.875 ´ 1010 V/s (b) id = ic = 0.15 A E 15. (a) c = 0 B0 \

E0 c w = 2pf

2

= E0 =

\

=

Now, or

18. (a)

or

1 æ e0 ö e0 ç ÷ c 2 è 2ø 2I e0c 2 ´ 2.5 ´ 1014 8.86 ´ 10-12 ´ 3 ´ 108

= 4.3 ´ 108 N/C E c= 0 B0 E0 B0 = c c l= f E0 c= B0 E0 B0 = c 2

w k

w c 2p l= k

k=

1 wC With decrease in frequency, X C will increase, so current will decrease.

16. X C =

id = ic ic has decreased, so id will also decrease. I 1 17. I E = = e0 E 2c 2 2

(b)

B0 =

c=v = \

Electromagnetic Waves — 435

(c) uE =

1 1 æE ö 1 e0E 2 = e0 ç 0 ÷ = e0E02 2 2 è 2ø 4

Similarly,

uB =

B02 4m 0

After substituting the values we get, uE = uB dq 19. id = ic = dt = - 2pq0 f sin (2pft )

30. Reflection of Light INTRODUCTORY EXERCISE

30.1 hmin 2 = 10 3 20 hmin = m 3 hmax 4 = 10 3 40 hmax = m 3

1. 2 m/s

\

2 m/s

O

I

vOI = 4 m/s

\

2.

3. i 2q 90°

h max 2m

h min

q

q 3m M

4m Person

2m 3 m M¢

q

i = 2q

Exercises LEVEL 1

For normal incidence, i = 0° \ d = 180°

Assertion and Reason 1. Convex mirror can make real images of virtual objects.

Objective Questions 1. Image is real and incident beam is convergent.

2. F O

I

20 cm

20 cm

Real object is in front of mirror, not at F. So, image is not at infinity. 4. Field of view of convex mirror is large. 5. m = - 2, means image is real, inverted an 2-times magnified.

6. i

i

d

d = 180° - 2 i

4.

1 1 1 + = v u f

…(i)

For virtual object u is positive and for concave mirror f is negative. Substituting these signs in Eq. (i), we can see that v is always negative or image is always real.

5. _ ¥

C

F

O

+¥

Reflection of Light — 437

Chapter 30 S.No.

u

v

1.

From O to F or from O From O to + ¥ to - f

2.

From F to C from - f to - 2 f

3.

From C to - ¥ or from From C to F or from - 2 f to - f - 2 f to - ¥

9. Magnified image is formed only by concave mirror. But this image may be real or virtual.

From - ¥ to C or from - ¥ to - 2 f x 3x

In the above example, we have seen that for virtual object (u = + ve), image is always real (v = - ve) corresponding graph is shown in Fig. (b).

For real image, 2x = 80 cm x = 40 cm 1 1 1 + = - 120 - 40 f

\

M

6.

1 q 2

20° 20°

90° – q

q

70° 70°

70° P

O

90° - q + 70° + 70° = 180° \

Solving, we get x = (n - 1) f 1 1 1 11. + = v - 60 - 24

q = 50°

7. A

Solving, we get f = - 30 cm Similarly, we can check for virtual image. 10. Let, u = - x x Then, v=+ n 1 1 1 + = (+ x / n) - x + f

\

B x

2x

R f = = 30 cm 2 1 1 1 + = + x - 2x + 30 \ Hence,

v = - 40 cm v (40) -2 m=- == u (- 60) 3

x = 15 cm AB = 3x = 45 cm

Image speed is m2-times the object speed and opposite to the direction of object velocity. 12. For real objects image formed by a convex mirror is always virtual, erect and diminished. 10 m/s

13. 4 cm/s P

8. M 1.4 m

2 cm/s A

0.8 m

O C

B

P x

x

MP AB = PC BC MP 0.8 \ = 2x x or MP = 1.6 m > 1.4 m Hence, the boy cannot see his feet.

vPM = 14 cm/s towards right vIM = 14cm/s towards left So, actual speed of image = 24 cm/s towards left \ vIO = 26 cm/s towards left.

Subjective Questions 1. Reflected rays are neither converging nor diverging. Hence, mirror is a plane.

438 — Optics and Modern Physics 2. Image distance from plane mirror = object

8.

distance. Lateral magnifications = 1 M

i

I

3. 15° 15°

I

75°

60°

15°

I = incident ray R = reflected ray Angle of incidence = 15° Angle between reflected ray and horizontal = 60° 4. Image from one mirror will behave like object for other mirror. where,

3b

3b

O I4

I2

M2

I1

M1

q–i

q

q–i 90 ° +

5.

i

90° – i

R

I3

N i–q

dTotal = dM + dN = (180° - 2i ) + [180° - 2 (q - i )] = 360° - 2q 1 1 1 9. Apply + = v u f (a) f = - 10 cm, u = - 25 cm Solving, we get v = - 16.7 cm Since, v is negative, image is in front of mirror. So, it is real. Similarly, we can solve for other parts. 10. (a) From O to F , image is real. From F to P, image is virtual.

b

b

b

O

b

OI 1 = OI 2 = 2b I 3 is the image of I 2 from mirror M 1 similarly I 4 is the image of I 1 from mirror M 2. OI 3 = OI 4 = 4 b 6. Given in the theory.

2m C 0.5 m F

30°

0.2 m

30° 30 °

7.

P

At O d 1.6 m

d 1 = tan 30° = 0.2 3 0.2 \ d= 3 1.6 N = =8 3 d = 13.85 Therefore, actual number of reflections required are 14.

0.5 m

\

1 1 1 + = v u f 1 1 1 + = v - 3.0 - 0.5

v = - 0.6 m (b) When object is at C and P, image coincides with object. Using 1 s = gt 2 2 2s or t= g

Reflection of Light — 439

Chapter 30 t=

At C

t=

At P

2´2 = 0.639 s 9.8 2´3 = 0.782 s 9.8

Object

16.5 cm

11. (a) P

C

P

From O to F or from O to + 0.25 m

From O to - ¥

From F to C or from + 0.25 m to + 0.50 m

From - ¥ to C or from + ¥ to + 0.50 m

From C to + ¥ or from + 0.50 cm to + ¥

From C to F or from + 0.50 m to + 0.25 m

15. O is placed at centre of curvature of concave

F

mirror (= 42 cm). Therefore, image from this mirror I 1 will coincide with object O.

11 cm Q'

22 cm

O

I1

1 1 1 (b) Apply + = v u f m=-

and

Image

I2 21 cm

v u

21 cm 42 cm

Now, plane mirror will make its image I 2 at the same distance from itself.

12. (a) Q

Q'

16. O

P

P' F

1 1 1 + = v u f 1 1 1 + = \ - ( y + f ) - (x + f ) - f Solving this equation, we get xy = f 2 Using

1 1 1 + = v u f m=-

and

v u

R = - 18 cm 2 Let u = - x cm 1 x Then, v = - cm for real image of th size. 9 9 1 1 1 Using, + = v u f

13. f =

We have,

1 1 1 + = (- x / 9) (- x ) - 18

Solving we get, x = 180 cm

14. +¥

C

F

0.25 m 0.5 m

O

x y

10cm

(b) Apply,

F

I

_¥

v

17. A

I1

I2 x 2R

For convex mirror 1 1 1 + = v - x + R/2 1 2 1 Rx or v = \ = + v R x R + 2x Now, applying mirror formula for concave mirror we have 1 1 1 + = - (2R - x ) - (2R + v ) - R / 2 Solving this equation, we can find value of x.

440 — Optics and Modern Physics AB = PB - PA = 2(a + L) - 2a = 2L = constant v (- 20) m=- ==-2 u (- 10)

18. (a) Image has to be taken on a screen. So, it should be real. Hence, mirror should be concave. x

(b)

3.

5m

| Dv | = m 2 | Du |

I

O

= (- 2)2 (0.1 cm) = 0.4 cm Object and image travel in opposite directions (along the axis).

Image is 5 times magnified. Hence, |v |= 5|u| or (5 + x ) = 5x Solving we get, x = 1.25 cm Now using, 1 1 1 2 + = = v u f R

3m

4. I2

I1

O x

3–x

We have, 1 1 2 = - 6.25 1.25 R

3–x

x

I 1I 2 = (3 - x ) + (3) + x =4m Solving we get, x = 1 m 5. Image of c will coincide with this.

Solving we get, R = - 2.08 m

M

LEVEL 2

C

Single Correct Option 1.

P

vI

10 cm

60° v0

\

æ k ö ÷ A v0 = vI = wA = çç ÷ è Mø vr = v02 + vI2 - 2v0 cos 120° = 3 A

k M L

2. h P

A a

10 cm

For P :

60°

a a+L

B a+L

1 1 1 + = v - 20 - 5 20 v=cm 3 v (- 20/ 3) 1 m=- ==u (- 20) 3

Length of image of PM : 1 10 I 1 = 10 ´ = cm 3 3 Length of image of PC : 20 10 I 2 = 10 = cm 3 3 I1 \ =1 I2 1 1 1 6. + = v - 15 - 10 \

v = - 30 cm

Reflection of Light — 441

Chapter 30 m=-

v (- 30) ==-2 u (- 15)

| D v | = m2 | Du | = (- 2)2 (2 mm) = 8 mm

7. Let i is the angle between incident ray and original reflected ray. Then, initial angle between them will be 180° - 2 i. When mirror is rotated by 20° , then reflected ray will rotate by 40°. \ 180° - 2 i ± 40° = 45° Solving we get, i = 47.5° or 87.5°

s2 = displacement of mirror 1 = - gt 2 2 \ Vertical distance of particle from mirror, s = s1 - s2 = 0.5 m Hence, distance between particle and its image = 25 = 1 m 13. vOM = vO - vM = (- $i - 3 k$ ) \ v = (- i$ + 3 k$ ) = v - v IM

B

8. 6 cm

C

M

I

M

vI = (- $i + 3 k$ ) + vM = (3 $i + 4 $j + 11 k$ )

or

14.

D x = 2 cm

x=0 P

53°

16° 53°

vI = 5 m /s

vO = 5 m /s

vI = (5 cos 16° ) i$ + (5 sin 16° ) $j = (4.8 $i + 1.4 $j) m/s

12. uy = 2 sin 45° = 1 m/s In vertical direction, s1 = displacement of particle 1 1 = (1) (0.5) - gt 2 = 0.5 - gt 2 2 2

æ 90° + i - q ö r=ç ÷ 2 è ø p = 180° - (90° - r) - 2i æ 90° + i - q ö = 180° - 90° + ç ÷ - 2i 2 è ø

\

q 90° – q

i

q

–q

11.

15. 90° - i q + 2r = 180°

90°

aRHS = a2 Net pulling force 2 mg = = Total mass 3m 2 = g 3 The relative acceleration is therefore : 3 2 17 g+ g= g 4 3 12 10. See the hint of Q.No-7 of subjective questions for Level 1.

m=+ 5 \ v should be positive and 5 times | u | .

°–

9. aLHS

10 cm

2 cm

90

A

PM = AC = AD + DC DB 6 = AD + = 164 + = 167 cm 2 2 Net pulling force 3 mg 3 = a1 = = = g Total mass 4m 4

x = 12 cm

r m i i r p q 90° – r

180° - 4 i + 90° + i - q 2 æ 270° - 3i - q ö =ç ÷ 2 è ø m = p - q = 180° - (90° - i ) - 2i æ 270° - 3i - q ö \ ç ÷ - q = 90° - i 2 è ø Substituting q = 20° , we get i = 30° 16. Ray passing through c is only correct. =

442 — Optics and Modern Physics More than One Correct Options

m=

1. For real image, Let u = - x , then v = - 2x 1 1 1 + = \ -2x - x - 20 Solving, we get x = 30 cm. For virtual image, Let u = - x , then v = + 2x 1 1 1 \ + = 2x - x - 20

Speed of image (in event -1) is m2 times and m times in event-2. 6. In event -1 For concave mirror, 1 1 1 + = v -3 f - f \ v = - 1.5 f For convex mirror, 1 1 1 + = v -3f + f 3 v = f = 0.75 f 4 For plane mirror, v=+ 3f In event-2 For concave mirror, 1 1 1 + = v - 1.5 f - f

x = 10 cm

or

2. Inverted and real image is formed by concave mirror. Let u = - x , then v = + x /2 1 1 1 \ + = + x/2 - x + f \ x=3f Erect and virtual image is formed by convex mirror. x Let u = - x, then v = + 2 1 1 1 + = -x / 2 -x + f

\ v=-3f For convex mirror, 1 1 1 + = v - 1.5 f + f 3f v= = 0.6 f \ 5 For plane mirror, v = + 1.5 f

x=+ f

3.

vI = v q q

v (- 60) ==-2 u ( - 30)

v0 = v

Comprehension Based Questions 1. vr = v 2 + v 2 - 2v × v cos 2 q = 2 v sin q 4. Ray diagram is as shown in figure. O F C O

1 1 1 + = v - 30 - 20 Solving we get,

10 cm

x

O M

10 cm

I

I

5.

y

v = - 60 cm

OM = MI Coordinates of I are (10 cm, - 10 cm). 2. Object is placed at centre at curvature of mirror. Hence, image is at the same point, real, inverted and of same size. Hence, coordinates are (20 cm, 0). 1 1 1 3. + = v - 20 + 10 20 v=+ cm \ 3

Chapter 30 m=-

v (+20/ 3) 1 ==+ u (- 20) 3

æ 1 ö 10 I = 10 ç ÷ = cm è 3ø 3 10 20 x = 10 = cm 3 3 20 80 and y = 20 + = cm 3 3 4. Plane mirror forms image at equal distance on opposite sides. Hence, x = 0, y = 40 cm

Reflection of Light — 443

\ x = 60 cm In the similar manner, other options can be solved.

\

Subjective Questions 1.

Q B 20 cm

S A

Match the Columns 1. (a) m = - 2 means image is real, inverted and 2-times magnified. So, mirror should be concave. Same logic can be given for other options also. 1 1 1 2. + (as u is positive for virtual objects) = v +u f 1 1 1 = \ v f u For plane mirror, f = ¥. So, v is always negative. Hence, image is always real. For concave mirror, f is negative. So, v is again negative. Therefore, image is always real. For convex mirror, f is positive. So, v may be positive or negative. Hence, image may be virtual or real. 3. (a) Image is inverted, real and diminished. Hence, mirror is concave. Same logic can be applied for other options too. 1 1 1 4. (a) + = v - 20 - 20 \

v=µ

v =¥ u Same formulae can be applied for other options too. 5. (a) See the hint of Q.No-1 of more than one correct options section. (b) Half size image is formed only in case of real image. x Let u = - x, then v = 2 1 1 1 Now, + = - x / 2 - x - 20

P

Insect can see the image of source S in the mirror, so far as it remains in field of view of image overlapping with the road. Shaded portion is the field of view, which overlaps with the road upto length PQ. By geometry we can see that, PQ = 3 AB = 60 cm Distance 60 Ans. \ t= = =6s Speed 10 æ1

2. Using mirror formula, çç + èv 1 1 -1 = v 50 25

\

1 1ö = ÷ u f ÷ø

v = - 50 cm m=-

v = -1 u

Ans.

M1 I1

Optic axis of M1

S

m=-

Optic axis of M2

I2

0.5 cm 0.5 cm 0.5 cm 0.5 cm

M2 50 cm

3. Using mirror formula

1 1 1 + = for concave v u f

mirror first, we have 1 1 1 = v 60 -40 or

v = - 120 cm

Rö æ çQ f = ÷ 2ø è

444 — Optics and Modern Physics First image I 1 at 120 cm from concave mirror will act as virtual object for plane mirror. Plane mirror will form real image of I 1 at S.

\ The required time is t = t1 - t2 = 1.7 s

I1

30 cm

60 cm

120 cm

4.

Ans.

5. Applying mirror formula for concave mirror first

30 cm

Ray diagram is shown in figure. Distance between two mirrors is 90 cm.

Time taken to move the boy from G to topmost point and then from topmost point to G will be 2v t2 = = 2.83 s g

Ans.

æ1 1 1ö çç + = ÷÷ we have, èv u f ø 1 1 1 = v 110 -100 v = - 1100 cm AI 1 = 100 cm. Therefore, final image will be real and at distance 100 cm below point A at I 2.

G

1100 cm

I B

100 cm

F

H

45°

E

S

D

A

AI2 = 100 cm

A

I1 I2

C

\

6. In 15 seconds, mirror will rotate 15° in clockwise direction. Hence, the reflected ray will rotate 30° in clockwise direction.

FG IH HS = = BF BH BH æ HS ö FG = ( BF ) ç ÷ è BH ø

A

æ 1.0 ö = (5) ç ÷ è 0.5 ø

3m

30° 30°

60°

= 10 m FC = 2 + 10 = 12 m The boy has dropped himself at point F. So, his velocity is 20 m/s in upward direction. Let us first find the time to move from F to topmost point and then from topmost point to point 1 C. From s = ut + at 2 , we have 2 1 - 12 = (20t ) + (- 10) t 2 2 Solving this equation we get, t1 = 4.53 s. Velocity of boy at point G , v = (20)2 - 2 ´ 10 ´ 10 = 14.14 m/s

(Q v 2 = u2 - 2gh)

3 Ö3 m At t = 15 s

P

3m q y

At time t

Chapter 30 y = 3 tan q dy dq …(i) \ = (3 sec2 q). dt dt dy dq Here, = vP , = 2° per second dt dt 2´p p rad per second = = 180 90 At t = 15 s and q = 60° Substituting the values in Eq. (i), we have ìpü vP = {3 sec2 60°} í ý î 90 þ p = 3´4 ´ 90 2p Ans. = m/s 15 7. (a) Differentiating the mirror formula, (with respect to time)

Reflection of Light — 445

or

1 u u- f = -1= m f f

\

m=

f u- f

Differentiating we have, f du æ dm ö . ç ÷=2 dt (u - f ) dt è ø

Using mirror formula to find u with magnification 1 we get, m= 10 1 1 1 - = u / 10 u 10 or u = 90 m (with sign u = - 90 m) Substituting in Eq. (ii) we have, dm (10) =(- 1) dt (- 90 - 10)2 = 10-3 per second

POLICE

…(ii)

Ans.

u is decreasing at a rate of v - 20 or (21 - 20 ) or 1 m/s du = - 1 m/s \ dt 8. (a) At t = t,

Note

v

20 m/s Thief

1 1 1 + = v u f We get velocity of image, vI = (m)2 vO Here, vO = relative velocity of object with respect to mirror vI = relative velocity of image m = linear magnification Here, vO = (v - 20) m/s vI = 1 cm/s = 0.01 m/s 1 m= 10 Substituting in Eq. (i) we have, 2

æ 1ö 0.01 = ç ÷ (v - 20) è 10 ø \ v = 21 m/s 1 1 1 (b) + = v u f Multiplying with u we get, u u + 1= v f

…(i)

u = - (2 f + x ) = - (2 f + f cos w t ) Using the mirror formula, 1 1 1 + = v u f We have, 1 1 -1 = v 2 f + f cos wt f æ 2 + cos wt ö ÷÷ f v = - çç è 1 + cos wt ø i.e. distance of image from mirror at time t is æ 2 + cos wt ö çç ÷÷ f Ans. è 1 + cos wt ø (b) Ball coincides with its image at centre of curvature, i.e. at x = 0. (c) At t = T /2 \

or wt = p, x = - f i.e. u = - f or ball is at focus. So, its image is at ¥. m=¥

446 — Optics and Modern Physics æ y - y1 ö çç ÷÷ = - tan b è x - x1 ø

9. Let the ray is incident at a point P = (x1 , y1 ) on the mirror. Then, slope at P, y

= - tan 2q = P

a

q q

\

b

q F

1 æ dy ö tan a = ç ÷ = è dx ø (x1 y1) 4 by1

x

a = 90° - q and b = 2q Now, the reflected ray is passing through P (x1 y1 ) and has a slope - tan b. Hence, the equation will be

2 cot a (x1 - x ) 1 + cot 2 a

…(ii)

x1 = 4 by12

…(iii)

At F , x=0 From Eq. (i) to Eq. (iv),

…(iv)

Further,

1 8b It shows that the coordinates of F are unique æ 1 ö ç , 0÷ . è 8b ø we get

…(i)

y - y1 =

-2 tan q 1 - tan 2 q

x=

Hence, the reflected ray passing through one focus 1 Hence proved. and the focal length is . 8b

31. Refraction of Light INTRODUCTORY EXERCISE

31.1

m 2 m 1 m 2- m 1 , we get = v u R 1 1.5 1 - 1.5 = v - 10 - 15

Using

1. 1m 2 ´ 2m 3´ 3m 1 = 1 4 3 1 ´ = = 1m 3 or 3 2 3m 1

\

1m 3

=2

8

3. m =

c c 3 ´ 10 = 1.67 = = v f l 6 ´ 1014 ´ 300 ´ 10-9

INTRODUCTORY EXERCISE

Solving, we get

3. O

m2 m1 m2 -m1 , we get = v u R 1.5 1.0 1.5 - 1.0 = v - 20 +6 Solving, we get v = + 45cm Similarly other parts can be solved.

31.2

(a) Using

m 2 sin i1 sin 60° = = = 3 m 1 sin i2 sin 30° l 2. 1m 2 = 1.5 = 1 l2

1. 1m 2 =

INTRODUCTORY EXERCISE

31.3

4.

10 cm

10 50 = cm 1.5 3 = 10 + (1.5)(10) = 25 cm

O

1. (a) dapp = 10 + (b) happ

INTRODUCTORY EXERCISE

31.4

æ æ 1ö 1ö 1. Total shift = çç1 - ÷÷ t1 + çç1 - ÷÷ t2 m1ø m2ø è è 2ö 1ö æ æ = ç1 - ÷ 10 + ç1 - ÷ 10 3ø 2ø è è 25 = cm 3 25 125 = cm \ Image distance = 50 3 3

INTRODUCTORY EXERCISE

v = - 8.57 cm

Using

m 2 m 1 m 2 -m 1 , we get = v u R 1 4 /3 1 - 4 /3 = v - 10 - 15

Solving, we get v = - 9.0 cm

5.

31.5

Using the equation m2 m1 m2- m1 , we get = v u R 1.44 1.0 1.44 - 1.0 = v ¥ + 1.25

1. All rays starting from centre pass undeviated as they fall normal to the surface.

2. +ve

\

10 cm

v = 4.0 cm

0

INTRODUCTORY EXERCISE

C

1.

æ 1 1 1 1 1ö - = = (m - 1) çç - ÷÷ v u f R R è 1 2ø

31.6

448 — Optics and Modern Physics \

- v -2 × dv + u-2 × du = 0

æ 1 1 1 1 ö ÷÷ = (1.65 - 1) çç - 20 - 60 è -R + Rø

1 1 1 - = - 50 u + 30

2.

æ v dv = çç - 2 ÷÷ × du è u ø

\

Solving we get, R = 39 cm

8. It is just like a concave mirror. | f | = 0.2 m \ | R | = 0.4 m Focal length of this equivalent mirror is

Solving, we get u = - 18.75 cm v (- 50) m= = = 2.67 u (- 18.75)

+ ve

R1

1 2 (m 2 /m 1 ) 2 (m 2 /m 1 - 1) (extra points) = F R2 R1 2 (4 / 3) 2 (4 / 3 - 1) = - 0.4 + 0.4

+ ve

R1

R2

R2 f2

f1

æ 1 1 1 ö ÷ = (m - 1) çç f1 + R R2 ÷ø è 1 æ 1 1 1 ö ÷ = (m - 1) çç f2 + R R1 ÷ø è 2

…(i) …(ii)

Solving these two equations, we can see that f1 = f2.

4. –¥

f = 10 cm

f = 10 cm 2F

2F

2F 20 cm

7. Q

20 cm

In the shown figure, object appears at distance d = m e(0.2) + 0.2 Now, for image to further coincide with the object, d = |R | Solving we get, m e =1.5 (Displacement method) 10. O = I 1I 2 = 6´

When object is moved from O to F1 , its virtual, erect and magnified image should vary from O to - ¥. 1 æ 1.3 ö æ 1 1 ö ÷ 5. (a) = ç - 1÷ ç f è 1.8 ø çè - 20 + 20 ÷ø \ f = + 36cm (b) Between O and F1 image is virtual. Hence, for real image. | m | < f or 36 cm

6.

or F = - 0.12 m or - 12 cm 9. | R | = 0.5 m (from first case)

+¥

O

F1

20 cm

Air

Water

I = m (O ) = 2.67 ´ 2 = 5.33cm

3.

(as f = constant) 2ö

20 cm

1 1 1 - = v u f

Differentiating this equation, we get

2 = 2 cm 3

11. Virtual, magnified and erect image is formed by convex lens. Let u = -x Then, v = - 3x 1 1 1 Now, = - 3x - x + 12 \ x = 8 cm Distance between object and image = 3x - x = 2x = 16 cm 12. Diminished erect image is formed by concave lens. x Let u = - x, then v = 2 Now, | u | - | v | = 20 cm x = 20 cm or x = 40 cm \ 2 \ u = - 40 cm and v = - 20 cm 1 1 1 = f -20 - 40 or

f = - 40 cm

Refraction of Light — 449

Chapter 31

æ 1ö = (1.6) ç ÷ = 0.8 è 2ø \ i = 53° P ray deviates from its original path by an angle, d = i - 30° = 23°

13. If an object is placed at focus of lens (= 10 cm), rays become parallel and fall normal on plane mirror. So, rays retrace their path.

INTRODUCTORY EXERCISE

1.

31.7

P

30° 60°

d

60° = i

q d

Q

\ Angle between two rays, q = 2d = 46° Critical angle = i = 60° = qC m m sin qC = R Þ sin 60° = mD 3

2. q

Solving we get, m = 1.5 c 3 ´108 2. m = = = 1.3 v 2.3 ´ 108 æ 1ö æ 1 ö qC = sin -1 ç ÷ = sin -1 ç ÷ = sin -1 (0.77) è 1.3 ø èm ø

i q

3. (a) m 1 sin i1 = m 2 sin i2 = m 2 sin qC æ 1.30 ö (1.6)sin q= (1.80) ç ÷ è 1.80 ø æ 13 ö q = sin -1 ç ÷ \ è 16 ø (b) If q is decreased, then i2 will decrease from the value qC . Hence, refraction will take place in medium-3.

INTRODUCTORY EXERCISE

1.

30° 60° i 30°

31.8

i = qC \

sin i = sin qC =

4. i1 = 0° Þ r1 = 0° or r2 = A r2 = qC = A 1 2 = m 3 æ 2ö ç ÷ è 3ø

\

sin A = sin qC =

or

A = sin -1

5. d = i1 + i2 - A

m =

\

1 2 = mg 3

Applying Snell’s law at point A, We have m w sin q = m g sin i 4 3 2 \ sin q = ´ 3 2 3 3 \ sin q = 4 æ A + dm ö sin ç ÷ 3. m = è 2 ø , dm = 30° sin ( A / 2) Now,

sin i sin 30° sin i = m sin 30°

B i

A

30° = 60° + i2 - 30° \

i2 = 0 or r2 = 0

Hence proved.

450 — Optics and Modern Physics Þ Now,

6.

r1 = A = 30° sin i1 sin 60° m = = = 3 sin r1 sin 30° sin i1 2= sin (i1 / 2) 2 sin (i1 / 2)cos (i1 / 2) = sin (i1 / 2)

Solving this, we get i1 = 90° and r1 =

8. ABC can be treated as a prism with angle of prism A = 90°. Condition of no emergence is

i1 = 45° 2

At minimum deviation, r2 = r1 = 45° A = r1+ r2 = 90° æ A + dm ö 7. From m = sin ç ÷ / sin ( A / 2) è 2 ø

or

We can see that given deviation is the minimum deviation. A

C

B

or

\

M

M

A

or

A ³ 2qC æ Aö sin qC £ sin ç ÷ è 2ø 1 £ sin 45° m 1 1 £ m 2 m ³ 2

\ æ 1ö èm ø

æ 1 ö ÷ = 38.7° è 1.6 ø

9. qC = sin -1 ç ÷ = sin -1 ç

N

B

C

At minimum deviation, MN is parallel to BC is ÐB = ÐC.

r2 = qC = 38.7° \ r1 = A - r2 = 45° - 38.7° = 6.3° sin i1 Now applying, m = sin r1 we can find i1.

Exercises LEVEL 1

5. m1m2 = 1

Assertion and Reason

6. Image is formed at second focus (not the first

1ö mø è is in the direction of ray of R S light. Hence, Ram appears nearer to Anoop by that much distance. 4. P can be assumed a slab of negligible thickness. Deviation is almost negligible. æ

1. Shift due to a slab = ç1 - ÷ t

P

focus). 7. A = 60° and dm = 30° Substituting the values, we get m = 2 1 1 1 8. = + F f1 f2 f1 = Focal length of convex lens (+ ve) f2 = Focal length of concave lens (- ve) If f1 > | f2 | , then F comes out to be negative. So, it becomes a diverging lens. Negative power of a diverging lens is just for its diverging lens, nothing else.

Chapter 31 9. By inserting a slab between the lens and the object, effective distance between object and lens decreases. So, object now comes between F and 2F or between F and O. F 2F

F

2F

8. Frequency does not change during refraction, but wavelength and speed decrease in a denser (here water) medium.

9. m 1 sin i1 = m 3 sin i3 \

O

d = 2f

Now,

In the first case, image is real and magnified and in the second case it is virtual and magnified. 10. If medium on both sides of the lens is same, then it doesn’t matter, which side the object is kept.

or

11. i = 45°

Þ or

1.414 Þ

qC < 45°

\ They get TIR on face AC. 12. For virtual object a concave lens can form a real image. 1 æ 1 1ö 13. = (m - 1) ç - ÷ f è R Rø f =¥ 1 P= =0 f

\

Objective Questions 2. m , A and Hence,

3. dapp.

B are dimensionless. l2 [ B ] = [ l2 ]= [ L2 ]

d = m

m of red is least. So, dapp. for red is maximum. So, they appear to be raised least. æ 1ö 4. qC = sin -1 ç ÷ èm ø m for violet is maximum, so qC for violet is least. æ 1 1 1ö 5. = P = (m - 1) çç - ÷÷ f R R è 1 2ø æ 1 1 ö ÷ = (1.6 - 1) çç 0.1 0.1÷ø è = + 12 D c 3 ´ 108 6. v = = = 2.25 ´ 108 m/s m (4 / 3)

7. After a certain angle all colours get total internally reflected.

sin i m 3 = sin r m 1

m 1 sin i = m 3 sin r or

10. r1 = i

For m > 2

Refraction of Light — 451

r2 = 90° - r1 = 90° - i

and

m R sin iR = m D sin iD mR sin iD sin i = sin qC = = mD sin iR sin (90° - i ) sin qC =

or

sin i = tan i cos i

qC = sin -1 (tan i )

11. At minimum deviation, r1 = r2 = Now using,

m =

A = 30° 2

sin i1 sin r1

or

2=

sin i1 sin 30°

We get i1 = 45° æ 1 1 1ö 12. = (1.5 - 1) çç - ÷÷ 0.2 R R è 1 2ø

…(i)

æ 1 1ö çç - ÷÷ …(ii) R R è 1 2ø Here, m = refractive index of medium or liquid. Dividing Eq. (i) by Eq. (ii), we get 1 æ 1.5 ö =ç - 1÷ - 0.5 è m ø

-5=

1 15 or m = (1.5/m ) - 1 8

13. From Snell’s law, \

m sin i = m 4 sin x m sin x = sin i m4

14. Focal length of any one part will be 2 f . \

1 1 1 1 1 f or F = = + + + F 2f 2f 2f 2f 2

15. In air, focal length is 100 100 = = 20 cm P 5 æ 1 1 1ö = (1.5 - 1) çç - ÷÷ 20 è R1 R2 ø æ 1.5 ö æ 1 1 1ö =ç - 1÷÷ çç - ÷÷ R R - 100 çè m e øè 1 2ø f =

…(i) …(ii)

452 — Optics and Modern Physics Dividing Eq. (i) by Eq. (ii), we get 0.5 -5= (1.5 / m e ) - 1 1.5 5 or me = = 0.9 3 1 1 1 d 16. = + F f1 f2 f1 f2

Originally, F=

When d is made 4 times. f1 f2 f1 f2 f f F¢ = = =- 1 2 ( f1 + f2 ) - 4 d 3d - 4 d d

To behave like concave lens, F should be negative. d 1 1 So, > + f1 f2 f1 f2 d f1 + f2 or > f1 f2 f1 f2 d > ( f1 + f2 ) or 30 cm

or

17. Condition of no emergence from opposite face is 18. r2 = 0, r1 = A sin i1 i1 i1 » = Þ sin r1 r1 A

i1 = mA

æ A + dm ö sin ç ÷ è 2 ø 19. m = sin ( A / 2) æ Aö m = cot ç ÷ è 2ø Solving, we get sin = 180° - 2A 20. At minimum deviation, A r1 = r2 = = 30° 2 sin i1 sin i1 Now, or 2 = m = sin r1 sin 30°

If d is doubled, focal length is doubled or denominator becomes half. 1 ( f1 + f2 ) - 2d = [( f1 + f2 ) - d ] \ 2 or ( f1 + f2 ) = 3d Substituting in Eq. (i), we have

R = 6 cm \ 2R = 12 cm 23. Lens formula: 1 1 1 = v1 -15 30

\ v3 = + 60 cm Hence, distance of final image from object = 60 - 15 = 45 cm

Given,

Solving, we get i1 = 45° 1 1 1 d 21. = + F f1 f2 f1 f2 f + f2 d = 1 f1 f2 f1 f2 1 ( f1 + f2 ) - d = F f1 f2 f1 f2 \ F= ( f1 + f2 ) - d

= - 2F 1 1 ö æ1 22. = (1.5 - 1) ç ÷ 24 è R 2R ø

Þ v1 = - 30 cm Mirror formula: v2 = - u2 = - (-45) = 45 cm Lens formula: 1 1 1 = v3 - 60 30

A > 2qC m =

f1 f2 f f f f = 1 2 = 1 2 ( f1 + f2 ) - d 3d - d 2d

60 cm

I3

I1

I2

O

15 cm 15 cm

45 cm

24. n1 sin i1 = n3 sin i3 \ (1) (sin 45° ) = 2 sin i3 1 Þ sin i3 = or i3 = 30° 2 æ A + dm ö sin ç ÷ è 2 ø 25. m = sin ( A / 2) …(i)

A A » 2 2 æ A + dm ö m A \ sin ç ÷= 2 è 2 ø æ A + dm ö With increase in m , sin ç ÷ will increase. è 2 ø Hence, dm will increase. For small angled prism, sin

Chapter 31 26. n =

sin i 2 sin (i / 2) cos (i / 2) = sin (i / 2) sin (i / 2)

33. m =

Refraction of Light — 453

sin i or sin 45°

3 sin i or i = 60° = 2 (1/ 2 )

Solving we get, i = 2 cos-1 (n/ 2) i

27. If object is placed at focus of convex lens, rays become parallel and they are incident normally on plane mirror. So, ray of light will retrace its path. sin i sin 45° 28. m = or r = 30° Þ 2= sin r sin r 45º

45º A

d

45º

\

d = 60° - 45° = 15°

C 30º

d

30º

q

30º 30º d

B

dTotal = dA - dB + dC = (45° - 30° ) - [180° - 2(30° )] + (45° - 30° ) = - 90° or | dTotal | = 90° æ A + dm ö sin ç ÷ è 2 ø 29. m = æ Aö sin ç ÷ è 2ø 1 æ 1 1 ö öæ 1 ÷ 30. =ç - 1÷ ç f è 1.5 ø çè 10 - 10 ÷ø Solving, we get f = - 15 cm 31. Image formed by convex lens I 1 should coincide at C 1, the centre of curvature C of convex mirror.

q = 2d = 30° 34. r2 = 0° and r1 = A = 30° sin i1 Applying, m = sin r1 sin i1 2= sin 30° Solving, we get i1 = 45° 1 1 1 1 1 35. = + = + F f1 f2 10 10 \

F = 5 cm 1 1 1 = v - 7.5 + 5

\

I1 O

C

12 cm

10 cm

2f

36. Applying,

For convex lens, 1 1 1 = + (10 + 2 f ) - 12 + 10

\

Solving this equation we get, f = 25 cm 32. For convex lens, 1 1 1 = Þ v = - 30 cm v - 12 + 20

or

This image I 1 is therefore, (30 + 10) cm or 40 cm towards left of plane mirror. Therefore, second image I 2 (by the plane mirror) will be formed 40 cm behind the mirror.

\ 1 æ 1.5 ö 37. = ç - 1÷ f è 1.4 ø \

v = + 15 cm v + 15 m= = =-2 u - 7.5 I = mO = (- 2) (1 cm) = - 2 cm m2 m1 m2 -m1 = v u R 1 3/ 2 1 - 3/ 2 = v + 30 + 20 1 1 3 = v 20 40 v = 40 cm æ 1 1 ö çç ÷÷ è - 30 - 50 ø f = - 1050 cm

38. Between 1 and 3 there is no deviation. Hence, m1 =m3

454 — Optics and Modern Physics MP = PQ tan r MN = 2 (MP ) = 2PQ × tan r = 2 ´ 6 ´ tan 28.8° = 6.6cm sin 45° 4 m = = sin r 3

Between 3 and 2 \

Ray to light binds towards normal. Hence, m 2 > m 3. æ A + dm ö sin ç ÷ è 2 ø 39. A = dm = 60° Þ m = sin ( A / 2)

5.

+ ve I1

PI1 = 4.3 cm

O Q

P

m2 m1 m2 -m1 two times = v u R 1.5 1.0 1.5 - 1.0 = v1 - 2.5 + 10

Applying

Subjective Questions 343 = 0.229 1498 (b) qC = sin -1 (0.229) = 13.2° l l m l m l 2. t1 = = = 1 Þ t2 = 2 v1 c/m 1 e c l \ t2 - t1 = (m 2 - m 1 ) c (1.63 - 1.47) (20) = 3 ´ 108

1. (a) m =

Solving, we get v1 » - 4.3 cm Again applying the same equation, we get 1.0 1.5 1.0 - 1.5 = v2 - 24.3 - 10 Solving, we get

-8

= 1.07 ´10 s l l nl 3. (a) t1 = = = 1 = minimum v1 c/ n1 c =

v2 » - 85 cm QI 2 = 85 cm

or

6.

(towards left)

+ ve

1.2 ´ 10-6 3 ´ 108

= 4 ´ 10-15 s l (b) l = 0 n Number of wavelengths across any film, l ln N = = l l0 sin i 4. m = sin r M

P r r Q

1.8 =

sin 60° sin r

Solving, we get r = 28.8°

N

m2 m1 m2 -m1 we get, = v u R 1.0 1.5 1.0 - 1.5 = - 1.0 - u - 2.0

Applying

Solving we get, u = 1.2 cm

7. For plane surface, happ = mh 3 ´ 10 = 15cm 2 This first image is at a distance, (15 + 3) cm from the plane mirror. So, mirror will make its second image at a distance 18 cm below the mirror or 21 cm below the plane surface. Now further applying, d 21 dapp = = = 14 cm m 1.5 =

below the plane surface.

Chapter 31 8.

Refraction of Light — 455

CM = R sec qC = I3 O

P

=

Q

8 cm

10 cm

Applying happ for first refraction, PI 1 = 8m (towards left) QI 1 = PQ + PI 1 = (6 + 8m ) \ QI 2 = QI 1 = (6 + 8m ), towards right. PI 2 = PQ + QI 2 = (12 + 8m ) d Applying dapp = for third and last refraction we m have, æ 12 + 8m ö ÷ = 16 PI 3 = ç è m ø m = 1.5

9.

i

11. I

II

III

IV

Q

P 5 cm 10 c

m

Apply We have \

m2 m1 m2 -m1 four times, = v u R 1.5 1.0 1.5 - 1.0 = v1 ¥ + 10 v1 = + 30 cm 1.0 1.5 1.0 - 1.5 = v2 + 25 - 5.0

r i

h

5cm = 6.7 cm cos 41.8°

\ PM = CM - CP = (6.7 - 5.0) = 1.7 cm

6 cm

\

R cos qC

v2 = - 25 cm 1.5 1.0 1.5 - 1.0 = v3 - 35 - 5.0 v3 = - 11.67 cm 1.0 1.5 1.0 - 1.5 = v4 - 16.67 - 10

2 cm 4 cm

m=

Applying,

sin i we get sin r

Þ v4 = - 25 cm \`Final image is at 25 cm to the left of P.

2

4 4 / 16 + h = 3 2 / 4 + h2

12. Applying

Solving this equation we get, h = 2.4 cm

m2 m1 m2 -m1 we have, = = v u R P +ve

10. qC

90°

R qC C

P

M

æ 1ö æ 2ö qC = sin -1 ç ÷ = sin -1 ç ÷ = 41.8° è nø è 3ø

1.0 1.6 1.0 - 1.6 = v - 3.0 - 5.0 Solving we get, v = - 2.42 cm This is distance from P (downwards). \ Distance from observer = 5 + 2.42 = 7.42 cm

456 — Optics and Modern Physics 13. x

1 1 1 = f + 15 + 10

or

Bird

Solving, we get f = - 30 cm L1

17.

y

L2 I1

Fish

dy = 4 cm/s dt Distance of bird as observed by fish 4 Z = y+mx = y+ x 3 - dZ - dy 4 æ - dx ö = + ç \ ÷ dt dt 3 è dt ø - dZ Given, = 16 cm/s dt Substituting in Eq. (i), we get - dx = 9 cm/s dt d 2 - x2 (displacement method) 14. f = 4d f = 16 cm, x = 60 cm Substituting the values we get, d = 100 cm æ 1 1 1ö 15. (a) = (m - 1) çç - ÷÷ f è R1 R2 ø Given, -

\ or

25 cm

I 1 is formed at second focus of L1 and first focus of L2.

…(i)

æ 1 ö ÷÷ f µ çç è m - 1ø f1 m 2 - 1 1.7 - 1.0 = = = 1.4 f2 m 1 - 1 1.5 - 1.0

(b) If refraction index of the liquid (or the medium) is greater than the refraction index of lens it changes its nature or converging lens behaves as diverging.

16.

15 cm

40 cm

L1

18.

L2 I2

O

I1

7.5 cm 30 cm

30 cm

30 cm

For L1, 1 1 1 = v - 30 + 20 \ For L2 ,

v = + 60cm 1 1 1 = v + 30 + 10

\

v = 7.5 cm

19. Object is placed at distance 2 f from the lens. Hence, image is also formed at distance 2 f on other side. For mirror, 40 cm 40 cm O

I2

I1

5 cm f = 20 cm f = 10 cm

O I

30 cm

1 1 1 + = v + 10 - 10 15 cm 10 cm

Using

1 1 1 = - , f v u

Ray diagram is as shown below. I2

I1

Refraction of Light — 457

Chapter 31 20. For second lens, 1 1 1 = Þ v - 5 - 20

v = - 4 cm 5 cm

10 cm

5 cm

æmR ö ÷÷ = sin -1 èm D ø

24. qC = sin -1 çç

æ 4 /3ö çç ÷÷ = sin -1 (8/ 9) è 3/ 2 ø d2

Water

90°

Glass

I1 f2 = –20 cm f3 = 9 cm

f1 = 10 cm

For third lens, 1 1 1 = v - (4 + 9) + 9 \ v=¥ 21. See the result of sample example 31.30 h R= 2 m -1 h = R m2 -1 = (1 cm ) (5/ 3)2 - 1 =

22. sin qC =

m 1 sin i1 = m 2 sin i2 m 1 sin qC = m 2 sin q æ 3ö æ 2ö æ 4 ö ç ÷ ç ÷ = ç ÷ sin q è 2ø è 3ø è 3 ø æ 3ö q = sin -1 ç ÷ è4ø

\ or

23.

4 cm 3

1 2 = m1 3

Now, or

n1

qc

M qc

N q1 90° – qc n2

\

d1 = 0°

\ d2 = 90° - qC 8 cos d2 = cos (90° - qC ) = sin qC = 9 -1 æ 8 ö d2 = cos ç ÷ \ è 9ø æ A + dm ö sin ç ÷ è 2 ø, 25. Using m = æ Aö sin ç ÷ è 2ø Given, dm = A A A 2 sin × cos sin A 2 2 3= = \ A A sin sin 2 2 Solving this equation, we get A = 60°. 26. Condition of no emergence is, A > 2qC æ 1 ö Amax = 2qC = 2 sin -1 ç ÷ \ è 1.5 ø = 84 ° 27. (a) At minimum deviation, A r1 = r2 = = 30° 2 sin i1 sin i1 or 1.5 = Applying m = sin r1 sin 30° i1 = 48.6°

We get

n1 should be less than n2 for TIR at M. n sin qC = 1 n2 At N , n1 sin q1 = n2 sin (90° - qC ) = n2 cos qC n sin q1 = 2 cos qC \ n1 n2 sin q1 = 1 - sin 2 qC n1 =

n22 - n12 n1

i = qC

i = 0°

60°

48.6°

P

Q

30° 30° 30° R

60° 48.6°

dTotal = dP + dQ + dR

458 — Optics and Modern Physics = (48.6°- 30° ) + (180°- 2 ´ 30° ) + (48.6° - 30° ) = 157.2° 4 (b) sin (48.6° ) = 1.5 sin r1 3 Solving we get, r1 = 41.8° r2 = 60° - r1 = 18.2° r3 = 60° - r2 = 41.8° \ i3 = 48.6° Hence, dTotal = (48.6° - 41.8° ) + (180° - 2 ´ 18.2° ) + (48.6° - 41.8° ) = 157.2° 28. Dispersion power, m -mr w= v my -1 w1 w2 29. (a) + =0 f1 f2 æ f ö \ w2 = çç - 2 ÷÷ w1 è f1 ø (- 30) (0.18) = 0.27 (+ 20) 1 1 1 1 1 (b) = + = F f1 f2 20 30 or w1 w2 30. + =0 f1 f2 \

31. For blue light, sin i1 sin r1 sin 65° 1.68 = sin r1 m =

or

Solving this equation, we get r1 = 32.6° r2 = A - r1 = 27.4 ° Again applying, sin i2 sin i2 or 1.68 = m= sin r2 sin 27.4 ° Solving this equation, we get i2 = 50.6° Now, dB = i1 + i2 - A or dB = 65° + 50.6° - 60° = 55.6° For red light, sin 65° 1.65 = sin r1 or \

…(i)

r1 = 33.3° r2 = A - r1 = 26.7° sin i2 1.65 = sin 26.7°

=-

Now,

F = + 60 cm

\ i2 = 47.8° \ dR = i1 + i2 - A or dR = 65° + 47.8° - 60° = 52.8° …(ii) From Eqs. (i) and (ii), we get dB - dR = 2.8°

w1 = w2

f1 f2

Let 1-stands for flint glass and 2-stands for crown glass. Then, f1 3 = or | f1 | = 1.5 | f2 | f2 2 Focal length of flint glass is more. So, its power is less. Combined focal length (and hence combined power) is positive. So, convex lens (converging lens) should be made up of crown glass (having more positive power). 1 1 1 Now, = + F f1 f2 1 1 1 \ = + 150 - 1.5 f f Solving this equation, we get f2 = f = 50 cm and f1 = - 1.5, f = - 75 cm

LEVEL 2 Single Correct Option 1. happ = mh =

4 8 ´2= m 3 3

11 m. So, mirror 3 will make image at same distance (= 11/ 3 m from itself). Now in third refraction, depth of second image, 11 14 d= +1= m 3 3 d 14 3 7 dapp = = ´ = m m 3 4 2 Distance from mirror = 1 + happ =

The desired distance is therefore, (dapp + h) 11 æ7 ö or m ç + 2÷ m or 2 è2 ø

Chapter 31 m 2 m1 m 2 - m 1 we get, = v u R 1.5 1 1.5 - 1.0 = v ¥ +R

7. i1 = r1 = 0°

2. Applying,

\

r2 = 90° - q = 53° æ 1ö qC = sin -1 ç ÷ = 37° èm ø

v = + 3R

3.

Refraction of Light — 459

Since, r2 > qC , T / R will take place on the face AB d = dAC + dAB = 0 + 180° - 2r2 = 180° - 2 ´ 53° = 74 °

60º 90º

30º 60º

8. m =

B

sin 60° = 3 sin 30°

x

4. Refraction from first surface,

y

m2 m1 m2 -m2 = v1 µ +R Refraction from second surface, m3 m2 m3 -m2 = f v1 -R

F

(as v2 = f )

…(i)

h2 = y + mx

…(ii)

Eq. (i) can be written as mx + y h h = h1 or 2 = h1 or m = 2 m m h1

Adding these two equations, we get m3 m2 -m1 m3 -m2 = f R R Lens becomes diverging if f is negative or m3 -m2 >m2 -m1 or m3 + m1 > 2m2 Same result is obtained if parallel beam of light is incident from RHS. 5. In first case, u = - 16 cm, then v = (+ 16n) cm 1 1 1 …(i) \ = 16n - 16 f In second case, u = - 6 cm, then v = - (6n) cm 1 1 1 = \ - 6n - 6 f Solving these two equations, we get f = 11 cm d 6. dapp = (n1 / n2 ) n d = 2 dapp \ n1 -

y m

h1 = x +

d n æ d ö æn ö (d ) = 2 ç dapp ÷ = çç 2 ÷÷ x dt n1 è dt ø è n1 ø 2 dv é d ù x pR n2 = A ê(d )ú = dt n1 ë dt û

…(ii)

9. MC = QC - QM = (R - 0.3) cm PC 2 = MC 2 + PM 2 R 2 = (R - 0.3)2 + (3)2 P R Q

C M

Solving this equation, we get R = 15 cm 10. See the hint of Q.No-5 of the same section. In that example, 16 + 6 f = = 11 cm 2 1 1 1 11. = + F1 f1 f2 1 1 1 d = + F2 f1 f2 f1 f2 F2 > F1 So, image of distant object will be formed to the right of P.

460 — Optics and Modern Physics æ A + dm ö sin ç ÷ è 2 ø 12. m = sin ( A / 2)

18. (a) Two images are formed in case (i).

m and A for both rays are same. Hence, value of dm is also same for both rays.

13.

qC = sin

-1

æ 1ö ç ÷ èm ø

æ 3ö ÷ = 40.9° = sin -1 çç ÷ è 7ø r2 = qC = 40.9° r1 = A - r2 = 19.1° sin i1 m = sin r1 7 sin i1 = 3 sin r1 Solving this equation, we get i1 = 30°

14. If object is placed at focus of plano-convex lens, then it will make rays parallel. Now, these rays fall normal on plane mirror. So, they retrace their path.

15. If object is placed at centre of the sphere, then all rays starting from C fall normal on spherical surface and pass understand. m 6/ 5 4 16. sin qC = R = = m D 3/ 2 5 r1 = 0° Þ r2 = < B = 90° - q Now, r2 > qc or sin r2 > sin qC 4 sin (90° - q) > \ 5 4 or …(i) cos q > 5 4 cos 37° = 5 From Eq. (i), we see that q < 37° 17. From the first refraction, rays should become parallel. Or, image should formed at infinity. Applying, m2 m1 m2 -m1 = u u R 3/ 2 1.0 3/ 2 - 1.0 = ¥ -x + 10

(b) One image is formed in case (ii). æ 1 1 1ö 2 (c) = (m - 1) çç - ÷÷ = - (m - 1) f2 R è - R Rø 1 1 1 = + f3 F1 F2

æ 1 1ö æ 1 1ö = (m - 1)ç - ÷ + (m - 1)çç - ÷÷ …(ii) ¥ R R ¥ è ø è ø From Eqs. (i) and (ii), we can see that f2 = f3. 1 1 1 1 1 19. = = + f OB - OA OB OA (OA ) (OB ) f = AB + OB (OA ) (OB ) …(i) f = \ AB Now, AB 2 = AC 2 + BC 2 or

(OA + OB )2 = AC 2 + BC 2

or OA 2 + OB 2 + 2 (OA ) (OB ) = AC 2 + BC 2 \ ( AC 2 - OC 2 ) + ( BC 2 - OC 2 ) + 2 (OA ) (OB ) = AC 2 + BC 2 Solving, we get (OA ) (OB ) = OC 2 Substituting in Eq. (i), we get OC 2 f = AB æ ö m 20. Shift = çç1 - medium ÷÷ t m slab ø è é 4 /3ù = ê1 ú ´ 36 = 4 cm 3/ 2 û ë h 21 h 21 21. or = = m 2 (4 / 3) 2 Solving, we get h = 14 cm. 2 cm 22. m = =-2 1cm Þ

|v | = m |u | y

Object

x = 50 cm x = – 40 cm

O

x = –10 cm

Solving, we get x = 20 cm

…(i)

30 cm

60 cm

Image

Chapter 31 23. Using the equation, 1 2 (m 2 /m 1 ) (m /m - 1) = -2 2 1 F R2 R1 we get 1 2 (1.5) 1 (1.5 - 1) = - 10 ¥ R1

1

2

Rö æ ç as F = ÷ 2ø è

Solving, we get R1 = 10 cm. 24. In air, object lies between F and 2F. In liquid, focal length will become 4 times. So, object will now lie between optical centre and focus. 25. Condition of no emergence, A > 2qC A or 45° \ qC < 2 1 1 < sin 45° or \ sin qC or m 2

æ 1ö YI2 = (5 mm) ç ÷ = 2.5 mm è 2ø = 0.25 cm 28. Ð i = 0° for all values of q, as the rays fall normal to sphere at all points. 29. Cavity of placed at centre. Hence, image of cavity is also formed at centre, as all rays fall normal to surface at all points. (so pass undedicated).

Now,

For no emergence, m > 2

26. t = nl =

n l0 m

l ö æ ç as l = 0 ÷ m ø è

30.

\ or

1 m m 2 t1 = m 1 t2 t m2 = 1 m1 t2 tµ

æ 5ö =ç ÷ è4ø

\

æ4ö 5 ç ÷= è 3ø 3

O

5 mm

I1

20 cm

10 cm

1 1 1 = =Þ v = - 5 cm v - 10 10 v (- 5 cm) 1 = m= = u (- 10 cm) 2 X I2 = 20 + 5 = 25 cm

(4 / 3) 1

=

4 3

æ1 mö ç ÷ ç1 m÷ è ø

f = + 20 cm 1 1 1 = v1 - 30 + 20 v1 = + 60 cm v + 60 m1 = 1 = =-2 u1 - 30

v2 = + 30 cm v (+ 30) 1 m2 = 2 = =u2 (- 60) 2

5 cm I2

=

æ vö ç ÷ è uø

For second lens, 1 1 1 = v2 - 60 + 20

27. For concave lens, y

æm ö m = çç 1 ÷÷ èm 2 ø

æ 1 1 1 ö ÷ = (1.5 - 1) çç ÷ f è 20 - 20 ø Þ

or

Refraction of Light — 461

m = m1 m2 = 1 Final image is 30 cm to the right of second lens or 150 cm to the right of first lens. m =1 Hence, image height = object height = 3 mm m2 m1 m2 -m1 31. Apply = v v R 1 2 1- 2 We get = v - 10 - 10 Solving we get,

v = - 10 cm

462 — Optics and Modern Physics 32.

O

I1

2F

2F

20 cm

If angle of q is less than this value, then angle of incidence at N will be greater than qC . Hence, TIR will take place at N . 37. Reflection from the concave mirror, I 1 I1

20 cm

2 cm

1 æ1 1ö 33. = (1.5 - 1) ç - ÷ 15 èR ¥ø

1 cm

\ R = 7.5 cm With polished surface, 1 2 (m 2 /m 1 ) 2 (m 2 /m 1 - 1) = F R2 R1 2 (1.5) 2 (1.5 - 1) = - 7.5 ¥ F = - 2.5 cm Now, using mirror formula, 1 1 1 + = v - 20 - 2.5 20 Solving, we get v = cm 7 2 1ö æ 34. Shift = ç1 - ÷ t = æç1 - ö÷ (6) = 2 cm 3ø mø è è For mirror, object distance = 50 - 2 = 48 cm. So, mirror will make image at a distance 48 cm behind it. In return journey of ray of light, we will have to farther take 2 cm shift in the direction of ray of light. So, image distance as observed by observer = (50 + 48) - 2 = 96 cm 1 1 1 35. Using - = , we get v u f 1 1 1 = + ( f + 40) - ( f + 10) + f

O

1 1 1 + = Þ v -1 - 2

1ö æ Shift due to slab = ç1 - ÷ t nø è = (1 - 2/ 3) (9 cm ) = 3 cm Now, slab will make next image at a distance 3 cm from I 1 in the direction of ray of light, i.e. at O itself but it is virtual, as the ray of light has crossed the slab and we are making image behind the slab.

38.

= n1 1 - sin 2 qC = n1 1 90°– q q N

M

m1

æ1 æ 1 1 ö 1 ö ÷ + (1.5 - 1) ç ÷ = (1.6 - 1) çç ÷ ç - 10 - - 20 ÷ ¥ 10 è ø è ø Solving we get, F = 28.57 cm

= n12 - n22

40. q

m2

m2 m1 m2 -m1 = v u -R m2 m1 -m2 m1 \ = + v R u If m 1 > m 2 and u is positive (i.e. virtual object), then v is always positive or image is always real. 1 1 1 39. = + F f1 f2

Solving this equation, we get f = + 20 cm 36. Applying Snell’s law at point M, we get sin q or sin q = n1 cos qC n1 = sin (90° - qC ) n22 n12

v = + 2 cm

Refraction of Light — 463

Chapter 31 1.5 1.0 1.5 - 1.0 = v -u +R

For n = 2m + 1, it is just like an identical prism of larger size.

1.5 - 1 1 = + v u 2R

or

For image to real (for negative value of u) v should 1 1 be positive. Hence, < or u > 2R. u 2R

2. 41.

¥

2F

F

O

sin i1 v = 1 sin i2 v2

vy sin r 1 = slope = = sin i 3 vx vx vy = \ 3 Speed of light in medium- y is less. So, it is denser. TIR takes place when ray of travels from denser to rarer medium. 3. (a) If vacuum speed of light of all colours is same. \

When object moves towards F to O virtual erect and magnified image moves from ¥ to O. 1 1 1 d2 42. = + F f1 f2 f1 / f2 f1 and f2 both are negative. Hence, F is also negative. Object is real so combined lens (having negative focal length) will always make, its virtual image. m m m -m1 43. Using 2 - 1 = 2 we get u u R 3/ 2 1.0 3/ 2 - 1 = µ -x + 60 \ x = 120 cm Hence for x = 120 cm, rays of light become parallel to principal axis and fall normal to polished surface. Hence, rays retrace their path. 1.6 1.0 1.6 - 1.0 44. = v1 - 2 + 1.0 Þ

\

v1 = 16 m 2.0 1.0 2.0 - 1.0 = v2 -2 + 1.0

(b)

¥

P

F

If object moves from ¥ to P, then its virtual, erect and diminished image move from F to P. d

(c) d 0

I1

i

I2

4. Displacement method of finding focal length of convex lens.

5. Deviation, d = (m - 1) A = (1.5 - 1) 4 ° = 2°

v2 = 4 m I 2 I 2 = v 1 - v2 = 12 m

More than One Correct Options 1. For n = 2m, it is just like slab.

2º

(a)

2º 2º

N

M

P

\

Deviation = 0

To rotate ray MP by 2° (to make it parallel to MN ) we will have to rotate the mirror only by 1°.

464 — Optics and Modern Physics (b) Without prism, ray of light was falling normal to plane mirror. So, ray of light was retracing its path prism has deviated it 2°. So, if we rotate the mirror by 2°, ray of light further falls normal to plan mirror and retraces its path.

Comprehension Based Questions 1. Using the mirror formula,

1 1 1 1 1 = + = f v u - 40 10

Solving we get, f = - 8 cm

2. Focal length of lenses is æ1 1 1 ö ÷ = (1.8 - 1) çç ÷ F è ¥ - 2R ø æ 1 1 ö ÷ + (1.2 - 1) çç 2 R R ÷ø è F = 2R 2R R

Now combined power of system,

or or

P = 2PL + Pm 1 - = 2PL + Pm f 1 æ 1 ö 2 =2ç ÷ + 8 è 2R ø R

or R = 24 cm \ Radius of curvature of common surface = 2R = 48 cm 3. Combined focal length of lens, æ 1 1 1 ö ÷ = (1.2 - 1) çç ÷ f è + 24 + 48 ø æ 1 1ö + (1.8 - 1) çç - ÷÷ + ¥ 48 è ø 1 = 48 Now, combined power of system æ 1ö 1 æ 1ö \ - = 2 çç ÷÷ + ¥ = 2 ç ÷ F f è 48 ø è ø \

F = - 24 cm

Match the Columns 1. (c) and (d)

1 1 1 - = v u f

\

1 1 1 = + v u f

Between O and F2 or between F2 and 2F2, u is positive. So, v is also positive. u is positive m= \ v Therefore, image is real (as v is positive) and erect (as m is also positive). 1 1 1 2. (a) - = v u - f 1 1 1 \ = u u f Between O and F1, u is positive and less than f . So, v is positive (therefore image is real). Further from v m= u We can see that m is allow positive. So, it is erect also. 1 1 1 (b) - = v u - f 1 1 1 = \ v u f Between F1 and 2F1 , m positive and greater than f . So, v negative (therefore image is virtual). Further from v m= u We can see that m is negative so, image is inverted. 3. (a) and (b) m v= 2u m1 Þ |v | < |u | as m1 >m2 i.e v and u are of same sign. Or they are on same side of plane surface. From plane surface, if object is real, image is virtual and vice-versa. (c) and (d) m v= 1u m2 Þ |v | > |u | Other explanations are same. m 1 m -1 4. (a) = v -u -R

Refraction of Light — 465

Chapter 31 m -1 m -1 = v u R Therefore, v is always negative. Or image is always virtual. m 1 m -1 (b) = v +u -R m 1 m -1 or = v u R So, v may be positive or negative. Hence, image may be real or virtual. Same logic can be applied for two options. For them R is positive. In option (c), u is negative and in option (d) u is positive. æ 1 1 1ö 5. or P1 = (1.5 - 1) çç …(i) - ÷÷ f1 R R 2ø è 1 æ 1.5 ö æ 1 1 1ö or P2 = çç …(ii) - 1÷÷ çç - ÷÷ f2 ème ø è R1 R2 ø Dividing Eq. (ii) by Eq. (i), we get æ 3 ö …(iii) P2 = çç - 2÷÷ P1 èm e ø \

2. The paths of the rays are shown in figure. Since, the virtual image is diminished, the lens is concave.

A

Subjective Questions 1. First draw a ray AA¢ until it intersects with the principal optical axis and find the centre of the lens C. Since, the virtual image is magnified, the lens is convex.

C

F

3. (a) OA = 8.0 cm \ AI 1 = (ng )(OA ) æ 8ö = ç ÷ (8.0) = 12.8 cm è 5ø For refraction at EG (R = ¥ ), using C

(a) m e = 1.4, then P2 is positive and less than P1. Other options can be checked from Eq. (iii). 6. Concave lens can make only virtual, erect and diminished images of real objects. Convex lens can make real, inverted and diminished size or real inverted and magnified or virtual, erect and magnified images of real objects. Rest type of cases are possible with virtual, objects.

B A¢

B

A

O

E

D

G

F

+ve

n2 n1 n2 - n1 - = v u R 4 /3 8/ 5 =0 BI 2 - (12.8 + 3)

\ \ \

BI 2 = - (15.8)(4 / 3)(5/ 8) = - 13.2 cm FI 2 = 13.2 + 6.8 = 20.0 cm

(b) For face EF C

E

A¢ B

A

O

C

A

B

F D

Draw a ray AB parallel to the principal optical axis. It is refracted by the lens so that it passes through its focus and its continuation passes through the virtual image. The ray A ¢ B intersects the principal optical axis at point F, the focus of the lens.

G +ve

8/ 5 4 / 3 =0 BI 1 -6.8 \

BI 1 = - (6.8)(8/ 5)(3/ 4 ) = - 8.16 cm

F

Ans.

466 — Optics and Modern Physics Therefore, we will have to shift the screen a distance x = v2 - v1 = 60 cm away from lens. Ans.

For face CD 1.0 8/ 5 =0 AI 2 -11.16 \ \

7.

AI 2 = - (11.16)(5/ 8) = - 6.975 cm FI 2 = 8 + 6.975 = 14.975 cm

æ 1 1 1ö ÷ = (1.5 - 1) çç + ÷ 40 è 120 R1 ø Solving we get, R2 = 24 cm Applying lens formula, for L2

Ans.

O

4. The system behaves like a mirror of focal length given by 1 2(n2 / n1 ) 2(n2 / n1 - 1) = F R2 R1

x n1 n2

L2

Substituting the values with proper sign.

or

10 cm

1 2 ´ 4 /3 = F -20

(Q R1 = ¥ )

F = - 7.5 cm

Ans.

i.e. system behaves as a concave mirror of focal length 7.5 cm. æ 1 1 1ö 5. = (n - 1) çç - ÷÷ f R R 2ø è 1 æ 1 1 1 ö R1 n R2 ÷ = (1.5 - 1) çç ÷ 10 è R1 -10 ø 1 2 \ R1 = + 10 cm Now using, 1 1 2(n2 / n1 ) 2(n2 / n1 - 1) + = v u R2 R1 Substituting the values, 1 1 2(1.5) 2(1.5 - 1) = v -15 -10 +10 \

v = - 2.14 cm 1 1 1 6. Using lens formula, - = v u f 1 1 1 = v1 -40 30 \

\ \ \

v2 = 180 cm

R1 R2

1 1 1 …(i) + = v1 x 20 m2 m1 m2 -m1 for unsilvered side of Using = v u R L1. 1.5 1.0 1.5 – 1.0 = -120 v1 - 10 24 Solving Eqs. (i) and (ii), we get x = 10 cm 1 1 1 8. Case I + = v u f 1 1 1 \ = v1 30 -10

…(ii) Ans.

\ Ans.

v1 = 120 cm Shift due to the slab, 1 ö 1ö æ æ Dx = ç1 - ÷ d = ç1 ÷ 9 = 4 cm 1.8 ø mø è è u ¢ = - (40 - Dx ) = - 36 cm 1 1 1 = v2 -36 30

L1

v1 = - 15 cm 1ö 1 ö æ æ Case II Shift = ç1 - ÷ t = ç1 ÷ 6 = 2 cm mø 1.5 ø è è 1 1 1 \ = v2 28 -10

\ \

v2 = - 15.55 cm Dv = 0.55 cm

Ans.

m m m -m1 9. Using 2 - 1 = 2 , twice with u = ¥, v u R we have 1.5 1.5 – 1.4 …(i) = v1 +20 1.6 1.5 1.6 – 1.5 = v2 v1 -20 Solving Eqs. (i) and (ii), we get f = v2 = ¥ ,

…(ii)

Refraction of Light — 467

Chapter 31 i.e. the system behaves like a glass plate.

10. First image will be formed by direct rays 1 and 2, etc. PI 1 =

PO 5 = = 3.33 cm m 1.5

Ans.

Second image will be formed by reflected rays 3 and 4, etc.

æ 1ö qC = sin -1 ç ÷ = 45° èm ø Applying sine law in DCPM, CP CM = sin qC sin (90° + r) CP R \ = (1/ 2 ) cos r \

2

1 O

5cm

3

3

Object is placed at the focus of the mirror. Hence, I 2 is formed at infinity. 11. (a) Applying Snell's law at D, 4 3 sin i = sin 30° 3 2 Ans. \ i = 34.2°

30°

A

30°

\ R = 20 cm m m m -m1 twice with the Applying 2 - 1 = 2 v u R condition that rays must fall normally on the concave mirror. 1.5 12 . 1.5 – 1.2 …(i) = v1 - 40 + 20 2.0 1.5 2.0 – 1.5 …(ii) = d - 80 v1 –20 Solving Eqs. (i) and (ii), we get d = 30 cm

E

D i

2 R 3

As we move away from C, angle PMC will 2 increase. Therefore, CP >| R. Same is the case 3 on left side of C. 1 1 ö æ1 13. = (1.5 - 1) ç ÷ 20 R Rø è

P 5cm

CP =

i 30°

C

(b) d = dD + dE = 2dD = 2(i - 30° ) = 8.4 ° sin 45° 12. 2 = sin r

90° I1

Ans.

I2 100 cm 40 cm 80 cm

30 cm

14. Using lens formula,

45° P

1 1 1 = 36 -45 f

r qC M

r = 30°

Ans.

and v1 = - 100 cm The ray diagram is as shown in figure.

B

C

[R = radius]

\ f = 20 cm In the second case, let m be the refractive index of the liquid, then

468 — Optics and Modern Physics 1 48

1 1 = 40 ö 20 æ - ç5 + ÷ m ø è Solving this, we get m = 1.37

Note A ray passing through O and then O¢ goes undeviated.

Ans.

15. As the angles are small we can take,

Therefore I 1 and I 2 both should be on this line, which is also the x-axis. That’s why for final image we have taken projection of O ¢ I 2 ¢ on x-axis. 17. Since, the vessel is cubical, ÐGDE = 45° GE = ED = h (say ) EF = ED - FD = (h - 10) 4 sin 45° = 3 sin r r = 32°

and then

A 90°

Further, A A

A + 1°15¢

\

2A

6°30¢

Eye C

A

B

45°

C

G

sin q » q A + 1°15¢ m = A 6°30¢ = 2A

Now,

r h

Solving this equation, we get Ans.

16. Using lens formula for L2 , P I2N

L2

y

60° I1

O

Ans.

ÐOBC = ÐBCO = r A

x

r i

ON f

PN

1 1 1 = v - f /2 f or v=- f This f length will be along PP¢ from point O ¢ (towards P). \ O ¢ I 2¢ = f On x-axis this distance will be f sec 60° = 2 f . Since, OO ¢ = 2 f , therefore image will be formed at origin.

(say)

i B

D

f

D

EF = tan r = tan 32° GE h - 10 \ = 0.62 h Solving this, we get h = 26.65 cm \

I2

F

18. BO = OC I1N f/2

L1

E

Now,

A = 2° m = 1.62

and

45°

B

r O

C

Let angle of incidence be i, (external angle) i = r + r = 2r sin i sin 2r 2r Ans. m = = » =2 \ sin r sin r r

19. dTotal = dRefraction + 2dReflection + dRefraction or

d = (i – r) + 2(180° - 2r) + (i - r) = 360° + 2i - 6r

Refraction of Light — 469

Chapter 31 æ sin i ö = 360° + 2i - 6 sin -1 ç ÷ è m ø dd For deviation to be minimum, =0 di By putting first derivative of d (w.r.t. i) equal to zero, we get the desired result. æ1 1 1ö 20. (a) For refraction at first half lens çç - = ÷÷ èv u f ø 1 1 1 = v -20 15 \ v = 60 cm v 60 Magnification, m = = = -3 u -20 The image formed by first half lens is shown in figure (a). AB = 2 mm, A1 B1 = 6 mm, AO1 = 20 cm, O1F = 15 cm and O1 A1 = 60 cm.

So, length of final image 1 A3 B3 = A2 B2 = 2 mm 3 Point B2 is 2 mm below the optic axis of second half lens. Hence, its image B3 is formed 2/3 mm above the principal axis. Similarly, point A2 is 8 mm below the principal axis. Hence, its image is 8/3 mm above it. Therefore, image is at a distance of 20 cm behind the second half lens and at a distance of 2/3 mm above the principal axis. The size of image is 2 mm and is inverted as compared to the given object. Image formed by second half lens is shown in figure (c). A3 B3

O¢

O ¢¢

B2

A2 B

A

(b) Ray diagram for final image is shown in figure (d).

A1

F O1

(c)

4 mm C1

2 mm

B1

(a)

A3

B A1 A

B3

B2

Point B1 is 6 mm below the principal axis of the lenses. Plane mirror is 4 mm below it.

B1

B2

A2 2 mm

(d)

C2

21. (i) PO = OQ 4 mm

\

A2

ÐOPQ = ÐOQP = r (say) i

(b)

P

h

Hence, 4 mm length of A1 B1 (i.e. A1 C 1) acts as real object for mirror. Mirror forms its virtual image A2 C 2. 2 mm length of A1 B1 (i.e. C 1 B1) acts as virtual object for mirror. Real image C 2 B2 is formed of this part. Image formed by plane mirror is shown in figure (b). 1 1 1 For the second half lens, = v -60 15 \ v = + 20 v 20 1 m= = =u -60 3

i R

Also, In DPOR, or Also,

Radius = 0.1 m

r r O

Q i

i = r + r = 2r h = OP sin i = 0.1 sin i = 0.1 sin 2r h = 0.2 sin r cos r sin i 2 sin r cos r 3= = sin r sin r

…(i)

470 — Optics and Modern Physics = 2 cos r \ r = 30° Substituting in Eq. (i), we get 1 3 h = 0.2 ´ ´ 2 2 = 0.086 m Hence, height from the mirror is 0.1 + 0.086 = 0.186 m (ii) Use the principle of reversibility.

Further,

1 m 1-m = BI 2 ( AI 1 - R ) -R

Solving this equation, we get 2R (4 m - 1) BI 2 = 3m - 1 +ve

A

O

Q O

S

B

R

i

Eye

2R

C M

\ \

22.

Vi V

i = 2r = 60° QS 1 Now, = cot i = cot 60° = MS 3 MS 0.1 QS = = 3 3 The desired distance, 2 ´ 0.1 OC = 2 ´ 0.1 + 3 = 0.315 r r 1 = S = = rL 2 r 2

\ Distance between the final image and object is 2R (4 m - 1) d = 3R 3m - 1 (m - 1)R Hence proved. = (3m - 1)

24. dTotal = dP + dQ

P i

i.e. half the sphere is inside the liquid. For the image to coincide with the object light should fall normally on the sphere. m m m -m1 twice, we have Using 2 - 1 = 2 v u R 3/ 2 1 3/ 2 - 1 = v1 -8 +2 \

v1 = 12 cm 4 /3 3/ 2 4 / 3 - 3/ 2 Further, = h - 10 8 -2 Solving this equation, we get h = 15 cm

Ans.

23. We have to see the image of O from the other side. Applying,

\

m2 m1 m2 -m1 twice, we have = v u R m 1 m -1 = AI 1 -2R -R AI 1 =

2 mR 1 - 2m

a

Ans.

Q r

b

r

i

O

\

a = (i - r) + (i - r) a …(i) or i-r= 2 Further, in DOPQ, r + r + b = 180° b …(ii) r = 90° \ 2 From Eq. (i), a æ a - bö …(iii) i=r+ = 90° + ç ÷ 2 è 2 ø é æ a - böù sin ê 90° + ç ÷ú sin i è 2 øû ë m = = bö æ sin r sin ç 90° - ÷ 2ø è

Chapter 31 Further, \ Hence proved.

25. q = 90° - i y

P (x, y) i x

tan q = cot i dy = cot i dx m 0 sin i0 = m P sin iP

or

1 m 1-m = v2 - (3R - v1 ) R/2

…(i)

v2 =

R (9 - 4m ) (10 m - 9)(m - 2)

(i) sin 90° = ( 1 + ay) sin i

I2

\

1 sin i = 1 + ay

\

cot i = ay =

\

y

ò0

x dy = ò dx or 0 ay

Substituting and We get

26. Applying,

\

0.3 m

q P q

M I1

d 6.0 m

y a

y = 2 m, a = 2.0 ´ 10-6 m -1 xmax = 2000 m = 2 km

m2 m1 m2 -m1 twice, we have = v u R m 1 m -1 = v1 -2R R v1 =

q

N

O

dy dx x=2

Ans.

Final image is real if, v2 > 0. As 10 m - 9 is always positive (m > 1). Therefore, for v2 > 0, either (9 - 4m ) and (m - 2) both should be greater than zero or both should be less than zero. For the first condition (when both > 0) 2 < m < 2.25 and for the second condition (when both < 0), m < 2 and m > 2.25 which is not possible. Hence, m should lie between 2 and 2.25. 27. For the lens, u = - 2.0 m, f = + 1.5 m 1 1 1 = \ v -2.0 1.5 or v = 6.0 m 6.0 m= = - 3.0 –2.0 90°– q

æb - a ö cos ç ÷ è 2 ø = æbö cos ç ÷ è 2ø b æb - a ö or cos ç ÷ = m cos 2 è 2 ø

Refraction of Light — 471

2 mR 2m - 3

Ans.

Therefore, y-coordinate of image formed by lens is m(0.1) = - 0.3 m. 0.3 = tan q = 0.3 NP \ NP = MP = 1.0 m or d = 6.0 – 1.0 Ans. = 5.0 m and

x-coordinate of final image I 2 is x = d - 1.0 = 4.0 m

Ans.

32. Interference and Diffraction of Light INTRODUCTORY EXERCISE æ I / I + 1ö I 9 I ÷ 1. 1 = , max = çç 1 2 I 2 16 I min è I 1 / I 2 - 1 ÷ø Amax A1 + A2 2. (a) = Amin A1 - A2 (b)

3.

I max æ Amax ö ÷ =ç I min çè Amin ÷ø

4. A1 = 3 A2

32.1

2

So,

I 1 = 9I 2

Let

A2 = A0

Then, and Now,

I 1 = 9I 0

I = I 1 + I 2 + 2 I 1I 2 cos f

= 10I 0 + 6I 0 cos f I max æI ö + 6 ç max ÷ cos f 16 è 16 ø 5 3 = I max + I max cos f 8 8 5 3 f ö æ = I max + I max ç 2 cos2 - 1÷ 8 8 2 ø è 1 3 f = I max + I max × cos2 4 4 2 I max æ f ö 2 = ç1 + 3 cos ÷ 4 è 2ø f I = I max cos2 2 = 10 ´

Ö2A0 3A0

Ö2A0

INTRODUCTORY EXERCISE

I max = 16I 0 = 9I 0 + I 0 + 2 9I 0 ´ I 0 cos f

l path difference is equivalent to 90° phase 4 difference Anet = 5 A0 Þ I max = 25 I 0

3. I = I max cos2

and

I2 = I0

Amax = ( A1 + A2 ) = 4 A0

2

45°

A1 = 3 A0

and

32.2 5. (a)

f 2

= I 0 cos2 30°

3 f f 3 or cos = I max = I max cos2 4 2 2 2 f p \ = 2 6 p æ 2p ö 2p æ yd ö f = = ç ÷ (Dx ) = \ ç ÷ l èDø 3 è lø lD (600 ´ 10-19 ) (1.2) y= = \ 6d (6) (0.25 ´ 10-2 )

=

(as f = 60° )

3 I0 4

(b) 60° phase difference is equivalent to

l path 6

difference. f 6. aR = 2a cos 2 (i) For aR = 2a, f = 0° (ii) For aR = 2a, f = 90° etc

= 48 ´ 10-6 m = 48 mm

Exercises LEVEL 1 Assertion and Reason 1. I = 4 I 0 cos2 f=

f 2

2p or 120° Þ 3

…(i) f = 60° 2

Substituting in Eq. (i), we get æ lö I = I 0 Þ Dx = ç ÷ f è 2p ø 2p l For f = , Dx = 3 3 2. The whole fringe pattern will shift upwards.

Chapter 32 3. No reflected ray reaches below O. 4. I max = ( I 1 + I 2 )2 and I min = ( I 1 - I 2 )

2

I 1 = I 2 = I 0 , then I max = 4 I 0 and I min = 0 When slit of one width is slightly increased, then intensity due to that slit becomes greater than I 0 . In that case, we can see that I max > 4 I 0 and I min > 0 6. Locus of points of equal path difference in the shown case is circle. 7. At centre, path difference is maximum and this is equal to S 1S 2. Then, path difference decreases as we move away on the screen. So, order of fringe also decreases. Hence, 11th order maxima occurs before 10th order maxima. 8. At points P and R | Dx | = S 1S 2 = 4 l , therefore maxima.

Interference and Diffraction of Light — 473 Objective Questions 1. A = (8)2 + (6)2 = 10 mm 6 mm

When

Q

R

S1

S2

A

Þ 4 mm

2.

12 mm

8 mm

I1 = b2 I2 So let,

I 2 = 1 unit, then I 1 = b I max = ( I 1 + I 2 )2 = (1 + b )2 I min = ( I 1 - I 2 )2 = (1 - b )2

= I max - I min = 4b I max + I min = 2(1 + b 2 ) \ The asked ratio is

3. d sin q = P

6 mm

\

2b . 1+ b 2

l 2 æ lö q = sin -1 ç ÷ = sin -1 è 2d ø

æ 5460 ´ 10-10 ö ç ÷ ç 2 ´ 0.1 ´ 10-3 ÷ è ø

= 0.16° S

At Q and S Dx = 0, therefore again maxima. Then, three maxima in each quadrant (between P and Q or between Q and R etc.) corresponding to | Dx | = l , 2l and 3l. Therefore, there are total 16 maxima. In each quadrant there are four minima corresponding to | Dx | = 0.5 l , 1.5 l , 2.5 l and 3.5l. Hence, there are total 16 minima. 9. d sin q = 2l (for second order maxima) l 1 1 =2´ = d 4 2 \ q = 30° For maxima, d sin q = nl d (for q = 90°) \ nmax = = 4 l So, there are total 7 maxima corresponding to n = 0, ± 1, ± 2 and ± 3. We cannot take n = 4, as it is for q = 90°, which is out of screen. (m - 1) tD is independent of l. 10. Shift = d sin q = 2

4. 6th dark fringe distance in vacuum = 10th bright fringe distance in liquid. \ 5.5 w = 10 w¢ w 10 or =m = = 1.81 w¢ 5.5 (m - 1)tD 5. Shift = d (1.5 - 1) (10 ´ 10-6 ) (1.0) = 2.5 ´ 10-3 = 2 ´ 10- 3 m = 2 mm w (lD / d ) lD 6. Distance = = = 2 2 2d 7. Path difference at centre is always zero. Hence, all wavelengths under all conditions always interfere constructively. lD 8. w = or w µ l d 6000 w will increase 4000 or 1.5 times. Hence, number of fringes in same distance will decrease 1.5 times.

474 — Optics and Modern Physics 9. w = Þ \

lD d

1m A

l (DD ) Dw = d (d ) (Dw) (10-3 ) (3 ´ 10-5 ) l= = DD 5 ´ 10-2 = 0.6 ´ 10-6 m = 6000 Å

10.

2 49 I max ( I 1 / I 2 + 1) = = 2 9 I min ( I 1 / I 2 - 1)

Solving, we get

11. \

I 1 25 = I2 4

3lD y= d yd (7.5 ´ 10-3 ) (0.2 ´ 10-3 ) l= = 3D (3) (1)

= 500 ´ 10- 9 m = 500 nm nl D nl D 12. 1 1 = 2 2 d d n1 l 1 5200 4 or = = = n2 l 2 6500 5 \ 4th maxima of l 1 coincides with 5th maxima of l 2. 4 l 1D ymin = d 4 ´ 6500 ´ 10-10 ´ 1.2 = 2 ´ 10-3 = 1.56 ´ 10- 3 m = 0.156 cm

13. Only fringe pattern will shift. Number of fringes on screen will remain unchanged.

4m P1

B

4m A

1m P2

l 2 l At P2 AP2 - BP2 = 3m = 2 c 3 ´ 108 3. Q l = = = 2.5 m f 120 ´ 106 BP1 - AP1 = 3m =

At P1

(9 – x)

x A

P

» 0.590 nm l 500 ´ 10-9 radian 5. q = = d 2.0 ´ 10-3 » 0.014 °

1. Amax = 5 + 3 = 8 units

6. Wavelength in water, l ¢ =

\

2. (a) At centre path difference is zero. Therefore, construction interference will be obtained. l (b) = 3 m. At a distance, where path difference 2 l is or 3 m destructive interference will be 2 obtained.

B

Dx = ( BP - AP ) = (9 - 2x ) = nl 9 - nl 9 - 2.5 n \ x= = 2 2 Now, substituting n = 1, 2,L etc. We can find different values of x. x1 = 3.25 m for n = 1 for n = 2 x2 = 2.0 m and x3 = 0.75 m for n = 3 Similarly, we will get three points at same distance from other point B. lD 4. Q w= d wd (2.82 ´ 10-3 ) (0.46 ´ 10-3 ) \ l= = D 2.2 = 0.589 ´ 106 m

Subjective Questions Amin = 5 - 3 = 2 units Amax =4 Amin I max = (4 )2 = 16 I min

B

l m l ¢D lD Fringe width, w = = d md =

(700 ´ 10- 9 )(0.48) (4 / 3) (0.25 ´ 10-3 )

= 10-3 m = 1 mm 3l D 3l D 7. Distance = 2 - 1 d d 3(l 2 - l 1 )D = d

Chapter 32 =

3 ´ (600 - 480) ´ 10-9 ´ 1.0 5.0 ´ 10-3

= 7.2 ´ 10-5 m = 0.072 mm 8. The required distance = one fringe width lD =w= d (500 ´ 10-9 ) (0.75) = (0.45 ´ 10-3 )

Interference and Diffraction of Light — 475 12. (a) Dx =

2p × Dx l (2p ) yd = lD (2 ´ 180) yd = degree lD

f=

and

= 1978°

= 0.83 mm w2 2 l 1D l 2D \ = d 2d Þ l 2 = 2l 1 = 1200 nm l 10. d sin q1 = 2 l 550 ´ 10-9 = sin q1 = 2d 2 ´ 1.8 ´ 10-6 \

\

\

q1 = 8.78° y1 = tan q1 D y1 = D tan q1 = 35tan 8.78° = 5.41 cm 3l d sin q2 = 2 3l 3 ´ 550 ´ 10-9 sin q2 = = 2d 2 ´ 1.8 ´ 10-6

\ \

q2 = 27.27° y2 = D tan q2 = 35tan 27.27° = 18 cm Dy = y2 - y1 = 12.6 cm

11. l(in Å) =

150 E (in eV)

(de-Broglie wavelength of electron) 150 = = 1.22 Å 100 lD (1.22 Å) (3m) w= = d (10 Å) = 0.366 m = 36.6 cm

360 ´ 0.3 ´ 10-3 ´ 10 ´ 10-3 546 ´ 10-9 ´ 1.0

=

= 8.33 ´ 10-4 m

9. y = w1 =

yd D

Now, I = I 0 cos2

f 2

= 2.97 ´ 10- 4 I 0 » 3.0 ´ 10- 4 I 0 lD d \ Number of fringes between central fringe and P y yd N = = w lD (b) w =

=

(10 ´ 10-3 ) (0.3 ´ 10-3 ) (546 ´ 10-9 ) (1.0)

= 5.49 So, bright fringes are five. (m - 1) tD 13. Shift = d S1 =

(1.6 - 1) (10 ´ 10- 6 ) (1.5) 1.5 ´ 10-3

= 0.6 ´ 10-2 m = 0.6 m (1.2 - 1) (15 ´ 10-6 ) (1.5) S2 = 1.5 ´ 10-3 = 0.3 ´ 10-2 m = 0.3 cm = DS = S 1 - S 2 = 0.3 cm = 3 mm l (2D ) (m - 1)tD 14. (fringe width = shift) = d d (m - 1)t \ l= 2 (1.6 - 1) (1.964 ´ 10-6 ) = m 2 = 0.5892 ´ 10-6 m = 589 nm

476 — Optics and Modern Physics 15. Q d = 2 cm -9

w=

lD (500 ´ 10 ) (100) = d 2 ´ 10-2

= 2.5 ´ 10-3 m = 2.5 mm

For n = 4 , l = 424 nm For n = 5, l = 329 nm Therefore, two wavelengths lying in the given range are 424 nm and 594 nm. (b) Ray-1 and ray-2 are in same phase.

S d

P

n

S¢

In this figure, P is in a dark fringe, as conditions of maxima and minima are interchanged. Hence, next dark fringe will be obtained at a distance w or 2.5 mm from P. 16. 1 and 2 both are reflected from denser medium. 1

2

n1

t

n2

Glass

Hence, 2n1t = or

tmin

l for first order minima 2 l 650 = = 4 n1 4 ´ 1.42

1

(for maximum intensity) Dx = 2nt = ml 2nt (2) (1.53) (485 nm) l= = \ m m æ 1484 ö =ç ÷ nm è m ø Substituting m = 1, 2, 3 L etc, l 1 = 1484 nm, l 2 = 742 nm, l 3 = 495 nm, l 4 = 371 nm etc. Therefore, only wavelength lying in the given range is 495 nm. 18. Ray-1 and ray-2 both are reflected from denser medium. Hence, they are in phase. 1

t

1.5

n

l for minima 2 l 600 t= = 4m 4 ´ 1.3

2 mt =

\ or

l Dx = 2nt = (2m - 1) for maximum intensity 2 4 nt l= \ (2m - 1) (4 ) (1.53) (485 nm) = 2m - 1 æ 2968 ö ÷÷ nm = çç è 2m - 1ø For m = 1, l = 2968 nm For m = 2, l = 989 nm For m = 3, l = 594 nm

2

1.3

2

t

2

Hence,

= 114 nm 17. (a) Ray-1 is reflected from a denser medium and ray-2 by a rarer medium. 1

t

= 1154 Å

19. Ray-1 is reflected from denser medium and ray-2 from denser medium. 1 Oil

m

2 t

Water

\ Dx = 2 mt = l = 800 nm for destruction interference.

Chapter 32 \ mt = 400 nm For constructive interference,

l=

l 2

4mt 1600 = 2n - 1 2n - 1

For n = 1, l = 1600 nm For n = 2, l = 533 nm For n = 3, l = 320 nm The only wavelength lying in the given range is 533 nm. 20. 2mt = l / 2 This is the condition for destructive interference. l 3.0 t= = = 0.5 cm \ 4m 4 ´ 1.5

21. Path difference produced by slab, Dx = (m - 1)t =

l 2

l path difference is equivalent to 180° phase 2 difference. Hence, maxima and minima interchange their positions. 22. Dx = d sin q = nl d sin q n= l d 4.0 ´ 10-6 nmax = = l 600 ´ 10-9 = 6.66 Highest integer is 6.

23. 45º

lD b = 4d 4 æf ö Further, I 0 = 4 I 0 cos2 ç 2 ÷ è 2ø 2p æ 2p ö æ 2p ö æ y d ö f2 = = ç ÷ (D x2 ) = ç ÷ ç 2 ÷ \ 3 è lø è løè D ø lD b y2 = = \ 3d 3 b Dy = y2 - y1 = 12 2. At path difference l, we get maximum intensity. \

Dx = 2 mt = (2n - 1) \

Interference and Diffraction of Light — 477

A (each)

y1 =

\

I max = I

æ fö I R = I max cos2 ç ÷ è 2ø I æ fö = I cos2 ç ÷ \ 4 è 2ø 1 æ fö or cos ç ÷ = ± 2 è 2ø f = 60° or 120° \ 2 2p 4p and \ f = 120° or 240° or 3 3 æ lö From the relation, Dx = ç ÷ × f è 2p ø We see that, l 2l Dx = and 3 3 3. Ray-1 is reflected from a denser medium (D f = p) while ray-2 comes after reflecting from a rarer medium (D f = 0° ). l \ Dx = 2mt = (2n - 1) for maximum intensity. 2 4 mt or l = (2n - 1) =

Anet = 0

4 ´1.5 ´ 500 æ 3000 ö ÷÷ nm = çç 2n - 1 è 2n - 1ø 1

2

LEVEL 2 Single Correct Option q1 ö ÷ è 2ø p æ 2p ö æ 2p ö æ y d ö f1 = = ç ÷ (D x1 ) = ç ÷ ç 1 ÷ 2 è lø è løè D ø

1. 2 I 0 = 4 I 0 cos2 æç \

m

t

Substituting n = 1, 2, 3 ××× etc, we get l = 3000 nm, 1000 nm, 600 nm etc. \ Answer is 600 nm.

478 — Optics and Modern Physics 4. D x =

yd D

7. Q q=

d D S1

P y

q

d S2 D

For destructive interference at P. yd l Dx = = (2n - 1) D 2 2 yd l= \ (2n - 1) D Substituting n = 1, 3, 5 ×××etc we get 2 yd 2 æ yd ö 2 æ yd ö l= , ç ÷ , ç ÷ etc D 3èDø 5èDø

Third minima, 2.5l æ lD ö y = ± 2.5 w = ± 2.5 ç ÷=± q è d ø

8. At points P and Q, R

…(i)

yd (2 ´ 10-3 ) (0.1´ 10-3 ) = D 1.0 = 2 ´ 10-7 m = 2000 Å Substituting in Eq. (i), we get l = 4000 Å, 2680 Å, 1600Å etc. So that answer is 4000 Å. 3 f 5. I max = I max cos2 æç ö÷ 4 è 2ø f p 5p and = \ 2 12 6 p f= \ 6 5p æ 2p ö æ 2p ö æ yd ö and = ç ÷ (Dx ) = ç ÷ ç ÷ 3 è lø è løèDø lD (6000 ´ 10-10 ) (1) y1 = = \ 12d (12) (10-3 ) Here,

= 0.05 ´ 10-3 m = 0.05 mm æ lD ö y2 = 5 ç ÷ = 5 ´ 0.05 mm è 12d ø = 0.25 mm D y = y2 - y1 = 0.2 mm (m - 1) tD 3.5 lD 6. Shift = = 3.5, w = d d 3.5l t= m -1 =

(3.5) (6000 ´ 10-10 ) 1.5 - 1

= 4.2 ´ 10-6 m = 4.2 mm

Q

S1

P

O S2

3a = 15l S

| Dx | = 15, therefore maxima At points R and S Dx = 0, therefore maxima. Between P and R (and similarly in other three quadrants), we will get 14 maxima corresponding to, Dx = l , 2l ××× 14 l. Therefore, total maximas are 60. yd d læ lD ö 9. Dx = = (w/ 4 ) = ç w = ÷ D D 4 è d ø p æ 2p ö f = ç ÷ Dx = or 90° 2 è lø 2 q I = I max cos 2 I 1 2 \ = cos 45° = I max 2 (m - 1) tD 10. Shift = d At m = 1,shift = 0 P

I0

Q Zero R

I0

Therefore, intensity at centre is maximum or I 0.

Chapter 32 As m increases fringes shift upwards as shown in figure. So, intensity at P first decreases to zero (as Q reaches at P), then it further increases to I 0 (as point R reaches to P). 3 f 11. I max = I max cos2 4 2 f p p æ 2p ö æ 2p ö \ = or f = = ç ÷ (Dx ) = ç ÷ (m - 1) t 2 6 3 è lø è lø \ t=

l 6000 ´ 10-10 = 6 (m - 1) 6 (1.5 - 1)

= 0.2 ´ 10-6 m = 0.2 mm

12. Dx1 = (m 1 - 1), Dx2 = (m 2 - 1)t \

Dx = (m 1 - m 2 ) t = (1.52 - 1.40) (10400 nm ) = 1248 nm For maximum intensity, Dx = 1248 = nl 1248 (n = 1, 2, 3×××) l= \ n For n = 2, l = 624 nm and for n = 3, l = 416 nm

Interference and Diffraction of Light — 479 5. Fringe pattern shifts in the direction of slab. But, (m - 1) tD d So, actual shift will depend on the values of m , t , D and d. 6. For overlapping of maxima n1l 1D n2l 2D = d d n1 l 2 7 14 or = = , ××× n2 l 1 5 10 Shift =

Þ 14th order maxima of l 1 will coincide with 10th order maxima of l 2. For overlapping of minima (2n1 - 1) l 1D (2n2 - 1) l 2D = 2d 2d 2n1 - 1 l 2 7 = = \ 2n2 - 1 l 1 5 (c) Option with n1 = 11 and n2 = 8 gives this ratio.

Comprehension Based Questions 1. l =

l0 lD l 0D , w= = m d md

More than One Correct Options

=

1. lv is least. Therefore wv is minimum (as w µ l). Hence, the fringe next to centre will be violet. At centre, Dx = 0 for all wavelengths. Hence, all wavelengths interfere constructively at centre. So, it is white. 2. (c) I max = ( I 1 + I 2 )2

= 0.63 ´ 10-3 m = 0.63 mm 2. Dy = 7w - 3w = 4 w = 4 ´ 0.63 mm = 2.52 mm

I min = ( I 1 - I 2 )2 I1 = I2 = I0 I max = 4 I 0 and I min = 0 I When I2 = 0 , 2 then I max < 4 I 0 and I min > 0. 1 3. Dx0 = d sin q = (10-3 ) æç ö÷ è 2ø When

= 5 ´ 10-4 m = (103 ) l Since, Dx0 is integer multiple of l, it will produce maximum intensity or 4 I 0 at O. lD (5 ´ 10-7 ) (2) w= = d (10-3 ) = 10-3 m = 1 mm At 4 mm, we will get 4th order maxima.

(6300 ´ 10-10 ) (1.33) 1.33 ´ 10-3

æm 2 ö l - 1÷÷ t = 2 m è 1 ø

3. Dx = çç \

t=

=

l æm 2 ö 2 çç - 1÷÷ m è 1 ø (6300 ´ 10-10 / 133 . ) æ 1.53 ö - 1÷ 2ç è 1.33 ø

= 1575 . ´ 10-3 m = 1575 . mm

4. Fringe width remains unchanged by the introduction of glass sheet.

480 — Optics and Modern Physics \ I 0 = 4 I S3 or I S4 = 3I Same explanations can be given for (c) and (d) options. 6. (a) Fringe pattern will shift in the direction of slab. (b) No interference pattern will be obtained due to single slit. \ I = I S2 = I 0 » uniform (c) Fringe pattern will shift in the direction of slab. (d) No interference pattern will be obtained due to two real incoherent sources. I = I S1 + I S2 » uniform

Match the Columns f 1. I = 4 I 0 cos2 æç ö÷ è 2ø 2p 2. f = Dx and then apply, l æ fö I = 4 I 0 cos2 ç ÷ è 2ø

3.

y6 y5 y4 y3 y2 y1

Y6

Y ® maxima y ® minima

Y5 Y4 Y3 Y2 Y1

Subjective Questions 1. I 1 = 0.1I 0 , I 2 = 0.081I 0

w

4. (a) Inclined rays and slab both will shift the fringe

2

pattern upwards. So, zero order maxima will definitely lie about O. (b) Inclined rays will shift the fringe pattern upwards, but slab will shift the fringe pattern downwards. Hence, zero order maxima may lie above O, below O or at O. (c) Same explanations can be given for this option.

(d)

S2

0.09I0 1

I

P O

DX p = 0, where S 1 P = S 2 P I 5. I = I max = 4 I 0 Þ I 0 = 4 lD yd l (a) For y= , Dx = 2d D 2 \ I S 3 = I S4 = 0 for Dx = l/2 I0 = 0 lD yd l (b) For y = , Dx = = 6d D 6 2p æ 2p ö f = ç ÷ (Dx ) = = 60° 6 è lø q Now, I S3 = I S4 = I max cos2 2 3 2 = I cos 30° = I 4

0.1I0 0.9I0

\

S1

0.081I0

\

I 1 10 = I2 9 I max æç I 1 / I 2 + 1 ö÷ = I min çè I 1 / I 2 – 1÷ø

2

= (19)2

2. Q \

= 361 sin i m = sin r 4 sin 53° 4 / 5 = = 3 sin r sin r

\

37° 5

53° 3

\ \

3 5 r = 37°

sin r =

4

Ans.

Chapter 32

Interference and Diffraction of Light — 481

Refer figure (a) 1 i i

2 i

B

A

S1

1

C

r

S

t

2 S2

r r 15 cm

D (a)

D

Distance between two slits, d = 1.5 mm, D = 30 cm lD Fringe width, w = d (5.0 ´ 10–7 ) (0.3) = = 10–4 m (1.5 ´ 10–3 )

1 E

= 0.1 mm

2

Ans.

4. l = 0.25 m, d = 2 m = 8l

i B

30 cm 60 cm

Dx1 = between 2 and 1 = 2 ( AD ) = 2BD sec r = 2t sec r Their optical path Dx1 = 2mt sec r. Refer figure (b)

A

0.5 mm 0.25 mm 0.25 mm 0.5 mm

C

B

r D (b)

C

S1

\ (Dx )net = Dx1 – Dx2 = 2 mt sec r – 2t (tan r) (sin i ) 4 5 3 4 32 = 2´ ´t ´ – 2´t ´ ´ = t 3 4 4 5 15 Phase difference between 1 and 2 is p. \ For constructive interference, 32 l 15l 15 ´ 0.6 or t = t= = 15 2 64 64 1 1 1 3. Applying lens formula, – = v u f 1 1 1 + = v 15 10 \

Ans.

D

At A and C , Dx = d = 8l, i.e. maximum intensity is obtained. At B and D, Dx = 0, i.e. again maximum intensity will be obtained. Between A and B seven maximas corresponding to Dx = 7l, 6l , 5l , 4 l , 3l , 2l and l will be obtained. Similarly, between B and C ,C and D , and D and A. \ Total number of maximas = 4 ´7+ 4 Ans. = 32

5. (a) Q

v 30 = =–2 u –15

Dx = d cos q cos q = 1 –

v = 30 cm m=

A

d

D x2 = AC sin i = (2t tan r) sin i

= 0.14 mm

S2

\

q2 2

æ q2 ö Dx = d çç1 – ÷÷ 2ø è

(When q is small)

482 — Optics and Modern Physics æ y2 ö ÷ = d çç1 – 2D 2 ÷ø è

d sin f = Dx1

7. (a)

= (50 ´ 10-4 ) sin 30° P y

q S1

q = 30°

f

q

C

O

S2

Dxnet = 0

d

= 2.5 ´ 10-3 cm

For nth maxima Dx = nl \ y = radius of nth bright ring =D

nl ö æ 2 ç1 – ÷ d ø è

Dx2 = (m - 1)t æ3 ö = ç - 1÷ (0.01) è2 ø

Ans.

= 5.0 ´ 10-3 cm

(b) d = 1000l At O , D x = d = 1000l i.e. at O, 1000th order maxima is obtained. Substituting n = 998 in, y=D

(c) n = 998

\ Central maxima will be obtained at q = 30° below C.

\ Ans.

= 50

6. (a) The optical path difference between the two

Ans.

(c) Number of fringes that will pass if we remove the slab Path difference due to slab = l

waves arriving at P is yd yd Dx = 1 + 2 D1 D2 (1) (10) (5) (10) + 103 2 ´ 103

=

= 3.5 ´ 10–2 mm = 0.035 mm As, Dx = 70l \ 70th order maxima is obtained at P. yd (b) At O , Dx = 1 = 10–2 mm D1 = 0.01 mm As Dx = 20l \ 20th order maxima is obtained at O. (c) (m – 1) t = 0.01 mm 0.01 t= = 0.02 mm = 20 mm \ 1.5 – 1

2.5 ´ 10-3 l 2.5 ´ 10-3 = 500 ´ 10-7

n=

Ans.

=

(also)

(b) At C DX net = 2.5 ´ 10-3 cm = nl

nl ö æ 2ç1 – ÷ d ø è

We get the radius of second closest ring r = 6.32 cm

Dx2 - Dx1 = 2.5 ´ 10-3 cm = Dx1

5 ´ 10-3 500 ´ 10-7

= 100

Ans.

8. (a) (Dx )net = 0

Ans.

Since, the pattern has to be shifted upwards, therefore, the film must be placed in front of S 1.

\

y1d y2d = D1 D2

\

d /2 y = 1.5 2.0

or

y=

d 1.5 6 = 1.5

= 4 mm

Ans.

Chapter 32 (b) At O , net path difference, yd Dx = 1 D1 (d / 2) (d ) = D1 =

= 120 ´ 10–7 m l = 6000 Å = 6 ´ 10–7 m As, Dx = 20l, therefore at O bright fringe of order 20 will be obtained. æ fö (c) I = I max cos2 ç ÷ è 2ø 3 æ fö I max = I max cos2 ç ÷ 4 è 2ø f p = 2 6

p æ 2p ö = ç ÷ (m – 1) t 3 è lø l t= 6 (m – 1) 6000 = 6 (1.5 – 1)

f= \

(6 ´ 10–3 )2 2 ´ 1.5

= 12 ´ 10–6 m

\

Interference and Diffraction of Light — 483

= 2000 Å æ

9. (a) ç1 – è

\

Ans.

1ö 3l ÷t= mø m 3l (m – 1) 3 ´ 0.78 = 1.3 – 1

t=

= 7.8 mm yd æ 1ö 4l (b) Upwards – ç1 – ÷ t = D è mø m Solving, we get y = 4.2 mm 1ö yd 4 l æ Downwards t ç1 – ÷ + = mø D m è Solving, we get y = 0.6 mm

Ans.

Ans.

Ans.

33. Modern Physics - I INTRODUCTORY EXERCISE

33.1

hc h ,P= l l 2. Number of photons emitted per second, Power of source P Pl N1 = = = Energy of one photon (hc/ l ) hc

2. En =

1. E =

4.

(qm)d = (qm)P

3. The expressions of kinetic energy, potential energy and total energy are me4 Kn = 2 2 2 Þ 8e0n h

ld = la

En =

unchanged while in electric field it will increase. Further, 1 lµ K h 6. l= mv 6.63 ´ 10-34 (a) l = (46 ´ 10-3 ) (30) = 4.8 ´ 10-34 m 6.63 ´ 10-34 9.31 ´ 10-31 ´ 107

= 7.12 ´ 1011 m

INTRODUCTORY EXERCISE

33.2

1. Z = 3 for doubly ionized atom E µ Z 2 Ionization energy of hydrogen atom is 13.6 eV. \ Ionisation energy of this atom = (3)2 (13.6) = 122.4 eV

- me4 4e20n2h2

- me4 8e20n2h2

Kn µ

1 n2

Þ Un µ Þ

En µ -

1 n2

and

1 n2

In the transition from some excited state to ground state, the value of n decreases, therefore kinetic energy increases, but potential and total energy decrease.

4.

5. Kinetic energy in magnetic field remains

(b) l =

Un =

1 ´2 = 2 1 ´1

h 1 Þ lµ 2Km m ma 4 = = 2 md 2

l=

n=3

æ hö Ln = n ç ÷ è 2p ø

Now,

At a distance r, these photons are falling on an area 4 pr2. \ Number of photons incident per unit area per unit time, N1 Pl = N2 = 2 4 pr (4 pr2 ) hc h 1 3. l = Þ lµ 2qVm qm lP = ld

- 13.6 = - 1.51 Þ (n)2

n=4 n=3

n=4 n=3

n=2

n=2

n=1

First line of Balmer series

n=1

Second line of Balmer series

For hydrogen or hydrogen type atoms, æ 1 ö ç - 1÷ 2 2 ç nf ni ÷ è ø In the transition from ni ¾® nf , 1 lµ \ æ 1 1ö Z2 ç 2 - 2 ÷ ç nf ni ÷ è ø æ ö 1 1 Z12 ç 2 - 2 ÷ ç nf ni ÷ l2 è ø1 \ = l1 æ 1 1ö Z22 ç 2 - 2 ÷ ç nf ni ÷ è ø2 1 = RZ 2 l

æ 1 1ö l 1Z12 ç 2 - 2 ÷ ç nf ni ÷ è ø1 l2 = æ ö 1 1 Z22 ç 2 - 2 ÷ ç nf ni ÷ è ø2

Modern Physics-I — 485

Chapter 33 Substituting the values, we have 1ö æ 1 (6561 Å) (1)2 ç 2 - 2 ÷ 2 3 è ø = 1215 Å = 1ö 2 æ 1 (2) ç 2 - 2 ÷ 4 ø è2 \ Correct option is (a). 5. The series in U-V region is Lyman series. Largest wavelength corresponds to minimum energy which occurs in transition from n = 2 to n = 1. 1 R …(i) 122 = \ 1ö æ1 ç 2 - 2÷ 2 ø è1 The smallest wavelength in the infrared region corresponds to maximum energy of Paschen series. 1 R …(ii) l= \ 1ö æ 1 ç 2 ÷ ¥ø è3 Solving Eqs. (i) and (ii), we get l = 823.5 nm \ Correct option is (b). 6. The first photon will excite the hydrogen atom (in ground state) to first excited state (as E2 - E1 = 10.2 eV). Hence, during de-excitation a photon of 10.2 eV will be released. The second photon of energy 15 eV can ionise the atom. Hence, the balance energy, i.e. (15 - 13.6) eV = 1.4 eV is retained by the electron. Therefore, by the second photon an electron of energy 1.4 eV will be released. \ Correct answer is (c). 7. In second excited state n = 3, So, while

æ hö lH = lLi = 3 ç ÷ è 2p ø E µ Z 2 and ZH = 1, ZLi = 3

So, | ELi | = 9 | EH | or | EH | < | ELi | 8. Energy of infrared radiation is less than the energy of ultraviolet radiation. In options (a), (b) and (c), energy released will be more, while in option (d) only, energy released will be less. 9. For hydrogen and hydrogen like atoms, En = - 13.6

(Z 2 ) eV (n2 )

Therefore, ground state energy of doubly ionized lithium atom (Z = 3, n = 1) will be

(3)2 (1)2 = - 122.4 eV \ Ionization energy of an electron in ground state of doubly ionized lithium atom will be 122.4 eV. 10. Shortest wavelength will correspond to maximum energy. As value of atomic number (Z) increases, the magnitude of energy in different energy states gets increased. Value of Z is maximum for doubly ionised lithium atom (Z = 3) among the given elements. Hence, wavelength corresponding to this will be least. \ Correct option is (d). E1 = (-13.6)

11. 5E - E = hf

Þ

E=

hf 4

…(i)

Between 5E and 4E 5E - 4 E = hf1 E f [from Eq. (i)] \ f1 = = h 4 Between 4E and E 4 E - E = hf2 3E æ fö 3 f2 = =3ç ÷ = f \` h è4ø 4

12. Longest wavelength means minimum energy. (DE )min = E3 - E2 13.6 13.6 =+ = 1.9 eV 9 4 12375 = 6513 l (in Å) = 1.9 or l » 651 nm

INTRODUCTORY EXERCISE

33.3

1. K a transition takes place from n1 = 2 to n2 = 1 é 1 1 1 ù = R (Z - b)2 ê 2 ú l (2)2 û ë (1) For K-series, b = 1 1 µ (Z - 1)2 \ l l Cu (ZMo - 1)2 (42 - 1)2 Þ = = lMo (ZCu - 1)2 (29 - 1)2 41 ´ 41 1681 = = = 2.144 28 ´ 28 784

\

2. Cut off wavelength depends on the applied voltage not on the atomic number of the target. Characteristic wavelengths depend on the atomic number of target.

486 — Optics and Modern Physics 3.

1 µ (Z - 1)2 l

2. K max = E - W = hf - hf0 2

2

l 1 æ Z2 - 1ö 1 æ Z2 - 1ö ÷÷ or ÷ \ = çç =ç l 2 è Z1 - 1ø 4 çè 11 - 1 ÷ø Solving this, we get Z2 = 6 \ Correct answer is (a). 4. Wavelength l k is independent of the accelerating voltage (V), while the minimum wavelength l c is inversely proportional to V. Therefore, as V is increased, l k remains unchanged whereas l c decreases or l k - l c will increase. 5. The continuous X-ray spectrum is shown in figure.

= h ( f - f0 ) \ K max µ ( f - f0 ) 3. K max = E - W 1.2 = E - W 4.2 = 1.5 E - W Solving these equations, we get W = 4.8 eV = hf0 4.8 ´ 1.6 ´ 10-19 \ f0 = 6.63 ´ 10-34

…(i) …(ii)

= 1.16 ´ 1015 Hz

4. Energy corresponding to 248 nm wavelength

El

1240 eV = 5 eV 248 Energy corresponding to 310 nm wavelength 1240 = eV = 4 eV 310 KE 1 u12 4 = = KE 2 u22 1 =

lmin

l

All wavelengths >l min are found, where 12375 Å l min = V (in volt ) Here, V is the applied voltage. æ 1 1ö 6. DE = hn = Rhc (Z - b)2 çç 2 - 2 ÷÷ è n1 n2 ø For K-series, b = 1 æ 1 1ö \ n = Rc (Z - 1)2 çç 2 - 2 ÷÷ n n è 1 2ø æ1 1ö 4.2 ´ 1018 = (1.1 ´ 107 ) (3 ´ 108 ) (Z - 1)2 ç - ÷ è1 4 ø (Z - 1)2 = 1697

or or

Z - 1 » 41 Z = 42

INTRODUCTORY EXERCISE

1. K min = 0 and

Þ

16 - 4W = 5 - W

33.4

11 = 3W 11 Þ W = = 3.67 eV 3 ~ 3.7 eV = 5. Saturation current is proportional to intensity while stopping potential increases with increase in frequency. Hence, fa = fb while I a < I b Therefore, the correct option is (a). 12375 6. l (in Å) = W (eV) 12375 = Å » 3093 Å 4.0 or l » 309.3 nm or » 310 nm

Note l(in Å) = K max = E - W 12375 = - 3.0 2000 » 3.19 eV

5 eV - W 4 eV - W

Þ

Substituting the values,

\

=

12375 hc comes from W = W ( eV) l

7. Stopping potential is the negative potential applied to stop the electrons having maximum kinetic energy. Therefore, stopping potential will be 4 V.

Exercises LEVEL 1

10. In X-ray spectrum, all wavelengths greater than l min are obtained.

Assertion and Reason h l 1 \ E and P µ l Speed of all wavelengths (in vacuum) is c.

2. E =

hc l

and

P=

3. Intensity = energy incident per unit area per unit time or I = n(hf ) . Here, n = number of photons incident per unit area per unit time. f = frequency of incident photons. Hence, I µ nf . Hence, intensity can be increased either by increasing n or f . But saturation current only depends on n (I s µ m). By increasing n and decreasing f , we can increase the saturation current even without increasing the intensity. 5. Let us take n = 3 as N = 1. Then, n = 6 means N = 4. So, total number of emission lines between N = 1 and N = 4 are N (N - 1) 4 ´ 3 = =6 2 2 6. eV0 = hn - hn 0 h or V0 = (n - n 0 ) e h hn 0 V0¢ = (2n - n 0 ) + e e hn 0 = 2V0 = > 2V0 e 7. Energy of X-ray > 13.6 eV

8. Dl = l Ka - l min Here, l min (in Å) =

12375 V (in volts)

If V is increased, l min decreases. Therefore, Dl increases. E2 = - 3.4 eVü 9. ý U 2 = - 6.8 eV þ E1 = - 13.6 eVü ý U 1 = - 27.2 eVþ E2 > E1 , similarly U 2 > U 1

Objective Questions 1. K max = hf - W K max versus f graph is a straight line of slope h (a universal constant) c 2. v1H = 2.19 ´ 106 m/s » 137 3. Let a-particles are n and b-particles are m. Then, …(i) 86 - 2n + m = 84 …(ii) 222 - 4 n = 210 Solving these two equations, we get n = 3 and b=4 12375 4. l min = in Å V (in volts) =

12375 = 0.62 Å 20 ´ 103

K max = 18 ´ 103 eV =

5. \

vmax =

1 2 mvmax 2

2 ´ 18 ´ 103 ´ 1.6 ´ 10-19 9.1 ´ 10-31

= 8 ´ 107 m/s

6. (2pr2 ) = 2l 2 (2pr3 ) = 3l 3 \ Now,

l 2 3r2 = l 3 2r3

…(i)

r µ n2 2

r2 æ 2 ö =ç ÷ r3 è 3 ø Substituting in Eq. (i), we get l2 2 = l3 3

\

7. E µ Z 2 \ \

(- 13.6) (Z 2 ) = - 122.4

Z=3 1 8. l min µ µ V -1 V % change in l min = (- 1) (% change in V ) for small % changes

488 — Optics and Modern Physics 9. E µ

Z2 n2

16. Q (- 13.2) ´ (Z )2 = - 13.6 (2)2

\ \

For

12375 V (in volts)

l min = 1Å

\

V = 12375 V » 12.4 kV h l 12375 12. l = = 7734 Å 1.6 12375 13. DE = = 11.4 eV 1085 Third Balmer line is corresponding to the transition, n = 5 to n = 2. E5 - E2 = 11.4 E1 E \ - 1 = 11.4 (5)2 (2)2

\

\

V0 = =

» 1.8 volt

f1

Slope µ

2

1 1 - 2 2 n1 n2

1 - 1/ 9 Slope1 32 = = Slope2 27 1 14 For K b , n1 = 1, n2 = 3 For K a , n1 = 1, n2 = 2 æ 1 1 1ö 1 1 5R 19. = R çç 2 - 2 ÷÷ = R æç - ö÷ = 4 9 36 l n n è ø è 1 2ø 36 l= \ 5R 20. Energy of electrons = 10000 eV

…(ii) \

\

hc (l 1 - l 2 ) el 1l 2 (6.63 ´ 10-34 ) (3 ´ 108 ) (110) 1.6 ´ 10-19 ´ 330 ´ 220 ´ 10-9

2

\

Subtracting Eq. (i) from Eq. (ii), we get hc (l 1 - l 2 ) eV0 = l 1l 2 \

æ 1 1ö f µ (Z - b) çç 2 - ÷÷ n n 2ø è 1

18.

increased it by 27.2 eV. It implies that we have increased it by 27.2 eV in all states. U 2 = - 6.8 eV \ U 2¢ = (- 6.8 + 27.2) eV = + 20.4 eV E2¢ = U 2¢ + K 2 = (20.4 + 3.4) eV = 23.8 eV hc …(i) 15. eV0 = -W l1 hc -W l2

æ Z - 1ö ÷÷ f2 = çç 2 è Z1 - 1ø

2

14. OtherwiseU 1 = - 27.2 eV. Therefore, we have

e (2V0 ) =

f µ (Z - 1)2

æ 51 - 1ö ÷÷ f = çç è 31 - 1ø 25 = f 9

E1 = - 54.28 eV | E1 | = 54.28 eV

Similarly ,

…(ii)

f2 (Z2 - 1)2 = f1 (Z1 - 1)2

11. P =

\

…(i)

Solving these equations, we get hc W = 3l 17. f µ (Z - 1) for K-series

Z=2

10. l min (in Å) =

hc -W l hc 4E = -W l/ 3 E=

21.

150 = 0122 . 10000 12375 l 2 (in Å) = » 1.2 10000 l1 » 0.1 l2 hc e (5V0 ) = -W l hc eV0 = -W 3l l 1 (in Å) =

…(i) …(ii)

Chapter 33 h 2Em where, \

Solving these equations, we get hc W = 6l 22. eV0 = 2hn 0 - hn 0

30. l =

\ hn 0 - eV0 In second condition, eV = 3hn 0 - hn 0 = 2hn 0 - 2eV0 or V = 2V0 23. Frequency µ energy and energy µ Z 2

24. E µ

Z2 n2 Z2 =1 n2 n=Z =3

\

or 1 25. mv12 = hn 1 - W 2 1 2 mv2 = hv2 - W 2 From these two equations, we can see that 2h v12 - v22 = (n 1 - n 2 ) m 26. Longest wavelength of Lyman series means, minimum energy corresponding n = 2 to n = 1. \ (E2 - E1 )H = (En - E2 )He+ - 13.6 13.6 - 13.6(Z )2 13.6(Z )2 + = + (2)2 (1)2 n2 (2)2

\

Putting Z = 2, we get n = 4. æ 1 1 1ö 27. = R (Z - 1)2 çç 2 - 2 ÷÷ l n n è 1 2ø For K a -line, n1 = 1 and n2 = 2

28.

La

Ka

\ \

29. Q l = \

l = constant Em = constant 1 or Eµ m 31. U 1 = - 27.2 eV U 1 is assumed zero. Therefore, it is increased by 27.2 eV at all points. Eµ = 0 under normal conditions. Hence, in charged conditions it is 27.2 eV. 32. Number of photons emitted per second Energy radiated per second P = = Energy of one photon hf 1000 = 6.63 ´ 10-34 ´ 880 ´ 103 = 1.7 ´ 1030 12375 33. E = = 4.125 eV 3000 Since, E < W no photoelectric effect will be observed. 1 34. E µ 2 n - 13.6 \ E5 = (5)2 = - 0.544 eV 35. K max = E - W = 2 eV = 3.2 ´ 10-19 J h 36. meve = l h \ ve = lme =

EKb = EKa + ELa hg 2 = hg 1 + hg 3 g2 = g1 + g3 h 1 or l µ 2qVm qm 2´4 1 ´1

6.63 ´ 10-34 (5200 ´ 10- 10 ) ´ 9.1 ´ 10-31

» 1400 m/s 1 2 37. K max = mvmax = E-W 2 2(E - W ) \ U max = m 12375 Here, E= = 4.125 eV 3000

Kb

l P ( qm )a = = l a ( qm )P

Modern Physics-I — 489

= =2 2

2 ´ 3.125 ´ 1.6 ´ 10-19 9.1 ´ 10-31

» 1.09 ´ 106 m/s

490 — Optics and Modern Physics Subjective Questions 1.

4. (a) p = La

Ka

Kb

ELa = EKb - EKa \

hc hc hc = l La l Kb l Ka

or

l La = =

l Ka l Kb l Ka - l Kb 0.71 ´ 0.63 = 5.59 nm 0.71 - 0.63

2. l Ka = 0.71nm = 7.1 Å 127375 = 1743 eV 7.1 \ E1 = E2 - 1743 = - 2870 - 1743 = - 4613 eV l Kb = 0.63 nm = 6.3 Å 12375 E3 - E1 = 6.3 = 1964 eV \ E3 = E1 + 1964 = - 4613 + 1964 = - 2649 eV 3. Number of photons incident per second Power = Energy of one photon P Pl = = (hc/ l ) hc E2 - E1 =

Number of electrons emitted per second = 0.1% of Pl Pl = hc 1000 hc \ Current = Charge (on photoelectrons per second) Ple = 1000 hc (1.5 ´ 10-3 ) (400 ´ 10-9 ) (1.6 ´ 10-19 ) = (1000) (6.63 ´ 10-34 ) (3 ´ 108 ) = 0.48 ´ 10-6 A = 0.48 mA

E c

Þ

E = pc

12375 E (in eV) 12375 = = 8035 Å 1.54 » 804 nm So, this wavelength lies in ultraviolet region. c 5. (a) f = l (b) Q Number of photons emitted per second Power of source P N1 = = Energy of one photon hc/l Pl N1 = hc (75) (600 ´ 10-9 ) = (6.63 ´ 10-34 ) (3 ´ 108 ) (b) l =

» 2.3 ´ 1020 photons/s E 6. (a) E = hf Þ f = h c (b) l= f p2 or K µ p2 2m If momentum is doubled, kinetic energy becomes four times. E or E µ p (for a photon) (b) p = c If p is doubled, E will also become two times. 8. (a) Number of photons incident per second = number of photons absorbed per second Power = Energy of one photon P Pl N = = hc/l hc

7. (a) K =

(b) Force = Rate of change of momentum = (Number of photons absorbed per second) ´ (momentum of one photon) æ Pl ö æ h ö P =ç ÷ç ÷= (Q P = power) è hc ø è l ø c

9. See the hint of above example. P . If c surface is perfectly reflecting, then force will be 2P . c If surface is perfectly absorbing, force is

Modern Physics-I — 491

Chapter 33 In this case, 0.7P 0.3 (2P ) 1.3P F= + = c c c 10. Force = Rate of change of momentum é hù = 2 [ N ] ê ú × cos 60° ë lû N = number of photons striking per second h = momentum of one photon. l h h 11. l = = p mv Since, wavelength is too short, it does not behave wave like property. h 12. l = mv h h 13. (a) l = Þ p= p l (b) l =

l Þ 2Km

K =

h2 2ml2

3RT = v (say) M h h M and l= = mv m 3RT Substituting the values, we get

14. vrms =

l=

6.63 ´ 10-34 (2/ 6.02 ´ 1023 ) ´ 10-3

2 ´ 10-3 3(8.31) (20 + 273)

= 1.04 ´ 10-4 m = 1.04 Å

15. KE = | Total energy | = 3.4 eV l (in Å) =

150 for an electron KE (in eV)

150 3.4 = 6.6Å =

E1 = - 13.6 eV

16. (a)

KE = | E1 | = 13.6 eV 150 for an electron l 1 (in Å) = KE (in eV)

l4 =

» 13.3 Å 2pr4 = 2p (n)2 r1

150 = 3.32Å 13.6 2pr1 = 2p (0.529Å) » 3.32 Å (b) E4 =

E1 - 13.6 = = - 0.85 eV (4 )2 (4)2

(as r µ n2)

= 53.15 Å » 4 l 4

17. Energy required to remove first electron is 24.6 eV. After removing first electrons from this atom, it will become He+ . E1 = - (13.6)(2)2 (as E µ Z 2 and Z = 2) = - 54.4 eV \ Energy required to remove this second electron will be 54.4 eV. \ Total energy required to remove both electrons = 24.6 + 54.4 = 79 eV 12375 18. En = E1 = 1023 - 13.6 or + 13.6 = 12.1 (n)2 Solving this equation, we get n = 3 n (n - 1) =3 \Total possible emission lines = 2 Longest wavelength means, minimum energy, which is corresponding to n = 3 to n = Z 12375 \ l max = E3 - E2 13.6 13.6 Here, E3 - E2 = + = 1.9 eV (3)2 (4 )2 12375 l max = = 6513 Å \ 1.9 19. In hydrogen atom, DE1 = E3 - E1 = 12.1 eV E µ Z2 \ For

\

=

150 as KE = | E4 | = 0.85 eV 0.85

\

Z = 3, DE2 = (3)2 DE1 = (3)2 (12.1) = 108.9Å 12375 l= » 113 Å 108.9

20. En - E1 = DE1 + DE2 12375 12375 é 13.6 ù \ ê - 2 + 13.6ú (Z )2 = + 1085 304 ë n û Putting Z = 2 for He+ , we get n = 5

492 — Optics and Modern Physics 21. (a) Let these two levels are n1 and n2. Then, - 0.85 = - 0.544 = -

13.6 (Z ) n12

25. F = -

2

13.6 (Z )2 n22

…(i) Now, …(ii)

Total number of lines between n2 and n1are given by (n2 - n1 ) (n2 - n1 + 1) …(iii) = 10 2 Solving these equations, we get Z = 4 , n1 = 16 and n2 = 20 (b) Smallest wavelength means maximum energy. DEmax = En2 - En1 = - 0.544 + 0.85 = 0.306 eV 12375 l min = = 40441Å 0.306 22. (a) Ionization energy = 15.6 eV \ Ionization potential = 15.6 V (b) Eµ - E2 = 5.3 eV 12375 l= = 2335Å \ 5.3 (c) E3 - E1 = 12.52 eV \ Excitation potential is 12.52 V. (d) E3 - E1 = 12.52 eV 12375 \ l= Å 12.52 = 988.41 Å 1 1 \ Wave number = = m -1 l 988.41 ´ 10-10 = 1.01 ´ 107 m -1 12375 eV = 1.44 eV 23. (a) Energy of photon = 8600 \ Internal energy after absorption = - 6.52 + 1.44 = - 5.08 eV 12375 (b) Energy of emitted photon = = 2.946 eV 4200 Internal energy after emission = - 2.68 - 2.946 = - 5.626 eV

24. K max (in eV) = E - W =

12375 - 4.3 2000

= 1.9 eV Therefore, stopping potential is 1.9 V.

or or Now, or

dU = - m2w2r dr mv 2 = | F | = m2w2r r v µr v = ar h m vr = n 2p h m (ar)r = n 2p

\

26.

rµ n

1 æ 1 1ö µç - ÷ l çè n12 n22 ÷ø l Kb

\

l Ka

=

(1/ n12 - 1/ n22 )Ka

=

(1/ n12 - 1/ n22 )Kb (1 - 1/ 4 ) 27 = (1 - 1/ 9) 32

27 27 l Ka = l0 32 32 27. DE1 = 50% of 50 keV = 25 keV \

l Kb =

\

l1 =

= 49.5 pm DE2 = 50% of 25 keV = 12.5 keV 12375 l2 = = 0.99Å 12.5 ´ 103

\

28.

1 l Ka or

12375 = 0.495 Å 25 ´ 103

= 99 pm æ 1 1ö = R (Z - 1)2 çç 2 - 2 ÷÷ n n è 1 2ø 1 1ö æ = 1.097 ´ 107 (Z - 1)2 ç1 - ÷ -10 4ø 0.76 ´ 10 è

Solving this equation, Z = 41

29. 26 pm = 0.26 Å Now,

0.26 =

12375 12375 V 1.5 V

30. Similar type of example is given in the theory. 31. Transition if from L-shell to K-shell. 1 = R (Z - 1)2 l

æ 1 1ö ç - ÷ ç n2 n2 ÷ è 1 2ø

Modern Physics-I — 493

Chapter 33 æ 1 1ö ç - ÷ ç n2 n2 ÷ è 1 2ø

or

f = R (Z - 1)2 c

or

4.2 ´ 108 = 1.1 ´ 107 (Z - 1)2 3 ´ 108

12375 = 6.875 eV 1800 K max or K = E - W = 4.875 eV 2Km r= Bq

38. E =

1ö æ ç1 - ÷ 4ø è

Solving we get, Z » 42

32. (a) P = Vi = (40 ´ 103 ) (10 ´ 10-3 ) = 400 W 1% of 400 W is 4 W. (b) Heat generated = 400 W - 4 W = 396 W = 396 J/s 33. Stopping potential = 10.4 V K max = 10.4 eV E = W + K max = 1.7 + 10.4 = 12.1 eV 12375 l= = 1022 Å 12.1 12.1 eV is the energy gap between n = 3 and n = 1 in hydrogen atom. (6.63 ´ 10-34 ) (1.5 ´ 1015 ) 34. E = hf = eV = 6.21 eV 1.6 ´ 10-19 K max = E - W = 6.21 - 3.7 = 2.51 eV 35. l 0 = 5000 Å 12375 W = = 2.475 eV 5000 Stopping potential is 3V. Therefore, K max = 3 eV E = W + K max = 5.475 eV 12375 l= = 2260Å 5.475 36. (a) f0 = fA = 10 ´ 1014 Hz = 1015 Hz

=

39. fhigher

5 ´ 10-5 ´ 1.6 ´ 10-19

= 0.148 m 8 ´ 1015 -1 = s 2p E=

\

(b) W = | K max |c = 4 eV W (c) W = hf0 Þ h = f0 12357 37. E1 = = 4.125 eV 3000 12357 E2 = = 2.0625 eV 6000 Maximum speed ratio is 3 : 1. Therefore, maximum kinetic ratio is 9 : 1. Now, …(i) 9 K max = 4.125 - W …(ii) K max = 2.0625 - W Solving these two equations, we get W » 1.81eV. and K max = 0.26 eV 1 2 we can find vmax . Here, m Putting K max = mvmax 2 is the mass of electron.

2 ´ 4.875 ´ 1.6 ´ 10-19 ´ 9.1 ´ 10-31

=

40.

hf eV 1.6 ´ 10-19 (6.63 ´ 10-34 ) (8 ´ 1015 ) 1.6 ´ 10-19 ´ 2p

= 5.27 eV K max = E - W = 3.27 eV w (1.57 ´ 107 ) c f = = 2p 2p hf E= eV 1.6 ´ 10-19 =

(6.63 ´ 10-34 ) (1.57 ´ 107 ) (3 ´ 108 ) eV 1.6 ´ 10-19 ´ 2 ´ p

= 3.1 eV K max = E - W = 1.2 eV

LEVEL 2 Single Correct Option mv 2 GMm …(i) = 2 r r h (for n = 1) …(ii) mvr = 2p Solving these two equations, we can find v and r. Then, 1 GMM E = mv 2 2 r M q 2. = = constant L 2m M µL nh and or L µ n L= 2p \ M µn

1.

494 — Optics and Modern Physics Momentum of photon = Momentum of hydrogen atom 3Rh 3Rh \ v= = Mv Þ 4M 4 15 9. wmax = 1.5 ´ 10 rad /s = 2pfmax

3. E2 - E1 = 40.8 eV E1 - E1 = 40.8 eV \ (2)2 3 or - E1 = 40.8 eV 4 or E1 = - 54.4 eV | E1 | = 54.4 eV v ö 4. i = qf = q æç ÷ 2p è rø v (1/ n) 1 or or i µ 3 iµ µ 2 r (n) n

=

3

5. nf = 6 nf (nf - 1) 2

6.63 ´ 10-34 ´ 1.5 ´ 1015 eV 2p ´ 1.6 ´ 10-19

» 1.0 eV Since, E < W , no photoemission can take place. n (n - 1) 10. f f = 6 Þ nf = 4 2

i1 æ n2 ö = ç ÷ = (2)3 = 8 i2 çè n1 ÷ø

Total emission lines =

1.5 ´ 1015 Hz 2p hfmax eV E= 1.6 ´ 10-19

fmax =

\

= 15

nf = 4

6. A = pr2 A µ r2 \

or

A µ n4

(as r µ n2)

ni = 2

An = (n)4 A1

æA ö ln çç n ÷÷ = 4 ln (n) è A1 ø æA ö Therefore, ln çç n ÷÷ versus ln n graph is a straight è A1 ø line of slope 4. m i 7. B = 0 2p i or Bµ r See the hint of Q.No. 4 of same section. 1 i µ 3 and r µ n2 n 1 \ B= 5 n 5

B1 æ n2 ö = ç ÷ = (2)5 = 32 B2 çè n1 ÷ø 1 1 3R 8. = R æç1 - ö÷ = l 4 4 è ø h \ p= l 3Rh = = momentum of photon 4 From conservation of linear momentum,

From ni = 2, energy of six emission lines is either greater than less or equal to the energy of absorption line. 2pr r (n2 / Z ) or T µ or T µ 11. T = v v (Z / n) T µ

\

n3 Z2 3

æZ ö æ T1 ö æ n1 ö çç ÷÷ = çç ÷÷ = çç 2 ÷÷ è Z1 ø è T2 ø è n2 ø 3

\

12. l Ka

2

2

æ 1ö æ Z ö 2=ç ÷ ç ÷ è 2ø è 1 ø Z=4 will depend on atomic number Z and values Z

is same of all three isotopes.

\

13. n=2 10.2 eV n=1

l2 ® 1 =

12375 » 1213 Å 10.2

Chapter 33 14. For K ³ 10.2 eV electrons can excite the hydrogen atom (as E2 - E1 = 10.2 eV). So, collision may be inelastic. - 13/ 6 15. E3 - E1 = + 13.6 = 12.1 eV (3)3 From momentum conservation, Momentum of photon = Momentum of hydrogen atom E = mv \ c 12.1 ´ 1.6 ´ 10-19 E or v= = mc 1.67 ´ 10-27 ´ 3 ´ 108 = 3.86 m/s 16. P = Vi = 150 ´ 103 ´ 10 ´ 10-3 = 1500 W 99 99% of this = 1500 ´ J/s = 1485 J/s 100 1485 = cal/s 4.18 » 355cal/s 17. E4 - E3 = 32.4 \

- 13.6 (Z )2 13.6 (Z )2 + = 32.4 (4 )2 (3)2

Solving, we get Z=7

18. Q l =

h 2qVm V =

h2 2qml2

1 and Ln µ n n2 M q (always) 20. = L 2m \ M µL nh L= Þ L µn 2p Hence, M µn Third excited state means n = 4. p2 21. K = 2m In collision, momentum p remains constant. 1 \ K µ mass After collision, mass has doubled. So, kinetic K K energy will remain . Hence, loss is also . 2 2

19. En µ

Modern Physics-I — 495

K = minimum excitation energy required. 2 = 10.2 eV Þ K = 20.4 eV 22. KE = | E | = 3.4 eV Now,

l (in Å) =

150 KE (in eV)

150 3.4 = 6.6 Å

=

More than One Correct Options 1. l Ka and l K b will remain unchanged. But, l 0 will 1ö æ decrease ç as l 0 µ ÷ . Vø è \ Dl 1 = l Ka - l 0 or Dl 2 = l Kb - l 0 will increase. 1 1 2. R µ n2 , v µ and E µ 2 n n

3. L µ n, r µ n2 and

T =

2pr v

T µ

or

r n2 µ v (1/ n)

T µ n3 h h h 4. l = = = p mv 2Km or

1 (if v is same) m 1 (if K is same) lµ m If change in potential energy is same, then change in kinetic energy is also same. But, this does not mean that kinetic energy is same. n (n - 1) 5. f f =6 2 Þ nf = 4 lµ

nf = 4

1 l0

ni = 2

2

5 3

4

l 1 and l 4 are longer than l 0. l 3 , l 5 and l6 belong to Lyman series.

6

496 — Optics and Modern Physics f µ (Z - b)

6.

…(i)

If f versus Z is a straight line. c f = l 1 Hence, versus Z is also a straight line. l

Comprehension Based Questions 12375 = 2.5 eV 4950 K 1 = maximum kinetic energy = 0.6 eV

1. E1 =

W = E1 - K 2 = 1.9 eV 2. K 2 = 1.1eV E2 = W + K 2 = 3.0 eV 12375 l2 = = 4125 Å 3.0

\

3. Magnetic field cannot change the kinetic energy of charged particle. 4. K max = E - W = 2 eV l= =

150 , for an electron KE (in eV) 150 » 8.6 Å 2

5. K max is 2 eV. Hence, stopping potential is 2V. Photoemission stops when potential of sphere becomes 2V. q 2= \ 4 pe0r \

q = 8pe0 r

6. Q q = ne = 8pe0r \

8pe0r e 2 ´ 8 ´10-3 = 9 ´ 109 ´ 1.6 ´ 10- 19

n=

= 1.11 ´ 107 So, this much number of electrons are required to be ejected from the sphere. Number of photons emitted per second, Power of source N1 = Energy of one photon =

3.2 ´ 10-3 5 ´ 1.6 ´ 10-19

= 4 ´ 1015

Number of photons incident on sphere per second, æ N ö N 2 = ç 1 2 ÷ (pr2 ) è 4 pR ø N = 12 r2 4R (4 ´ 1015 ) (8 ´ 10-3 )2 = 4 (0.8)2

r R

= 1011 Number of photoelectrons emitted per second, N N 3 = 26 = 105 10 Now, N 3t = n n \ t= = 111 s N3

Match the Columns 2. For He+ , Z = 2 Z2 n2 E2 = - 3.4 eV E1 = - 13.6 eV Eµ

————— ————— H-atom

Ionisation energy from first, excited state of H-atom (given) = | E2 | = 3.4 eV = E (a) | E1 | = (13.6 eV) (2)2 = 16 (3.4 eV) = 16E (2)2 (b) U 2 = 2E2 = 2 (- 13.6) 2 (2) = - 8 (3.4 eV) = - 8E (c) K 1 = | E1 | = 16E (d) | E2 | = (13.6)

(2)2 (2)2

= 4 (3.4 eV) = 4 E

3. K max = hf - W \ K max versus f graph is a straight line of slope h and intercept - W . eV0 = hf = W W æ hö \ V0 = ç ÷ f e è eø \ V0 versus f graph is again a straight line of slope h W and intercept . e e

Chapter 33 4. T =

2pr v

\

or

T µ T µ

m has become half, so l will become two times or 1312 nm or 1.31mm. 1 (b) E µ Z 2 \ l µ 2 Z For singly ionized helium atom Z = 2 1 \ l is th or 164 nm 4 v 2. (a) f = 2pr (2.2 ´ 106 ) f1 = (2p ) (0.529 ´ 10-10 )

r n2 / Z µ v Z /n

n3 Z2

nh or L µ n 2p Z n2 and r µ vµ n Z 1 æ 1 1ö 5. µ çç 2 - 2 ÷÷ l è n1 n2 ø L=

or

l 2 (1/ n12 - 1/ n22 )i = l 1 (1/ n12 - 1/ n22 ) f

or

l2 =

(1/ 4 - 1/16)l (1/ n12 - 1/ n22 ) f

é ù 1 ê 2 2 ú êë (1/ n1 - 1/ n2 ) f úû (a) For first line of Balmer series, æ 27 ö n1 = 2, n2 = 3 \ l 2 = ç ÷ l è 20 ø (b) For third line of Balmer series, æ 25 ö n1 = 2, n2 = 5 \ l 2 = ç ÷ l è 28 ø (c) For first line of Lyman series, l n1 = 1, n2 = 2 \ l 2 = 4 (d) For second line of Lyman series, æ 27 ö n1 = 2, n2 = 3 \ l 2 = ç ÷l è 128 ø 7. See the hint of Q.No. 3 of section Assertion and Reason. Stopping potential increases with increase in maximum kinetic energy of photoelectrons of frequency of incident light. With increase in distance between cathode and anode, f remains unchanged. K max = hf - W = eV0 If W is decreased, K max and V0 both will increase. æ 3 ö = ç l÷ è 16 ø

Subjective Questions 1. (a) Reduced mass of positronium and electron is where, m = mass of electron E µm \ l µ

Modern Physics-I — 497

1 m

m , 2

» 6.6 ´ 1015 Hz 1 and r µ n2 vµ n v 1 f µ Þ f µ 3 r n f \ f2 = 1 8 (b) DE = E2 - E1 = 10.2 eV = hf 10.2 ´ 1.6 ´ 10-19 \ f = 6.6 ´ 10-34 » 2.46 ´ 1015 Hz (c) In option (a), we have found that f2 = 0.823 ´ 1015 Hz 1 T2 = f2 t N = = tf2 = (10-8 ) (0.823 ´ 1015 ) T2 = 8.23 ´ 106 revolutions 1 3. (a) r µ m (rH ) 0.529 ´ 10-10 \ r= 1 = m 207 -13 = 2.55 ´ 10 m (b) E µ m \ Ionization energy of given atom = (m) (ionization energy of hydrogen atom) = (207) (13.6 eV) = 2815.2 eV = 2.81 keV 4. (a) En - E1 = 12.5 13.6 \ - 2 + 13.6 = 12.5 n Solving we get, n = 3.51

498 — Optics and Modern Physics Hence, electron jumps to n = 3. So, possible lines are between n = 3 to n = 2, n = 3 to n = 1and between n = 2 to n = 1. For n = 3 to n = 2, 13.6 13.6 DE = E3 - E2 = + 9 4 = 1.9 eV 12375 \ l= Å = 6513 Å 1.9 » 651 nm Similarly, other wavelengths can also be obtained. (b) n = 3.51(in option-a) A photon always transfers its energy completely. So, it cannot excite the ground state electrons to n = 3 (like and electrons excited it in part-a). 5. (a) E = hf = (6.6 ´ 10-34 )(5.5 ´ 1014 ) -20

= 36.3 ´ 10

J

= 2.27 eV

Ans.

(b) Number of photons leaving the source per second, 0.1 P n= = E 36.3 ´ 10-20 = 2.75 ´ 1017

Ans.

(c) Number of photons falling on cathode per second, 0.15 n1 = ´ 2.75 ´ 1017 100 = 4.125 ´ 1014 Number of photoelectrons emitting per second,

\

n2 =

6 ´ 10-6 = 3.75 ´ 1013 1.6 ´ 10-19

%=

n2 3.75 ´ 1013 ´ 100 = ´ 100 n1 4.125 ´ 1014

= 9%

Ans.

K 6. 1 = 5 K2 E4 = –0.85 eV E3 = –1.51 eV E2 = –3.4 eV

E1 = –13.6 eV

\

DE1 - W =5 DE2 - W

Here, DE1 = E4 - E1 = 12.75 eV and DE2 = E3 - E1 = 12.09 eV Substituting in Eq. (i) and solving, we get W = 11.925 eV

…(i)

Ans.

7. For shorter wavelength DE = E4 - E3 =

(-13.6)(3)2 é (-13.6)(3)2 ù -ê ú (4 )2 (3)2 ë û

= 5.95 eV W = E - K max = (5.95 - 3.95) eV = 2 eV For longer wavelength (-13.6)(3)2 é (-13.6)(3)2 ù DE = E5 - E4 = -ê ú (5)2 (4 )2 ë û = 2.754 eV \ K max = E - W = 0.754 eV or stopping potential is 0.754 V. Ans. e 8. Magnetic moment, m = NiA = æç ö÷ (pr2 ) èT ø æ e ö evr 2 ÷÷ (pr ) = or …(i) m = çç 2 è 2pr / v ø nh …(ii) We know that mvr = 2p Solving Eqs. (i) and (ii), we get neh Ans. m = 4 pm m i m e Magnetic induction, B = 0 = 0 2r 2rT m 0ev m ev …(iii) or B= = 0 2 (2r)(2pr) 4 pr e2 mv 2 From Newton's second law, = 2 r 4pe0r 2 e or …(iv) v2 = 4 pe0mr Solving all these equations, we get m pm2e7 B= 0 5 5 8e0h n

Ans.

9. Energy of electron in ground state of hydrogen atom is – 13.6 eV. Earlier it had a kinetic energy of 2 eV. Therefore, energy of photon released during formation of hydrogen atom, DE = 2 – (– 13.6) = 15.6 eV 12375 12375 l= = \ DE 15.6 Ans. = 793.3 Å

Chapter 33 10. U = – 1.7 eV

Modern Physics-I — 499

12375 Ans. = 28.43 Å 1ö æ (13.6) (6)2 ç1 – ÷ 9ø è (d) Ionization energy = (13.6) (6)2 = 489.6 eV Ans. 12375 Ans. l= = 25.3 Å 489.6 vT 12. Pitch of helical path, p = (v cos q) T = . 2 (as q = 60°) 2pm 2p qö æ T = = ça = ÷ Bq Ba mø è pv p= \ Ba Bap …(i) or v= p 1 KE = mv 2 = E – W 2 1 …(ii) \ W = E – mv 2 2 Substituting value of v from Eq. (i) in Eq. (ii), we get 1 W = 4.9 – 2 9.1 ´ 10–31 ´ (2.5 ´ 10–3 )2 (1.76 ´ 1011 )2 =

–13.6 U \ E = = – 0.85 eV = 2 n2 \ n=4 Ejected photoelectron will have minimum de-Broglie wavelength corresponding to transition from n = 4 to n = 1. DE = E4 – E1 = – 0.85 – (– 13.6) = 12.75 eV K max = DE – W = 10.45 eV 150 (for an electron) \ l= Å 10.45 Ans. = 3.8 Å n (n – 1) 11. (a) =3 2 \ n=3 i.e after excitation atom jumps to second excited state. Hence, nf = 3. So, ni can be 1 or 2. If ni = 1, then energy emitted is either equal to or less than the energy absorbed. Hence, the emitted wavelength is either equal to or greater than the absorbed wavelength. Hence, ni ¹ 1.

(2.7 ´ 10–3 )2

´

2

p ´ 1.6 ´ 10

–19

= (4.9 – 0.4 ) eV = 4.5 eV

ni = 1

Ans.

æ

1 ö ÷ 1 – 1/ p2 ÷ø è

13. (a) l = 1500 çç

l max corresponds to least energetic photon with p = 2. æ 1 ö ÷÷ = 2000 Å Ans. \ l max = 1500 çç è 1 – 1/ 4 ø ni = 2

If ni = 2, then Ee ³ Ea . Hence, l e £ l b \ ni = 2

l min corresponds to most energetic photon with p = ¥ Ans. \ l min = 1500 Å Ans.

(b) E3 – E2 = 68 eV \ \ (c) l min

æ 1 1ö (13.6) (Z 2 ) ç – ÷ = 68 è 4 9ø Z=6 12375 = E3 – E1

(b) l ¥ – 1 = 1500 Å E3 = – 0.95 eV E2 = – 2.05 eV

Ans. E1 = – 8.25 eV

500 — Optics and Modern Physics \ \

\

12375 eV 1500 = 8.25 eV E1 = – 8.25 eV l 2 – 1 = 2000 Å

E¥ – E1 =

E2 – E1 =

(as E¥ = 0)

(c) Energy of photon in first case, 12375 = 3000 = 4.125 eV or E1 = 6.6 ´ 10–19 J Rate of incident photons P 10–3 = 1 = E1 6.6 ´ 10–19

12375 eV 2000

= 6.2 eV E2 = – 2.05 eV æ 1 ö ÷÷ Similarly, l 31 = 1500 çç è 1 – 1/ 9 ø

\

= 1687.5 Å 12375 \ E3 – E1 = eV = 7.3 eV 1687.5 \ E3 = – 0.95 eV (c) Ionization potential = 8.25 V 12375 14. (a) K 1 = –W 3000 12375 K2 = –W 1650 v2 = 2v1 \ K 2 = 4 K 1

= 1.52 ´ 1015 per second Number of electrons ejected 4.8 ´ 10–3 per second = 1.6 ´ 10–19 = 3.0 ´ 1016 per second

Ans.

\ Efficiency of photoelectrons generation 1.52 ´ 1015 = ´ 100 3.0 ´ 1016 = 5.1%

…(i) …(ii)

15. Balmer Series l 32 =

…(iii)

Solving these equations, we get

12375 = E3 - E2

12375 æ1 (13.6) ç è4

1ö ÷ 9ø

= 6551 Å

W = 3 eV \ Threshold wavelength, 12375 l0 = 3

l 42

Ans. = 4125 Å 12375 (b) E2 = = 7.5 eV = 12 ´ 10–19 J 1650 Therefore, number of photons incident per second P 5.0 ´ 10–3 n2 = 2 = E2 12 ´ 10–19 = 4.17 ´ 1015 per second Number of electrons emitted per second (h = 5.1%) 5.1 = ´ 4.17 ´ 1015 100 = 2.13 ´ 1014 per second \ Saturation current in second case i = (2.13 ´ 1014 ) (1.6=´3.4 10–19 ´ 10 ) A–5 A = 34 mA

Ans.

Ans.

= 655.1 nm 12375 = = E4 - E2

12375 æ1 1ö (13.6) ç - ÷ è 4 16 ø

= 4853 Å l 52

= 485.3 nm 12375 = = E5 - E2

12375 1ö æ1 (13.6) ç - ÷ è 4 25 ø

= 4333 Å = 433.3 nm First two lie in the given range. Of these l 42 corresponds to more energy. æ1 1ö E = E4 - E2 = (13.6) ç - ÷ è 4 16 ø \

= 2.55 eV K max = E - W = (2.55 - 2.0) eV = 0.55 eV

Ans.

16. From the theory of standing wave, we can say that l = (2.5 - 2.0) = 0.5 Å 2

Chapter 33 l =1Å

or

mv 2 k = r r According to Bohr’s assumption, h mvr = n 2p Solving Eqs. (i) and (ii), we get nh r= 2p mk

\

2Å

2.5 Å

Therefore, least value of d required will correspond to a single loop. l Ans. dmin = = 0.5 Å \ 2 Further for de-Broglie wavelength of an electron, 150 l= Å K (in eV) l =1Å \

K = 150 eV

Ans.

17. (a) Reduced mass m =

(1837me) (207me ) m1m2 = m1 + m2 1837me + 207me

= 186me = 186 ´ 9.1 ´ 10-31 = 1.69 ´ 10-28 kg

(c) E2 = 186 (- 3.4 ) eV = - 632.4 eV \ DE21 = 1897.2 eV 12375 \ l 21 = Å DE21 12375 Å 1897.2 = 6.53 Å » 0.653 nm

v=

and

…(i)

…(ii)

k m

1 k k mv 2 = k ln r – + = k ln r 2 2 2 nh Thus, rn = 2p mk ì nh ü and Ans. En = k ln í ý î 2p mk þ mv …(i) 19. (a) r= Be nh …(ii) mvr = 2p Solving these two equations, we get nh r= 2pBe \ E =U +

Ans.

(b) En µ m Here, reduced mass is 186 times mass of electron. Hence, ground state energy will also be 186 times that of hydrogen atom. \ E1 = 186 (- 13.6) eV Ans. = - 2529.6 eV » - 2.53 keV

and

v=

nhBe 2pm2

1 nhBe mv 2= 2 4pm æ eö (c) M = iA = ç ÷ (pr2 ) èT ø e evr = (pr2 ) = æ 2pr ö 2 ç ÷ è v ø (b) K =

Ans.

e nh nhBe 2 2pBe 2pm2 nhe = 4pm U = – MB cos 180° nheB = 4pm =

=

Ans.

18. Force of interaction between electron and proton is dU – k = dr r Force is negative. It means there is an attraction between the particles and they are bound to each other. This force provides the necessary centripetal force for the electron. F =–

Modern Physics-I — 501

Note Angle between M and B will be 180°. Think why? nheB 2pm nh 2 (e) | f | = Bpr = 2e

(d) E = U + K =

502 — Optics and Modern Physics 20. (a) and (b) When hydrogen atom is excited, then æ1 1 ö eV = E0 ç – 2 ÷ è1 n ø When ion is excited, é 1 1ù eV = E0Z 2 ê 2 – 2 ú n 2 ë 1û Wavelength of emitted light, hc æ1 1 ö = E0 ç – 2 ÷ l1 è1 n ø æ1 1 ö hc = E0 Z 2 çç – 2 ÷÷ l2 è 1 n1 ø Further it is given that l1 5 = l2 1 Solving the above equations, we get Z = 2, n = 2, n1 = 4 and V = 10.2 V

…(i)

…(ii)

…(iii)

12375 –W 4950 1.1 =

…(v)

Ans.

Ans. …(i)

12375 –W l

i = ne = 1.0 ´ 1013 ´ 1.6 ´ 10–19 = 1.6 ´ 10–6 A

and by the ion = E4 – E1

21. 0.6 =

= 1.0 ´ 1013 per second \

…(iv)

(c) Energy of emitted photon by the hydrogen atom = E2 – E1 Ans. = 10.2 eV 1ö æ = (13.6) (2)2 ç1 – ÷ 16 ø è = 51 eV

Solving above two equations, we get W = 1.9 eV and Ans. l = 4125 Å 12375 22. E = = 3.1 eV 4000 Number of photoelectrons emitted per second, ö 5 æ 1 öæ ÷ n = ç 6 ÷ çç –19 ÷ è 10 ø è 3.1 ´ 1.6 ´ 10 ø

…(ii)

Ans. = 1.6 mA 12375 23. (a) E = = 3.1 eV 4000 Energy of electron after first collision E1 = 90% of E = 2.79 eV (as 10% is lost) Energy of electron after second collision E2 = 90% of E1 = 2.51 eV KE of this electron after emitting from the metal surface Ans. = (2.51 – 2.2) eV = 0.31 eV (b) Energy after third collision, E3 = 90% of E2 = 2.26 eV Similarly, E4 = 90% of E3 = 2.03 eV So, after four collisions it becomes unable for the electrons to come out of the metal.

34. Modern Physics-II INTRODUCTORY EXERCISE

34.1

2. Penetrating power is maximum for g -rays, then of b-particles and then a -particles because basically it depends on the velocity. However, ionization power is in reverse order. 3. Both the beta rays and the cathode rays are made up of electrons. So, only option (a) is correct. (b) Gamma rays are electromagnetic waves. (c) Alpha particles are doubly ionized helium atoms and (d) Protons and neutrons have approximately the same mass. Therefore, (b), (c) and (d) are wrong options. 4. During b-decay, a neutron is transformed into a proton and an electron. This is why atomic number (Z = number of protons) increases by one and mass number (A = number of protons + neutrons) remains unchanged during b-decay. 5. Following nuclear reaction takes place 1 1 0 0 n ¾¾® 1H + -1e + n n is antineutrino. n 1 6. From R = R0 æç ö÷ è 2ø

1 (l N ) 2 2 2

or

l1 N = 2 l 2 2N 1

or

T1 2N 1 = T2 N2

ln 2 ö æ çT = half - life = ÷ l ø è

N 1 = 2N 2 T1 \ =4 T2 \ Correct option is (a). 1 8. After two half-lives th fraction of nuclei will 4 3 remain un-decayed. Or, th fraction will decay. 4 Given,

n

…(i) è 2ø Here, R = activity of radioactive substance after n R half-lives (given) = 0 16 Substituting in Eq. (i), we get n = 4 \ t = (n) t1/ 2 = (4 ) (100 ms) = 400 ms 11. Using N = N 0e- lt l=

where,

ln 2 ln 2 = t1/ 2 3.8 ln 2

n

l1 N 1 =

1 10. R = R0 æç ö÷

t N0 = N 0e 3.8 20 Solving this equation with the help of given data we find t = 16.5 days \ Correct option is (b).

\

æ 1ö we have, 1 = 64 ç ÷ è 2ø or n = 6 = number of half-lives \ t = n ´ tt1/ 2 = 6 ´ 2 = 12 h 1 (activity of S 2) 7. Activity of S 1 = 2 or

Hence, the probability that a nucleus decays in two 3 half-lives is . 4 9. Activity reduces from 6000 dps to 3000 dps in 140 days. It implies that half-life of the radioactive sample is 140 days. In 280 days (or two half-lives) 1 activity will remain th of the initial activity. 4 Hence, the initial activity of the sample is 4 ´ 6000 dps = 24000 dps Therefore, the correct option is (d).

n

1 12. 1000 = æç ö÷ 8000

è 2ø \ n = 3 = number of half-lives

These half-lives are equivalent to 9 days. Hence, one half-life is 3 days. tav = 1.44 t1/ 2 = 1.44 ´ 3 = 4.32 days R 13. R0 = lN 0 Þ N 0 = 0 l ln 2 where, l= t1/ 2 N = N 0 e- lt Find

N 1 = N 0e- lt1

and

N 2 = N 0e- lt2

\ Number of nuclei decayed in given time = N 1 - N 2

504 — Optics and Modern Physics 14. (a) R = R0e-lt

2. Heavy water is used as moderators in nuclear reactors to slow down the neutrons.

R0 = 20 MCI R = 8 MCI t = 4.0 h

4. During fusion process two or more lighter nuclei combine to form a heavy nucleus. Hence, the correct option is (c). 5. (a) m(c)2 = P ´ t P ´t \ m= 2 c (109 ) (24 ) (3600) = (3 ´ 108 )2

Find l. (b) R0 = lN 0.

Find

N0 =

R0 l

(c) Find R = R0 e- lt

15. R0 = lN 0 6.0 ´ 1011 = l (1015 )

= 9.6 ´ 10-4 kg

l = 6.0 ´ 10-4 s

\

t1/ 2 =

(b) Number of fissions required per second Energy required per second = Energy released in one fission

ln 2 0.693 = l 6.0 ´ 10-4

= 1.16 ´ 103 s

=

16. In 200 minute time, n1 = number of half-lives of X 200 = =4 50 n2 = number of half-lives of Y 200 = =2 100

109 200 ´ 10 ´ 1.6 ´ 10-19 6

= 3.125 ´ 1019

6. Mass defect Smi - Sm f = Dm = (238.050784) - (234.043593 + 4.002602) = 4.589 ´ 10-3 u Energy released = Dm ´ 931.48 MeV = 4.27 MeV 8. Q = (Dm in atomic mass unit) ´ 931.4 MeV

N X N 0 (1/ 2)4 1 = = N Y N 0 (1/ 2)2 4

= (2 ´ mass of 1H2 - mass of 2He4 ) ´ 9314 . MeV

INTRODUCTORY EXERCISE

34.2

1. 4 (2 He4 ) = 8 O16 Mass defect, Dm = {4 (4.0026) - 15.9994} = 0.011 amu \ Energy released per oxygen nuclei = (0.011) (931.48) MeV = 10.24 MeV \ Correct answer is (c).

= (2 ´ 2.0141 - 4.0024) ´ 931.4 MeV Q » 24 MeV 9. 2 1H2 ¾® 2He4 Binding energy of two deuterons, E1 = 2 [ 2 ´ 1.1] = 4.4 MeV Binding energy of helium nucleus, E2 = 4 (7.0) = 28.0 MeV \ Energy released DE = E2 - E1 = (28 - 4.4 ) MeV = 23.6 MeV

Exercises LEVEL 1 Assertion and Reason 1. Huge amount of energy is involved in any nuclear process, which cannot be increased or decreased by pressure or temperature.

2. Some lighter nuclei are also radioactive.

3. By emission of one a -praticle, atomic number decreases by 2 and mass number by 4. But by the emission of one b-particle, atomic number increases by 1 and mass number remains unchanged.

4. In moving from lower energy state to higher energy state electromagnetic waves are absorbed.

Chapter 34 5. Neutrino or antineutrino is also produced during b-decay. important for stability.

10. a -particles are heaviest. Hence, its ionizing power is maximum.

11. In binding energy per nucleon versus mass number graph binding energy per nucleon of daughter nuclei should increase (for release of energy) or the daughter nuclei should lie towards the peak of the graph.

Objective Questions 1. Nuclear density is constant hence, mass µ volume mµV

2. Radius of a nucleus is given by (where, R0 = 1.25 ´ 10-15 m)

= 1.25 A 1/ 3 ´ 10-15 m Here, A is the mass number and mass of the uranium nucleus will be m » Amp , where mp = mass of proton = A (1.67 ´ 10-27 kg) mass m = \ Density, r = volume 4 pR 3 3 =

\ n = number of half-lives = 2 Two half-lives are equivalent to 15 min. Therefore, one half-life is 7.5 min. ln 2 ln 2 12. l = = day -1 t1/ 2 4 Now, apply R = R0e- lt Þ 5 = 100e- lt

A (1.67 ´ 10-27 kg) A (1.25 ´ 10-15 m )3

r » 2.0 ´ 1017 kg/m 3

or

1 æ 1ö =ç ÷ 8 è 2ø \ n = number of half-lives =3 3 half-lives = 8 s 8 1 half-life = s \ 3 10. Number of atoms disintegrated, N = N 0 (1 - e- lt ) N = 1 - e-lt \ N0 1 At t = one mean life = l N 1 = 1 - e-1 = 1 N0 e 3 11. Decayed fraction is th. Therefore, left fraction is 4 1 th. 4 n n N 1 æ 1ö æ 1ö N = N 0 ç ÷ or =ç ÷ = N 0 è 2ø 4 è 2ø Now,

9. (1 amu )(c2 ) = 931.48 MeV

R = R0 A 1/ 3

1 8

9. Number of nuclei left = th n

6. Total binding energy per nucleon is more

or

5. Let n - a particles and m - b particles are emitted.

Substituting value of l, we can find t.

13. Let N nuclei decay per second. Then,

Then, 90 - 2n + m = 80 200 - 4 n = 168 Solving Eqs. (i) and (ii), we get n = 8 and m = 6

…(i) …(ii)

N (200 ´ 1.6 ´ 10-13 ) = 1.6 ´ 106 Solving we get N = 5 ´ 1016 per sec

14. Nuclear density is independent of A. It is of the

7. By emitting one a -particle, atomic number decreases by 2. By emitting two b-particles, atomic number increases by 2. Hence, ZA = ZC or A and C are isotopes.

8. Binding energy = (Dm) ´ 931.5 MeV = [2 ´ 1.00785 + 2 ´ 1.00866 - 4.00274]

order of 107 kg/m 3. ln 2 0.693 1 15. l= = = sec-1 t1/ 2 6.93 10 N = N 0 e- lt \

æ 1 ö

- ç ÷ (10) N = e-lt = e è 10 ø N0 = e-1 » 0.63

16. Activity of atoms is 6.25% after four half-lives. ´ 931.5

= 28.2 MeV

Modern Physics - II — 505

\ Four half-lives » 2 h = 120 min \ One half-life is 30 min.

506 — Optics and Modern Physics = 1 - e- lt = 1 - e-0.1 ´ 5

17. Probability of survival, - lt

Number of nuclei left N e = 0 Initial number of nuclei N0 1 t = one mean life = l 1 -1 P=e = e

P= At

= 0.39 3 U238 6. = Pb206 1 N0 = 3 + 1= 4 N =3 N = N 0e- lt

Subjective Questions

l=

1. (a) R = R0e-lt R = 2700 per minute, R0 = 4750 per minute t = 5 min Find l. ln 2 (b) t1/ 2 = l 2. R = lN 6 ´ 1011 = 1.0 ´ 1015 l l = 6 ´ 10-4 s ln 2 0.693 = s t1/ 2 = l 6 ´ 10-4

\

= 1155 s = 19.25 min

3. Q R = lN \ N =

R R Rt = = 1/ 2 l (ln 2)/ t1/ 2 ln 2

(8 ´ 3.7 ´ 1010 ) (5.3) (365) (24 ) (3600) = 0.693 = 7.14 ´ 1019 7.14 ´ 109 m= ´ 60 g 6.02 ´ 1023 = 7.11 ´ 10-3 g æ ln 2 ö ÷÷ N 4. R = lN = çç è t1/ 2 ø 0.693 æ 1 ö 23 = ç ÷ (6.02 ´ 10 ) 9 4.5 ´ 10 ´ 365 ´ 24 ´ 3600 è 238 ø = 1.23 ´ 104 dps 1 5. = 10 days l \ l = 0.1 day -1 Probability of decay Number of atoms decayed = Initial number of atoms =

N 0 (1 - e- lt ) N0

…(i)

ln 2 t1/ 2

…(ii)

From Eqs. (i) and (ii), we get t = 1.88 ´ 109 yr ln 2 T1 ln 2 l2 = T2

7. l 1 =

(T = half-life)

R01 + R02 = 8 mCi \ l 1 (4 N 0 ) + l 2 (N 0 ) = 8 mCi From here we can find number after t = 60 yr R = R1 + R2

(given)

= (4 l 1N 0 )e- l1t + (l 2N 0 )e-l 2 t

9. (a) 82 + 10 = 92 , 206 + 10 + 20 = 236 So, this reaction is possible. (b) 82 + 16 - 6 = 92 , 206 + 32 = 238 But antineutrino is also emitted with b -1 (or electron) decay. 10. Binding energy = Dm ´ 931.5 MeV = (7 ´ 1.00783 + 7 ´ 1.00867 - 14.00307) 931.5 = 104.72 MeV

11. Energy released = binding energy = Dm ´ 931.5 MeV = (8 ´ 1.007825 + 8 ´ 1.008665 - 15.994915) ´ 931.5 = 127.62 MeV 12. (a) Number of nuclei in 1 kg of U235, æ 1 ö 26 N =ç ÷ (6.02 ´ 10 ) è 235 ø \ Total energy released = (N ´ 200) MeV æ 1 ö -13 26 =ç ÷ (6.02 ´ 10 ) (200) (1.6 ´ 10 ) è 235 ø = 8.19 ´ 1013 J

Chapter 34 m=

(b)

8.19 ´ 1013 g 30 ´ 103 9

= 2.73 ´ 10 g = 2.73 ´ 106 kg

13. From momentum of conservation, p1 = p2 2K 1m1 = 2K 2m2

\

K 1m1 (6.802) (4) = m2 208 = 0.1308 MeV 14. Number of nuclei in 1 kg of uranium, æ 1 ö 26 N =ç ÷ (6.02 ´ 10 ) è 235 ø æ 1 ö 26 -13 Now, ç ÷ (6.02 ´ 10 ) (185 ´ 1.6 ´ 10 ) è 235 ø = (100 ´ 106 )t \

K2 =

t = 7.58 ´ 105 s

\

=

7.58 ´ 105 day 60 ´ 60 ´ 24

= 8.78 days 15. Q-value = (Dm) (931.5) MeV (a) Q-value = (2 ´ 2.014102 - 3.016049 - 1.007825) ´ 931.5 = 4.05 MeV Similarly, Q-value of other parts can also be obtained. 16. Q-value = (Dm) ´ 931.5 MeV = (2 ´ 4.0026 - 8.0053) ´ 931.5 = - 0.0931 MeV = - 93.1 keV

Modern Physics - II — 507

2. l = l 1 + l 2 ln 2 ln 2 ln 2 = + (T = Half-life) T T1 T1 TT T = 1 2 = 20 y T1 + T2

\ or

1 th sample remains after 2 half-lives or 40 y. 4 3. Q-value = Final binding energy Initial binding energy = E2N 2 + E3N 3 - E1N 1 4. Energy released = Final binding energy - initial binding energy = 110 ´ 8.2 + 90 ´ 8.2 - 200 ´ 7.4 = 160 MeV 5. It means we are getting only 100 MeV of energy by the fission of one uranium nucleus. Number of nuclei per second Energy required per second = Energy obtained by one fission =

16 ´ 106 = 1018 100 ´ 1.6 ´ 10-13

6. When the rate production = rate of disintegration, number of nuclei or maximum. lN = A \ ln 2 AT or N = A or N = = maximum T ln 2

7. R0 = 15 ´ 200 = 3000 decay/min from 200 g carbon. Using

æ 1ö R = R0 ç ÷ è 2ø

n

n

LEVEL 2 Single Correct Option 1.

n

æ 1ö ç ÷ \ n=4 è 2ø So, 3t times is equivalent to four half-lives. Hence, 3t one half-life is equal to . 4 11 9 The given time t - t = t is equivalent to 6 2 2 half-lives. 6 N æ 1ö N = N0 ç ÷ = 0 \ 64 è 2ø N0 = N0 16

æ 1ö 375 = 3000 ç ÷ è 2ø \ n = number of half-lives = 3 \ t = 5730 ´ 3 = 17190 y 8. AP = A0e-lt1 \

AQ = A0e- lt2 l t1 = ln ( A0 / AP ) 1 t1 = ln ( A0 / AP ) = T ln ( A0 / AP ) \ l Similarly, t2 = T ln ( A0 / AQ ) æ AQ ö ÷÷ t1 - t2 = T ln çç \ è AP ø

508 — Optics and Modern Physics 9. Combining two given equations,

\

we have, 31 H2 = 2He4 + p + n Dm = 3 ´ 2.014 - 4.001- 1.007 - 1.008 = 0.026 u Energy released by 3 deuterons = 0.026 ´ 931.5 ´ 1.6 ´ 10-13 J = 3.9 ´ 10-12 J æ 10 40 ö ÷ (3.9 ´ 10-12 ) Now, (1016 ´ t ) = çç ÷ è 3 ø Solving we get, t » 1.3 ´ 1012 s

R01e

= R02e

- l1t

l 1N 0 e

= l 2N 0 e- l 2 t ln 2 0.693 l1 = = t1 t1 0.693 l2 = t2

From conservation of momentum, pa = pg = 2K a ma = 2K g mg

\

K a mg 228 = = K g ma 4 æ 228 ö ÷÷ KTotal K a = çç è 228 + 4 ø æ 228 ö =ç ÷ KTotal è 232 ø

…(i)

ln 2 ×x T x 1 = = constant y l x T x or = > T (as ln 2 = 0.693) y ln 2 y

Further, xy = x (lx ) = lx 2 After one half-life, x remains half. Hence, x 2 1 remains th. 4 3. By the emission of an a-particle, atomic number decreases by 2 and by the emission of two particles atomic number increases by 2. Hence, net atomic number remains unchanged. 4. At t = 4T Number of half-lives of first n1 = 4 and number of half-lives of second n2 = 2 N1 N (1/ 2)4 1 =x= 0 = N2 N 0 (1/ 2)2 4

12. From momentum of conservation, momentum of photon = photon of nucleus E = 2Km \ c E2 \ K = 2mc2 (7 ´ 1.6 ´ 10-13 )2 keV = 2 ´ 24 ´ 1.67 ´ 10-27 ´ (3 ´ 108 )2 ´ 1.6 ´ 10-16 = 1.1 keV 13. Activity A µ Number of atoms -lt1

A1 = A0e

æ A ö T ln ç 2 ÷ ln 2 çè 2 A1 ÷ø

More than One Correct Options

- lt

Substituting these values in Eq. (i), we can get the t. 11. A = 232 - 4 = 228

or

æ 2A ö T ln çç 0 ÷÷ ln 2 è A2 ø T æ A0 A ö çç t1 - t2 = ´ 2 ÷÷ ln 2 è A1 2 A0 ø t2 =

1. y = lx = - l1t

1 æ A0 ö T A ln ç ÷ = ln 0 l çè A1 ÷ø ln 2 A1

A2 = 2 A 0 e-lt2

=

10. R1 = R2 \

t1 =

R1 l 1 N 0 (1/ 2)4 = R2 l 2 N 0 (1/ 2)2 l T = 1 = 2 4 l 2 4T1 2T 1 = = 4T 2 or R µ A 1/ 3

y=

6. R = R0 A 1/ 3

Comprehension Based Questions 1. Energy released = (Dm) (931.48) MeV = [2 ´ 2.01102 - 3.0160 - 1.007825] ´ 931.5 = 4.03 MeV » 4 MeV

Chapter 34 2. Let N number of fusion reactions are required, then N ´ 4 ´ 1.6 ´ 10-13 = 103 ´ 3600

(c) l to X-rays is of the order of 1Å - 100 Å é 12375 ù E=ê ú in eV ë l (in Å) û

N = 5.625 ´ 1018

3. In one fusion reaction two

2 1 H nuclei

or 1.125 ´ 1019 Mass in kg æ 1.125 ´ 1019 ö ÷ (2) kg = çç 26 ÷ è 6.02 ´ 10 ø = 3.7 ´ 10-8 kg

dt ø

y = lx or

l=

y x

ln 2 = (ln 2) (x / y) l R = R0e- lt

t1/ 2 =

\

109 = 7.43 ´ 1013 0.693 14.3 ´ 3600 N t dN dN Now, = ò dt = q - lN or ò 0 q - lN 0 dt q \ N = (1 - e- lt ) l Substituting the values,

1. N =

dN ö 1. æç ÷ = lN \

where, l is in Å. (d) l is of the order of 4000 Å - 7000 Å 12375 Now, E (in eV) = l (in Å)

Subjective Questions

Match the Columns è

é 12.375 ù in keV =ê ë l úû

are used.

Hence, total number of 21 H nuclei are 2N .

Modern Physics - II — 509

= ye- l (1/ l) = y/ e y R = = lN e y y x N = = = el e ( y/ x ) e

2. Energy is released when daughter nuclei lie towards peak of this graph, so that binding energy per nucleon or total binding energy in the nuclear process increases. 3. A will continuously decrease, but C will increase. ( A + B ) will continuously decrease as C is formed only from A and B. (C + B ) = Total - A A is continuously decreasing. Hence, (C + B ) will continuously increase. 4. (a) Z ¢ = Z - Z + 1 = Z - 1 A¢ = A - 4 (b) Z ¢ = Z - 2 ´ 2 + 1 = Z - 3

R = l

7.43 ´ 1013 =

Solving this equation we get, t = 14.3 h

2. (a) At t = 0 At t

or

B 0 N2

A N0 N1

dN 2 = l (N 1 - N 2 ) dt dN 2 = lN 0e- lt - lN 2 dt

or

dN 2 + lN 2dt = lN 0e-lt

\

elt dN 2 + lN 2elt dt = lN 0dt

or

d (N 2elt ) = lN 0dt

\

A¢ = A - 4 (d) Z ¢ = Z - 2 ´ 2 + 2 ´ 2 = Z - 2

\

Ans. C 0 N3

N 1 = N 0e-lt

Here,

A¢ = A - 2 ´ 4 = A - 8 (c) Z ¢ = Z - 2 + 2 ´ 1 = Z

A¢ = A - 2 ´ 4 = A - 8 5. (a) E is of the order of kT , where k = Boltzmann constant and T » 300 K

2 ´ 109 . / 14. 3 ´ 3600) t ] [1 - e-(0693 0.693 14.3 ´ 3600

N 2elt = lN 0t + C

At t = 0, N 2 = 0, C =0 N 2 = lN 0 (te- lt )

\ (b) Activity of B is

R2 = lN 2 = l2N 0 (te- lt )

…(i)

510 — Optics and Modern Physics dR2 =0 dt 1 Ans. t= \ l lN 0 Ans. \ Rmax = e 3. (a) Let at time t, number of radioactive nuclei are N . Net rate of formation of nuclei of A dN dN = a - lN or = dt dt a - lN For maximum activity,

t dN = ò dt 0 0 a - N

N

òN

or

…(i) ln 2 = l

3 N0 2 (ii) Substituting a = 2lN 0 and t ® ¥ in Eq. (i), we get a N = = 2N 0 l or N = 2N 0 4. (a) Let RA0 and RB0 be the initial activities of A and B. Then, …(i) RA0 + RB 0 = 1010 dps in Eq. (i), we get N =

Activity of A after time t = 20 days (two half-lives of A) is 2

æ 1ö RA = ç ÷ RA0 = 0.25 RA0 è 2ø Similarly, activity of B after t = 20 days (four half-lives of B) is 4

æ 1ö RB = ç ÷ RB 0 = 0.0625 RB 0 è 2ø Now, it is given that RA + RB = 20% of 1010 or 0.25 RA0 + 0.0625 RB 0 = 0.2 ´ 1010 dps …(ii) Solving Eqs. (i) and (ii), we get RA0 = 0.73 ´ 1010 dps RB 0 = 0.27 ´ 1010 dps

and (b)

RA0 RB 0

=

l AN A0 lB N B0

=

(t1/ 2 )B N A0 × (t1/ 2) A N B 0

N A0 N B0

æ RA ö (t ) æ 0.73 ö æ 10 ö = ç 0 ÷ 1/ 2 A = ç ç RB ÷ (t1/ 2 )B è 0.27 ÷ø çè 5 ÷ø è 0ø = 5.4

5. Let N be the number of radio nuclei at any time t. Then, net rate of formation of nuclei at time t is dN = a - lN dt N t dN or ò0 a - lN = ò0 dt a or N = (1 - e- lt ) l N

Solving this equation, we get 1 N = [ a - (a - l N 0 )e-lt ] l (b) (i) Substituting a = 2lN 0 and t = t1/ 2

\

Rate of formation = a

Rate of decay = lN

Number of nuclei formed in time t = at and number of nuclei left after time a t = (1 - e-lt ) l Therefore, number of nuclei disintegrated in time a t = at - (1 - e-lt ) l \ Energy released till time, a é ù t = E0 ê at - (1 - e-lt )ú l ë û But only 20% of it is used in raising the temperature of water. a é ù So, 0.2 E0 ê at - (1 - e-lt )ú = Q l ë û where, Q = msDq Q \ Dq = increase in temperature of water = ms a é ù 0.2 E0 ê at - ( 1 - e-lt )ú l ë û \ Dq = ms dN B 6. We have for B = P - l 2N B dt NB t dN Þ ò0 P - l2BN B = ò0 dt Þ Þ

æ P - l 2N B ö ln ç ÷ = - l 2t P è ø NB =

P (1 - e- l 2 t ) l2

The number of nuclei of A after time t is N A = N 0e- l1t

Modern Physics - II — 511

Chapter 34

Þ Þ

7.

210 84 Po

R1 1 = R 9 and probability of getting b -particles R 8 = 2= R 9 (ii) R01 = R02 N 01 N 02 \ = T1 T2 N 01 1 \ = N 02 4

dN c = l 1N A + l 2N B dt dN c = l 1N 0e-l1t + P (1 - e-l 2 t ) dt æ e-l 2 t - 1ö ÷ N c = N 0 (1 - e-l1t ) + P çç t + l 2 ÷ø è

Thus,

¾®

206 82 Pb

=

+ 42He

Dm = 0.00564 amu Energy liberated per reaction = (Dm)931 MeV = 8.4 ´ 10-13 J Electrical energy produced = 8.4 ´ 10-14 J Let m g of energy.

210

Po is required to produce the desired

m ´ 6 ´ 1023 210 0.693 l= = 0.005 per day t1/ 2

N =

\ Electrical energy produced per day (0.005)(6 ´ 1023 m ) ´ 8.4 ´ 10-14 J 210 This is equal to 12 (given) . ´ 107 J =

\

m = 10 g

Ans.

Activity at the end of 693 days is n

0.005 ´ 6 ´ 1023 ´ 10 1021 æ 1ö = per day = R0 ç ÷ 210 7 è 2ø 693 Here, n = number of half-lives = =5 138.6 1021 \ R0 = R (2)5 = 32 ´ = 4.57 ´ 1021 per day 7 Ans. 8. (i) At t = 0, probabilities of getting a and b particles are same. This implies that initial activity of both is equal. Say it is R0. Activity after t = 1620 s, R=

and

æ 1ö R1 = R0 ç ÷ è 2ø

1620/ 405

æ 1ö R2 = R0 ç ÷ è 2ø

1620/ 1620

=

R0 16

R = 0 2

9 R0 16 Probability of getting a-particles,

Total activity, R = R1 + R2 =

Let N 0 be the total number of nuclei at t = 0. N 4N 0 Then, N 01 = 0 and N 02 = 5 5 N0 Given that N 1 + N 2 = 2 N0 5

or

(0.005)(6 ´ 1023 m ) æ dN ö per day ç÷ = lN = 210 è dt ø

Ans.

æ 1ö ç ÷ è 2ø

t / 405

+

4N 0 5

æ 1ö ç ÷ è 2ø

t / 1620

=

N0 2

t / 1620

æ 1ö =x ç ÷ è 2ø Then, above equation becomes x 4 + 4 x - 2.5 = 0 Let

\

x = 0.594

or

æ 1ö ç ÷ è 2ø

t/ 1620

Solving, it we get t = 1215 s.

= 0.594 Ans.

10-3 9. N = ´ 6.02 ´ 1023= 2.87 ´ 1018 210 During one mean life period 63.8% nuclei are decayed. Hence, energy released E = 0.638 ´ 2.87 ´ 1018 ´ 5.3 ´ 1.6 ´ 10-13 J = 1.55 ´ 106 J

Ans.

0.693 10. R0 = lN = ´ 6.02 ´ 1023 per sec 14.3 ´ 3600 ´ 24 = 3.37 ´ 1017 per sec After 70 hours activity, R = R0e- lt = (3.37 ´ 1017 ) e- (0.693/ 14.3 ´ 24)(70) = 2.92 ´ 1017 per sec In fruits activity was observed 1mCi or 3.7 ´ 104 per sec. Therefore, percentage of activity transmitted from root to the fruit. =

3.7 ´ 104 ´ 100 2.92 ´ 1017

= 126 . ´ 10-11 %

Ans.

35. Semiconductors INTRODUCTORY EXERCISE

35.1

1. Eg Carbon = 5.4 eV Eg Silicon = 1.1 eV Eg Germanium = 0.7 eV \ (Eg )C > (Eg )Si > (Eg )Ge

INTRODUCTORY EXERCISE

B

Y1

Y2

Y

0

1

1

0

1

1

1

0

0

1

Clearly output resembles an ‘OR’ gate.

35.2

3. For given amplifier, V0 = 2 V, R0 = 2 kW b ac = 100, Ri = 1 kW We have, output voltage, V0 = IC R0 Þ IC = collector current V 2 = 0 = = 10-3 A R0 2 ´ 103

1. Hole diffusion from p to n side can be viewed as “electron diffusion” from n to p side. Diffusion occurs due to difference in concentrations in different regions. An electron (or hole) diffuses where its concentration is less. 2. Due to forward biasing depletion layer thickness decreases, potential barrier is reduced and diffusion of electrons from n to p side occurs.

INTRODUCTORY EXERCISE

A

= 1 mA Also, current amplication b =

35.3 Þ

1. In a transistor, base must be very thin and lightly doped so that all of the charge carriers are not combined in base and majority of them passes the reverse bias layer to collector side.

2. Voltage gain is maximum and constant for mid frequency range but is less for both low and high frequencies.

iB =

iC 10-3 = A b 100

= 10-5 A = 10 ´ 10-6 A = 10 mA. and voltage amplification bR V AV = 0 = 0 Ri Vi Þ

Av (Voltage gain)

iC iB

Vi =

2 ´ 1 ´ 103 V0Ri = bR0 100 ´ 2 ´ 103

= 10-2 volts = 10 mV w (Input frequency)

INTRODUCTORY EXERCISE

1. Input waveforms are as shown

(b) We construct truth table to see the logical operation. A

Y1

NAND

NAND

B

NAND

35.4

Y

Y2

A¢ O B¢ O

t1

t2

t3

t4

t5

These are fed to an NAND gate,

A

B

Y1

Y2

Y

A

0

0

1

1

0

B

1

0

0

1

1

Output waveform is as shown

Y

t6

Chapter 35 t1

t2

t3

t4

t5

t6

1 Y: 0

2. Truth table for the circuit given is A

NAND

B

Y

NAND

Y1

Semiconductors — 513

A

B

Y1

Y

0

0

1

0

1

0

1

0

0

1

1

0

1

1

0

1

As the output resembles output of a AND gate so, given circuit behaves like an ‘AND’ gate.

Exercises 1. With increase in temperature, number of electrons reaching conduction band increases but, mean relation time decreases. Effect of decrease in relation time is much less than that of increase in number density of charge carriers.

2. For diode D1, Vp - Vn = - 10 - 0 = - 10V So, D1 is in reverse bias. For diode D2, Vp - Vn = 0 - (-10) = 10 V So, D2 is in forward bias.

3. Hole is a vacancy created by an electron moving from valance band to conduction band.

4. Capacitor is charged to maximum potential difference V = Vmax = Vrms ´ 2 \ = 220 2 volts

5. Electrons take the energy and move up to neat higher energy level both in conduction and valance band. So, a hole move downwards due to movement of electron.

6. In an n- p-n transistor, E

IE

B

C

95 100 IE Þ IE = ´ IC 100 95 100 = ´ 10 mA = 10.53 mA 95 So, I B = 0.53 mA 8. Depletion zone is formed due to diffusion of efrom n to p side. Due to diffusion positive immobile ions exists on n side and negative ions on p-side. Recombination of e- and holes takes place on p-side. This results in (n-side more positive than p-side) formation of a junction field and potential barrier across the depletion zone. 9. When applied voltage approaches zener potential diode breakdown to conduct excess current. This causes a change of current in series resistance. 10. Reverse breakdown may be “Avlanche break down” (breakdown due to high velocity collision of minority carrier) or it may be a zener breakdown (breakdown of bonds due to strong field). As IC =

11. No, it cannot be measured by using a voltmeter. Barrier potential is less than 0.2 V for germanium diode and is less than 0.7 V for silicon. Also there are no free charge carriers which provide ‘current’ for working of voltmeter. 12. An “OR” gate may be made to operate motor relays for garage gates. Pick up signal to relay

IB -

+

-

+

Electrons move from emitter to base and electrons which are not combined with holes in base region, crosses to collector side. 7. In an n- p-n transistor, I E = I B + IC

+ –

Sensor output from gate

514 — Optics and Modern Physics 13. Such combination is called “Cascade”

19. Supply voltage may be 7 V and zener breakdown

combination. AV = AV1 ´ AV2 (net )

= 10 ´ 20 = 200 V0 = output voltage = AV ´ Vi

\

(net )

= 200 ´ 0.01 = 2V hc 1240 eV× nm Eg = Þ Eg = l 600 nm

14. We use,

» 2 eV So, photon energy is less than that required (2.8 eV). So, detector is not able to detect this wavelength. 15. (i) Diode is a rectifier diode. (ii) Point P represents zener breakdown potential.

16. Given is a p - n - p CE configuration, when R1 is increased, base current decreases as a result current also decreases. Hence, both ammeter and voltmeter readings decreases.

17. A NOT gate can be formed by proper biasing of a transistor as shown. VCC=V(1)

at 5 V. So, voltage drop across resistance R = 7 - 5 = 2V. Now power rating of zener = 1W. So, current through zener must not exceed P 1 I = = A V 5 Hence, series resistance V 2 R = = = 10 W. 1 i 5

20. Current flows only in branches AB and EF as diode of CD branch is in reverse bias. (\ I 3 = 0) So, given circuit is equivalent to A E

I1

I4 25 W

125 W

25 W

125 W

B F

I2

G 5V

25 W

Current through cell is V 5 i1 = = = 0.05 A RTotal 100 Also by symmetry, 0.05 = 0.025 A 2 21. Given circuit is equivalent to I2 = I4 =

Y

C B A

10 V 3 kW

VB=V(1)

If input (A) is low (0), then the transistor is in cut off stage and output (Y) is same as VCC = V (1). If input (A) is high, then transistor current is in saturation and the net voltage at output (Y) is V (0) or is in state 0.

18. Given is an AND gate, its truth table is A

B

Y

0

0

0

1

0

0

0

1

0

1

1

1

H

400 W 10 V

In base emitter loop, VB = I B RB V 10 Þ IB = B = RB 400 ´ 103 = 25 ´ 10-6 A = 25 mA In emitter-collector loop, VC = IC RC

Semiconductors — 515

Chapter 35 Þ

IC =

b DC

and

10 3 ´ 103

23. Consider the figure to solve this question,

= 3.33 mA I 3.33 mA = C = IB 25 mA

I E = IC + I B and IC = bI B

K(i)

IC RC + VCE + I E RE = VCC

K(ii)

RI B + VBE + I E RE = VCC

K(iii)

I E = IC = b I B

= 133

VCC = 12 V A

22. Consider the figure given here to solve this [As base current is very small] RC = 7.8 kW From the figure, IC (RC + RE ) + VCE = 12 (RE + RC ) ´ 1 ´ 10-3 + 3 = 12

C H

I

= 0.085 mA 12 - 1.7 10.3 Resistance, RB = = IC 0.01 + 0.085 + 0.085 b

IC

IC + I RB

D

From Eq. (iii), Þ

(R + bRE ) I B = VCC - VBE V - VBE I B = CC R + bRE

RC = 7.8 kW

J

VCE = 3 V

IB VBE = 0.5V E IE R = 200 kW RE D

=

12 - 0.5 80 + 1.2 ´ 100

=

11.5 mA 200

From Eq. (ii), VCC = 12 V A

C H

E IE RE = 1 kW

J

[Given, b = 100] = 108 kW

VCE = 3 V

IB VBE = 0.5V R = 20 kW

RE = 9 - 7.8 = 1.2 kW VE = I E ´ RE = 1 ´ 10-3 ´ 1.2 ´ 103 = 1.2 V Voltage, VB = VE + VBE = 1.2 + 0.5 = 1.7 V VB 1.7 Current, I = = 20 ´ 103 20 ´ 103

RC

RC =100 kW

RE + RC = 9 ´ 103 = 9 kW

I

IC

IB + I

problem IC = I E

(RC + RE ) =

VCE - VBE IC

=

VCC - VCE bI B

(RC + RE ) =

2 (12 - 3) kW 11.5

= 1.56 kW RC + RE = 1.56 RC = 1.56 - 1 = 0.56 kW

(Q IC = bI B )

36. Communication System Exercises Single Correct Option

= 2 ´ 6.4 ´ 106 ´ 240

1. Range of frequencies is as follows :

» 55 ´ 103 m

Ground wave : 300 Hz to 300 kHz (it may go upto 3 MHz) Sky wave : 300 kHz to 3 MHz Space wave : 3 MHz to 300 GHz. l 2. Length of antenna ³ 4 l l³ Þ 4 Þ l £ 4 ´l or l £ 400 m

= 55 km

3. Sideband frequencies are :

c 3 ´ 108 = 20 ´ 103 m = F 15 ´ 103

So, size of antenna required l = 4 = 5 ´ 103 m = 5 km

(at least)

Also, effective power radiated by antenna is very less. = 1.5 ´ 106 - 3 ´ 103

= (1000 - 3) ´ 103 Hz

= 1497 kHz USB = wc + wm = 1503 kHz Bandwidth required = USB - LSB = 2wm = 2 ´ 3 kHz = 6 kHz 12. Due to over modulation m > 1, sidebands overlaps and fading (loss of information) may occur. Also when m » 1, distortion in carrier waveform occurs. 1 13. Frequency of signal obtained = = 1 MHz 2p LC 1 LC = Þ (2p ´ 106 )2 and

= 2997 kHz and = 2.997 MHz Upper sideband (USB) = wc + wm = 1 ´ 106 Hz + 3 ´ 103 Hz = 1003 kHz = 1.003 MHz 4. Frequency of amplitude modulated wave is same as that of carrier wave. 5. Basic communication system is

From source

l=

11. LSB = wc = wm

Lower sideband (LSB) = wc - wm = 1 ´ 106 Hz - 3 ´ 103 Hz

Transmitter Channel

10. Wavelength of signal is

Receiver To user

6. Beyond horizon, a signal can reach via ionospheric reflection or sky wave mode. Frequency range suitable is 3 MHz to 30 MHz. 7. UHF band is in range of 150 to 900 MHz. So, suitable mode of communication is “space wave mode”. 8. Digital signals employs a discrete values of amplitudes which are coded using binary system. 9. In LOS communication maximum distance upto which a signals can be received from tower is d = 2Rh

Þ

LC = 0.25 ´ 10-13 s2

Þ

LC = 2.5 ´ 10-14 s2

14. m = modulation index = Þ

Am Ac

Am = m ´ Ac = 0.75 ´ 12 = 9 volts 15. Vibrating tuning fork and string produces all possible values of displacement hence, these signals are analog. A light pulse and output of NAND gate gives only discrete values of output. So, signals are digital these.

Chapter 36 16. Penetrating power of signal a frequency of signal. So, 3 MHz signal travels longer distance in ionosphere. 17. Modulation index m, A - Amin = max Amax + Amin (A = amplitude of modulated wave) 15 - 3 12 2 m = = = 15 + 3 18 3

\

Communication System — 517

(iii) Percentage increase in area covered A - A1 = 2 ´ 100 A1 3608 - 804 = ´ 100 804 » 349 % 21. That can be done by using six antennas, 1

18. Man made noises and atmospheric interferences

6

affect only amplitude of a signal. So, an AM signal is more noisy than a FM signal. 19. In LOS communication, it is not necessary that transmitting antenna and receiving antenna are at same height. Only requirement is that there must not be any obstacle in between.

2

R R

3

5 4 1

R

R

6

T

5

2

3 4

Given, h = 81 m \ Distance upto which transmission can be made = d = 2Rh and, area covered in broadcast = pd 2 = 2pRh = 2 ´ p ´ 6.4 ´ 106 ´ 81 (m 2 ) = 3258 km 2

20. (i) When receiver is at ground level, then service area covered = pd 2 = p ( 2Rh )2 = 2pRh = 2 ´ p ´ 6.4 ´ 106 ´ 20

From above figure, side of triangle = R + h Also, altitude of triangle 3 = ´ side = R 2 3 Þ (R + h) = R 2 3 (R + h) = 2R Þ 2 Þ R+ h= R 3 2 h= R - R » 0.15 R Þ 3 1

22. From fmax = 9(N max ) 2 Þ

N max =

Þ

N max F =

» 804 km 2 (ii) When receiver is at height of 25 m, area covered = pd12 + pd22

1

» 3608 km 2

(5 ´ 106 )2 81 = 0.3086 ´ 1012 m -3

= 3.086 ´ 1011 m -3

= p ( 2RhT )2 + p ( 2RhR )2 = 2pR (hT + hR ) = 2 ´ p ´ 6.4 ´ 106 ´ 45 (m 2 )

2 fmax 81

and

(8 ´ 106 )2 81 = 7.9 ´ 1011 m -3

N max F = 2

518 — Optics and Modern Physics 23. Let the receiver is at point A and source is at B.

x

A

d

h3

x

d

or

3 ´ 108 ´ 4.04 ´ 10-3 2 = 6.06 ´ 105 = 606 km

x=

Using Pythagoras theorem, d 2 = x 2 - hs2 = (606)2 - (600)2 = 7236 B

Velocity of waves = 3 ´ 108 m/s Time to reach a receiver = 4.04 ms = 4.04 ´ 10-3 s Let the height of satellite is hs = 600 km Radius of earth = 6400 km Size of transmitting antenna = hT We know that Distance travelled by wave = Velocity of waves Time 2x = 3 ´ 108 4.04 ´ 10-3

or d = 85.06 km So, the distance between source and receiver = 2d = 2 ´ 85.06 = 170 km The maximum distance covered on ground from the transmitter by emitted EM waves d = 2RhT d2 = hT 2R 7236 or size of antenna hT = 2 ´ 6400 or

= 0.565 km = 565m

JEE Main and Advanced Previous Years’ Questions (2018-13) JEE Main 1. An EM wave from air enters a medium. The electric fields are é æz öù E1 = E01x$ cosê2pnç - t ÷ú in air and c è øû ë E2 = E02x$ cos[k( 2z - ct )] in medium, where the wave number k and frequency n refer to their values in air. The medium is non-magnetic. If e r1 and e r2 refer to relative permittivities of air and medium respectively, which of the following options is correct? (2018) (a)

er1 er2

= 4 (b)

er1 er2

= 2 (c)

er1 er2

=

er 1 1 (d) 1 = 4 er2 2

2. Unpolarised light of intensity I passes through an ideal polariser A. Another identical polariser B is placed behind A. The intensity of light beyond B is found to I be . Now, another identical polariser C is 2 placed between A and B. The intensity 1 beyond B is now found to be . The angle 8 between polariser A and C is (2018) (a) 0°

(b) 30°

(c) 45°

(d) 60°

3. The angular width of the central maximum in a single slit diffraction pattern is 60°. The width of the slit is 1 mm. The slit is illuminated by monochromatic plane waves. If another slit of same width is made near it, Young’s fringes can be observed on a screen placed at a distance 50 cm from the slits. If the observed fringe width is 1 cm, what is slit separation distance? (i.e. distance between the centres of each slit.) (2018) (a) 25 mm

(b) 50 mm

(c) 75 mm

(d) 100 mm

4. An electron from various excited states of hydrogen atom emit radiation to come to the ground state. Let ln , l g be the de-Broglie wavelength of the electron in the nth state and the ground state, respectively. Let Ln be the wavelength of the emitted photon in the transition from the nth state to the ground state. For large n, (A, B are constants) (2018) B

(a) Ln » A +

(b) Ln » A + Bl2n

l2n

(c) L2n » A + Bl2n

(d) L2n » l

5. If the series limit frequency of the Lyman series is n L , then the series limit frequency of the Pfund series is (2018) (a) 25 nL

(b) 16 nL

(c)

nL 16

(d)

nL 25

6. It is found that, if a neutron suffers an elastic collinear collision with deuterium at rest, fractional loss of its energy is Pd ; while for its similar collision with carbon nucleus at rest, fractional loss of energy is Pc. The values of Pd and Pc are respectively (2018) (a) (.89, .28) (c) (0, 0)

(b) (.28, .89) (d) (0, 1)

7. The reading of the ammeter for a silicon diode in the given circuit is 200W

3V

(a) 0 (c) 11.5 mA

(b) 15 mA (d) 13.5 mA

(2018)

2

Optics & Modern Physics

9. A diverging lens with magnitude of focal length 25 cm is placed at a distance of 15 cm from a converging lens of magnitude of focal length 20 cm. A beam of parallel light falls on the diverging lens. The final image formed is (2017) (a) virtual and at a distance of 40 cm from convergent lens (b) real and at a distance of 40 cm from the divergent lens (c) real and at a distance of 6 cm from the convergent lens (d) real and at a distance of 40 cm from convergent lens

(d)

log 13 . log 2 T (c) t = log 13 . (a) t = T

11. Some energy levels of a molecule are shown in the figure. The ratio of the wavelengths r = l1 / l2 is given by (2017)

log V

(b) t = T log 13 . (d) t =

T log 2 2 log 13 .

14. A particle A of mass m and initial velocity v collides with a particle B of mass m/ 2 which is at rest. The collision is held on, and elastic. The ratio of the de-Broglie wavelengths lA to lB after the collision is (2017)

l (a) A = 2 lB l 1 (c) A = lB 2

l 2 (b) A = lB 3 lA 1 (d) = lB 3

15. In a common emitter amplifier circuit using an n-p-n transistor, the phase difference between the input and the output voltages will be (2017)

–E l2 l1

(a) 90°

–2E

(b) 135°

(c) 180°

(d) 45°

16. A transparent slab of thickness d has a

–3E

(b) r =

(c)

log V

decays into a nucleus B. At t = 0, there is no nucleus B. After sometime t, the ratio of the number of B to that of A is 0.3. Then, t is given by (2017)

(a) 7.8 mm (b) 9.75 mm(c) 15.6 mm(d) 1.56 mm

2 3

log V

13. A radioactive nucleus A with a half-life T ,

are separated by 0.5 mm and the screen is placed 150 cm away. A beam of light consisting of two wavelengths, 650 nm and 520 nm, is used to obtain interference fringes on the screen. The least distance from the common central maximum to the point where the bright fringes due to both the wavelengths coincide, is (2017)

(a) r =

(b)

log V

10. In a Young’s double slit experiment, slits

–4/3E

(a)

loglmin

(b) 2 ´ 104 (d) 2 ´ 106

potential difference V to hit a metallic target to produce X-rays. It produces continuous as well as characteristic X-rays. If lmin is the smallest possible wavelength of X-rays in the spectrum, the variation of log lmin with log V is correctly represented in (2017)

loglmin

(2018)

(a) 2 ´ 103 (c) 2 ´ 105

12. An electron beam is accelerated by a

loglmin

working at carrier frequency of 10 GHz. Only 10% of it is utilised for transmission. How many telephonic channels can be transmitted simultaneously, if each channel requires a bandwidth of 5 kHz?

loglmin

8. A telephonic communication service is

3 4

(c) r =

1 3

(d) r =

4 3

refractive index n ( z ) that increases with z. Here, z is the vertical distance inside the slab, measured from the top. The slab is placed between two media with uniform

3

Previous Years’ Questions (2018-13) refractive indices n1 and n 2( > n1 ), as shown in the figure. A ray of light is incident with angle q i from medium 1 and emerges in medium 2 with refraction angle q f with a (2016) lateral displacement l. n1=constant z

qi

20. Radiation of wavelength l, is incident on

1

a photocell. The fastest emitted electron has speed v. If the wavelength is changed 3l to , the speed of the fastest emitted 4 electron will be (2016)

n (z) d

n2=constant

l

qf

2

1/ 2

Which of the following statement(s) is (are) true? (a) l is independent on n (z) (b) n1 sin qi = (n2 - n1) sin qf (c) n1 sin qi = n2 sin qf (d) l is independent of n2

refractive index of glass of a prism by i-d plot, it was found that a ray incident at an angle 35° suffers a deviation of 40° and that it emerges at an angle 79°. In that case, which of the following is closest to the maximum possible value of the refractive index? (2016) (b) 1.6

(c) 1.7

(d) 1.8

18. An observer looks at a distant tree of height 10 m with a telescope of magnifying power of 20. To observer the tree appears (2016) (a) 10 times taller (c) 20 times taller

1/ 2

æ4ö (a) > v ç ÷ è3ø 1/ 2 æ4ö (c) = v ç ÷ è3ø

æ4ö (b) < v ç ÷ è3ø 1/ 2 æ3ö (d) = v ç ÷ è4ø

21. Half-lives of two radioactive elements A

17. In an experiment for determination of

(a) 1.5

æ 2 l2 ö ÷ lL and bmin = çç ÷ è L ø (c) a = lL and bmin = 4lL l2 (d) a = and bmin = 4lL L (b) a =

(b) 10 times nearer (d) 20 times nearer

and B are 20 min and 40 min, respectively. Initially, the samples have equal number of nuclei. After 80 min, the ratio of decayed numbers of A and B nuclei will be (2016) (a) 1 : 16

(b) 4 : 1

(c) 1 : 4

(d) 5 : 4

22. The temperature dependence of resistances of Cu and undoped Si in the temperature range 300-400 K, is best described by (2016) (a) linear increase for Cu, linear increase for Si (b) linear increase for Cu, exponential increase for Si (c) linear increase for Cu, exponential decrease for Si (d) linear decrease for Cu, linear decrease for Si

23. Identify the semiconductor devices whose characteristics are as given below, in the order (a),(b),(c),(d). (2016) I

I

19. The box of a pin hole camera of length L, has a hole of radius a. It is assumed that when the hole is illuminated by a parallel beam of light of wavelength l the spread of the spot (obtained on the opposite wall of the camera) is the sum of its geometrical spread and the spread due to diffraction. The spot would then have its minimum size (say bmin ) when (2016) (a) a =

æ 2 l2 ö l2 ÷ and bmin = çç ÷ L è L ø

(a)

V

(b)

I

I Dark

(c)

V

Resistance V

Illuminated

(d) Intensity of light

V

4

Optics & Modern Physics (a) Simple diode, Zener diode, Solar cell, Light dependent resistance (b) Zener diode, Simple diode, Light dependent resistance, Solar cell (c) Solar cell, Light dependent resistance, Zener diode, Simple diode (d) Zener diode, Solar cell, Simple diode, Light dependent resistance

24. Monochromatic light is incident on a glass prism of angle A. If the refractive index of the material of the prism is m, a ray incident at an angle q, on the face AB would get transmitted through the face AC of the prism provided (2015) é ì æ 1 ö üù (a) q < cos- 1 êm sin íA + sin- 1 çç ÷÷ ýú êë è m ø þúû î é ì æ 1 ö üù (b) q < sin- 1 êm siníA - sin- 1 çç ÷÷ ýú êë è m ø þúû î é ì æ 1 ö üù (c) q > cos- 1 êm siníA + sin- 1 çç ÷÷ ýú êë è m ø þúû î é ì æ 1 ö üù (d) q > sin- 1 êm siníA - sin- 1 çç ÷÷ ýú êë è m ø þúû î

25. Assuming human pupil to have a radius of 0.25 cm and a comfortable viewing distance of 25 cm, the minimum separation between two objects that human eye can resolve at 500 nm wavelength is (2015) (a) 30 mm

(b) 1 mm

(c) 100 mm (d) 300 mm

26. On a hot summer night, the refractive index of air is smallest near the ground and increases with height from the ground. When a light beam is directed horizontally, the Huygens principle leads us to conclude that as it travels, the light beam (2015) (a) becomes narrower (b) goes horizontally without any deflection (c) bends upwards (d) bends downwards

27. A signal of 5 kHz frequency is amplitude modulated on a carrier wave of frequency 2MHz. The frequencies of the resultant signal is/are (2015)

(a) 2 MHz only (b) 2005 kHz 2000 kHz and 1995 kHz (c) 2005 kHz and 1995 kHz (d) 2000 kHz and 1995 kHz

28. A red LED emits light at 0.1 W uniformly around it. The amplitude of the electric field of the light at a distance of 1 m from the diode is (2015) (a) 2.45 V/m (c) 5.48 V/m

(b) 1.73 V/m (d) 7.75 V/m

29. As an electron makes a transition from an excited state to the ground state of a hydrogen like atom/ion (2015) (a) kinetic energy, potential energy and total energy decrease (b) kinetic energy decreases, potential energy increases but total energy remains same (c) kinetic energy and total energy decrease but potential energy increases (d) its kinetic energy increases but potential energy and total energy decrease

30. Match Column I (fundamental experiment) with Column II (its conclusion) and select the correct option from the choices given below the list. (2015) Column I

Column II

A

Franck-Hertz experiment

1. Particle nature of light

B

Photo-electric experiment

2. Discrete energy levels of atom

C

Davisson-Germe r experiment

3. Wave nature of electron 4. Structure of atom

A (a) 1 (c) 2

B 4 4

C 3 3

A (b) 2 (d) 4

B 1 3

C 3 2

31. A green light is incident from the water to the air-water interface at the critical angle (q). Select the correct statement. (2014) (a) The entire spectrum of visible light will come out of the water at an angle of 90° to the normal (b) The spectrum of visible light whose frequency is less than that of green light will come out of the air medium

5

Previous Years’ Questions (2018-13) (c) The spectrum of visible light whose frequency is more than that of green light will come out to the air medium (d) The entire spectrum of visible light will come out of the water at various angles to the normal

32. A thin convex lens made from crown glass

magnetic field of 3 ´ 10-4 T. If the radius of the largest circular path followed by these electrons is 10.0 mm, the work function of the metal is close to (2014) (a) 1.8 eV (b) 1.1 eV (c) 0.8 eV (d) 1.6 eV

36. The forward biased diode connection is

(m = 3/ 2) has focal length f. When it is

measured in two different liquids having refractive indices 4/3 and 5/3. It has the focal lengths f1 and f2, respectively. The correct relation between the focal length is (2014) (a) (b) (c) (d)

f1 = f2 < f f1 > f and f2 becomes negative f2 > f and f1 becomes negative f1 and f2 both become negative

(c) 1

(d) 1/3

34. Hydrogen (1H1 ), deuterium (1H 2 ), singly

ionised helium ( 2He4 )+ and doubly ionised lithium (3 Li8 )+ + all have one electron around the nucleus. Consider an electron transition from n = 2 to n = 1. If the wavelengths of emitted radiation are l1 , l2 , l3 and l4, respectively for four elements, then approximately which one of the following is correct? (2014)

(a) (b) (c) (d)

–3V

–3V

(c)

2V

4V

(d)

–2V

+2V

(b)

electromagnetic waves in a medium,

light with mutually perpendicular planes of polarisation are seen through a polaroid. From the position when the beam A has maximum intensity (and beam B has zero intensity), a rotation of polaroid through 30° makes the two beams appear equally bright. If the initial intensities of the two beams are I A and I B (2014) respectively, then I A / I B equals (b) 3/2

(a)

–2V

37. During the propagation of

33. Two beams, A and B, of plane polarised

(a) 3

(2014)

+2V

4 l1 = 2 l2 = 2 l3 = l4 l1 = 2 l2 = 2 l3 = l4 l1 = l2 = 4 l3 = 9l4 l1 = 2 l2 = 3 l3 = 4 l4

35. The radiation corresponding to 3 ® 2 transition of hydrogen atom falls on a metal surface to produce photoelectrons. These electrons are made to enter a

(2014)

(a) electric energy density is double of the magnetic energy density (b) electric energy density is half of the magnetic energy density (c) electric energy density is equal to the magnetic energy density (d) Both electric and magnetic energy densities are zero

38. Match List I (Electromagnetic wave type) with List II (Its association/application) and select the correct option from the choices given below the lists. (2014) List I

List II

A.

Infrared waves

1.

To treat strain

B.

Radio waves

2.

For broadcasting

C.

X-rays

3.

To detect fracture of bones

D.

Ultraviolet

4.

Absorbed by the ozone layer of the atmosphere

Codes A (a) 4 (b) 1 (c) 3 (d) 1

B 3 2 2 2

C 2 4 1 3

D 1 3 4 4

muscular

6

Optics & Modern Physics

39. The graph between angle of deviation ( d) and angle of incidence (i) for a triangular prism is represented by (2013) d

d

frequency of radiation emitted is proportional to 1 (a) n

(b)

1

(c)

n2

1

(2013)

1

(d)

n4

n3

44. The I-V characteristic of an LED is (a)

(b)

(2013)

R YG B O

O

i

d

i

d

(c)

R G (b) Y R

(a) I O

(d)

O

V

V V

O

O

i

(c) I

i

40. Two coherent point sources S1 and S 2 are separated by a small distance d as shown. The fringes obtained on the screen will be (2013)

Screen d S1

(d)

O

V

R Y G B

I

O - Red - Yellow - Green - Blue

45. The anode voltage of a photocell is kept fixed. The wavelength l of the light falling on the cathode is gradually changed. The plate current I of photocell varies as follows (2013)

S2

I

I

D

(a)

(a) points (c) semi-circle

is passed through a polaroid A and then through another polaroid B which is oriented so that its principal plane makes an angle of 45° relative to that of A. The intensity of the emergent light is (2013) (b) I 0 / 2 (d) I 0 / 8

42. Diameter of a plano-convex lens is 6 cm and thickness at the centre is 3 mm. If speed of light in material of lens is 2 ´ 108 m/s, the focal length of the lens is (2013) (a) 15 cm (c) 30 cm

(b) 20 cm (d) 10 cm

43. In a hydrogen like atom electron makes transition from an energy level with quantum number n to another with quantum number ( n - 1). If n >> 1, the

l

l I

41. A beam of unpolarised light of intensity I 0

(a) I 0 (c) I 0 / 4

(b)

(b) straight lines (d) concentric circles

I

(c)

(d) l

l

46. A diode detector is used to detect an amplitude modulated wave of 60% modulation by using a condenser of capacity 250 pico farad in parallel with a load resistance 100 kW. Find the maximum modulated frequency which could be detected by it. (2013) (a) 10.62 MHz (c) 5.31 MHz

(b) 10.62 kHz (d) 5.31 kHz

47. The magnetic field in a travelling electromagnetic wave has a peak value of 20 nT. The peak value of electric field strength is (2013) (a) 3 V/m

(b) 6 V/m

(c) 9 V/m

(d) 12 V/m

Answer with Explanations 1. (c) Speed of progressive wave is given by, v =

w k

As electric field in air is, æ 2 pnz ö E1 = E 01x cos ç - 2 pnt ÷ è c ø 2 pn =c \ Speed in air = æ 2 pn ö ç ÷ è c ø 1 Also, c= m 0er1 e0

…(i)

In medium, E 2 = E 02xcos (2 kz - kct ) kc c \ Speed in medium = = 2k 2 c 1 Also, = 2 m 0er2 e0

…(ii)

As medium is non-magnetic medium, m medium = m air On dividing Eq. (i) by Eq. (ii), we have er2 er 1 2 = Þ 1 = er1 er2 4 C I/2

de-excitation is L n; n

B

n

I 8

I a

4. (a) If wavelength of emitted photon in


45°

(d)

¥

3. In a radioactive decay chain, 232 90 Th nucleus decays to 212 82 Pb nucleus. Let N a and N b be the number of a and bparticles respectively, emitted in this decay process. Which of the following statements is (are) true? (More than One Correct Option, 2018)

(a) N a = 5 (c) N b = 2

electrons. The value of n is ........... . (Take, mass of the electron, me = 9 ´ 10-31kg and eV = 1.6 ´ 10-19 J) (Numerical Value, 2018)

5. Consider a hydrogen-like ionised atom with atomic number Z with a single electron. In the emission spectrum of this atom, the photon emitted in the n = 2 to n = 1 transition has energy 74.8 eV higher than the photon emitted in the n = 3 to n = 2 transition. The ionisation energy of the hydrogen atom is 13.6 eV. The value of Z is ............ . (Numerical Value, 2018)

6. For an isosceles prism of angle A and

(b) 0>1. Which of the following statement(s) is (are) true? (More than One Correct Option, 2016)

(a) Relative change in the radii of two consecutive orbitals does not depend on Z (b) Relative change in the radii of two consecutive orbitals varies as 1/ n (c) Relative change in the energy of two consecutive orbitals varies as 1 / n 3 (d) Relative change in the angular momenta of two consecutive orbitals varies as 1/n

18. A hydrogen atom in its ground state is irradiated by light of wavelength 970Å. Taking hc / e = 1.237 ´ 10-6 eVm and the ground state energy of hydrogen atom as - 13.6 eV, the number of lines present in the emission spectrum is (Single Integer Type, 2016)

19. An accident in a nuclear laboratory resulted in deposition of a certain amount of radioactive material of half-life 18 days inside the laboratory. Tests revealed that

uniformly distributed throughout a spherical nucleus of radius R is given by 3 Z ( Z - 1)e2 The measured masses of E= 5 4pe 0R the neutron, 11H , 157 N and 158 O are 1.008665 u, 1.007825 u, 15.000109 u and 15.003065 u, respectively. Given that the radii of both the 157 N and 158 O nuclei are same, 1 u = 931.5 MeV/c 2 (c is the speed of light) and e2 /( 4pe 0 ) = 1.44 MeV fm. Assuming that the difference between the binding energies of 157 N and 158 O is purely due to the electrostatic energy, the radius of either of the nuclei is (1fm = 10- 15 m) (Single Correct Option, 2016) (a) 2.85 fm (c) 3.42 fm

(b) 3.03 fm (d) 3.80 fm

21. A Young’s double slit interference arrangement with slits S1 and S 2 is immersed in water (refractive index = 4/ 3) as shown in the figure. The positions of maxima on the surface of water are given by x 2 = p2m 2l2 - d 2, where l is the wavelength of light in air (refractive index = 1 ), 2d is the separation between the slits and m is an integer. The value of (Single Integer Type, 2015) p is S1 d d S2

x

Air Water

17

Previous Years’ Questions (2018-13) 22. Consider a concave mirror and a convex lens (refractive index = 1.5) of focal length 10 cm each, separated by a distance of 50 cm in air (refractive index = 1) as shown in the figure. An object is placed at a distance of 15 cm from the mirror. Its erect image formed by this combination has magnification M1.When the set-up is 7 kept in a medium of refractive index , 6 the magnification becomes M 2. The ½M ½ magnitude½ 2½is ½M1½ (Single Integer Type, 2015)

Passage (Q. Nos. 25-26) Light guidance in an optical fibre can be understood by considering a structure comprising of thin solid glass cylinder of refractive index n1 surrounded by a medium of lower refractive index n2. The light guidance in the structure takes place due to successive total internal reflections at the interface of the media n1 and n2 as shown in the figure. All rays with the angle of incidence i less than a particular value im are confined in the medium of refractive index n1. The numerical aperture (NA) of the structure is defined as sin im. (Passage Type, 2015)

25. For two structures namely S1 with 45 3 8 and n 2 = , and S 2 with n1 = 4 2 5 7 and n 2 = and taking the refractive index 5 4 of water to be and that to air to be 1, 3 the correct option(s) is/are n1 =

15 cm 50 cm

23. Two identical glass rods S1 and S 2

(refractive index = 1.5) have one convex end of radius of curvature 10 cm. They are placed with the curved surfaces at a distance d as shown in the figure, with their axes (shown by the dashed line) aligned. When a point source of light P is placed inside rod S1 on its axis at a distance of 50 cm from the curved face, the light rays emanating from it are found to be parallel to the axis inside S 2. The distance d is (Single Correct Option, 2015) S1

S2

P 50 cm

(a) 60 cm (c) 80 cm

d

(b) 70 cm (d) 90 cm

24. A monochromatic beam of light is q incident at 60° on 60° one face of an equilateral prism of refractive index n and emerges from the opposite face making an angle q with the normal (see figure). For n = 3 the value of q is 60° dq and = m. The value of m is dn (Single Integer Type, 2015)

Air q i

n1>n2

Cladding Core

n2 n1

(a) NA of S1 immersed in water is the same as that of S2 immersed in a liquid of refractive 16 index 3 15 (b) NA of S1 immersed in liquid of refractive index 6 is the same as that of S2 immersed in 15 water (c) NA of S1 placed in air is the same as that S2 4 immersed in liquid of refractive index 15 (d) NA of S1 placed in air is the same as that of S2 placed in water

26. If two structures of same cross-sectional area, but different numerical apertures NA1 and NA2( NA2 < NA1 ) are joined longitudinally, the numerical aperture of the combined structure is (a)

NA1NA2 NA1 + NA2

(c) NA1

(b) NA1 + NA2 (d) NA2

18

Optics & Modern Physics

27. A nuclear power plant supplying electrical power to a village uses a radioactive material of half life T years as the fuel. The amount of fuel at the beginning is such that the total power requirement of the village is 12.5% of the electrical power available from the plant at that time. If the plant is able to meet the total power needs of the village for a maximum period of nT years, then the value of n is (Single Integer Type, 2015)

31. A fission reaction is given by 236 92 U

® 140 54 Xe +

94 38 Sr

where x and y are two particles. Considering 236 92 U to be at rest, the kinetic energies of the products are denoted by K Xe , K Sr , K x (2 MeV) and K y (2 MeV), respectively. Let the binding energies per 140 94 nucleon of 236 92 U, 54 Xe and 38 Sr be 7.5 MeV, 8.5 MeV and 8.5 MeV, respectively. Considering different conservation laws, the correct options is/are

28. Match the nuclear processes given in Column I with the appropriate option(s) in Column II. Column I

Column II

A. Nuclear fusion

P. absorption of thermal neutrons by 235 92 U

B. Fission in a nuclear reactor

Q.

C. b-decay

R. Energy production in stars via hydrogen conversion to helium

D. g-ray emission

S. Heavy water T.

60 27 Co

nucleus

Neutrino emission (Matching Type, 2015)

29. For a radioactive material, its activity A and rate of change of its activity R are dN dA defined as A = and R = , where dt dt N ( t ) is the number of nuclei at time t. Two radioactive source P(mean life t) and Q (mean life 2t) have the same activity at t = 0. Their rate of change of activities at t = 2t are RP and RQ , respectively. If RP n = , then the value of n is RQ e (Single Integer Type, 2015)

30. An electron in an excited state of Li 2+ ion 3h . The de 2p Broglie wavelength of the electron in this state is ppa0 (where a0 is the Bohr radius). The value of p is

(Single Correct Option, 2015)

(a) x = n, y = n, K Sr = 129 MeV, K Xe = 86 MeV

(b) x = p, y = e - , K Sr = 129 MeV, K Xe = 86 MeV (c) x = p, y = n, K Sr = 129 MeV, K Xe = 86 MeV (d) x = n, y = n, K Sr = 86 MeV, K Xe = 129 MeV

32. A light source, which emits two wavelengths l1 = 400 nm and l2 = 600 nm, is used in a Young’s double-slit experiment. If recorded fringe widths for l1 and l2 are b1 and b2 and the number of fringes for them within a distance y on one side of the central maximum are m1 and m2, respectively, then (More than One Correct Option, 2014)

(a) b 2 > b1 (b) m1 > m2 (c) from the central maximum, 3rd maximum of l2 overlaps with 5th minimum of l1 (d) the angular separation of fringes of l1 is greater than l2

33. A transparent thin film of uniform thickness and refractive index n1 = 1.4 is coated on the convex spherical surface of radius R at one end of a long solid glass cylinder of refractive index n 2 = 1.5, as shown in the figure. n1

has angular momentum

(Single Integer Type, 2015)

+ x + y,

Air

n2

Rays of light parallel to the axis of the cylinder traversing through the film from

19

Previous Years’ Questions (2018-13) air to glass get focused at distance f1 from the film, while rays of light traversing from glass to air get focused at distance f2 from the film. Then (More than One Correct Option, 2014)

(a) | f1 | = 3R

(b) | f1 | = 2.8R

(c) | f2 | = 2R

(d) | f2 | = 1.4R

34. Four combinations of two thin lenses are given in Column I.The radius of curvature of all curved surfaces is r and the refractive index of all the lenses is 1.5. Match lens combinations in Column I with their focal length in Column II and select the correct answer using the code given below the columns. Column I

Column II

P.

1.

2r

Q.

2.

r/2

R.

3.

-r

36. If lCu is the wavelength of K a , X-ray line of copper (atomic number 29) and lMO is the wavelength of the K a , X-ray line of molybdenum (atomic number 42), then the ratio lCu / lMo is close to

(Single Correct Option, 2014)

(a) 1.99

(b) 2.14

(c) 0.50 (d) 0.48

37. The image of an object, formed by a plano-convex lens at a distance of 8 m behind the lens, is real and is one-third the size of the object. The wavelength of 2 light inside the lens is times the 3 wavelength in free space. The radius of the curved surface of the lens is (Single Correct Option, 2013)

(a) 1 m

(b) 2 m

(c) 3 m

(d) 6 m

38. A ray of light travelling in the direction 1 ( i + 3 j) is incident on a plane mirror. 2 After reflection, it travels along the 1 direction ( i - 3 j). The angle of 2 incidence is (Single Correct Option, 2013) (a) 30°a

(b) 45°

(c) 60°

(d) 75°

39. In the Young’s double slit experiment S.

4.

r

(Matching Type, 2014)

Codes P Q R S (a) 1 2 3 4 (c) 4 1 2 3

l 2 l (c) (2n + 1) 8

(a) (2n + 1) P Q R S (b) 2 4 3 1 (d) 2 1 3 4

35. A metal surface is illuminated by light of two different wavelengths 248 nm and 310 nm. The maximum speeds of the photoelectrons corresponding to these wavelengths are u1 and u 2, respectively. If the ratio u1 : u 2 = 2 : 1 and hc = 1240 eV nm, the work function of the metal is nearly (Single Correct Option, 2014) (a) 3.7 eV (c) 2.8 eV

using a monochromatic light of wavelength l the path difference (in terms of an integer n) corresponding to any point having half the peak intensity is (Single Correct Option, 2013)

(b) 3.2 eV (d) 2.5 eV

l 4 l (d) (2n + 1) 16 (b) (2n + 1)

40. A right angled prism of refractive index m 1 is placed in a rectangular block of refractive index m 2 , which is surrounded by a medium of refractive index m 3 , as shown in the figure, A ray of light ‘e’ enters the rectangular block at normal incidence. Depending upon the relationships between m 1 , m 2 and m 3 , it takes one of the four possible paths ‘ef’, ‘eg’, ‘eh’ or ‘ei’.

20

Optics & Modern Physics 43. A freshly prepared sample of a

f 45°

e

radioisotope of half-life 1386 s has activity 103 disintegrations per second. Given that ln 2 = 0.693, the fraction of the initial number of nuclei (expressed in nearest integer percentage) that will decay in the first 80 s after preparation of the sample is (Single Integer Type, 2013)

g

m1

h

i m2

m3

Match the paths in Column I with conditions of refractive indices in Column II and select the correct answer using the codes given below the columns.

44. The radius of the orbit of an electron in a Hydrogen-like atom is 4.5 a0 where a0 is the Bohr radius. Its orbital angular 3h momentum is . It is given that h is 2p Planck constant and R is Rydberg constant. The possible wavelength(s), when the atom de-excites, is (are)

(Matching, Type 2013)

Column I

Column II

P.

e ®f

1.

m1 > 2 m 2

Q.

e ®g

2.

m 1 > m 1 and m 2 > m 3

R.

e ®h

3.

m1 = m 2

S.

e ®i

4.

m 2 < m 1 < 2 m 2 and m 2 >m 3

Codes P Q R S (a) 2 3 1 4 (c) 4 1 2 3

P Q R (b) 1 2 4 (d) 2 3 4

S 3 1

41. A point source S is placed at the bottom of a transparent block of height 10 mm and refractive index 2.72. It is immersed in a lower refractive index liquid as shown in the figure. It is found that the light emerging from the Liquid block to the liquid forms a circular Block bright spot of S diameter 11.54 mm on the top of the block. The refractive index of the liquid is (Single Correct Option, 2013)

(a) 1.21

(b) 1.30

(c) 1.36

(d) 1.42

42. A pulse of light of duration 100 ns is absorbed completely by a small object initially at rest. Power of the pulse is 30 mV and the speed of light is 3 ´ 108 ms-1. The final momentum of the object is (Single Correct Option, 2013) (a) 0.3 ´ 10-17 kg-ms-1 (b) 10 . ´ 10-17 kg-ms-1 (c) 3.0 ´ 10-17 kg -ms-1 (d) 9.0 ´ 10-17 kg-ms-1

(More than One Correct Option, 2013)

9 (a) 32R 9 (c) 5R

9 16R 4 (d) 3R

(b)

Passage (Q. Nos. 45-46) The mass of a nucleus AZ X is less that the sum of the masses of ( A - Z ) number of neutrons and Z number of protons in the nucleus. The energy equivalent to the corresponding mass difference is known as the binding energy of the nucleus. A heavy nucleus of mass M can break into two light nuclei of masses m1 and m2 only if (m1 + m2) < M . Also two light nuclei of masses m3 and m4 can undergo complete fusion and form a heavy nucleus of mass M ¢ only if (m3 + m4 ) > M ¢ . The masses of some neutral atoms are given in the table below : (Passage Type, 2013) 1 1H 6 3Li

1.007825u

2 1H

2.014102u

3 1H

3.016050u 4.002603u

6.01513u

7 3Li

7.016004u

4 2 He

152 64 Gd

151.919803u

206 82 Pb

205.974455u

70 30 Zn

69.925325u

82 34 Se

81.916709u

209 83 Bi

208.980388u

201 84 Po

205.974455u

45. The correct statement is (a) The nucleus 63 Li can emit an alpha particle (b) The nucleus 210 84 Po can emit a proton (c) Deuteron and alpha particle can undergo complete fusion 82 (d) The nuclei 70 30 Zn and 34 Se can undergo complete fusion

21

Previous Years’ Questions (2018-13) 46. The kinetic energy (in keV) of the alpha particle, when the nucleus undergoes alpha decay, is (a) 5316

(b) 5422

210 84 Po at

(c) 5707

rest

(d) 5818

47. The isotopes 125 B having a mass 12.014 u undergoes b-decay to 126 C. 126 C has an

excited state of the nucleus (126 C* ) at 4.041 MeV above its ground state. If 125 B decays to 125 C *, the maximum kinetic energy of the b-particle in units of MeV is (1 u = 931.5MeV / c2, where c is the speed of light in vacuum) (Single Integer Type, 2013)

Answer with Explanations 2

1. (130.0 kW / m )

IPQ =

Þ

f æ PQ = ç x è

f ö æ 2 ( AB) x ö ÷ ÷ç xø è f ø 2( AB) x ù é Q PQ = f ûú ëê

I0 I'0 A0

A'0 2cm

20 cm

3. (a, c)

2

A 0¢ æ 2 ö 1 =ç ÷ = Þ A 0 è 20 ø 100

A 0¢ =

A0 100

P = I0 A 0 = I0 ¢ A 0 ¢ I A I0¢ = 0 0 = 100I0 = 130 kW/m 2 A0 100

Þ

2. (d) Image of point A Q x

P

C

For A :

PQ =

1 1 1 + = v [- ( f /2 )] - f

2 ( AB) x f

Þ

IAB v f Þ =- =Þ AB u æ fö ç- ÷ è 2ø For height of PQ, 1 1 1 + = v - [( f - x )] - f

212 82 Pb.

Change in mass number ( A ) = 20 20 \Number of a-particle emitted = =5 4 Due to 5 a-particles, Z will change by 10 units. Since, given change is 8, therefore number of b-particles emitted is 2.

4. (24) \ Power = nhf

Þ

n=

200 16 . ´ 10 -19 ´ 625 .

As photon is just above threshold frequency KEmax is zero and they are accelerated by potential difference of 500 V.

f/2

PQ AB = Þ x f /2

converting into

200 = nW = n [625 . ´ 16 . ´ 10 -19 ]

f/2 A

232 90 Th is

(where, n = number of photons incident per second) Since, KE = 0, hf = work-funcition W

B

O

IPQ = 2 AB (Size of image is independent of x. So, final image will be of same height terminating at infinity)

KE f = qDV

\

P2 = qDV Þ 2m

v =f IAB = 2 AB

P = 2 mq DV

Since, efficiency is 100%, number of electrons emitted per second = number of photons incident per second. As, photon is completely absorbed, force exerted = n( mV ) = nP = n 2 mqDV =

200 6.25 ´ 1.6 ´ 10 -19

´ 2 ( 9 ´ 10 -31 ) ´ 1.6 ´ 10 -19 ´ 500

= 24

Þ

1 1 1 = v ( f - x) f

Þ

IPQ v f( f - x ) æfö =- = =ç ÷ PQ u x[( f - x )] è x ø

Þ

f( f - x ) v = x

é

1ù

é 3ù

5. (3) DE 2-1 = 13.6 ´ Z 2 ê1 - ú = 13.6 ´ Z 2 ê ú 4û ë ë 4û é 1 1ù é 5ù DE 3- 2 = 13.6 ´ Z 2 ê - ú = 13.6 ´ Z 2 ê ú ë 4 9û ë 36 û

22

Optics & Modern Physics DE 2 = DE 3- 2 + 74.8

\

é 5ù é 3ù 13.6 ´ Z 2 ê ú = 13.6 ´ Z 2 ê ú + 74.8 ë 36 û ë 4û 5ù é3 = 74.8 13.6 ´ Z 2 ê ë 4 36 úû Z2 = 9 Z =3

\

6. (a,b,c) The minimum deviation produced by a prism dm = 2 i - A = A \ i 1 = i 2 = A and r1 = r2 = A /2 Þ r1 = i 1 /2 Now, using Snell’s law sin A = m sin A /2 Þ m = 2 cos ( A /2 ) A

at P1 Dx = 0 at P2 Dx = 1. 8 mm = nl Number of maximas will be Dx 1. 8 mm n= = = 3000 l 600 nm at P2 Dx = 3000l Hence, bright fringe will be formed. At P2, 3000 th maxima is formed. For (a) option Dx = d sin q Þ dDx = d cos dq Rl Rl = d cos qRdq Þ Rdq = d cos q As we move from P1 to P2 q ­ cos q ¯ Rdq ­

9. (5) 1131 ¾T¾ ¾ ¾ ¾® Xe 131 + b = 8 Days 1/ 2

i1

r1

r2

A 0 = 2.4 ´ 10 5 Bq = lN0

i2

C

B

For this prism when the emergent ray at the second surface is tangential to the surface i 2 = p /2 Þ r2 = qc Þ r1 = A - qc so, sin i 1 = m sin( A - qc ) ù A -1 é so, i 1 = sin êsin A 4 cos 2 - 1 - cos A ú 2 ë û For minimum deviation through isosceles prism, the ray inside the prism is parallel to the base of the prism if ÐB = ÐC. But it is not necessarily parallel to the base if, ÐA = ÐB or ÐA = ÐC

Let the volume is V, t = 0 A 0 = lN0 Þ t = 11. 5 h, A = lN l æN ö 115 = lç ´ 2 . 5 ÷ 115 = ´ 2 . 5 ´ ( N0e - lt ) V èV ø ln2

(11.5 h) ( N0l) ´ (2 . 5) ´ e 8 day V (2 . 4 ´ 10 5 ) 115 = ´ (2 . 5) ´ e -1/ 24 V 2 . 4 ´ 10 5 1ù é V = ´ 2 . 5 ê1 115 24 úû ë

115 =

= =

7. (8) But this value of refractive index is not possible. 1. 6 sin q = ( n - mDn)sin 90° 1. 6 sin q = n - mDn

1. 6 ´

1 = 1. 6 - m( 0 .1) Þ 0 . 8 = 1. 6 - m ( 0 .1) 2 m ´ 0 .1 = 0 . 8 Þ m = 8

8. (c,d) P1 Dx= d sinq

10 5 ´ 23 ´ 25 115 ´ 10 2

= 5 ´ 10 3 ml = 5 L

10. (d) According to photoelectric effect equation KEmax =

hc p2 hc = - f0 - f0 Þ 2m l l

[KE = p2 /2 m]

( h / ld )2 hc - f0 [ p = h / l] = 2m l Assuming small changes, differentiating both sides, h2 æç 2dld ö hc dld l3 - 3 ÷÷ = - 2 dl Þ µ d2 2 m çè ld ø dl l l

11. (a) According to question,

d sinq q S1

2 . 4 ´ 10 5 é 23 ù ´ 2 . 5ê ú 115 ë 24 û

S2

P

P2

q

d

a M

l = 600 nm

Q

r1

N r2 R

23

Previous Years’ Questions (2018-13)

Radius of curvature of convex surface is 30 cm. Faint image is erect and virtual. Focal length of lens is 20 cm.

Applying Snell's law at M, sin a sin 45° n= Þ 2 = sin r1 sin r1 sin 45° 1 / 2 1 = = r1 = 30° 2 2 2 1 1 Þ qc = 45° sin qc = = n 2

Þ

13. (a,c,d) From Snell's law, n sin q = constant

sin r1 =

\ n1 sin q i = n2 sin q f Further, l will depend on n1 and n( z). But it will be independent of n2.

Let us take r2 = qc = 45° for just satisfying the condition of TIR. In DPNM,

14. (d) 50 4–Ö3

q + 90°+ r1 + 90°- r2 = 180° or q = r2 - r1 = 45°- 30°= 15° Note If a > 45° (the given value). Then, r1 >30° (the obtained value) \ r2 > qc (as r2 - r1 = q or r2 = q + r1 ) or TIR will take place. So, for taking TIR under all conditions a should be greater than 45° or this is the minimum value of a.

12. (a,d) Case 1

n

50 4 – Ö3 Ö3

r a 30°

O

25 30°

50

25Ö3 2

For Lens I

l2

Þ

1 1 1 - = Þ v u f v =

v =

uf u + f

( -50)( 30) = 75 cm - 50 + 30

1 1 1 uf + = Þ v = v u f u-f æ 25 3 ö ç ÷ ç 2 ÷ ( 50) - 50 3 è ø cm = v = Þ 4- 3 25 3 - 50 2 æ - 50 3 ö ç ÷ ç 4 - 3 ÷ 25 v h m = - = 2 Þ h2 = - ç ÷× 2 u h1 ç 25 3 ÷ ç ÷ 2 è ø + 50 cm h2 = 4- 3 For Mirror

30 cm 60 cm Using lens formula, 1 1 1 1 1 2 + = Þ = + Þ f1 = 20 cm 60 30 f1 f1 60 60

Further,

1 æ1 1ö = ( n - 1) ç - ÷ f1 èR ¥ø

Þ

f1 =

R = + 20 cm n-1

Case 2 10 cm n

I

30 cm

Using mirror formula, 1 1 1 3 1 1 2 = Þ = = 10 30 f 2 30 30 f2 30 R Þ R = 30 Þ R = 30 cm 2 R 30 = +20 cm = n-1 n-1

The x-coordinate of the images = 50 - v cos 30° + h2 cos 60° » 25 cm The y-coordinate of the images = v sin 30° + h2 sin 60° » 25 3 cm

15. (a,b)

f2 = 15 =

Þ 2 n - 2 = 3 Þ f1 = + 20 cm Refractive index of lens is 2.5.

d S1

S2

O

Path difference at point O = d = 0.6003 mm = 600300 nm.

I1 25/2

24

Optics & Modern Physics lö æ This path difference is equal to ç 1000 l + ÷. 2ø è

n

Þ Minima is formed at point O. Line S1S 2 and screen are ^ to each other so fringe pattern is circular (semi-circular because only half of screen is available) hc (f = work function) 16. (b) - f = eV0 l hc …(i) - f = 2e 0.3 ´ 10 -6 hc 0.4 ´ 10 -6

…(ii)

- f = 1e

Subtracting Eq. (ii) from Eq. (i) 1 ö 6 æ 1 æ 0.1 ö ´ 10 6 ÷ = e hc ç ÷10 = e Þ hc ç è 0.3 0.4 ø è 0.12 ø

æ1ö 1 = 64 ç ÷ è2 ø Solving we get, n = 6 Now, t = n(t 1/ 2 ) = 6(18 days) = 108 days

20. (c) Electrostatic energy = Binding energy of N – Binding energy of O = {[7MH + 8Mn - MN ] - [8MH + 7Mn - MO ]} ´ C 2 = [- MH + Mn + MO -MN ] C 2 = [- 1007825 + 1008665 + 15.003065 . . - 15.000109] ´ 9315 . = + 3.5359 MeV . ´ 8 ´ 7 3 1. 44 ´ 7 ´ 6 3 144 DE = ´ - ´ 5 5 R R = 3.5359 MeV 3 ´ 144 . ´ 14 R = = 3.42 fm 5 ´ 3.5359

h = 0.64 ´10 -33 = 6.4 ´ 10 -34 J-s

17. (a,b,d) As radius r µ 2

n2 z 2

æ n + 1ö æ nö ç ÷ -ç ÷ Dr è z ø è z ø = 2n + 1 » 2 µ 1 Þ = 2 r n n n2 æ nö ç ÷ è zø Energy E µ

z

z2

2

n2

Þ

DE n2 = E

-

( n + 1)2 - n2

× ( n + 1)2 Þ

d S2

m(S 2P ) - S1P = ml Þ

m d 2 + x 2 - d 2 + x 2 = ml

Þ

(m - 1) d 2 + x 2 = ml

æ4 ö Þ ç - 1÷ d 2 + x 2 = ml or è3 ø

12375 = 12.7 eV 970

x 2 = 9m2l2 - d 2 Þ p2 = 9 or

p= 3

22. (7) Case I Reflection from mirror 1 1 1 = + Þ f v u

1 1 1 = + -10 v -15

v = - 30

Þ

After excitation, let the electron jumps to nth state, then - 13.6 = - 13.6 + 12.7 n2 Solving this equation, we get n = 4 \Total number of lines in emission spectrum, n( n - 1) 4 ( 4 - 1) = = =6 2 2 n æ1ö 19. (c) Using the relation, R = R 0 ç ÷ è2 ø Here, R is activity of radioactive substance, R 0 initial activity and n is number of half lives.

d 2 + x 2 = 3ml

Squaring this equation we get,

18. (6) Energy of incident light (in eV) E=

P

x

O

( n + 1)2 z2

DE 2 n + 1 ~ 2 n 1 = - 2 µ n E n2 n nh Angular momentum L = 2p ( n + 1)h nh DL 2p = 1 µ 1 Þ = 2p nh L n n 2p n2 × ( n + 1)2

S1 d

z2

( n + 1)2 =

21. (3)

air

O 15 30

For lens

air I2

I1

20

20

1 1 1 1 1 1 = - Þ = Þ v = 20 f v u 10 v -20

| M1| =

v1 v 2 æ 30 ö æ 20 ö =ç ÷ç ÷ u1 u 2 è 15 ø è 20 ø

= 2 ´1= 2

(in air)

25

Previous Years’ Questions (2018-13) Case II For mirror, there is no change. v = - 30 1.5 medium medium (7/6) O (7/6)

60° 60°

N

M r

q

60°– r 30

I1 20

140

1 æ 3/2 ö =ç - 1÷ fair è 1 ø

For lens,

dr + sin r = 0 dn - tan r sin r dr or == dn n cos r n dq æ - tan r ö = - n cos ( 60°- r ) ç Þ cos q ÷ + sin ( 60°- r ) dn è n ø dq 1 = [cos( 60°- r )tan r + sin ( 60°- r )] dn cos q Differentiating Eq. (i), n cos r

æ 1 1 ö ÷÷ çç è R1 R 2 ø

1 ö 1 æ 3/2 öæ 1 ÷ =ç - 1÷ çç fmedium è 7 / 6 ø è R1 R 2 ÷ø with We get

fair = 10 cm 4 1 1 4 = cm-1 Þ = fmedium 70 v -20 70 1

Form Eq. (i), r = 30° for n = 3 dq 1 = (cos 30 ´ tan 30 + sin 30) = 2 dn cos 60

1 1 4 1 4 1 æ 2 öæ 2 ö + = ç ÷ç ÷ = Þ = v 20 è 7 ø è 10 ø 70 v 70 20 v = 140, |M2| =

v1 v 2 æ 30 ö =ç ÷ u1 u 2 è 15 ø

æ 140 ö ç ÷, è 20 ø

25. (a)

½M ½ 14 æ 140 ö = (2 ) ç =7 ÷ = 14 Þ ½ 2½ = è 20 ø ½M1½ 2

sin qc =

Þ

Þ

Þ

m 2 m1 m 2 - m1 two times = v u R 1 1.5 1 - 1.5 = v -50 -10 1 1.5 0.5 + = v 50 10 2.5 - 15 . . 1 0.5 15 = = v 10 50 50

æ n2 ö çç ÷÷ è n1 ø

2

4 sin i = 3

15 16 8 Þ sin i = sin( 90 - qc ) 8 5 3 15 9 Simplifying we get, sin i = 16 (a) is correct.

26. (d) sin i m = n1 sin ( 90 - qc ) Þ

sin i m = n1 cos qc Þ NA = n1 1 - sin2 qc = n1 1 -

n22 n12

=

n12 - n22

Substituting the values we get, 3 15 3 and NA 2 = NA1 = = Þ NA 2 < NA1 4 5 4

15 . -1 15 . 1 = ¥ - (d - 50) 10 1 1 = Þ d = 70 d - 50 20

Therefore, the numerical aperture of combined structure is equal to the lesser of the two numerical aperture, which is NA 2.

24. (2) Applying Snell’s law at M and N, sin 60° = n sin r sin q = n sin ( 60 - r ) Differentiating we get dr dq = - n cos ( 60 - r ) + sin ( 60 - r ) cos q dn dn

n2 Þ cos qc = 1 n1

Þ cos qc =

v = 50 MN = d , MI1 = 50 cm Þ NI1 = (d - 50) cm +ve I1 O N M

Again,

45 45 sin( 90 - q c ) = cos qc 4 4

45 3 9 ; sin i = 4 16 45 n 7 In second case, sin qc = 2 = n1 8

23. (b) R = 10 cm Applying

4 sin i = 3

...(i) ...(ii)

27. (3) Let initial power available from the plant is P0. After n

æ1ö time t = nT or n half lives, this will become ç ÷ P0. è2 ø

26

Optics & Modern Physics Number of fringes in a given width Y 1 or m µ Þ m2 < m1 as b 2 > b1 m= b b

Now, it is given that, n

æ1ö . ) P0 ç ÷ P0 = 12.5% of P0 = ( 0125 è2 ø Solving this equation we get, n = 3

28. A-(p, q), B- (p, r), C-(p, s) D- (p, q, r) 29. (2) Let initial numbers are N1 and N2. l1 t 2t T =2 = 2 = 2 = l2 t1 t T1

(T = Half life)

- dN = lN dt Initial activity is same \ l1N1 = l2N2 Activity at time t, A = lN = lN0 e - lt Þ A1 = l1N1 e - l1t A=

Þ

R1 = -

dA1 = l21 N1 e - l1 dt

Similarly, R 2 = After t = 2 t

…(i)

l22

N2 e

- l2t

l1t =

1 1 (t ) = (2 t ) = 2 t1 t

l2t =

1 1 (t ) = 1 = (2 t ) = 1 t2 2t

l2 N e -l1t l æ e -2 ö 2 RP R = 12 1 -l t Þ P = 1 çç -1 ÷÷ = l1 è e ø e RQ RQ l2 N2 e 2 æ h ö æ h ö 30. (2) Angular momentum = n ç ÷ = 3ç ÷ 2 p è ø è 2p ø \ n=3 n2 ( 3)2 Now, rn µ Þ r3 = ( a0 ) = 3a0 z 3 æ h ö æ h ö Now, mv 3r3 = 3 ç ÷ Þ mv 3( 3a0 ) = 3ç ÷ 2 p è ø è 2p ø h h or = 2 pa0 Þ = 2 pa0 ( P = mv ) mv 3 P3 hö æ or l3 = 2 pa0 çl = ÷ Pø è \ Answer is 2.

Distance of 3rd maximum of l2 from central maximum 3l2D 1800D = = d d Distance of 5th minimum of l1 from central maximum 9l D 1800D = 1 = 2d d So, 3rd maximum of l2 will overlap with 5th minimum of l1 l Angular separation (or angular fringe width) = µ l d Þ Angular separation for l1 will be lesser. 1 æ1 1ö 33. (a,c) = ( n1 - 1) ç - ÷ Þ ffilm = ¥ (infinite) ffilm èR Rø \ There is no effect of presence of film. From Air to Glass n 1 n -1 Using the equation 2 - = 2 v u R 1.5 1 1.5 - 1 Þ v = 3R Þ f1 = 3R = v ¥ R From Glass to Air Again using the same equation 1 n2 1 - n2 1 1.5 1 - 1.5 = Þ = v u -R v ¥ -R

34. (b) (P) Þ Þ (Q) Þ (R)

31. (d) From conservation laws of mass number and atomic number, we can say that x = n, y = n ( x = 10n, y = 10n) \ Only (a) and (d) options may be correct. From conservation of momentum,| Pxe| = | Psr| P2 1 From K = ÞK µ 2m m Ksr m = xe Þ K sr = 129 MeV Þ K xe = 86 MeV K xe msr lD or b µ l Þ l2 > l1 32. (a, b, c) Fringe width b = d So b 2 > b1

v = 2R Þ 1 æ3 ö = ç - 1÷ f è2 ø

Þ

f =r r 1 1 1 2 Þ feq = = + = 2 feq f f r 1 æ3 ö æ 1ö = ç - 1÷ ç ÷ Þ f = 2 r f è2 ø èrø 1 1 2 1 Þ feq = r + = = f f f r 1 = f

(S)

1 æ3 ö æ 1ö ç - 1÷ ç - ÷ = 2 r 2 r è øè ø f = - 2r

Þ Þ

f2 = 2 R æ 1 1ö 1 ç + ÷= rø r èr

1 1 1 2 Þ feq = - r = + =feq f f 2r Þ

1 1 1 1 Þfeq = 2 r = + = feq r -2 r 2 r

35. (a) Energy corresponding to 248 nm wavelength =

1240 eV = 5 eV 248

27

Previous Years’ Questions (2018-13) Energy corresponding to 310 nm wavelength 1240 = eV = 4 eV 310 2 KE1 u1 4 5 eV - W = = = KE 2 u 22 1 4 eV - W Þ Þ

40. (d) For e ® i 45° > qc Þ sin 45° > sin qc 1 m > 2 Þ m 1 > 2 m 2 , (s-1) 2 m1 For e ® f

16 - 4W = 5 - W Þ 11 = 3W 11 W = = 3.67 eV ~ = 3.7 eV 3

angle of refraction is lesser than angle of incidence, so (P-2) m 2 > m 1 and then m 2 > m 3 For e ® g , m 1 = m 2 (Q-3) for e ® h, m 2 < m 1 < 2m 2 and m 2 > m 3 (R-4)

36. (b) K a transition takes place from n1 = 2 to n2 = 1

41. (c) At point Q angle of incidence is critical angle qC ,

é 1 1 1 ù = R ( Z - b )2 ê 2 ú l (2 )2 û ë (1) 1 For K-series, b = 1 Þ µ ( Z - 1)2 l ( zMo - 1)2 ( 42 - 1)2 lCu Þ = = lMo ( zCu - 1)2 (29 - 1)2

\

where

Q

qC h

41 ´ 41 1681 = = 2.14 28 ´ 28 784 lair 1 3 37. (c) m = = = lmedium (2 /3) 2 1 v Further, | m| = = 3 u |u| (Real object) | v| = Þ u = - 24 m \ 3 (Real image) \ v = + 8m 1 1 1 1ö æ 1 Now, - = = (m - 1) ç - ÷ v u f è+R ¥ ø =

\

…(i)

Imax 2 \From Eqs. (i) and (ii), we have p 3p 5p , f= , 2 2 2 æ l ö Or path difference, Dx = ç ÷×f è 2p ø l 3l 5l æ 2n + 1ö Dx = , , Kç \ ÷l 4 4 4 è 4 ø Given,

I=

In DPQS,sin qC = Þ ml =

r r 2 + h2

2

r + h ´ 2.72 =

2

Þ

ml = m block

r + h2

5.77 ´ 2.72 = 1.36 11.54

42. (b) Final momentum of object = 30 ´ 10 -3 ´ 100 ´ 10 -9 3 ´ 10 8

r 2

Power ´ time Speed of light

= 1.0 ´ 10 -17 kg-m/s -1

Nd = N0(1 - e - lt )

3 and component perpendicular to the plane = 2 æ1ö ç ÷ 1 2 \ tan i = è ø = æ 3ö 3 ç ÷ ç 2 ÷ è ø \ i = 30° = angle of incidence 2

ml m block r

sin qC =

43. (4) Number of nuclei decayed in time t ,

1 2

39. (b) I = I cos 2 f max

Denser medium QP = r

S

=

1 1 æ3 öæ 1 ö + = ç - 1÷ ç ÷ Þ R = 3 m 8 24 è 2 øè R ø

38. (a) Component along the plane =

Rarer medium

P

qC

…(ii)

æN ö \ % decayed = çç d ÷÷ ´ 100 = (1 - e - lt d ) ´ 100 …(i) è N0 ø 0.693 Here, l= = 5 ´ 10 -4 s -1 1386 \% decayed » ( lt ) ´ 100 = ( 5 ´ 10 -4 ) ( 80) (100) = 4 æ h ö 44. (a, c) L = 3 ç ÷ è 2p ø n=3 l1 l2 n=2 l3 n=1

n2 æ h ö \ n = 3, as L = n ç ÷ Þ rn µ z è 2p ø r3 = 4.5a0 \ z = 2 1 1 ö æ 1 æ1 1ö = Rz2 ç 2 - 2 ÷ = 4R ç - ÷ l1 3 ø è2 è4 9ø

28

Optics & Modern Physics 9 5R

46. (a)

\

l1 =

Þ

1 1 ö 1ö æ1 æ = Rz2 ç 2 - 2 ÷ = 4R ç 1 - ÷ l2 9ø 3 ø è1 è l2 =

Þ

1 1 ö 1ö æ1 æ = Rz2 ç 2 - 2 ÷ = 4R ç 1 - ÷ l3 4ø 2 ø è1 è

Þ

1 l3 = 3R

210 ® 83Bi 209 + 1P1 84 Po

Dm is negative so reaction is not possible. (c) 1H 2 ® 2He 4 + 3Li 6 Dm is positive so reaction is possible. 30 Zn

+

34 Se

206

Q = ( Dm) ( 931.48) MeV = 5.4193 MeV = 5419 keV

82

152

® 64 Gd

Dm is positive so reaction is not possible.

Pb

From conservation of linear momentum, pPb = pa \

\

= [6.01513 - 4.002603 - 3.016050] = - 1.003523 u Dm is negative so reaction is not possible.

(d)

82Pb

Mass defect Dm = ( mPo - MHe - mPb ) = 0.005818 u

Þ

Dm = [MLi - MHe - MH 3 ]

70

¾® 2He 4 +

a

45. (a) 3Li 7 ® 2He 4 + 1H 3

(b)

210

\

9 32 R

Þ

84 Po

2 mPb kPb = 2 ma k a ka m 206 = Pb = k Pb ma 4 æ 206 ö ÷ ( k total ) k a = çç ÷ è 206 + 4 ø æ 206 ö =ç ÷ ( 5419) = 5316 keV è 210 ø

47. (9) 125 B ¾® 126 C + 0-1e + n Mass of 126 C = 12.000 u (by definition of 1 a.m.u.) Q-value of reaction, Q = ( M B - MC ) ´ c 2 . = 13.041 MeV = (12.014 - 12.000) ´ 9315 4.041 MeV of energy is taken by 126 C * Þ Maximum KE of b-particle is (13.041 - 4.041) = 9 MeV